See our solution for Question 3E from Chapter 1.1 from Lay's Linear Algebra and Its Applications, 5th Edition. Step 1 Step
2 Step 3: $R_2 = R_2 - R_1$ Step 4: ${R_2} = -{R_2}/7 $ Step 5: Write the reduced form in system of equations Step 6 ANSWERS \[\left[ {\begin{array}{*{20}{l}}{{x_1} = \dfrac{4}{7}}\\{{x_2} = \dfrac{9}{7}}\end{array}} \right]\] See our solution for Question 3E from Chapter 1.9 from Lay's Linear Algebra and Its Applications, 5th Edition. Step 1 Step 2: The Rotation Angle Step 3: The
Transformation Matrix ANSWER \[A = \left[ {\begin{array}{*{20}{c}}0&1\\{ - 1}&0\end{array}} \right]\] Table of contents : INSTRUCTOR’S SOLUTIONS MANUAL JUDI J. MCDONALD Washington State University LINEAR ALGEBRA AND ITS APPLICATIONS FIFTH EDITION GLOBAL EDITION David C. Lay University of Maryland Steven R. Lay Lee University Judi J. McDonald Washington State University Boston Columbus Hoboken Indianapolis New York San Francisco Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto Delhi Mexico City São Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo Copyright © 2016 Pearson Education Limited All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise without the prior written permission of the publisher. Many of the designations used by manufacturers and sellers to distinguish their products are claimed as trademarks. Where those designations appear in this book, and the publisher was aware of a trademark claim, the designations have been printed in initial caps or all caps. ISBN-13: 978-1-292-09227-0 ISBN-10: 1-292-09227-0 Contents Chapter 1 1-1 Chapter 2 2-1 Chapter 3 3-1 Chapter 4 4-1 Chapter 5 5-1 Chapter 6 6-1 Chapter 7 7-1 Chapter 8 8-1 iii Copyright © 2016 Pearson Education, Ltd. 1.1 SOLUTIONS Notes: The key exercises are 7 (or 11 or 12), 19–22, and 25. For brevity, the symbols R1, R2,…, stand for row 1 (or equation 1), row 2 (or equation 2), and so on. Additional notes are at the end of the section. 1. x1 + 5 x2 = 7 −2 x1 − 7 x2 = −5 1 −2 5 −7 7 −5 Replace R2 by R2 + (2)R1 and obtain: x1 + 5 x2 = 7 3 x2 = 9 x1 + 5 x2 = 7 Scale R2 by 1/3: x2 = 3 x1 Replace R1 by R1 + (–5)R2: = −8 x2 = 3 1 0 5 3 7 9 1 0 5 1 7 3 1 0 0 1 −8 3 1 5 2 7 −2 11 1 0 2 −3 1 0 2 1 −2 −7 1 0 0 1 12 −7 The solution is (x1, x2) = (–8, 3), or simply (–8, 3). 2. 2 x1 + 4 x2 = −4 5 x1 + 7 x2 = 11 2 5 4 7 −4 11 Scale R1 by 1/2 and obtain: Replace R2 by R2 + (–5)R1: x1 + 2 x2 = −2 5 x1 + 7 x2 = 11 x1 + 2 x2 = −2 −3 x2 = 21 x1 + 2 x2 = −2 Scale R2 by –1/3: x2 = −7 Replace R1 by R1 + (–2)R2: x1 = 12 x2 = −7 −2 21 The solution is (x1, x2) = (12, –7), or simply (12, –7). Copyright © 2016 Pearson Education, Ltd. 1-1 1-2 CHAPTER 1 • Linear Equations in Linear Algebra 3. The point of intersection satisfies the system of two linear equations: x1 + 5 x2 = 7 5 7 1 x1 − 2 x2 = −2 1 −2 −2 x1 + 5 x2 = 7 Replace R2 by R2 + (–1)R1 and obtain: −7 x2 = −9 x1 + 5 x2 = 7 Scale R2 by –1/7: x2 = 9/7 x1 Replace R1 by R1 + (–5)R2: = 4/7 x2 = 9/7 1 0 5 −7 7 −9 1 0 5 1 7 9/7 1 0 0 1 4/7 9/7 1 0 −5 8 1 2 1 0 1 −5 1 1/4 1 0 0 1 The point of intersection is (x1, x2) = (4/7, 9/7). 4. The point of intersection satisfies the system of two linear equations: x1 − 5 x2 = 1 1 −5 1 3 −7 5 3 x1 − 7 x2 = 5 x1 − 5 x2 = 1 Replace R2 by R2 + (–3)R1 and obtain: 8 x2 = 2 x1 − 5 x2 = 1 Scale R2 by 1/8: x2 = 1/4 x1 Replace R1 by R1 + (5)R2: = 9/4 x2 = 1/4 9/4 1/4 The point of intersection is (x1, x2) = (9/4, 1/4). 5. The system is already in “triangular” form. The fourth equation is x4 = –5, and the other equations do not contain the variable x4. The next two steps should be to use the variable x3 in the third equation to eliminate that variable from the first two equations. In matrix notation, that means to replace R2 by its sum with 3 times R3, and then replace R1 by its sum with –5 times R3. 6. One more step will put the system in triangular form. Replace R4 by its sum with –3 times R3, which 4 0 −1 1 −6 0 2 −7 0 4 . After that, the next step is to scale the fourth row by –1/5. produces 0 0 1 2 −3 0 0 −5 15 0 7. Ordinarily, the next step would be to interchange R3 and R4, to put a 1 in the third row and third column. But in this case, the third row of the augmented matrix corresponds to the equation 0 x1 + 0 x2 + 0 x3 = 1, or simply, 0 = 1. A system containing this condition has no solution. Further row operations are unnecessary once an equation such as 0 = 1 is evident. The solution set is empty. Copyright © 2016 Pearson Education, Ltd. 1.1 • Solutions 1-3 8. The standard row operations are: 1 0 0 −4 1 0 0 1 0 ~ 0 0 0 9 7 2 −4 1 0 0 1 0 ~ 0 0 0 9 7 1 −4 1 0 0 0 1 0 1 0 ~ 0 0 0 0 1 0 0 0 1 0 0 0 The solution set contains one solution: (0, 0, 0). 9. The system has already been reduced to triangular form. Begin by scaling the fourth row by 1/2 and then replacing R3 by R3 + (3)R4: 1 0 0 0 −1 1 0 0 0 −3 1 0 0 0 −3 2 −4 1 −7 0 ~ −1 0 4 0 −1 1 0 0 0 −3 1 0 0 0 −3 1 −4 1 −7 0 ~ −1 0 2 0 −1 1 0 0 0 −3 1 0 0 0 0 1 −4 −7 5 2 Next, replace R2 by R2 + (3)R3. Finally, replace R1 by R1 + R2: −1 1 0 0 1 0 ~ 0 0 0 0 1 0 0 0 0 1 −4 1 8 0 ~ 5 0 2 0 0 1 0 0 0 0 1 0 0 0 0 1 4 8 5 2 The solution set contains one solution: (4, 8, 5, 2). 10. The system has already been reduced to triangular form. Use the 1 in the fourth row to change the –4 and 3 above it to zeros. That is, replace R2 by R2 + (4)R4 and replace R1 by R1 + (–3)R4. For the final step, replace R1 by R1 + (2)R2. 1 0 0 0 −2 1 0 0 0 0 1 0 3 −4 0 1 −2 1 7 0 ~ 6 0 −3 0 −2 1 0 0 0 0 1 0 0 0 0 1 7 1 −5 0 ~ 6 0 −3 0 0 1 0 0 0 0 1 0 −3 −5 6 −3 0 0 0 1 The solution set contains one solution: (–3, –5, 6, –3). 11. First, swap R1 and R2. Then replace R3 by R3 + (–3)R1. Finally, replace R3 by R3 + (2)R2. 0 1 3 1 3 7 4 5 7 −5 1 −2 ~ 0 6 3 3 1 7 5 4 7 −2 1 −5 ~ 0 6 0 3 1 −2 5 4 −8 −2 1 −5 ~ 0 12 0 3 1 0 5 4 0 −2 −5 2 The system is inconsistent, because the last row would require that 0 = 2 if there were a solution. The solution set is empty. 12. Replace R2 by R2 + (–3)R1 and replace R3 by R3 + (4)R1. Finally, replace R3 by R3 + (3)R2. 4 −4 1 −3 4 −4 1 −3 4 −4 1 −3 3 −7 7 −8 ~ 0 2 −5 4 ~ 0 2 −5 4 6 −1 7 0 −6 15 −9 0 0 0 3 −4 The system is inconsistent, because the last row would require that 0 = 3 if there were a solution. The solution set is empty. Copyright © 2016 Pearson Education, Ltd. 1-4 CHAPTER 1 • Linear Equations in Linear Algebra 13. Replace R2 by R2 + (–2)R1. Then interchange R2 and R3. Next replace R3 by R3 + (–2)R2. Then divide R3 by 5. Finally, replace R1 by R1 + (–2)R3. 8 1 0 −3 8 1 0 −3 8 1 0 1 0 −3 2 2 9 7 ~ 0 2 15 −9 ~ 0 1 5 −2 ~ 0 1 0 1 5 −2 0 1 5 −2 0 2 15 −9 0 0 8 1 0 0 1 0 −3 5 ~ 0 1 5 −2 ~ 0 1 0 3 . The solution is (5, 3, –1). 0 0 1 −1 0 0 1 −1 −3 5 5 8 −2 −5 14. Replace R2 by R2 + R1. Then interchange R2 and R3. Next replace R3 by R3 + 2R2. Then divide R3 by 7. Next replace R2 by R2 + (–1)R3. Finally, replace R1 by R1 + 3R2. 1 −1 0 −3 1 1 0 5 1 5 1 2 ~ 0 0 0 1 ~ 0 0 −3 1 0 0 0 1 5 1 −1 ~ 0 1 0 −3 −2 1 0 5 1 5 1 7 ~ 0 0 0 −3 1 −2 0 1 0 0 0 1 2 −1 . The solution is (2, –1, 1). 1 0 1 5 5 1 0 ~ 0 7 0 −3 1 0 0 1 7 5 1 0 ~ 0 7 0 −3 1 0 0 1 1 5 0 1 15. First, replace R4 by R4 + (–3)R1, then replace R3 by R3 + (2)R2, and finally replace R4 by R4 + (3)R3. 1 0 0 3 1 0 ~ 0 0 2 1 3 0 ~ 1 0 −5 0 0 1 −2 0 3 0 3 0 0 −3 2 7 0 1 0 0 3 0 3 0 0 2 −3 3 . −4 7 −5 10 0 1 −2 0 3 0 3 −9 0 −3 2 7 2 1 3 0 ~ 1 0 −11 0 0 1 0 0 3 0 3 −9 0 −3 −4 7 2 3 7 −11 The resulting triangular system indicates that a solution exists. In fact, using the argument from Example 2, one can see that the solution is unique. 16. First replace R4 by R4 + (2)R1 and replace R4 by R4 + (–3/2)R2. (One could also scale R2 before adding to R4, but the arithmetic is rather easy keeping R2 unchanged.) Finally, replace R4 by R4 + R3. 0 −2 −3 1 0 0 −2 −3 1 0 0 −2 −3 1 0 0 −2 −3 1 0 0 2 2 0 0 0 2 2 0 0 0 2 2 0 0 0 2 2 0 0 ~ ~ ~ 0 0 1 3 1 0 0 1 3 1 0 0 1 3 1 0 0 1 3 1 1 5 0 3 2 −3 −1 0 0 −1 −3 −1 0 0 0 0 0 −2 3 2 The system is now in triangular form and has a solution. The next section discusses how to continue with this type of system. Copyright © 2016 Pearson Education, Ltd. 1.1 • Solutions 1-5 17. Row reduce the augmented matrix corresponding to the given system of three equations: 1 1 −4 1 1 −4 1 1 −4 2 −1 −3 ~ 0 7 −5 ~ 0 7 −5 −1 −3 4 0 −7 5 0 0 0 The system is consistent, and using the argument from Example 2, there is only one solution. So the three lines have only one point in common. 18. Row reduce the augmented matrix corresponding to the given system of three equations: 1 4 1 2 1 4 1 2 1 4 1 2 0 1 −1 1 ~ 0 1 −1 1 ~ 0 1 −1 1 1 3 0 0 0 1 −1 −4 0 0 0 −5 The third equation, 0 = –5, shows that the system is inconsistent, so the three planes have no point in common. h 4 1 h 4 1 19. ~ Write c for 6 – 3h. If c = 0, that is, if h = 2, then the system has no 3 6 8 0 6 − 3h −4 solution, because 0 cannot equal –4. Otherwise, when h ≠ 2, the system has a solution. h −3 1 h −3 1 20. ~ . Write c for 4 + 2h. Then the second equation cx2 = 0 has a 6 0 4 + 2h 0 −2 4 solution for every value of c. So the system is consistent for all h. 3 −2 1 3 −2 1 21. ~ . Write c for h + 12. Then the second equation cx2 = 0 has a 8 0 h + 12 0 −4 h solution for every value of c. So the system is consistent for all h. 2 −3 h 2 22. ~ 9 5 0 −6 only if h = –5/3. −3 0 h . The system is consistent if and only if 5 + 3h = 0, that is, if and 5 + 3h 23. a. True. See the remarks following the box titled “Elementary Row Operations”. b. False. A 5 × 6 matrix has five rows. c. False. The description given applies to a single solution. The solution set consists of all possible solutions. Only in special cases does the solution set consist of exactly one solution. Mark a statement True only if the statement is always true. d. True. See the box before Example 2. 24. a. True. See the box preceding the subsection titled “Existence and Uniqueness Questions”. b. False. The definition of row equivalent requires that there exist a sequence of row operations that transforms one matrix into the other. c. False. By definition, an inconsistent system has no solution. d. True. This definition of equivalent systems is in the second paragraph after equation (2). Copyright © 2016 Pearson Education, Ltd. 1-6 CHAPTER 1 −4 3 5 1 25. 0 −2 • Linear Equations in Linear Algebra 7 −5 −9 g 1 h ~ 0 k 0 −4 3 −3 7 −5 5 −4 3 0 g 1 h ~ 0 k + 2 g 0 7 −5 0 g h k + 2 g + h Let b denote the number k + 2g + h. Then the third equation represented by the augmented matrix above is 0 = b. This equation is possible if and only if b is zero. So the original system has a solution if and only if k + 2g + h = 0. 26. A basic principle of this section is that row operations do not affect the solution set of a linear system. Begin with a simple augmented matrix for which the solution is obviously ( –2, 1, 0 ) , and then perform any elementary row operations to produce other augmented matrices. Here are three examples. The fact that they are all row equivalent proves that they all have the solution set ( –2, 1, 0) . 1 0 0 0 1 0 0 0 1 −2 1 1 ~ 2 0 0 0 1 0 0 0 1 −2 1 −3 ~ 2 0 2 0 1 0 0 0 1 −2 −3 −4 27. Study the augmented matrix for the given system, replacing R2 by R2 + (–c)R1: 3 f 1 3 f 1 c d g ~ 0 d − 3c g − cf . This shows that shows d – 3c must be nonzero, since f and g are arbitrary. Otherwise, for some choices of f and g the second row would correspond to an equation of the form 0 = b, where b is nonzero. Thus d ≠ 3c. 28. Row reduce the augmented matrix for the given system. Scale the first row by 1/a, which is possible since a is nonzero. Then replace R2 by R2 + (–c)R1. a c b d f 1 ~ g c b/a d f / a 1 ~ g 0 b/a d − c(b / a ) f /a g − c( f / a ) The quantity d – c(b/a) must be nonzero, in order for the system to be consistent when the quantity g – c( f /a) is nonzero (which can certainly happen). The condition that d – c(b/a) ≠ 0 can also be written as ad – bc ≠ 0, or ad ≠ bc. 29. Swap R1 and R2; swap R1 and R2. 30. Multiply R2 by –1/2; multiply R2 by –2. 31. Replace R3 by R3 + (–4)R1; replace R3 by R3 + (4)R1. 32. Replace R3 by R3 + (3)R2; replace R3 by R3 + (–3)R2. 33. The first equation was given. The others are: T2 = (T1 + 20 + 40 + T3 )/4, or 4T2 − T1 − T3 = 60 T3 = (T4 + T2 + 40 + 30)/4, or 4T3 − T4 − T2 = 70 T4 = (10 + T1 + T3 + 30)/4, or 4T4 − T1 − T3 = 40 Copyright © 2016 Pearson Education, Ltd. 1.1 • Solutions 1-7 Rearranging, 4T1 − −T1 + − T2 4T2 −T2 − T3 + 4T3 − T3 −T1 T4 − T4 + 4T4 = 30 = 60 = 70 = 40 34. Begin by interchanging R1 and R4, then create zeros in the first column: 4 −1 0 −1 −1 4 −1 0 −1 0 −1 4 0 −1 4 −1 30 −1 60 −1 ~ 70 0 40 4 −1 −1 4 0 0 4 −1 −1 4 0 −1 −1 40 −1 60 0 ~ 70 0 30 0 −1 0 4 −4 0 4 −1 −1 4 40 −4 20 −1 70 15 190 Scale R1 by –1 and R2 by 1/4, create zeros in the second column, and replace R4 by R4 + R3: 1 0 ~ 0 0 0 1 −1 −1 −4 −1 −1 15 1 0 4 −4 −40 1 5 0 ~ 70 0 190 0 0 1 0 0 1 0 4 −4 −4 −1 −2 14 −40 1 5 0 ~ 75 0 195 0 0 1 0 0 1 0 4 0 −4 −1 −2 12 −40 5 75 270 Scale R4 by 1/12, use R4 to create zeros in column 4, and then scale R3 by 1/4: 1 0 ~ 0 0 0 1 0 0 1 0 4 0 −4 −1 −2 1 −40 1 5 0 ~ 75 0 22.5 0 0 1 0 0 1 0 4 0 0 0 0 1 50 1 27.5 0 ~ 120 0 22.5 0 0 1 0 0 1 0 1 0 0 0 0 1 50 27.5 30 22.5 The last step is to replace R1 by R1 + (–1)R3: 1 0 ~ 0 0 0 1 0 0 0 0 1 0 0 0 0 1 20.0 27.5 . The solution is (20, 27.5, 30, 22.5). 30.0 22.5 Notes: The Study Guide includes a “Mathematical Note” about statements, “If … , then … .” This early in the course, students typically use single row operations to reduce a matrix. As a result, even the small grid for Exercise 34 leads to about 25 multiplications or additions (not counting operations with zero). This exercise should give students an appreciation for matrix programs such as MATLAB. Exercise 14 in Section 1.10 returns to this problem and states the solution in case students have not already solved the system of equations. Exercise 31 in Section 2.5 uses this same type of problem in connection with an LU factorization. For instructors who wish to use technology in the course, the Study Guide provides boxed MATLAB notes at the ends of many sections. Parallel notes for Maple, Mathematica, and ssome calculators appear in separate appendices at the end of the Study Guide. The MATLAB box for Section 1.1 describes how to access the data that is available for all numerical exercises in the text. This feature has the ability to save students time if they regularly have their matrix program at hand when studying linear algebra. The MATLAB box also explains the basic commands replace, swap, and scale. These commands are included in the text data sets, available from the text web site, www.pearsonhighered.com/lay. Copyright © 2016 Pearson Education, Ltd. 1-8 CHAPTER 1 1.2 • Linear Equations in Linear Algebra SOLUTIONS Notes: The key exercises are 1–20 and 23–28. (Students should work at least four or five from Exercises 7–14, in preparation for Section 1.5.) 1. Reduced echelon form: a and b. Echelon form: d. Not echelon: c. 2. Reduced echelon form: a. Echelon form: b and d. Not echelon: c. 1 3. 4 6 2 5 7 1 ~ 0 0 1 4. 3 5 2 1 0 3 5 7 1 ~ 0 0 4 1 7 ~ 0 9 0 3 6 8 3 2 0 3 1 0 4 1 3 ~ 0 0 0 7 1 9 ~ 0 1 0 5 7 9 5 2 0 7 1 3 ~ 0 1 0 5. 0 * , 0 * 0 , 0 0 1 7. 3 3 9 7 1 ~ 6 0 4 7 2 −3 −5 0 1 0 3 −4 −8 3 1 0 3 −6 −10 −1 2 0 5 −8 −16 5 2 0 4 1 −9 ~ 0 −15 0 7 1 −12 ~ 0 −34 0 0 1 0 ~ 0 1 0 0 1 0 7 1 ~ −15 0 Corresponding system of equations: 1 4 6 2 5 7 5 2 −16 −1 2 0 0 0 . Pivot cols 1, 2, and 4 1 3 0 * , 0 0 0 4 1 x1 + 3 x2 x3 * 0 0 , 0 0 0 7 1 ~ 3 0 4 7 9 3 6 8 7 1 3 ~ 0 −34 0 3 1 −8 6. 0 0 4 −5 4 3 −15 3 2 −10 −2 3 . Pivot cols 1 and 2. 0 0 3 0 2 1 −5 3 1 0 1 3 5 7 3 −10 5 2 0 3 5 7 5 7 9 7 9 1 0 0 3 0 0 1 −5 3 = −5 = 3 The basic variables (corresponding to the pivot positions) are x1 and x3. The remaining variable x2 is free. Solve for the basic variables in terms of the free variable. The general solution is x1 =−5 − 3 x2 x2 is free x = 3 3 Note: Exercise 7 is paired with Exercise 10. Copyright © 2016 Pearson Education, Ltd. 1.2 1 2 8. 4 7 0 0 7 1 ~ 10 0 4 −1 7 1 ~ −4 0 0 0 Corresponding system of equations: 4 1 0 0 7 1 ~ 4 0 0 1 0 0 • Solutions 1-9 −9 4 = x1 x2 −9 = 4 The basic variables (corresponding to the pivot positions) are x1 and x2. The remaining variable x3 is free. Solve for the basic variables in terms of the free variable. In this particular problem, the basic variables do not depend on the value of the free variable. x1 = −9 General solution: x2 = 4 x is free 3 Note: A common error in Exercise 8 is to assume that x3 is zero. To avoid this, identify the basic variables first. Any remaining variables are free. (This type of computation will arise in Chapter 5.) 0 9. 1 1 −2 −6 7 5 1 ~ −6 0 Corresponding system: −2 1 x1 x2 7 −6 −6 1 ~ 5 0 − 5 x3 − 6 x3 0 1 −5 −6 4 5 = 4 = 5 x1= 4 + 5 x3 Basic variables: x1, x2; free variable: x3. General solution: x2= 5 + 6 x3 x is free 3 1 10. 3 −2 −6 −1 −2 −2 0 3 1 ~ 2 0 Corresponding system: −1 1 3 1 ~ −7 0 x1 − 2 x2 −2 0 0 1 −4 −7 = −4 x3 = −7 x1 =−4 + 2 x2 Basic variables: x1, x3; free variable: x2. General solution: x2 is free x = −7 3 3 11. −9 −6 −4 12 8 2 −6 −4 0 3 0 ~ 0 0 0 x1 Corresponding system: −4 0 0 2 0 0 − 4 x2 3 0 1 0 ~ 0 0 0 + 2 x3 3 0 0 −4 / 3 0 0 2/3 0 0 0 0 0 0 = = 0 = 0 Copyright © 2016 Pearson Education, Ltd. 1-10 CHAPTER 1 • Linear Equations in Linear Algebra 4 2 x1 x2 − x3 = 3 3 Basic variable: x1; free variables x2, x3. General solution: x2 is free x is free 3 1 12. 0 −1 −7 0 7 0 1 −4 5 1 −3 ~ 0 7 0 6 −2 2 −7 0 0 0 1 −4 x1 − 7 x2 x3 Corresponding system: 5 1 −3 ~ 0 12 0 6 −2 8 −7 0 0 0 1 0 6 −2 0 5 −3 0 + 6 x4 = 5 − 2 x4 = −3 0 = 0 5 7 x2 − 6 x4 x1 =+ x is free Basic variables: x1 and x3; free variables: x2, x4. General solution: 2 x3 =−3 + 2 x4 x4 is free 1 0 13. 0 0 −3 1 0 0 0 0 0 0 −1 0 1 0 −2 1 1 0 ~ 4 0 0 0 0 −4 9 0 −3 1 0 0 x1 Corresponding system: x2 x4 0 0 0 0 0 0 1 0 − 3 x5 − 4 x5 + 9 x5 0 2 1 1 0 ~ 4 0 0 0 9 −4 9 0 0 1 0 0 0 0 0 0 = 5 = 1 = 4 = 0 x1= 5 + 3 x5 x = 1 + 4x 5 2 Basic variables: x1, x2, x4; free variables: x3, x5. General solution: x3 is free x = 4 − 9x 5 4 x5 is free Note: The Study Guide discusses the common mistake x3 = 0. 1 0 14. 0 0 2 1 0 0 −5 −6 0 0 −6 −3 0 0 0 0 1 0 −5 1 2 0 ~ 0 0 0 0 0 1 0 0 7 −6 0 0 0 −3 0 0 0 0 1 0 −9 2 0 0 Copyright © 2016 Pearson Education, Ltd. 0 0 1 0 −3 −4 9 0 5 1 4 0 1.2 x1 x2 Corresponding system: + 7 x3 − 6 x3 − 3 x4 • Solutions 1-11 = −9 = 2 x5 = 0 0 = 0 x1 =−9 − 7 x3 x =+ 2 2 6 x3 + 3 x4 Basic variables: x1, x2, x5; free variables: x3, x4. General solution: x3 is free x is free 4 x5 = 0 15. a. The system is consistent, with a unique solution. b. The system is inconsistent. The rightmost column of the augmented matrix is a pivot column. 16. a. The system is consistent, with a unique solution. b. The system is consistent. There are many solutions because x2 is a free variable. 2 17. 4 3 6 h 2 ~ 7 0 3 0 h The system has a solution only if 7 – 2h = 0, that is, if h = 7/2. 7 − 2h −3 −2 1 −3 −2 1 If h +15 is zero, that is, if h = –15, then the system has no 18. ~ h −7 0 h + 15 3 5 solution, because 0 cannot equal 3. Otherwise, when h ≠ −15, the system has a solution. h 2 1 h 2 1 19. ~ 4 8 k 0 8 − 4h k − 8 a. When h = 2 and k ≠ 8, the augmented column is a pivot column, and the system is inconsistent. b. When h ≠ 2, the system is consistent and has a unique solution. There are no free variables. c. When h = 2 and k = 8, the system is consistent and has many solutions. 1 20. 3 3 h 2 1 ~ k 0 3 h−9 2 k − 6 a. When h = 9 and k ≠ 6, the system is inconsistent, because the augmented column is a pivot column. b. When h ≠ 9, the system is consistent and has a unique solution. There are no free variables. c. When h = 9 and k = 6, the system is consistent and has many solutions. 21. a. b. c. d. e. False. See Theorem 1. False. See the second paragraph of the section. True. Basic variables are defined after equation (4). True. This statement is at the beginning of “Parametric Descriptions of Solution Sets”. False. The row shown corresponds to the equation 5x4 = 0, which does not by itself lead to a contradiction. So the system might be consistent or it might be inconsistent. Copyright © 2016 Pearson Education, Ltd. 1-12 CHAPTER 1 • Linear Equations in Linear Algebra 22. a. False. See the statement preceding Theorem 1. Only the reduced echelon form is unique. b. False. See the beginning of the subsection “Pivot Positions”. The pivot positions in a matrix are determined completely by the positions of the leading entries in the nonzero rows of any echelon form obtained from the matrix. c. True. See the paragraph after Example 3. d. False. The existence of at least one solution is not related to the presence or absence of free variables. If the system is inconsistent, the solution set is empty. See the solution of Practice Problem 2. e. True. See the paragraph just before Example 4. 23. Yes. The system is consistent because with three pivots, there must be a pivot in the third (bottom) row of the coefficient matrix. The reduced echelon form cannot contain a row of the form [0 0 0 0 0 1]. 24. The system is inconsistent because the pivot in column 5 means that there is a row of the form [0 0 0 0 1] in the reduced echelon form. Since the matrix is the augmented matrix for a system, Theorem 2 shows that the system has no solution. 25. If the coefficient matrix has a pivot position in every row, then there is a pivot position in the bottom row, and there is no room for a pivot in the augmented column. So, the system is consistent, by Theorem 2. 26. Since there are three pivots (one in each row), the augmented matrix must reduce to the form x1 = a 1 0 0 a 0 1 0 b and so x2 = b 0 0 1 c x3 = c No matter what the values of a, b, and c, the solution exists and is unique. 27. “If a linear system is consistent, then the solution is unique if and only if every column in the coefficient matrix is a pivot column; otherwise there are infinitely many solutions. ” This statement is true because the free variables correspond to nonpivot columns of the coefficient matrix. The columns are all pivot columns if and only if there are no free variables. And there are no free variables if and only if the solution is unique, by Theorem 2. 28. Every column in the augmented matrix except the rightmost column is a pivot column, and the rightmost column is not a pivot column. 29. An underdetermined system always has more variables than equations. There cannot be more basic variables than there are equations, so there must be at least one free variable. Such a variable may be assigned infinitely many different values. If the system is consistent, each different value of a free variable will produce a different solution. 30. Example: x1 + x2 2 x1 + 2 x2 + x3 + 2 x3 = 4 = 5 31. Yes, a system of linear equations with more equations than unknowns can be consistent. Copyright © 2016 Pearson Education, Ltd. 1.2 x1 + x2 x2 Example (in which x1 = x2 = 1): x1 − 3 x1 + 2 x2 • Solutions 1-13 = 2 = 0 = 5 32. According to the numerical note in Section 1.2, when n = 30 the reduction to echelon form takes about 2(30)3/3 = 18,000 flops, while further reduction to reduced echelon form needs at most (30)2 = 900 flops. Of the total flops, the “backward phase” is about 900/18900 = .048 or about 5%. When n = 300, the estimates are 2(300)3/3 = 18,000,000 phase for the reduction to echelon form and (300)2 = 90,000 flops for the backward phase. The fraction associated with the backward phase is about (9×104) /(18×106) = .005, or about .5%. 33. For a quadratic polynomial p(t) = a0 + a1t + a2t2 to exactly fit the data (1, 12), (2, 15), and (3, 16), the coefficients a0, a1, a2 must satisfy the systems of equations given in the text. Row reduce the augmented matrix: 1 1 1 1 2 3 1 ~ 0 0 1 12 1 4 15 ~ 0 9 16 0 0 0 7 1 0 6 0 1 −1 1 1 12 1 1 3 3 ~ 0 2 8 4 0 1 1 0 1 3 2 12 1 3 ~ 0 −2 0 1 1 0 1 3 1 12 1 3 ~ 0 −1 0 1 1 0 0 0 1 13 6 −1 The polynomial is p(t) = 7 + 6t – t 2. 34. [M] The system of equations to be solved is: a0 + a1 ⋅ 0 + a2 ⋅ 02 + a3 ⋅ 03 + a4 ⋅ 04 + a5 ⋅ 05 = 0 a0 + a1 ⋅ 2 + a2 ⋅ 2 2 + a3 ⋅ 2 3 + a4 ⋅ 2 4 + a5 ⋅ 2 5 = 2.90 a0 + a1 ⋅ 4 + a2 ⋅ 42 + a3 ⋅ 43 + a4 ⋅ 44 + a5 ⋅ 45 = 14.8 a0 + a1 ⋅ 6 + a2 ⋅ 62 + a3 ⋅ 63 + a4 ⋅ 64 + a5 ⋅ 65 = 39.6 a0 + a1 ⋅ 8 + a2 ⋅ 8 + a3 ⋅ 8 + a4 ⋅ 8 + a5 ⋅ 8 = 74.3 a0 + a1 ⋅ 10 + a2 ⋅ 102 2 3 + a3 ⋅ 103 4 + a4 ⋅ 104 5 + a5 ⋅ 105 = 119 The unknowns are a0, a1, …, a5. Use technology to compute the reduced echelon of the augmented matrix: 1 0 1 2 1 4 1 6 1 8 1 10 0 4 16 36 64 0 8 64 216 512 0 16 256 1296 4096 0 32 1024 7776 32768 102 103 104 105 0 1 2.9 0 14.8 0 ~ 39.6 0 74.3 0 119 0 0 2 0 0 0 0 0 4 8 24 48 80 0 8 48 192 480 960 Copyright © 2016 Pearson Education, Ltd. 0 16 224 1248 4032 9920 0 32 960 7680 32640 99840 0 2.9 9 30.9 62.7 104.5 1-14 CHAPTER 1 • Linear Equations in Linear Algebra 1 0 0 ~ 0 0 0 0 2 0 0 0 0 0 4 8 0 0 0 0 8 48 48 192 480 1 0 0 ~ 0 0 0 0 2 0 0 0 0 0 4 8 0 0 0 0 8 48 48 0 0 0 16 224 576 384 0 0 32 960 4800 7680 3840 1 0 0 ~ 0 0 0 0 2 0 0 0 0 0 4 8 0 0 0 0 8 48 48 0 0 0 16 224 576 384 0 0 0 0 0 0 1 0 16 224 576 2688 7680 0 32 960 4800 26880 90240 0 1 2.9 0 9 0 ~ 3.9 0 8.7 0 14.5 0 0 1 2.9 0 9 0 ~ 3.9 0 −6.9 0 10 0 0 2 0 0 0 0 0 4 8 0 0 0 0 2 0 0 0 0 0 4 8 0 0 0 0 8 48 48 0 0 0 1 0 2.8167 0 6.5000 ~3~ −8.6000 0 0 −26.900 .002604 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 8 16 48 224 48 576 0 384 0 1920 0 16 224 576 384 0 0 0 0 1 0 0 0 0 0 0 1 0 0 32 960 4800 7680 42240 0 2.9 9 3.9 −6.9 −24.5 0 0 32 2.9 960 9 4800 3.9 7680 −6.9 1 .0026 0 0 0 0 0 1 0 1.7125 −1.1948 .6615 −.0701 .0026 Thus p(t) = 1.7125t – 1.1948t2 + .6615t3 – .0701t4 + .0026t5, and p(7.5) = 64.6 hundred lb. Notes: In Exercise 34, if the coefficients are retained to higher accuracy than shown here, then p(7.5) = 64.8. If a polynomial of lower degree is used, the resulting system of equations is overdetermined. The augmented matrix for such a system is the same as the one used to find p, except that at least column 6 is missing. When the augmented matrix is row reduced, the sixth row of the augmented matrix will be entirely zero except for a nonzero entry in the augmented column, indicating that no solution exists. Exercise 34 requires 25 row operations. It should give students an appreciation for higher-level commands such as gauss and bgauss, discussed in Section 1.4 of the Study Guide. The command ref (reduced echelon form) is available, but I recommend postponing that command until Chapter 2. The Study Guide includes a “Mathematical Note” about the phrase, “if and only if,” used in Theorem 2. 1.3 SOLUTIONS Notes: The key exercises are 11–14, 17–22, 25, and 26. A discussion of Exercise 25 will help students understand the notation [a1 a2 a3], {a1, a2, a3}, and Span{a1, a2, a3}. −1 −3 −1 + (−3) −4 1. u + v= + = = . 2 −1 2 + (−1) 1 Using the definitions carefully, −1 −3 −1 (−2)(−3) −1 + 6 5 = = u − 2= v + (−2) = + , or, more quickly 2 −1 2 (−2)(−1) 2 + 2 4 Copyright © 2016 Pearson Education, Ltd. 1.9 • Solutions −1 −3 −1 + 6 5 u − 2 v = − 2 = = . The intermediate step is often not written. 2 −1 2 + 2 4 3 2 3 + 2 5 2. u + v= + = = . 2 −1 2 + (−1) 1 Using the definitions carefully, 3 2 3 (−2)(2) 3 + (−4) −1 u − 2 v= + (−2) = = = + , or, more quickly, 2 −1 2 (−2)(−1) 2 + 2 4 3 2 3 − 4 −1 u − 2v = − 2 = = . The intermediate step is often not written. 2 −1 2 + 2 4 3. 4. 6 −3 1 6 x1 −3 x2 1 7 , 5. x1 −1 + x2 4 = −7 , − x1 + 4 x2 =− 5 0 −5 5 x1 0 −5 6 x1 − 3 x2 = 1 6 x1 − 3 x2 1 − x1 + 4 x2 = −7 5 x1 −5 − x1 + 4 x2 = −7 5 x1 = −5 Usually the intermediate steps are not displayed. −2 x1 8 x2 x3 0 −2 8 1 0 , 6. x1 + x2 + x3 = = + + , 3 5 −6 0 3 x1 5 x2 −6 x3 0 −2 x2 + 8 x2 + x3 = 0 3 x1 + 5 x2 − 6 x3 = 0 Usually the intermediate steps are not displayed. Copyright © 2016 Pearson Education, Ltd. −2 x1 + 8 x2 + x3 0 3x + 5 x − 6 x = 0 2 3 1 1-15 1-16 CHAPTER 1 • Linear Equations in Linear Algebra 7. See the figure below. Since the grid can be extended in every direction, the figure suggests that every vector in 2 can be written as a linear combination of u and v. To write a vector a as a linear combination of u and v, imagine walking from the origin to a along the grid "streets" and keep track of how many "blocks" you travel in the u-direction and how many in the v-direction. a. To reach a from the origin, you might travel 1 unit in the u-direction and –2 units in the vdirection (that is, 2 units in the negative v-direction). Hence a = u – 2v. b. To reach b from the origin, travel 2 units in the u-direction and –2 units in the v-direction. So b = 2u – 2v. Or, use the fact that b is 1 unit in the u-direction from a, so that b = a + u = (u – 2v) + u = 2u – 2v c. The vector c is –1.5 units from b in the v-direction, so c = b – 1.5v = (2u – 2v) – 1.5v = 2u – 3.5v d. The “map” suggests that you can reach d if you travel 3 units in the u-direction and –4 units in the v-direction. If you prefer to stay on the paths displayed on the map, you might travel from the origin to –3v, then move 3 units in the u-direction, and finally move –1 unit in the v-direction. So d = –3v + 3u – v = 3u – 4v Another solution is d = b – 2v + u = (2u – 2v) – 2v + u = 3u – 4v Figure for Exercises 7 and 8 8. See the figure above. Since the grid can be extended in every direction, the figure suggests that every vector in R2 can be written as a linear combination of u and v. w. To reach w from the origin, travel –1 units in the u-direction (that is, 1 unit in the negative u-direction) and travel 2 units in the v-direction. Thus, w = (–1)u + 2v, or w = 2v – u. x. To reach x from the origin, travel 2 units in the v-direction and –2 units in the u-direction. Thus, x = –2u + 2v. Or, use the fact that x is –1 units in the u-direction from w, so that x = w – u = (–u + 2v) – u = –2u + 2v y. The vector y is 1.5 units from x in the v-direction, so y = x + 1.5v = (–2u + 2v) + 1.5v = –2u + 3.5v z. The map suggests that you can reach z if you travel 4 units in the v-direction and –3 units in the u-direction. So z = 4v – 3u = –3u + 4v. If you prefer to stay on the paths displayed on the “map,” you might travel from the origin to –2u, then 4 units in the v-direction, and finally move –1 unit in the u-direction. So z = –2u + 4v – u = –3u + 4v Copyright © 2016 Pearson Education, Ltd. 1.9 9. 4 x1 − x1 + 5 x3 − x3 − 8 x3 x2 + 6 x2 + 3 x2 0 = 0, = 0 = • Solutions 1-17 x2 + 5 x3 0 4x + 6x − x = 0 2 3 1 − x1 + 3 x2 − 8 x3 0 0 1 5 0 0 x1 4 + x2 6 + x3 −1 = −1 3 −8 0 0 x2 5 x3 0 4 x + 6 x + − x = 0 , 3 1 2 − x1 3 x2 −8 x3 0 Usually, the intermediate calculations are not displayed. Note: The Study Guide says, “Check with your instructor whether you need to “show work” on a problem such as Exercise 9.” 4 x1 + x2 10. x1 − 7 x2 8 x1 + 6 x2 + 3 x3 − 2 x3 − 5 x3 = 9 = 2 , = 15 4 x1 + x2 + 3 x3 9 x − 7x − 2x = 2 2 3 1 8 x1 + 6 x2 − 5 x3 15 4 1 3 9 2 x1 1 + x2 −7 + x3 −2 = 8 6 −5 15 4 x1 x2 3 x3 9 x + −7 x + −2 x = 2 , 2 3 1 8 x1 6 x2 −5 x3 15 Usually, the intermediate calculations are not displayed. 11. The question Is b a linear combination of a1, a2, and a3? is equivalent to the question Does the vector equation x1a1 + x2a2 + x3a3 = b have a solution? The equation 1 0 5 2 x1 −2 + x2 1 + x3 −6 =−1 0 2 8 6 ↑ a1 ↑ a2 ↑ a3 (*) ↑ b has the same solution set as the linear system whose augmented matrix is 1 −2 M= 0 0 1 2 5 −6 8 2 −1 6 Row reduce M until the pivot positions are visible: 1 M ~ 0 0 0 1 2 5 4 8 2 1 3 ~ 0 6 0 0 1 0 5 4 0 2 3 0 The linear system corresponding to M has a solution, so the vector equation (*) has a solution, and therefore b is a linear combination of a1, a2, and a3. Copyright © 2016 Pearson Education, Ltd. 1-18 CHAPTER 1 • Linear Equations in Linear Algebra 12. The equation 1 0 2 −5 11 x1 −2 + x2 5 + x3 0 = 2 5 8 −7 ↑ ↑ ↑ ↑ a1 a2 a3 (*) b has the same solution set as the linear system whose augmented matrix is 1 0 2 −5 M = −2 5 0 11 2 5 8 −7 Row reduce M until the pivot positions are visible: 1 M ~ 0 0 0 5 5 2 4 4 −5 1 1 ~ 0 3 0 0 5 0 2 4 0 −5 1 2 The linear system corresponding to M has no solution, so the vector equation (*) has no solution, and therefore b is not a linear combination of a1, a2, and a3. 13. Denote the columns of A by a1, a2, a3. To determine if b is a linear combination of these columns, use the boxed fact on page 30. Row reduce the augmented matrix until you reach echelon form: 2 3 1 −4 2 3 1 −4 0 3 5 −7 ~ 0 3 5 −7 −2 8 −4 −3 0 0 0 3 The system for this augmented matrix is inconsistent, so b is not a linear combination of the columns of A. 1 −2 −6 11 1 −2 −6 11 3 7 −5 ~ 0 3 7 −5 . The linear system corresponding to this 14. [a1 a2 a3 b] = 0 1 −2 5 9 0 0 11 −2 matrix has a solution, so b is a linear combination of the columns of A. 15. Noninteger weights are acceptable, of course, but some simple choices are 0∙v1 + 0∙v2 = 0, and 7 1∙v1 + 0∙v2 = 1 , 0∙v1 + 1∙v2 = −6 −5 3 0 2 1∙v1 + 1∙v2 = 4 , 1∙v1 – 1∙v2 = −6 12 −2 −6 16. Some likely choices are 0∙v1 + 0∙v2 = 0, and Copyright © 2016 Pearson Education, Ltd. 1.9 3 1∙v1 + 0∙v2 = 0 , 0∙v1 + 1∙v2 = 2 • Solutions 1-19 −2 0 3 1 5 1∙v1 + 1∙v2 = 0 , 1∙v1 – 1∙v2 = 0 5 −1 4 1 −2 4 1 1 −2 4 1 −2 5 −15 ~ 0 1 −3 ~ 0 17. [a1 a2 b] = 4 −3 1 ~ 0 7 h 0 3 h + 8 0 3 h + 8 0 −2 b is in Span{a1, a2} when h + 17 is zero, that is, when h = –17. h 1 −3 h 1 −3 1 −3 1 −5 ~ 0 1 1 −5 ~ 0 18. [v1 v2 y] = 0 8 −3 0 2 −3 + 2h 0 0 −2 Span{v1, v2} when 7 + 2h is zero, that is, when h = –7/2. −2 1 0 4 −3 . The vector h + 17 h −5 . The vector y is in 7 + 2h 19. By inspection, v2 = (3/2)v1. Any linear combination of v1 and v2 is actually just a multiple of v1. For instance, av1 + bv2 = av1 + b(3/2)v2 = (a + 3b/2)v1 So Span{v1, v2} is the set of points on the line through v1 and 0. Note: Exercises 19 and 20 prepare the way for ideas in Sections 1.4 and 1.7. 20. Span{v1, v2} is a plane in 3 through the origin, because neither vector in this problem is a multiple of the other. Every vector in the set has 0 as its second entry and so lies in the xz-plane in ordinary 3-space. So Span{v1, v2} is the xz-plane. h h 2 2 h 2 2 21. Let y = . Then [u v y] = ~ . This augmented matrix k −1 1 k 0 2 k + h / 2 corresponds to a consistent system for all h and k. So y is in Span{u, v} for all h and k. 22. Construct any 3×4 matrix in echelon form that corresponds to an inconsistent system. Perform sufficient row operations on the matrix to eliminate all zero entries in the first three columns. 23. a. False. The alternative notation for a (column) vector is (–4, 3), using parentheses and commas. −5 b. False. Plot the points to verify this. Or, see the statement preceding Example 3. If were on 2 −5 −2 −2 the line through and the origin, then would have to be a multiple of , which is 2 5 5 not the case. c. True. See the line displayed just before Example 4. d. True. See the box that discusses the matrix in (5). Copyright © 2016 Pearson Education, Ltd. 1-20 CHAPTER 1 • Linear Equations in Linear Algebra e. False. The statement is often true, but Span{u, v} is not a plane when v is a multiple of u, or when u is the zero vector. 24. a. True. See the beginning of the subsection “Vectors in n ”. b. True. Use Fig. 7 to draw the parallelogram determined by u – v and v. c. False. See the first paragraph of the subsection “Linear Combinations”. d. True. See the statement that refers to Fig. 11. e. True. See the paragraph following the definition of Span{v1, …, vp}. 25. a. There are only three vectors in the set {a1, a2, a3}, and b is not one of them. b. There are infinitely many vectors in W = Span{a1, a2, a3}. To determine if b is in W, use the method of Exercise 13. 4 1 1 0 −4 0 3 −2 1 ~ 0 −2 6 3 −4 0 ↑ ↑ ↑ ↑ a1 a 2 a3 b −4 −2 −5 0 3 6 4 1 1 ~ 0 4 0 −4 −2 −1 0 3 0 4 1 2 The system for this augmented matrix is consistent, so b is in W. c. a1 = 1a1 + 0a2 + 0a3. See the discussion in the text following the definition of Span{v1, …, vp}. 26. a. [a1 a2 a3 2 b] = −1 1 0 6 8 5 −2 1 10 1 3 ~ −1 3 1 0 3 8 5 −2 1 5 1 3 ~ 0 3 0 0 3 8 8 −2 −2 5 1 8 ~ 0 −2 0 5 0 3 8 8 8 0 0 0 Yes, b is a linear combination of the columns of A, that is, b is in W. b. The third column of A is in W because a3 = 0∙a1 + 0∙a2 + 1∙a3. 27. a. 5v1 is the output of 5 days’ operation of mine #1. 150 b. The total output is x1v1 + x2v2, so x1 and x2 should satisfy x1 v1 + x2 v 2 = 2825 . 30 150 1 0 1.5 20 c. [M] Reduce the augmented matrix ~ . 550 500 2825 0 1 4.0 Operate mine #1 for 1.5 days and mine #2 for 4 days. (This is the exact solution.) 28. a. The amount of heat produced when the steam plant burns x1 tons of anthracite and x2 tons of bituminous coal is 27.6x1 + 30.2x2 million Btu. b. The total output produced by x1 tons of anthracite and x2 tons of bituminous coal is given by the 27.6 30.2 vector x1 3100 + x2 6400 . 250 360 27.6 30.2 162 23,610 . c. [M] The appropriate values for x1 and x2 satisfy x1 3100 + x2 6400 = 250 360 1,623 Copyright © 2016 Pearson Education, Ltd. 1.9 27.6 To solve, row reduce the augmented matrix: 3100 250 • Solutions 162 1.000 23610 ~ 0 1623 0 30.2 6400 360 0 1.000 0 1-21 3.900 1.800 . 0 The steam plant burned 3.9 tons of anthracite coal and 1.8 tons of bituminous coal. 29. The total mass is 2 + 5 + 2 + 1 = 10. So v = (2v1 +5v2 + 2v3 + v4)/10. That is, 5 4 −4 −9 10 + 20 − 8 − 9 1 1 v= 2 −4 + 5 3 + 2 −3 + 8 = −8 + 15 − 6 + 8= 10 10 6 − 10 − 2 + 6 3 −2 −1 6 1.3 .9 0 m m1 1 + mk v k= v1 + + k v k . (m1 v1 + ) m m m The second expression displays v as a linear combination of v1, …, vk, which shows that v is in Span{v1, …, vk}. v 30. Let m be the total mass of the system. By definition, = 31. a. The center of mass is 8 2 10 / 3 1 0 1 ⋅ + 1 ⋅ + 1 ⋅ = . 3 1 1 4 2 b. The total mass of the new system is 9 grams. The three masses added, w1, w2, and w3, satisfy the 0 8 2 2 1 equation ( w1 + 1) ⋅ + ( w2 + 1) ⋅ + ( w3 + 1) ⋅ = , which can be rearranged to 9 1 1 4 2 0 8 2 18 0 8 2 8 ( w1 + 1) ⋅ 1 + ( w2 + 1) ⋅ 1 + ( w3 + 1) ⋅ 4 = 18 and w1 ⋅ 1 + w2 ⋅ 1 + w3 ⋅ 4 = 12 . The condition w1 + w2 + w3 = 6 and the vector equation above combine to produce a system of three equations whose augmented matrix is shown below, along with a sequence of row operations: 1 1 1 6 1 0 8 2 8 ~ 0 1 1 4 12 0 1 0 0 3.5 ~ 0 1 0 .5 0 0 1 2 1 8 0 1 2 3 6 1 8 ~ 0 6 0 1 8 0 1 2 1 6 1 8 ~ 0 2 0 1 8 0 0 0 1 4 1 4 ~ 0 2 0 0 8 0 0 0 1 3.5 4 2 Answer: Add 3.5 g at (0, 1), add .5 g at (8, 1), and add 2 g at (2, 4). Extra problem: Ignore the mass of the plate, and distribute 6 gm at the three vertices to make the center of mass at (2, 2). Answer: Place 3 g at (0, 1), 1 g at (8, 1), and 2 g at (2, 4). 32. See the parallelograms drawn on the figure from the text. Here c1, c2, c3, and c4 are suitable scalars. The darker parallelogram shows that b is a linear combination of v1 and v2, that is c1v1 + c2v2 + 0∙v3 = b. The larger parallelogram shows that b is a linear combination of v1 and v3, that is, c4v1 + 0∙v2 + c3v3 = b. Copyright © 2016 Pearson Education, Ltd. 1-22 CHAPTER 1 • Linear Equations in Linear Algebra So the equation x1v1 + x2v2 + x3v3 = b has at least two solutions, not just one solution. (In fact, the equation has infinitely many solutions.) 33. a. For j = 1,…, n, the jth entry of (u + v) + w is (uj + vj) + wj. By associativity of addition in , this entry equals uj + (vj + wj), which is the jth entry of u + (v + w). By definition of equality of vectors, (u + v) + w = u + (v + w). b. For any scalar c, the jth entry of c(u + v) is c(uj + vj), and the jth entry of cu + cv is cuj + cvj (by definition of scalar multiplication and vector addition). These entries are equal, by a distributive law in . So c(u + v) = cu + cv. 34. a. For j = 1,…, n, uj + (–1)uj = (–1)uj + uj = 0, by properties of . By vector equality, u + (–1)u = (–1)u + u = 0. b. For scalars c and d, the jth entries of c(du) and (cd )u are c(duj) and (cd )uj, respectively. These entries in are equal, so the vectors c(du) and (cd)u are equal. Note: When an exercise in this section involves a vector equation, the corresponding technology data (in the data files on the web) is usually presented as a set of (column) vectors. To use MATLAB or other technology, a student must first construct an augmented matrix from these vectors. The MATLAB note in the Study Guide describes how to do this. The appendices in the Study Guide give corresponding information about Maple, Mathematica, and the TI and HP calculators. 1.4 SOLUTIONS Notes: Key exercises are 1–20, 27, 28, 31 and 32. Exercises 29, 30, 33, and 34 are harder. Exercise 34 anticipates the Invertible Matrix Theorem but is not used in the proof of that theorem. 1. The matrix-vector product Ax is not defined because the number of columns (2) in the 3×2 matrix 3 −4 2 1 6 does not match the number of entries (3) in the vector −2 . 7 0 1 Copyright © 2016 Pearson Education, Ltd. 1.9 • Solutions 2. The matrix-vector product Ax is not defined because the number of columns (1) in the 3×1 matrix 2 6 does not match the number of entries (2) in the vector 5 . −1 −1 5 6 5 12 −15 −3 2 −3 =2 −4 − 3 −3 = −8 + 9 = 1 , and −3 7 6 14 −18 −4 6 6 3. Ax = −4 7 6 Ax = −4 7 8 4. Ax = 5 5 2 −3 = −3 6 3 1 6 ⋅ 2 + 5 ⋅ (−3) (−4) ⋅ 2 + (−3) ⋅ (−3) = 7 ⋅ 2 + 6 ⋅ (−3) −3 1 −4 1 −4 8 3 −4 8 + 3 − 4 7 1 = 1⋅ + 1⋅ + 1⋅ = = , and 2 5 1 2 5 + 1 + 2 8 1 1 8 3 −4 8 ⋅ 1 + 3 ⋅ 1 + (−4) ⋅ 1 7 Ax = 1 = = 2 5 ⋅ 1 + 1 ⋅ 1 + 2 ⋅ 1 8 5 1 1 5. On the left side of the matrix equation, use the entries in the vector x as the weights in a linear combination of the columns of the matrix A: 5 1 −8 4 −8 5 ⋅ − 1 ⋅ + 3 ⋅ − 2 ⋅ = −2 −7 3 −5 16 6. On the left side of the matrix equation, use the entries in the vector x as the weights in a linear combination of the columns of the matrix A: 7 −3 1 2 1 −9 −2 ⋅ − 5 ⋅ = 9 −6 12 −3 2 −4 7. The left side of the equation is a linear combination of three vectors. Write the matrix A whose columns are those three vectors, and create a variable vector x with three entries: 4 −5 7 −1 3 −8 = A = 7 −5 0 −4 1 2 4 −1 7 −4 −5 3 −5 1 7 x1 −8 and x = x2 . Thus the equation Ax = b is 0 x3 2 Copyright © 2016 Pearson Education, Ltd. 1-23 1-24 CHAPTER 1 −5 3 −5 1 4 −1 7 −4 • Linear Equations in Linear Algebra 7 6 x1 −8 −8 x = 0 2 0 x3 2 −7 For your information: The unique solution of this equation is (5, 7, 3). Finding the solution by hand would be time-consuming. Note: The skill of writing a vector equation as a matrix equation will be important for both theory and application throughout the text. See also Exercises 27 and 28. 8. The left side of the equation is a linear combination of four vectors. Write the matrix A whose columns are those four vectors, and create a variable vector with four entries: 4 −4 −5 3 4 A = −2 5 4 0 −2 4 is −2 −4 5 −5 4 −4 5 −5 4 z1 z 3 2 . Then the equation Az = b = z , and z 3 0 z4 z1 3 z2 4 = . 0 z3 13 z4 For your information: One solution is (7, 3, 3, 1). The general solution is z1 = 6 + .75z3 – 1.25z4, z2 = 5 – .5z3 – .5z4, with z3 and z4 free. 9. The system has the same solution set as the vector equation 3 1 −5 9 x1 + x2 + x3 = 0 1 4 0 and this equation has the same solution set as the matrix equation 3 1 0 1 x −5 1 9 x = 4 2 0 x3 10. The system has the same solution set as the vector equation 8 −1 4 1 x1 5 + x2 4 = 1 −3 2 and this equation has the same solution set as the matrix equation 8 5 1 −1 4 x1 4 = 1 x2 −3 2 Copyright © 2016 Pearson Education, Ltd. 1.9 • Solutions 1-25 11. To solve Ax = b, row reduce the augmented matrix [a1 a2 a3 b] for the corresponding linear system: 1 0 −2 2 1 −4 4 5 −3 −2 1 2 ~ 0 9 0 2 1 0 4 5 5 −2 1 2 ~ 0 5 0 2 1 0 4 5 1 −2 1 2 ~ 0 1 0 x1 = 0 The solution is x2 = −3 . As a vector, the solution is x = x = 1 3 2 1 0 x1 x = 2 x3 0 0 1 −6 1 −3 ~ 0 1 0 0 1 0 0 0 1 0 −3 1 0 −3 . 1 12. To solve Ax = b, row reduce the augmented matrix [a1 a2 a3 b] for the corresponding linear system: 1 −3 0 2 −1 5 1 2 3 0 1 1 ~ 0 −1 0 2 5 5 1 5 3 0 1 1 ~ 0 −1 0 1 ~ 0 0 2 5 0 0 0 1 −1 1 −4 ~ 0 1 0 2 1 0 0 0 1 −1 1 −4 / 5 ~ 0 1 0 2 5 0 1 5 −2 0 1 0 0 0 1 0 1 1 ~ 0 −2 0 2 5 0 1 5 1 0 1 1 3/ 5 −4 / 5 1 x1 = 3/ 5 The solution is x2 = −4 / 5 . As a vector, the solution is x = x = 1 3 x1 x = 2 x3 3/ 5 . −4 / 5 1 13. The vector u is in the plane spanned by the columns of A if and only if u is a linear combination of the columns of A. This happens if and only if the equation Ax = u has a solution. (See the box preceding Example 3 in Section 1.4.) To study this equation, reduce the augmented matrix [A u] 1 4 1 1 4 1 1 4 3 −5 0 1 −2 6 4 ~ −2 6 4 ~ 0 8 12 ~ 0 8 12 1 1 4 3 −5 0 0 −8 −12 0 0 0 The equation Ax = u has a solution, so u is in the plane spanned by the columns of A. For your information: The unique solution of Ax = u is (5/2, 3/2). 14. Reduce the augmented matrix [A u] to echelon form: 7 2 1 3 0 2 1 3 0 2 1 3 0 2 5 8 0 1 −1 −3 ~ 0 1 −1 −3 ~ 0 1 −1 −3 ~ 0 1 −1 −3 1 3 0 2 5 8 7 2 0 −7 7 −8 0 0 0 −29 The equation Ax = u has no solution, so u is not in the subset spanned by the columns of A. b1 2 −1 b1 2 −1 , which is row equivalent to 15. The augmented matrix for Ax = b is 3 b2 0 b2 + 3b1 −6 0 . This shows that the equation Ax = b is not consistent when 3b1 + b2 is nonzero. The set of b for Copyright © 2016 Pearson Education, Ltd. 1-26 CHAPTER 1 • Linear Equations in Linear Algebra which the equation is consistent is a line through the origin–the set of all points (b1, b2) satisfying b2 = –3b1. b1 1 −3 −4 b . 16. Row reduce the augmented matrix [A b]: A = 2 6 , b = 2 −3 b3 5 −1 −8 1 −3 −4 −3 2 6 5 −1 −8 1 −3 −4 = 0 −7 −6 0 0 0 b1 1 b2 ~ 0 b3 0 −3 −7 14 −4 −6 12 b1 1 b2 + 3b1 ~ 0 b3 − 5b1 0 −3 −7 0 −4 b1 −6 b2 + 3b1 0 b3 − 5b1 + 2(b2 + 3b1 ) b1 b2 + 3b1 b1 + 2b2 + b3 The equation Ax = b is consistent if and only if b1 + 2b2 + b3 = 0. The set of such b is a plane through the origin in 3 . 17. Row reduction shows that only three rows of A contain a pivot position: 1 −1 A= 0 2 3 −1 −4 0 0 −1 2 3 3 1 1 0 ~ −8 0 −1 0 3 2 −4 −6 0 −1 2 3 3 1 4 0 ~ −8 0 −7 0 3 2 0 0 0 −1 0 0 3 1 4 0 ~ 0 0 5 0 3 2 0 0 0 −1 0 0 3 4 5 0 Because not every row of A contains a pivot position, Theorem 4 in Section 1.4 shows that the equation Ax = b does not have a solution for each b in 4 . 18. Row reduction shows that only three rows of B contain a pivot position: 1 0 B= 1 −2 3 1 2 −8 −2 1 −3 2 2 1 −5 0 ~ 7 0 −1 0 3 1 −1 −2 −2 1 −1 −2 2 1 −5 0 ~ 5 0 3 0 3 1 0 0 −2 1 0 0 2 1 −5 0 ~ 0 0 −7 0 3 1 0 0 −2 1 0 0 2 −5 −7 0 Because not every row of B contains a pivot position, Theorem 4 in Section 1.4 shows that the equation Bx = y does not have a solution for each y in 4 . 19. The work in Exercise 17 shows that statement (d) in Theorem 4 is false. So all four statements in Theorem 4 are false. Thus, not all vectors in 4 can be written as a linear combination of the columns of A. Also, the columns of A do not span 4 . 20. The work in Exercise 18 shows that statement (d) in Theorem 4 is false. So all four statements in Theorem 4 are false. Thus, not all vectors in 4 can be written as a linear combination of the columns of B. The columns of B certainly do not span 3 , because each column of B is in 4 , not 3 . (This question was asked to alert students to a fairly common misconception among students who are just learning about spanning.) 21. Row reduce the matrix [v1 v2 v3] to determine whether it has a pivot in each row. Copyright © 2016 Pearson Education, Ltd. 1.9 1 0 −1 0 0 −1 0 1 1 1 0 0 ~ 0 0 −1 0 0 −1 0 1 1 1 0 0 ~ 1 0 −1 0 0 −1 0 0 1 1 0 0 ~ 1 0 −1 0 0 1 0 0 • Solutions 1-27 1 0 . 1 0 The matrix [v1 v2 v3] does not have a pivot in each row, so the columns of the matrix do not span 4 , by Theorem 4. That is, {v1, v2, v3} does not span 4 . Note: Some students may realize that row operations are not needed, and thereby discover the principle covered in Exercises 31 and 32. 22. Row reduce the matrix [v1 v2 v3] to determine whether it has a pivot in each row. 0 0 −2 0 −3 8 4 −2 −1 ~ 0 −5 0 8 −3 0 −5 −1 4 The matrix [v1 v2 v3] has a pivot in each row, so the columns of the matrix span 3 , by Theorem 4. That is, {v1, v2, v3} spans 3 . 23. a. b. c. d. e. f. False. See the paragraph following equation (3). The text calls Ax = b a matrix equation. True. See the box before Example 3. False. See the warning following Theorem 4. True. See Example 4. True. See parts (c) and (a) in Theorem 4. True. In Theorem 4, statement (a) is false if and only if statement (d) is also false. 24. a. True. This statement is in Theorem 3. However, the statement is true without any “proof” because, by definition, Ax is simply a notation for x1a1 + ⋅ ⋅ ⋅ + xnan, where a1, …, an are the columns of A. b. True. See Example 2. c. True, by Theorem 3. d. True. See the box before Example 1. Saying that b is not in the set spanned by the columns of A is the same a saying that b is not a linear combination of the columns of A. e. False. See the warning that follows Theorem 4. f. True. In Theorem 4, statement (c) is false if and only if statement (a) is also false. 25. By definition, the matrix-vector product on the left is a linear combination of the columns of the matrix, in this case using weights –3, –1, and 2. So c1 = –3, c2 = –1, and c3 = 2. 26. The equation in x1 and x2 involves the vectors u, v, and w, and it may be viewed as x1 v ] = w. By definition of a matrix-vector product, x1u + x2v = w. The stated fact that x2 3u – 5v – w = 0 can be rewritten as 3u – 5v = w. So, a solution is x1 = 3, x2 = –5. [u 27. Place the vectors q1, q2, and q3 into the columns of a matrix, say, Q and place the weights x1, x2, and x3 into a vector, say, x. Then the vector equation becomes Copyright © 2016 Pearson Education, Ltd. 1-28 CHAPTER 1 • Linear Equations in Linear Algebra Qx = v, where Q = [q1 q2 x1 q3] and x = x2 x3 Note: If your answer is the equation Ax = b, you need to specify what A and b are. 28. The matrix equation can be written as c1v1 + c2v2 + c3v3 + c4v4 + c5v5 = v6, where −3 5 −4 9 c1 = –3, c2 = 2, c3 = 4, c4 = –1, c5 = 2, and = v1 = , v 2 = , v3 = , v4 , 5 8 1 −2 7 8 = v 5 = , v6 . −4 −1 29. Start with any 3×3 matrix B in echelon form that has three pivot positions. Perform a row operation (a row interchange or a row replacement) that creates a matrix A that is not in echelon form. Then A has the desired property. The justification is given by row reducing A to B, in order to display the pivot positions. Since A has a pivot position in every row, the columns of A span 3 , by Theorem 4. 30. Start with any nonzero 3×3 matrix B in echelon form that has fewer than three pivot positions. Perform a row operation that creates a matrix A that is not in echelon form. Then A has the desired property. Since A does not have a pivot position in every row, the columns of A do not span 3 , by Theorem 4. 31. A 3×2 matrix has three rows and two columns. With only two columns, A can have at most two pivot columns, and so A has at most two pivot positions, which is not enough to fill all three rows. By Theorem 4, the equation Ax = b cannot be consistent for all b in 3 . Generally, if A is an m×n matrix with m > n, then A can have at most n pivot positions, which is not enough to fill all m rows. Thus, the equation Ax = b cannot be consistent for all b in 3 . 32. A set of three vectors in cannot span 4 . Reason: the matrix A whose columns are these three vectors has four rows. To have a pivot in each row, A would have to have at least four columns (one for each pivot), which is not the case. Since A does not have a pivot in every row, its columns do not span 4 , by Theorem 4. In general, a set of n vectors in m cannot span m when n is less than m. 33. If the equation Ax = b has a unique solution, then the associated system of equations does not have any free variables. If every variable is a basic variable, then each column of A is a pivot column. So 1 0 0 0 1 0 . the reduced echelon form of A must be 0 0 1 0 0 0 Note: Exercises 33 and 34 are difficult in the context of this section because the focus in Section 1.4 is on existence of solutions, not uniqueness. However, these exercises serve to review ideas from Section 1.2, and they anticipate ideas that will come later. Copyright © 2016 Pearson Education, Ltd. 1.9 • Solutions 1-29 34. If the equation Ax = b has a unique solution, then the associated system of equations does not have any free variables. If every variable is a basic variable, then each column of A is a pivot column. So 1 0 0 the reduced echelon form of A must be 0 1 0 . Now it is clear that A has a pivot position in each 0 0 1 row. By Theorem 4, the columns of A span 3 . 35. Given Ax1 = y1 and Ax2 = y2, you are asked to show that the equation Ax = w has a solution, where w = y1 + y2. Observe that w = Ax1 + Ax2 and use Theorem 5(a) with x1 and x2 in place of u and v, respectively. That is, w = Ax1 + Ax2 = A(x1 + x2). So the vector x = x1 + x2 is a solution of w = Ax. 36. Suppose that y and z satisfy Ay = z. Then 4z = 4Ay. By Theorem 5(b), 4Ay = A(4y). So 4z = A(4y), which shows that 4y is a solution of Ax = 4z. Thus, the equation Ax = 4z is consistent. −5 −5 2 8 7 2 8 7 2 −5 8 7 −5 −3 4 −9 0 −11/ 7 3/ 7 −23/ 7 0 −11/ 7 −23/ 7 3/ 7 ~ ~ 37. [M] 6 10 −2 7 0 58 / 7 16 / 7 1/ 7 0 0 50 /11 −189 /11 −3 11 23 0 0 0 0 −7 9 2 15 0 7 0 or, approximately 0 0 −5 .429 4.55 0 8 −3.29 , to three significant figures. The original matrix does −17.2 0 not have a pivot in every row, so its columns do not span 4 , by Theorem 4. 2 −1.57 0 0 9 5 −7 −4 9 5 −7 −4 9 5 −7 −4 6 −8 −7 5 0 2 / 5 −11/ 5 −29 / 5 0 2 / 5 −11/ 5 −29 / 5 ~ ~ 38. [M] 4 −4 −9 −9 0 8 / 5 −29 / 5 −81/ 5 0 0 3 7 7 0 −8 / 5 44 / 5 116 / 5 0 0 * * −9 11 16 MATLAB shows starred entries for numbers that are essentially zero (to many decimal places) and in fact they are zero. So, with pivots only in the first three rows, the original matrix has columns that do not span 4 , by Theorem 4. 12 −9 39. [M] −6 4 12 0 ~ 0 0 −7 4 11 −6 −7 −5 / 4 0 0 11 −8 −7 10 11 1/ 4 0 28 / 5 −9 7 3 −5 5 12 −3 0 ~ −9 0 12 0 −9 1/ 4 0 −41/15 −7 −5 / 4 15 / 2 −11/ 3 5 12 3/ 4 0 ~ −2 0 122 /15 0 11 1/ 4 −3/ 2 19 / 3 −7 −5 / 4 0 0 −9 1/ 4 −3/ 2 −2 11 1/ 4 28 / 5 0 5 3/ 4 −13/ 2 31/ 3 −9 1/ 4 −41/15 0 5 3/ 4 122 /15 −2 The original matrix has a pivot in every row, so its columns span 4 , by Theorem 4. Copyright © 2016 Pearson Education, Ltd. 1-30 CHAPTER 1 8 −7 40. [M] 11 −3 8 0 ~ 0 0 • Linear Equations in Linear Algebra 11 −8 7 4 11 13/ 8 0 0 −6 5 −7 1 −6 −1/ 4 0 0 −7 6 −9 8 −7 −1/ 8 0 6 13 8 −9 0 ~ −6 0 7 0 11 13/ 8 −65 / 8 65 / 8 13 8 19 / 8 0 ~ −12 0 0 0 −6 −1/ 4 5/ 4 −5 / 4 11 13/ 8 0 0 −7 −1/ 8 5/8 43/ 8 −6 −1/ 4 0 0 13 19 / 8 −191/ 8 95 / 8 −7 −1/ 8 6 0 13 19 / 8 0 −12 The original matrix has a pivot in every row, so its columns span 4 , by Theorem 4. 41. [M] Examine the calculations in Exercise 39. Notice that the fourth column of the original matrix, say A, is not a pivot column. Let C be the matrix formed by deleting column 4 of A, let B be the echelon form obtained from A, and let D be the matrix obtained by deleting column 4 of B. The sequence of row operations that reduces A to B also reduces C to D. Since D is in echelon form, it shows that C has a pivot position in each row. Therefore, the columns of C span 4 . It is possible to delete columns 1, 2.or 3 of A instead of column 4. In this case, the fourth column of A becomes a pivot column of C, as you can see by looking at what happens when column 3 of B is deleted. For later work, it is desirable to delete a nonpivot column. Note: Exercises 41 and 42 help to prepare for later work on the column space of a matrix. (See Section 2.9 or 4.6.) The Study Guide points out that these exercises depend on the following idea, not explicitly mentioned in the text: when a row operation is performed on a matrix A, the calculations for each new entry depend only on the other entries in the same column. If a column of A is removed, forming a new matrix, the absence of this column has no affect on any row-operation calculations for entries in the other columns of A. (The absence of a column might affect the particular choice of row operations performed for some purpose, but that is not being considered here.) 42. [M] Examine the calculations in Exercise 40. The third column of the original matrix, say A, is not a pivot column. Let C be the matrix formed by deleting column 3 of A, let B be the echelon form obtained from A, and let D be the matrix obtained by deleting column 3 of B. The sequence of row operations that reduces A to B also reduces C to D. Since D is in echelon form, it shows that C has a pivot position in each row. Therefore, the columns of C span 4 . It is possible to delete column 1or 2 of A instead of column 3. (See the remark for Exercise 41.) However, only one column can be deleted. If two or more columns were deleted from A, the resulting matrix would have fewer than four columns, so it would have fewer than four pivot positions. In such a case, not every row could contain a pivot position, and the columns of the matrix would not span 4 , by Theorem 4. Notes: At the end of Section 1.4, the Study Guide gives students a method for learning and mastering linear algebra concepts. Specific directions are given for constructing a review sheet that connects the basic definition of “span” with related ideas: equivalent descriptions, theorems, geometric interpretations, special cases, algorithms, and typical computations. I require my students to prepare such a sheet that reflects their choices of material connected with “span”, and I make comments on their sheets to help them refine their review. Later, the students use these sheets when studying for exams. The MATLAB box for Section 1.4 introduces two useful commands gauss and bgauss that allow a student to speed up row reduction while still visualizing all the steps involved. The command B = gauss(A,1) causes MATLAB to find the left-most nonzero entry in row 1 of matrix A, and use Copyright © 2016 Pearson Education, Ltd. 1.9 • Solutions 1-31 that entry as a pivot to create zeros in the entries below, using row replacement operations. The result is a matrix that a student might write next to A as the first stage of row reduction, since there is no need to write a new matrix after each separate row replacement. I use the gauss command frequently in lectures to obtain an echelon form that provides data for solving various problems. For instance, if a matrix has 5 rows, and if row swaps are not needed, the following commands produce an echelon form of A: B = gauss(A,1), B = gauss(B,2), B = gauss(B,3), B = gauss(B,4) If an interchange is required, I can insert a command such as B = swap(B,2,5) . The command bgauss uses the left-most nonzero entry in a row to produce zeros above that entry. This command, together with scale, can change an echelon form into reduced echelon form. The use of gauss and bgauss creates an environment in which students use their computer program the same way they work a problem by hand on an exam. Unless you are able to conduct your exams in a computer laboratory, it may be unwise to give students too early the power to obtain reduced echelon forms with one command—they may have difficulty performing row reduction by hand during an exam. Instructors whose students use a graphing calculator in class each day do not face this problem. In such a case, you may wish to introduce rref earlier in the course than Chapter 4 (or Section 2.8), which is where I finally allow students to use that command. 1.5 SOLUTIONS Notes: The geometry helps students understand Span{u, v}, in preparation for later discussions of subspaces. The parametric vector form of a solution set will be used throughout the text. Figure 6 will appear again in Sections 2.9 and 4.8. For solving homogeneous systems, the text recommends working with the augmented matrix, although no calculations take place in the augmented column. See the Study Guide comments on Exercise 7 that illustrate two common student errors. All students need the practice of Exercises 1–14. (Assign all odd, all even, or a mixture. If you do not assign Exercise 7, be sure to assign both 8 and 10.) Otherwise, a few students may be unable later to find a basis for a null space or an eigenspace. Exercises 29–34 are important. Exercises 33 and 34 help students later understand how solutions of Ax = 0 encode linear dependence relations among the columns of A. Exercises 35–38 are more challenging. Exercise 37 will help students avoid the standard mistake of forgetting that Theorem 6 applies only to a consistent equation Ax = b. 1. Reduce the augmented matrix to echelon form and circle the pivot positions. If a column of the coefficient matrix is not a pivot column, the corresponding variable is free and the system of equations has a nontrivial solution. Otherwise, the system has only the trivial solution. 2 −2 4 −5 −7 2 8 1 7 0 2 0 ~ 0 0 0 −5 −12 12 8 9 −9 0 2 0 ~ 0 0 0 −5 −12 0 8 9 0 0 0 0 The variable x3 is free, so the system has a nontrivial solution. 1 2. −2 1 −3 1 2 7 −4 9 0 1 0 ~ 0 0 0 −3 7 −5 10 5 2 0 1 0 ~ 0 0 0 −3 7 −5 10 0 12 0 0 0 There is no free variable; the system has only the trivial solution. 5 −7 0 −3 5 −7 0 −3 3. . The variable x3 is free; the system has nontrivial ~ 1 0 0 −3 15 0 −6 7 solutions. An alert student will realize that row operations are unnecessary. With only two equations, Copyright © 2016 Pearson Education, Ltd. 1-32 CHAPTER 1 • Linear Equations in Linear Algebra there can be at most two basic variables. One variable must be free. Refer to Exercise 31 in Section 1.2 7 9 0 1 −2 6 0 1 −2 6 0 −5 . x3 is a free variable; the system has 4. ~ ~ 7 9 0 0 −3 39 0 1 −2 6 0 −5 nontrivial solutions. As in Exercise 3, row operations are unnecessary. 1 5. −4 0 x1 3 −9 −3 1 2 −6 0 1 0 ~ 0 0 0 3 3 −3 0 1 0 ~ 0 0 0 1 6 −6 0 3 0 −5 6 0 0 1 0 ~ 0 0 0 0 1 0 −5 2 0 0 0 0 − 5 x3 = 0 x2 + 2 x 3 = 0 . The variable x3 is free, x1 = 5x3, and x2 = –2x3. 0 = 0 x1 5 x3 5 In parametric vector form, the general solution is x = x3 −2 . x2 = −2 x3 = x3 x3 1 1 6. 1 −3 x1 −5 −8 9 3 4 −7 0 1 0 ~ 0 0 0 3 1 2 −5 −3 −6 0 1 0 ~ 0 0 0 3 1 0 −5 −3 0 0 1 0 4 −3 0 0 0 0 + 4 x3 = 0 x2 − 3 x 3 = 0 . The variable x3 is free, x1 = –4x3, and x2 = 3x3. 0 = 0 x1 In parametric vector form, the general solution is x2 = x = x3 1 7. 0 0 1 0 ~ 0 0 0 3 1 −3 −4 7 5 0 1 ~ 0 0 0 1 9 −4 −8 5 −4 x3 −4 3 x= x 3 . 3 3 x3 1 + 9 x3 − 8 x4 = 0 0 x1 . 0 x2 − 4 x3 + 5 x4 = 0 The basic variables are x1 and x2, with x3 and x4 free. Next, x1 = –9x3 + 8x4, and x2 = 4x3 – 5x4. The general solution is x1 −9 x3 + 8 x4 −9 x3 8 x4 −9 8 x 4 x − 5 x 4 x −5 x 4 −5 2 3 4 4 3 x = + = +x x= = x3 x3 0 3 1 4 0 x3 x4 0 1 x4 0 x4 1 8. 0 −2 1 −9 2 5 −6 0 1 ~ 0 0 0 1 −5 2 −7 −6 − 5 x3 − 7 x4 = 0 0 x1 . 0 x2 + 2 x3 − 6 x4 = 0 Copyright © 2016 Pearson Education, Ltd. 1.9 • Solutions 1-33 The basic variables are x1 and x2, with x3 and x4 free. Next, x1 = 5x3 + 7x4 and x2 = –2x3 + 6x4. The general solution in parametric vector form is x1 5 x3 + 7 x4 5 x3 7 x4 5 7 x −2 x + 6 x −2 x 6 x −2 6 2 3 4 4 3 x= = = + =x +x x3 x3 0 3 1 4 0 x3 x4 0 1 x4 0 x4 0 6 0 1 −3 2 0 1 −3 2 0 x1 − 3 x2 + 2 x3 = 3 −9 9. . ~ ~ 3 −2 0 3 −9 6 0 0 0 0 0 0 = 0 −1 The solution is x1 = 3x2 – 2x3, with x2 and x3 free. In parametric vector form, 3 x2 − 2 x3 3 x2 −2 x3 3 −2 x = x2 = x2 + 0 =x2 1 + x3 0 . x3 0 x3 0 1 − 4 x4 = 0 1 3 0 −4 0 1 3 0 −4 0 x1 − 3 x2 10. . ~ 0 0 0 = 0 2 6 0 −8 0 0 0 0 The only basic variable is x1, so x2, x3, and x4 are free. (Note that x3 is not zero.) Also, x1 = 3x2 + 4x4. x1 3 x2 + 4 x4 3 x2 0 4 x4 3 0 4 x x2 = x2 + 0 + 0 =x 1 + x 0 + x 0 . The general solution is x = 2 = 2 x3 0 x3 0 0 3 1 4 0 x3 x4 0 0 1 x4 0 0 x4 11. 1 0 0 0 −4 0 0 0 −2 1 0 0 x1 − 4 x2 x3 0 0 0 0 3 0 1 0 −5 −1 −4 0 0 1 0 0 ~ 0 0 0 0 −4 0 0 0 −2 1 0 0 0 0 0 0 0 0 1 0 7 −1 −4 0 0 1 0 0 ~ 0 0 0 0 −4 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 5 −1 −4 0 0 + 5 x6 = 0 − x6 = . The basic variables are x1, x3, and x5. The remaining variables are 0 x5 − 4 x6 = 0 = 0 free. In particular, x4 is free (and not zero as some may assume). The solution is x1 = 4x2 – 5x6, x3 = x6, x5 = 4x6, with x2, x4, and x6 free. In parametric vector form, Copyright © 2016 Pearson Education, Ltd. 0 0 0 0 1-34 CHAPTER 1 • Linear Equations in Linear Algebra x1 4 x2 − 5 x6 4 x2 0 −5 x6 4 0 −5 x x x 0 0 1 0 0 2 2 2 x3 0 0 x6 0 0 1 x6 x= = = = x2 + x4 + x6 + + x4 x4 0 x4 0 0 1 0 x5 4 x6 0 0 4 x6 0 0 4 x6 0 0 1 x6 0 0 x6 ↑ ↑ ↑ u v w Note: The Study Guide discusses two mistakes that students often make on this type of problem. 1 0 12. 0 0 5 0 0 0 x1 + 5 x2 2 1 0 0 −6 −7 0 0 9 4 0 0 0 −8 1 0 + 8 x4 + x5 x3 − 7 x4 + 4 x5 0 1 0 0 ~ 0 0 0 0 5 0 0 0 2 1 0 0 −6 −7 0 0 9 4 0 0 0 0 1 0 0 1 0 0 ~ 0 0 0 0 5 0 0 0 0 1 0 0 8 −7 0 0 1 4 0 0 0 0 1 0 = 0 = 0 x6 = 0 0 = 0 . The basic variables are x1, x3, and x6; the free variables are x2, x4, and x5. The general solution is x1 = –5x2 – 8x4 – x5, x3 = 7x4 – 4x5, and x6 = 0. In parametric vector form, the solution is x1 −5 x2 − 8 x4 − x5 −5 x2 −8 x4 − x5 −5 −8 −1 x x2 2 x2 0 0 1 0 0 x3 7 x4 − 4 x5 0 7 x4 −4 x5 0 7 −4 x2 + x4 + x5 x= = = + + = x4 x4 0 x4 0 0 1 0 x5 1 x5 x 0 0 0 0 5 0 0 0 0 0 0 0 x6 13. To write the general solution in parametric vector form, pull out the constant terms that do not involve the free variable: x1 5 + 4 x3 5 4 x3 5 4 x = x2 =−2 − 7 x3 =−2 + −7 x3 =−2 + x3 −7 =p + x3q. 1 x3 x3 0 x3 0 ↑ ↑ p q 5 Geometrically, the solution set is the line through −2 in the direction of 0 Copyright © 2016 Pearson Education, Ltd. 4 −7 . 1 0 0 0 0 1.9 • Solutions 14. To write the general solution in parametric vector form, pull out the constant terms that do not involve the free variable: x1 3 x4 0 3 x4 0 3 x 8 + x 8 x 8 1 2 4 4 =+ =+x = x= = p + x4q x3 2 − 5 x4 2 −5 x4 2 4 −5 x4 x4 0 x4 0 1 ↑ p ↑ q The solution set is the line through p in the direction of q. 15. Row reduce the augmented matrix for the system: 1 −4 0 3 −9 −3 1 2 −6 1 ~ 0 0 3 1 0 1 2 0 1 1 −1 ~ 0 −3 0 1 1 1 ~ 0 0 0 0 1 0 3 3 −3 −5 2 0 1 6 −6 1 1 3 ~ 0 −3 0 3 3 0 1 6 0 1 3 0 − 5 x3 = −2 −2 x1 x2 + 2 x3 = 1. 1 . 0 0 = 0 Thus x1 = –2 + 5x3, x2 = 1 – 2x3, and x3 is free. In parametric vector form, −2 5 x3 −2 5 1 + −2 x= 1 + x −2 3 3 0 x3 0 1 −2 The solution set is the line through 1 , parallel to the line that is the solution set of the 0 x1 x x= = 2 x3 −2 + 5 x3 1 − 2x = 3 x3 homogeneous system in Exercise 5. 16. Row reduce the augmented matrix for the system: 1 1 −3 x1 3 4 −7 −5 −8 9 4 1 7 ~ 0 −6 0 3 1 2 −5 −3 −6 4 1 3 ~ 0 6 0 3 1 0 −5 −3 0 4 1 3 ~ 0 0 0 0 1 0 4 −3 0 −5 3 0 + 4 x3 = −5 x2 − 3 x3 = 3 . Thus x1 = –5 – 4x3, x2 = 3 + 3x3, and x3 is free. In parametric vector form, 0 = 0 x1 −5 − 4 x3 −5 −4 x3 −5 −4 x = x2 = 3 + 3 x3 = 3 + 3 x3 = 3 + x3 3 x3 x3 0 x3 0 1 Copyright © 2016 Pearson Education, Ltd. 1-35 1-36 CHAPTER 1 • Linear Equations in Linear Algebra −5 The solution set is the line through 3 , parallel to the line that is the solution set of the 0 homogeneous system in Exercise 6. 17. Solve x1 + 9x2 – 4x3 = –2 for the basic variable: x1 = –2 – 9x2 + 4x3, with x2 and x3 free. In vector form, the solution is x1 −2 − 9 x2 + 4 x3 −2 −9 x2 4 x3 −2 −9 4 x2 x = x2 = = 0 + x2 + 0 = 0 + x2 1 + x3 0 x3 0 0 x3 0 0 1 x3 The solution of x1 + 9x2 – 4x3 = 0 is x1 = –9x2 + 4x3, with x2 and x3 free. In vector form, x1 −9 x2 + 4 x3 −9 x2 4 x3 −9 4 x = x2 = x2 = x2 + 0 =x2 1 + x3 0 = x2u + x3v x3 x3 0 x3 0 1 The solution set of the homogeneous equation is the plane through the origin in 3 spanned by u and v. The solution set of the nonhomogeneous equation is parallel to this plane and passes through −2 the point p = 0 . 0 18. Solve x1 – 3x2 + 5x3 = 4 for the basic variable: x1 = 4 + 3x2 – 5x3, with x2 and x3 free. In vector form, the solution is x1 4 + 3 x2 − 5 x3 4 3 x2 −5 x3 4 3 −5 x2 x = x2 = = 0 + x2 + 0 = 0 + x2 1 + x3 0 x3 0 0 x3 0 0 1 x3 The solution of x1 – 3x2 + 5x3 = 0 is x1 = 3x2 – 5x3, with x2 and x3 free. In vector form, x1 3 x2 − 5 x3 3 x2 −5 x3 3 −5 x= x2 1 + x3 0 = x2u + x3v x2 = x2 = x2 + 0 = x3 x3 0 x3 0 1 The solution set of the homogeneous equation is the plane through the origin in 3 spanned by u and v. The solution set of the nonhomogeneous equation is parallel to this plane and passes through 4 the point p = 0 . 0 19. The line through a parallel to b can be written as x = a + t b, where t represents a parameter: x1 =−2 − 5t x1 −2 −5 x = = + t , or x2 0 3 x2 = 3t 20. The line through a parallel to b can be written as x = a + tb, where t represents a parameter: x1= 3 − 7t x1 3 −7 x = = + t , or x2 −4 8 x2 =−4 + 8t Copyright © 2016 Pearson Education, Ltd. 1.9 • Solutions 1-37 2 −3 21. The line through p and q is parallel to q – p. So, given p = = and q , form −5 1 −3 − 2 −5 2 −5 , and write the line as x = p + t(q – p) = + t . = q−p = 1 − (−5) 6 −5 6 −6 0 22. The line through p and q is parallel to q – p. So, given p = = and q , form 3 −4 0 − (−6) 6 −6 6 , and write the line as x = p + t(q – p) = + t q−p = = −4 − 3 −7 3 −7 Note: Exercises 21 and 22 prepare for Exercise 27 in Section 1.8. 23. a. True. See the first paragraph of the subsection titled “Homogeneous Linear Systems”. b. False. The equation Ax = 0 gives an implicit description of its solution set. See the subsection entitled “Parametric Vector Form”. c. False. The equation Ax = 0 always has the trivial solution. The box before Example 1 uses the word nontrivial instead of trivial. d. False. The line goes through p parallel to v. See the paragraph that precedes Fig. 5. e. False. The solution set could be empty! The statement (from Theorem 6) is true only when there exists a vector p such that Ap = b. 24. a. False. A nontrivial solution of Ax = 0 is any nonzero x that satisfies the equation. See the sentence before Example 2. b. True. See Example 2 and the paragraph following it. c. True. If the zero vector is a solution, then b = Ax = A0 = 0. d. True. See the paragraph following Example 3. e. False. The statement is true only when the solution set of Ax = b is nonempty. Theorem 6 applies only to a consistent system. 25. Suppose p and w satisfy Ax = b. Then Ap = b and Aw = b . Set vh = w − p. Then Avh = A(w – p) = Aw – Ap = b – b = 0 So vh satisfies Ax = 0. 26. (Geometric argument using Theorem 6.) Since Ax = b is consistent, its solution set is obtained by translating the solution set of Ax = 0, by Theorem 6. So the solution set of Ax = b is a single vector if and only if the solution set of Ax = 0 is a single vector, and that happens if and only if Ax = 0 has only the trivial solution. (Proof using free variables.) If Ax = b has a solution, then the solution is unique if and only if there are no free variables in the corresponding system of equations, that is, if and only if every column of A is a pivot column. This happens if and only if the equation Ax = 0 has only the trivial solution. 27. When A is the 3×3 zero matrix, every x in 3 satisfies Ax = 0. So the solution set is all vectors in 3 . 28. No. If the solution set of Ax = b contained the origin, then 0 would satisfy A0= b, which is not true since b is not the zero vector. Copyright © 2016 Pearson Education, Ltd. 1-38 CHAPTER 1 • Linear Equations in Linear Algebra 29. a. When A is a 3×3 matrix with three pivot positions, the equation Ax = 0 has no free variables and hence has no nontrivial solution. b. With three pivot positions, A has a pivot position in each of its three rows. By Theorem 4 in Section 1.4, the equation Ax = b has a solution for every possible b. The term “possible” in the exercise means that the only vectors considered in this case are those in 3 , because A has three rows. 30. a. When A is a 3×3 matrix with two pivot positions, the equation Ax = 0 has two basic variables and one free variable. So Ax = 0 has a nontrivial solution. b. With only two pivot positions, A cannot have a pivot in every row, so by Theorem 4 in Section 1.4, the equation Ax = b cannot have a solution for every possible b (in 3 ). 31. a. When A is a 3×2 matrix with two pivot positions, each column is a pivot column. So the equation Ax = 0 has no free variables and hence no nontrivial solution. b. With two pivot positions and three rows, A cannot have a pivot in every row. So the equation Ax = b cannot have a solution for every possible b (in 3 ), by Theorem 4 in Section 1.4. 32. a. When A is a 2×4 matrix with two pivot positions, the equation Ax = 0 has two basic variables and two free variables. So Ax = 0 has a nontrivial solution. b. With two pivot positions and only two rows, A has a pivot position in every row. By Theorem 4 in Section 1.4, the equation Ax = b has a solution for every possible b (in 2 ). −2 −6 33. Look at x1 7 + x2 21 and notice that the second column is 3 times the first. So suitable values −3 −9 3 for x1 and x2 would be 3 and –1 respectively. (Another pair would be 6 and –2, etc.) Thus x = −1 satisfies Ax = 0. 34. Inspect how the columns a1 and a2 of A are related. The second column is –3/2 times the first. Put 3 another way, 3a1 + 2a2 = 0. Thus satisfies Ax = 0. 2 Note: Exercises 33 and 34 set the stage for the concept of linear dependence. 35. Look for A = [a1 a2 a3] such that 1∙a1 + 1a2 + 1∙a3 = 0. That is, construct A so that each row sum (the sum of the entries in a row) is zero. 36. Look for A = [a1 a2 a3] such that 1∙a1 – 2∙a2 + 1∙a3 = 0. That is, construct A so that the sum of the first and third columns is twice the second column. 37. Since the solution set of Ax = 0 contains the point (4,1), the vector x = (4,1) satisfies Ax = 0. Write this equation as a vector equation, using a1 and a2 for the columns of A: 4∙a1 + 1∙a2 = 0 Then a2 = –4a1. So choose any nonzero vector for the first column of A and multiply that column by 1 −4 – 4 to get the second column of A. For example, set A = . 1 −4 Copyright © 2016 Pearson Education, Ltd. 1.9 • Solutions 1-39 Finally, the only way the solution set of Ax = b could not be parallel to the line through (1,4) and the origin is for the solution set of Ax = b to be empty. This does not contradict Theorem 6, because that theorem applies only to the case when the equation Ax = b has a nonempty solution set. For b, take any vector that is not a multiple of the columns of A. Note: In the Study Guide, a “Checkpoint” for Section 1.5 will help students with Exercise 37. 38. No. If Ax = y has no solution, then A cannot have a pivot in each row. Since A is 3×3, it has at most two pivot positions. So the equation Ax = z for any z has at most two basic variables and at least one free variable. Thus, the solution set for Ax = z is either empty or has infinitely many elements. 39. If u satisfies Ax = 0, then Au = 0. For any scalar c, Theorem 5(b) in Section 1.4 shows that A(cu) = cAu = c∙0 = 0. 40. Suppose Au = 0 and Av = 0. Then, since A(u + v) = Au + Av by Theorem 5(a) in Section 1.4, A(u + v) = Au + Av = 0 + 0 = 0. Now, let c and d be scalars. Using both parts of Theorem 5, A(cu + dv) = A(cu) + A(dv) = cAu + dAv = c0 + d0 = 0. Note: The MATLAB box in the Study Guide introduces the zeros command, in order to augment a matrix with a column of zeros. 1.6 SOLUTIONS 1. Fill in the exchange table one column at a time. The entries in a column describe where a sector's output goes. The decimal fractions in each column sum to 1. Distribution of Output From: Goods Services Purchased by: output ↓ input ↓ .2 .7 Goods → .8 .3 Services → Denote the total annual output (in dollars) of the sectors by pG and pS. From the first row, the total input to the Goods sector is .2 pG + .7 pS. The Goods sector must pay for that. So the equilibrium prices must satisfy income pG = expenses .2pG + .7 pS From the second row, the input (that is, the expense) of the Services sector is .8 pG + .3 pS. The equilibrium equation for the Services sector is income expenses pS = .8pG + .3 pS Copyright © 2016 Pearson Education, Ltd. 1-40 CHAPTER 1 • Linear Equations in Linear Algebra Move all variables to the left side and combine like terms: .8 pG −.8 pG − + .7 pS .7 pS = 0 = 0 .8 −.7 0 .8 −.7 0 1 −.875 0 Row reduce the augmented matrix: ~ ~ .7 0 0 0 0 0 0 0 −.8 The general solution is pG = .875 pS, with pS free. One equilibrium solution is pS = 1000 and pG = 875. If one uses fractions instead of decimals in the calculations, the general solution would be written pG = (7/8) pS, and a natural choice of prices might be pS = 80 and pG = 70. Only the ratio of the prices is important: pG = .875 pS. The economic equilibrium is unaffected by a proportional change in prices. 2. Take some other value for pS, say 200 million dollars. The other equilibrium prices are then pC = 188 million, pE = 170 million. Any constant nonnegative multiple of these prices is a set of equilibrium prices, because the solution set of the system of equations consists of all multiples of one vector. Changing the unit of measurement to, say, European euros has the same effect as multiplying all equilibrium prices by a constant. The ratios of the prices remain the same, no matter what currencyis used. 3. a. Fill in the exchange table one column at a time. The entries in a column describe where a sector’s output goes. The decimal fractions in each column sum to 1. Distribution of Output From: Purchased Chemicals Fuels Machinery by: output ↓ ↓ ↓ input .2 .8 .4 Chemicals .3 .1 .4 Fuels .5 .1 .2 Machinery b. Denote the total annual output (in dollars) of the sectors by pC, pF, and pM. From the first row of the table, the total input to the Chemical & Metals sector is .2 pC + .8 pF + .4 pM. So the equillibrium prices must satisfy income pC = expenses .2pC + .8 pF + .4 pM From the second and third rows of the table, the income/expense requirements for the Fuels & pF = .3 pC + .1 pF + .4 pM Power sector and the Machinery sector are, respectively, pM = .5 pC + .1 pF + .2 pM .8 pC – .8 pF – .4 pM = 0 0 Move all variables to the left side and combine like terms: –.3 pC + .9 pF – .4 pM = –.5 pC – .1 pF + .8 pM = 0 c. [M] You can obtain the reduced echelon form with a matrix program. Actually, hand calculations are not too messy. To simplify the calculations, first scale each row of the augmented matrix by 10, then continue as usual. Copyright © 2016 Pearson Education, Ltd. 1.9 • Solutions 1-41 −8 9 −1 −4 0 1 −1 −.5 0 1 −1 −.5 0 −4 0 ~ −3 9 −4 0 ~ 0 6 −5.5 0 8 0 −5 −1 8 0 0 −6 5.5 0 −1 −.5 0 1 0 −1.417 0 The number of decimal 1 −.917 0 ~ 0 1 −.917 0 places displayed is 0 0 0 0 0 0 0 somewhat arbitrary. 8 −3 −5 1 ~ 0 0 The general solution is pC = 1.417 pM, pF = .917 pM, with pM free. If pM is assigned the value 100, then pC = 141.7 and pF = 91.7. Note that only the ratios of the prices are determined. This makes sense, for if the were converted from, say, dollars to yen or Euros, the inputs and outputs of each sector would still balance. The economic equilibrium is not affected by a proportional change in prices. 4. a. Fill in the exchange table one column at a time. The entries in each column must sum to 1. Distribution of Output From: output A ↓ .65 .10 .25 0 E ↓ .30 .10 .35 .25 M ↓ .30 .15 .15 .40 T ↓ .20 .10 .30 .40 Purchased by : input → → → → A E M T b. Denote the total annual output of the sectors by pA, pE, pM, and pT, respectively. From the first row of the table, the total input to Agriculture is .65pA + .30pE + .30pM + .20 pT. So the equilibrium prices must satisfy income expenses pA = .65 pA + .30 pE + .30 pM + .20 pT From the second, third, and fourth rows of the table, the equilibrium equations are .10 pA + .10 pE + .15 pM + .10 pT pE = pM = p= T .25 pA + .35 pE + .15 pM + .30 pT .25 pE + .40 pM + .40 pT Move all variables to the left side and combine like terms: .35 pA − .30 pE − .30 pM − .20 pT = 0 0 −.10 pA + .90 pE − .15 pM − .10 pT = 0 −.25 pA − .35 pE + .85 pM − .30 pT = 0 −.25 pE − .40 pM + .60 pT = Use gauss, bgauss, and scale operations to reduce the augmented matrix to reduced echelon form .35 0 0 0 −.3 .81 0 0 −.3 −.24 1.0 0 −.2 −.16 −1.17 0 0 .35 0 0 ~ 0 0 0 0 −.3 .81 0 0 0 0 1 0 −.55 −.43 −1.17 0 0 .35 0 0 ~ 0 0 0 0 Copyright © 2016 Pearson Education, Ltd. 0 1 0 0 0 0 1 0 −.71 −.53 −1.17 0 0 0 0 0 1-42 CHAPTER 1 • Linear Equations in Linear Algebra Scale the first row and solve for the basic variables in terms of the free variable pT, and obtain pA = 2.03pT, pE = .53pT, and pM = 1.17pT. The data probably justifies at most two significant figures, so take pT = 100 and round off the other prices to pA = 200, pE = 53, and pM = 120. 5. The following vectors list the numbers of atoms of boron (B), sulfur (S), hydrogen (H), and oxygen (O): 2 3 B2S3 : , H 2 O: 0 0 0 0 , H 3 BO3 : 2 1 1 0 , H 2S: 3 3 0 1 2 0 boron sulfur hydrogen oxygen The coefficients in the equation x1⋅B2S3 + x2⋅H20 → x3⋅H3BO3 + x4⋅H2S satisfy 2 0 1 0 3 0 0 1 x1 + x2 = x3 + x4 0 2 3 2 0 1 3 0 Move the right terms to the left side (changing the sign of each entry in the third and fourth vectors) and row reduce the augmented matrix of the homogeneous system: 2 3 0 0 2 0 ~ 0 0 −1 0 −3 −3 0 0 2 1 0 1 0 0 −1 −3 1 3 0 −1 −2 0 0 2 0 0 ~ 0 0 0 0 0 0 −2 / 3 −2 0 0 2 1 0 2 0 0 ~ 0 0 0 0 −1 3/ 2 −3 −3 0 1 0 0 0 −1 −2 0 0 0 1 0 0 2 0 0 ~ 0 0 0 0 −2 / 3 −2 −2 / 3 0 0 1 0 2 0 1 0 0 ~ 0 0 0 0 −1 −3 3/ 2 −3 0 1 0 0 0 0 −1 −2 0 0 1 0 0 2 0 0 ~ 0 0 0 0 −1/ 3 −2 −2 / 3 0 0 1 0 0 −1 −3 3/ 2 3 0 0 −1 −2 0 0 0 0 0 0 0 0 The general solution is x1 = (1/3) x4, x2 = 2x4, x3 = (2/3) x4, with x4 free. Take x4 = 3. Then x1 = 1, x2 = 6, and x3 = 2. The balanced equation is B2S3 + 6H20 → 2H3BO3 + 3H2S. 6. The following vectors list the numbers of atoms of sodium (Na), phosphorus (P), oxygen (O), barium (Ba), and nitrogen(N): 3 1 Na 3 PO 4 : 4 , Ba(NO3 ) 2 : 0 0 0 0 6 , Ba 3 (PO 4 ) 2 : 1 2 0 2 8 , NaNO3 : 3 0 1 0 3 0 1 sodium phosphorus oxygen barium nitrogen The coefficients in the equation x1⋅Na3PO4 + x2⋅Ba(NO3)2 → x3⋅Ba3(PO4)2 + x4⋅NaNO3 satisfy Copyright © 2016 Pearson Education, Ltd. 1.9 • Solutions 1-43 3 0 0 1 1 0 2 0 x1 4 + x2 6 = x3 8 + x4 3 0 1 3 0 0 2 0 1 Move the right terms to the left side (changing the sign of each entry in the third and fourth vectors) and row reduce the augmented matrix of the homogeneous system: 3 1 4 0 0 1 0 ~ 0 0 0 0 0 6 1 2 0 −2 −8 −3 0 0 1 0 0 0 −2 −3 18 6 6 −1 0 −3 0 −1 0 0 −3 −1 −1 0 1 0 3 0 ~ 4 0 0 0 0 −2 0 −8 −3 0 0 0 6 1 2 0 1 0 0 0 ~ 0 0 0 0 0 0 1 0 0 0 −2 −3 1 0 0 0 −1 −3 0 −1 0 1 0 0 0 ~ 0 0 0 0 0 0 0 −1/ 6 0 0 0 0 6 1 2 0 1 0 0 0 ~ 0 0 0 0 0 −2 6 0 −3 0 0 1 0 0 0 0 −1 −3 0 −1 0 0 1 0 0 0 1 0 0 0 ~ 0 0 0 0 0 −1/ 3 −1/ 2 −1/ 6 0 0 0 1 6 0 2 −2 −3 0 6 0 0 0 −3 −1 −1 0 0 0 0 0 0 0 0 0 0 The general solution is x1 = (1/3)x4, x2 = (1/2)x4, x3 = (1/6)x4, with x4 free. Take x4 = 6. Then x1 = 2, x2 = 3, and x3 = 1. The balanced equation is 2Na3PO4 + 3Ba(NO3)2 → Ba3(PO4)2 + 6NaNO3 7. The following vectors list the numbers of atoms of sodium (Na), hydrogen (H), carbon (C), and oxygen (O): 1 1 NaHCO3 : , H 3C6 H 5O7 : 1 3 0 8 , Na 3C6 H 5O7 : 6 7 3 0 0 5 2 , H 2 O : , CO 2 : 0 6 0 1 7 1 2 sodium hydrogen carbon oxygen The order of the various atoms is not important. The list here was selected by writing the elements in the order in which they first appear in the chemical equation, reading left to right: x1 · NaHCO3 + x2 · H3C6H5O7 → x3 · Na3C6H5O7 + x4 · H2O + x5 · CO2. The coefficients x1, …, x5 satisfy the vector equation 1 0 3 0 0 1 8 5 2 0 x1 + x2 = x3 + x4 + x5 1 6 6 0 1 3 7 7 1 2 Move all the terms to the left side (changing the sign of each entry in the third, fourth, and fifth vectors) and reduce the augmented matrix: Copyright © 2016 Pearson Education, Ltd. 1-44 CHAPTER 1 1 1 1 3 0 8 6 7 −3 −5 −6 −7 • Linear Equations in Linear Algebra 0 −2 0 −1 0 0 −1 −2 0 1 0 0 ~ ⋅⋅⋅ ~ 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 −1 −1/ 3 −1/ 3 −1 0 0 0 0 The general solution is x1 = x5, x2 = (1/3)x5, x3 = (1/3)x5, x4 = x5, and x5 is free. Take x5 = 3. Then x1 = x4 = 3, and x2 = x3 = 1. The balanced equation is 3NaHCO3 + H3C6H5O7 → Na3C6H5O7 + 3H2O + 3CO2 8. The following vectors list the numbers of atoms of potassium (K), manganese (Mn), oxygen (O), sulfur (S), and hydrogen (H): 1 0 0 0 2 0 potassium 1 1 0 1 0 0 manganese KMnO 4 : 4 , MnSO 4 : 4 , H 2 O: 1 , MnO 2 : 2 , K 2SO 4 : 4 , H 2SO 4 : 4 oxygen 0 1 0 0 1 1 sulfur 0 0 2 0 0 2 hydrogen The coefficients in the chemical equation x1⋅KMnO4 + x2⋅MnSO4 + x3⋅H2O → x4⋅MnO2 + x5⋅K2SO4 + x6⋅H2SO4 satisfy the vector 1 0 0 0 2 0 1 1 0 1 0 0 equation x1 4 + x2 4 + x3 1 = x4 2 + x5 4 + x6 4 0 1 0 0 1 1 0 0 2 0 0 2 Move the terms to the left side (changing the sign of each entry in the last three vectors) and reduce the augmented matrix: 1 1 4 0 0 0 1 4 1 0 0 0 1 0 2 0 −1 −2 0 0 −2 0 −4 −1 0 0 0 −4 −1 −2 0 1 0 0 0 ~ 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 −1.0 −1.5 −1.0 −2.5 −.5 0 0 0 0 0 The general solution is x1 = x6, x2 = (1.5)x6, x3 = x6, x4 = (2.5)x6, x5 = .5x6, and x6 is free. Take x6 = 2. Then x1 = x3 = 2, and x2 = 3, x4 = 5, and x5 = 1. The balanced equation is 2KMnO4 + 3MnSO4 + 2H2O → 5MnO2 + K2SO4 + 2H2SO4 9. [M] Set up vectors that list the atoms per molecule. Using the order lead (Pb), nitrogen (N), chromium (Cr), manganese (Mn), and oxygen (O), the vector equation to be solved is 1 0 3 0 0 0 6 0 0 0 0 1 x1 0 + x2 1 = x3 0 + x4 2 + x5 0 + x6 0 0 2 0 0 1 0 0 8 4 3 2 1 lead nitrogen chromium manganese oxygen Copyright © 2016 Pearson Education, Ltd. 1.9 • Solutions 1-45 The general solution is x1 = (1/6)x6, x2 = (22/45)x6, x3 = (1/18)x6, x4 = (11/45)x6, x5 = (44/45)x6, and x6 is free. Take x6 = 90. Then x1 = 15, x2 = 44, x3 = 5, x4 = 22, and x5 = 88. The balanced equation is 15PbN6 + 44CrMn2O8 → 5Pb3O4 + 22Cr2O3 + 88MnO2 + 90NO 10. [M] Set up vectors that list the atoms per molecule. Using the order manganese (Mn), sulfur (S), arsenic (As), chromium (Cr), oxygen (O), and hydrogen (H), the vector equation to be solved is 1 0 0 1 0 0 0 manganese 1 0 1 0 0 3 0 sulfur 0 2 0 0 1 0 0 arsenic x1 + x2 + x3 = x4 + x5 + x6 + x7 0 10 0 0 0 1 0 chromium 0 35 4 4 0 12 1 oxygen 3 0 0 2 1 0 2 hydrogen In rational format, the general solution is x1 = (16/327)x7, x2 = (13/327)x7, x3 = (374/327)x7, x4 = (16/327)x7, x5 = (26/327)x7, x6 = (130/327)x7, and x7 is free. Take x7 = 327 to make the other variables whole numbers. The balanced equation is 16MnS + 13As2Cr10O35 + 374H2SO4 → 16HMnO4 + 26AsH3 + 130CrS3O12 + 327H2O Note that some students may use decimal calculation and simply "round off" the fractions that relate x1, ..., x6 to x7. The equations they construct may balance most of the elements but miss an atom or two. Here is a solution submitted by two of my students: 5MnS + 4As2Cr10O35 + 115H2SO4 → 5HMnO4 + 8AsH3 + 40CrS3O12 + 100H2O Everything balances except the hydrogen. The right side is short 8 hydrogen atoms. Perhaps the students thought that the 4H2 (hydrogen gas) escaped! 11. Write the equations for each node: Node A B C Total flow: Flow in Flow out x1 + x3 =20 x2 = x3 + x4 80 = x1 + x2 80 = x 4 + 20 Rearrange the equations: x1 x1 + x2 x2 + − x3 x3 − x4 x4 = 20 = 0 = 80 = 60 Reduce the augmented matrix: 1 0 1 0 0 1 1 0 1 −1 0 0 0 −1 0 1 20 1 0 0 ~ ⋅⋅⋅ ~ 0 80 60 0 0 1 0 0 1 −1 0 0 0 0 1 0 20 60 60 0 For this type of problem, the best description of the general solution uses the style of Section 1.2 rather than parametric vector form: Copyright © 2016 Pearson Education, Ltd. 1-46 CHAPTER 1 • Linear Equations in Linear Algebra 20 − x3 x= 1 x= 60 + x 2 3 . Since x1 cannot be negative, the largest value of x3 is 20. x is free 3 x 4 = 60 12. Write the equations for each intersection: Intersection A B C D Total flow: Flow in Flow out = x1 x3 + x4 + 40 = x1 + x2 200 x2 + x3 = x5 + 100 x4 + x5 = 60 200 = 200 Rearrange the equations: x1 x1 + x2 x2 − x3 + x3 − x4 x4 − + x5 x5 40 1 200 0 ~ 100 0 60 0 0 1 0 0 −1 1 0 0 = 40 = 200 = 100 = 60 Reduce the augmented matrix: 1 1 0 0 0 1 1 0 −1 0 1 0 −1 0 0 1 0 0 −1 1 0 0 1 0 1 100 −1 100 1 60 0 0 The general solution (written in the style of Section 1.2) is x1 = 100 + x3 − x5 x = 100 − x + x 3 5 2 x is free 3 x= 60 − x 5 4 x is free 5 40 + x3 x= 1 = x2 160 − x3 b. When x4 = 0, x5 must be 60, and x3 is free x = 0 4 x5 = 60 c. The minimum value of x1 is 40 cars/minute, because x3 cannot be negative. 13. Write the equations for each intersection: Copyright © 2016 Pearson Education, Ltd. 1.9 Intersection A B C D E Total flow: Flow in x2 + 30 x3 + x5 x6 + 100 x4 + 40 x1 + 60 230 • Solutions Flow out x1 + 80 x2 + x4 x5 + 40 x6 + 90 x3 + 20 230 = = = = = = Rearrange the equations: x1 − x2 x2 − x3 + x4 − x5 x5 x4 − x1 − − = −50 = 0 = 60 = 50 −40 x6 x6 x3 Reduce the augmented matrix: 1 0 0 0 1 −1 1 0 0 0 1 0 ~ ⋅⋅⋅ ~ 0 0 0 0 −1 0 0 −1 0 1 0 0 0 0 1 0 1 0 0 −1 1 0 0 0 0 −1 −1 0 −1 −1 0 0 0 0 0 1 0 0 0 0 0 1 0 −50 1 0 0 60 ~ ⋅⋅⋅ ~ 0 50 0 0 −40 0 0 −1 −1 0 −1 1 0 0 0 0 −1 0 0 0 0 1 1 0 0 0 −1 0 1 0 0 0 −1 −1 0 −50 0 50 60 0 −40 10 50 60 0 x3 − 40 x= 1 x= x + 10 3 2 x3 is free a. The general solution is x6 + 50 4 x= x= x6 + 60 5 x6 is free b. To find minimum flows, note that since x1 cannot be negative, x3 > 40. This implies that x2 > 50. Also, since x6 cannot be negative, x4 > 50 and x5 > 60. The minimum flows are x2 = 50, x3 = 40, x4 = 50, x5 = 60 (when x1 = 0 and x6 = 0). Copyright © 2016 Pearson Education, Ltd. 1-47 1-48 CHAPTER 1 • Linear Equations in Linear Algebra 14. Write the equations for each intersection. Intersection Flow in Flow out A x1 = x2 + 100 B x2 + 50 = x3 C x3 = x4 + 120 D x4 + 150 =x5 E x5 = x6 + 80 F x6 + 100 =x1 Rearrange the equations: − x2 x1 x2 − x3 x3 − x4 x4 − − x1 x5 x5 − + x6 x6 = 100 = − 50 = 120 = −150 = 80 = −100 Reduce the augmented matrix: 1 0 0 0 0 −1 1 0 0 ~ ⋅⋅⋅ ~ 0 0 0 −1 1 0 0 0 0 0 1 0 0 0 0 0 −1 1 0 0 0 0 0 1 0 0 0 100 1 −50 0 0 120 ~ ⋅⋅⋅ ~ −150 0 80 0 −100 0 −1 1 0 0 0 0 0 0 −1 1 0 0 0 0 0 −1 1 0 0 0 0 0 −1 1 0 0 0 0 1 0 −1 −1 −1 −1 −1 0 100 0 50 . The general solution is −70 80 0 0 0 0 1 0 0 0 −1 1 0 0 0 0 0 −1 1 0 0 0 0 0 −1 1 0 0 0 0 0 −1 0 100 −50 120 −150 80 0 x1 100 + x6 = x = x 6 2 50 + x6 x= 3 . −70 + x6 x4 = x= 80 + x6 5 x6 is free Since x4 cannot be negative, the minimum value of x6 is 70. Note: The MATLAB box in the Study Guide discusses rational calculations, needed for balancing the chemical equations in Exercises 9 and 10. As usual, the appendices cover this material for Maple, Mathematica, and the TI and HP graphic calculators. 1.7 SOLUTIONS Note: Key exercises are 9–20 and 23–30. Exercise 30 states a result that could be a theorem in the text. There is a danger, however, that students will memorize the result without understanding the proof, and Copyright © 2016 Pearson Education, Ltd. 1.9 • Solutions 1-49 then later mix up the words row and column. Exercises 37 and 38 anticipate the discussion in Section 1.9 of one-to-one transformations. Exercise 44 is fairly difficult for my students. 1. Use an augmented matrix to study the solution set of x1u + x2v + x3w = 0 (*), where u, v, and w are 7 9 0 5 7 9 0 5 the three given vectors. Since 0 2 4 0 ~ 0 2 4 0 , there are no free variables. So 0 −6 −8 0 0 0 4 0 the homogeneous equation (*) has only the trivial solution. The vectors are linearly independent. 2. Use an augmented matrix to study the solution set of x1u + x2v + x3w = 0, where u, v, and w are the 0 −3 0 2 −8 1 0 0 three given vectors. Since 0 5 4 0 ~ 0 5 4 0 , there are no free variables. So 2 −8 1 0 0 0 −3 0 the homogeneous equation has only the trivial solution. The vectors are linearly independent. 3. Use the method of Example 3 (or the box following the example). By comparing entries of the vectors, one sees that the second vector is –3 times the first vector. Thus, the two vectors are linearly dependent. −1 −2 4. From the first entries in the vectors, it seems that the second vector of the pair , may be 2 4 −8 times the first vector. But there is a sign problem with the second entries. So neither of the vectors is a multiple of the other. The vectors are linearly independent. 5. Use the method of Example 2. Row reduce the augmented matrix for Ax = 0: 5 0 1 −3 2 0 1 −3 2 0 1 −3 2 0 1 −3 2 0 −8 3 −7 4 0 3 −7 4 0 0 2 −2 0 0 2 −2 0 0 2 −2 ~ ~ ~ ~ −1 5 −4 0 −1 5 −4 0 0 2 −2 0 0 0 0 0 0 0 −3 2 0 0 −8 5 0 0 −8 5 0 0 0 −3 0 0 0 0 1 −3 There are no free variables. The equation Ax = 0 has only the trivial solution and so the columns of A are linearly independent. 6. Use the method of Example 2. Row reduce the augmented matrix for Ax = 0: −4 0 1 5 −3 −1 0 4 0 4 3 6 0 1 0 0 ~ 0 −4 0 5 0 −1 −3 4 3 4 0 6 0 1 0 0 ~ 0 0 0 0 0 −1 −3 4 3 4 12 −9 0 1 0 0 ~ 0 0 0 0 0 −1 0 0 3 4 0 7 0 1 0 0 ~ 0 0 0 0 0 −1 0 0 3 4 7 0 0 0 0 0 There are no free variables. The equation Ax = 0 has only the trivial solution and so the columns of A are linearly independent. Copyright © 2016 Pearson Education, Ltd. 0 0 0 0 1-50 CHAPTER 1 • Linear Equations in Linear Algebra 7. Study the equation Ax = 0. Some people may start with the method of Example 2: 1 −2 −4 4 −7 −5 −3 5 7 0 1 5 0 1 4 0 ~ 0 1 0 0 11 −3 −1 −5 0 1 0 ~ 0 0 0 0 1 5 −3 −1 6 4 1 0 0 1 −6 0 0 0 But this is a waste of time. There are only 3 rows, so there are at most three pivot positions. Hence, at least one of the four variables must be free. So the equation Ax = 0 has a nontrivial solution and the columns of A are linearly dependent. 8. Same situation as with Exercise 7. The (unnecessary) row operations are 1 −3 0 −3 7 1 3 −1 −4 −2 2 3 0 1 0 ~ 0 0 0 −3 −2 1 3 8 −4 −2 −4 3 0 1 0 ~ 0 0 0 −3 −2 0 3 8 0 −2 −4 1 0 0 0 Again, because there are at most three pivot positions yet there are four variables, the equation Ax = 0 has a nontrivial solution and the columns of A are linearly dependent. 9. a. The vector v3 is in Span{v1, v2} if and only if the equation x1v1 + x2v2 = v3 has a solution. To find out, row reduce [v1 v2 v3], considered as an augmented matrix: 1 −3 2 −3 9 −6 5 1 −7 ~ 0 h 0 −3 0 0 5 8 h − 10 At this point, the equation 0 = 8 shows that the original vector equation has no solution. So v3 is in Span{v1, v2} for no value of h. b. For {v1, v2, v3} to be linearly independent, the equation x1v1 + x2v2 + x3v3 = 0 must have only the trivial solution. Row reduce the augmented matrix [v1 v2 v3 0]: 1 −3 2 −3 9 −6 5 −7 h −3 0 0 0 1 0 ~ 0 0 0 5 8 h − 10 0 1 0 ~ 0 0 0 −3 0 0 5 8 0 0 0 0 For every value of h, x2 is a free variable, and so the homogeneous equation has a nontrivial solution. Thus {v1, v2, v3} is a linearly dependent set for all h. 10. a. The vector v3 is in Span{v1, v2} if and only if the equation x1v1 + x2v2 = v3 has a solution. To find out, row reduce [v1 v2 v3], considered as an augmented matrix: 1 −5 −3 −2 10 6 2 1 −9 ~ 0 h 0 −2 0 0 2 1 h + 6 At this point, the equation 0 = 1 shows that the original vector equation has no solution. So v3 is in Span{v1, v2} for no value of h. b. For {v1, v2, v3} to be linearly independent, the equation x1v1 + x2v2 + x3v3 = 0 must have only the trivial solution. Row reduce the augmented matrix [v1 v2 v3 0]: Copyright © 2016 Pearson Education, Ltd. 1.9 1 −5 −3 −2 10 6 2 −9 h 0 1 0 ~ 0 0 0 −2 0 0 2 1 h+6 0 1 0 ~ 0 0 0 −2 0 0 • Solutions 1-51 0 0 0 2 1 0 For every value of h, x2 is a free variable, and so the homogeneous equation has a nontrivial solution. Thus {v1, v2, v3} is a linearly dependent set for all h. 11. To study the linear dependence of three vectors, say v1, v2, v3, row reduce the augmented matrix [v1 v2 v3 0]: 1 −1 4 3 −5 7 −1 5 h 0 1 0 ~ 0 0 0 3 −2 −5 −1 4 h+4 0 1 0 ~ 0 0 0 3 −2 0 −1 4 h−6 0 0 0 The equation x1v1 + x2v2 + x3v3 = 0 has a nontrivial solution if and only if h – 6 = 0 (which corresponds to x3 being a free variable). Thus, the vectors are linearly dependent if and only if h = 6. 12. To study the linear dependence of three vectors, say v1, v2, v3, row reduce the augmented matrix [v1 v2 v3 0]: 2 −4 1 −6 7 −3 8 h 4 0 2 0 ~ 0 0 0 −6 −5 0 8 h + 16 0 0 0 0 The equation x1v1 + x2v2 + x3v3 = 0 has a free variable and hence a nontrivial solution no matter what the value of h. So the vectors are linearly dependent for all values of h. 13. To study the linear dependence of three vectors, say v1, v2, v3, row reduce the augmented matrix [v1 v2 v3 0]: 1 5 −3 −2 −9 6 3 h −9 0 1 0 ~ 0 0 0 −2 1 0 3 h − 15 0 0 0 0 The equation x1v1 + x2v2 + x3v3 = 0 has a free variable and hence a nontrivial solution no matter what the value of h. So the vectors are linearly dependent for all values of h. 14. To study the linear dependence of three vectors, say v1, v2, v3, row reduce the augmented matrix [v1 v2 v3 0]: 1 −1 3 −5 7 8 1 1 h 0 1 0 ~ 0 0 0 −5 2 23 1 2 h−3 0 1 0 ~ 0 0 0 −5 1 0 1 1 h − 26 0 0 0 The equation x1v1 + x2v2 + x3v3 = 0 has a nontrivial solution if and only if h − 26 = 0 (which corresponds to x3 being a free variable). Thus, the vectors are linearly dependent if and only if h = 26. 15. The set is linearly dependent, by Theorem 8, because there are four vectors in the set but only two entries in each vector. 16. The set is linearly dependent because the second vector is 3/2 times the first vector. Copyright © 2016 Pearson Education, Ltd. 1-52 CHAPTER 1 • Linear Equations in Linear Algebra 17. The set is linearly dependent, by Theorem 9, because the list of vectors contains a zero vector. 18. The set is linearly dependent, by Theorem 8, because there are four vectors in the set but only two entries in each vector. 19. The set is linearly independent because neither vector is a multiple of the other vector. [Two of the entries in the first vector are – 4 times the corresponding entry in the second vector. But this multiple does not work for the third entries.] 20. The set is linearly dependent, by Theorem 9, because the list of vectors contains a zero vector. 21. a. b. c. d. False. A homogeneous system always has the trivial solution. See the box before Example 2. False. See the warning after Theorem 7. True. See Fig. 3, after Theorem 8. True. See the remark following Example 4. 22. a. True. See Fig. 1. 1 2 b. False. For instance, the set consisting of −2 and –4 is linearly dependent. See the warning 3 6 after Theorem 8. c. True. See the remark following Example 4. d. False. See Example 3(a). * * 23. 0 * 0 0 * 0 0 24. , , 0 0 0 0 0 0 0 * 0 0 0 0 and 25. 0 0 0 0 0 0 0 0 * * 0 * . The columns must be linearly independent, by Theorem 7, because the first column is 26. 0 0 0 0 0 not zero, the second column is not a multiple of the first, and the third column is not a linear combination of the preceding two columns (because a3 is not in Span{a1, a2}). 27. All five columns of the 7×5 matrix A must be pivot columns. Otherwise, the equation Ax = 0 would have a free variable, in which case the columns of A would be linearly dependent. 28. If the columns of a 5×7 matrix A span 5 , then A has a pivot in each row, by Theorem 4. Since each pivot position is in a different column, A has five pivot columns. 29. A: any 3×2 matrix with two nonzero columns such that neither column is a multiple of the other. In this case the columns are linearly independent and so the equation Ax = 0 has only the trivial solution. B: any 3×2 matrix with one column a multiple of the other. Copyright © 2016 Pearson Education, Ltd. 1.9 • Solutions 1-53 30. a. n b. The columns of A are linearly independent if and only if the equation Ax = 0 has only the trivial solution. This happens if and only if Ax = 0 has no free variables, which in turn happens if and only if every variable is a basic variable, that is, if and only if every column of A is a pivot column. 31. Think of A = [a1 a2 a3]. The text points out that a3 = a1 + a2. Rewrite this as a1 + a2 – a3 = 0. As a matrix equation, Ax = 0 for x = (1, 1, –1). 32. Think of A = [a1 a2 a3]. The text points out that a1 + 2a2 = a3. Rewrite this as a1 + 2a2 – a3 = 0. As a matrix equation, Ax = 0 for x = (1, 2, –1). 33. True, by Theorem 7. (The Study Guide adds another justification.) 34. True, by Theorem 9. 35. False. The vector v1 could be the zero vector. 36. False. Counterexample: Take v1, v2, and v4 all to be multiples of one vector. Take v3 to be not a multiple of that vector. For example, 1 1 ,v v1 = = 1 2 1 2 2 = ,v 2 3 2 1 0 = ,v 0 4 0 4 4 4 4 37. True. A linear dependence relation among v1, v2, v3 may be extended to a linear dependence relation among v1, v2, v3, v4 by placing a zero weight on v4. 38. True. If the equation x1v1 + x2v2 + x3v3 = 0 had a nontrivial solution (with at least one of x1, x2, x3 nonzero), then so would the equation x1v1 + x2v2 + x3v3 + 0×v4 = 0. But that cannot happen because {v1, v2, v3, v4} is linearly independent. So {v1, v2, v3} must be linearly independent. This problem can also be solved using Exercise 37, if you know that the statement there is true. 39. If for all b the equation Ax = b has at most one solution, then take b = 0, and conclude that the equation Ax = 0 has at most one solution. Then the trivial solution is the only solution, and so the columns of A are linearly independent. 40. An m×n matrix with n pivot columns has a pivot in each column. So the equation Ax = b has no free variables. If there is a solution, it must be unique. −3 4 −2 −1 8 −9 41. [M] A = 6 5 8 0 ~ 0 0 −3 5/8 0 0 0 5 0 0 0 5 2 7 −7 25 / 8 0 0 −7 11 −4 0 2 8 −7 0 ~ 4 0 10 0 2 8 −19 / 4 0 ~ 22 / 5 0 77 / 5 0 −3 5/8 1/ 4 7/8 −3 5/8 0 0 0 5 2 7 0 5 0 0 −7 25 / 8 5/ 4 35 / 8 −7 25 / 8 0 0 2 −19 / 4 5/ 2 35 / 4 2 −19 / 4 22 / 5 0 Copyright © 2016 Pearson Education, Ltd. 1-54 CHAPTER 1 • Linear Equations in Linear Algebra −3 4 −2 −1 8 −9 The pivot columns of A are 1, 2, and 5. Use them to form B = 6 5 8 −9 Other likely choices use columns 3 or 4 of A instead of 2: 6 5 2 −7 . 4 10 2 8 −7 −9 , 4 6 10 5 0 5 2 7 −7 11 −4 0 2 −7 . 4 10 Actually, any set of three columns of A that includes column 5 will work for B, but the concepts needed to prove that are not available now. (Column 5 is not in the two-dimensional subspace spanned by the first four columns.) 42. [M] 12 −7 9 −4 8 10 −6 9 −3 7 −6 4 −9 1 −5 −3 7 −5 6 −9 7 −9 5 −8 11 10 12 5 0 −1 ~ ⋅⋅⋅ ~ 0 9 0 −8 0 10 −1/ 6 0 0 0 −6 1/ 2 0 0 0 −3 21/ 4 89 / 2 0 0 7 −59 /12 −89 / 2 0 0 12 −7 The pivot columns of A are 1, 2, 4, and 6. Use them to form B = 9 −4 8 10 −6 9 −3 7 10 65 / 6 89 3 0 −3 7 −5 6 −9 10 5 −1 . 9 −8 Other likely choices might use column 3 of A instead of 2, and/or use column 5 instead of 4. 43. [M] Make v any one of the columns of A that is not in B and row reduce the augmented matrix [B v]. The calculations will show that the equation Bx = v is consistent, which means that v is a linear combination of the columns of B. Thus, each column of A that is not a column of B is in the set spanned by the columns of B. 44. [M] Calculations made as for Exercise 43 will show that each column of A that is not a column of B is in the set spanned by the columns of B. Reason: The original matrix A has only four pivot columns. If one or more columns of A are removed, the resulting matrix will have at most four pivot columns. (Use exactly the same row operations on the new matrix that were used to reduce A to echelon form.) If v is a column of A that is not in B, then row reduction of the augmented matrix [B v] will display at most four pivot columns. Since B itself was constructed to have four pivot columns, adjoining v cannot produce a fifth pivot column. Thus the first four columns of [B v] are the pivot columns. This implies that the equation Bx = v has a solution. Note: At the end of Section 1.7, the Study Guide has another note to students about “Mastering Linear Algebra Concepts.” The note describes how to organize a review sheet that will help students form a mental image of linear independence. The note also lists typical misuses of terminology, in which an Copyright © 2016 Pearson Education, Ltd. 1.9 • Solutions 1-55 adjective is applied to an inappropriate noun. (This is a major problem for my students.) I require my students to prepare a review sheet as described in the Study Guide, and I try to make helpful comments on their sheets. I am convinced, through personal observation and student surveys, that the students who prepare many of these review sheets consistently perform better than other students. Hopefully, these students will remember important concepts for some time beyond the final exam. 1.8 SOLUTIONS Notes: The key exercises are 17–20, 25 and 31. Exercise 20 is worth assigning even if you normally assign only odd exercises. Exercise 25 (and 27) can be used to make a few comments about computer graphics, even if you do not plan to cover Section 2.6. For Exercise 31, the Study Guide encourages students not to look at the proof before trying hard to construct it. Then the Guide explains how to create the proof. Exercises 19 and 20 provide a natural segue into Section 1.9. I arrange to discuss the homework on these exercises when I am ready to begin Section 1.9. The definition of the standard matrix in Section 1.9 follows naturally from the homework, and so I’ve covered the first page of Section 1.9 before students realize we are working on new material. The text does not provide much practice determining whether a transformation is linear, because the time needed to develop this skill would have to be taken away from some other topic. If you want your students to be able to do this, you may need to supplement Exercises 29, 30, 32 and 33. If you skip the concepts of one-to-one and “onto” in Section 1.9, you can use the result of Exercise 31 to show that the coordinate mapping from a vector space onto n (in Section 4.4) preserves linear independence and dependence of sets of vectors. (See Example 6 in Section 4.4.) 2 1. T(u) = Au = 0 0 1 2 , T(v) = = 2 −3 −6 .5 2. T(u) = Au = 0 0 0 .5 0 1 3. [ A b ] = −2 3 1 ~ 0 0 0 1 0 −2 2 1 −3 1 0 0 0 1 0 a 2a = 2 b 2b 0 1 .5 0 0 = 0 , T(v) = .5 −4 −2 −2 6 −5 −1 1 7 ~ 0 −3 0 −1 1 5 ~ 0 2 0 −3 1 −5 1 4. [ A= b ] 0 3 1 ~ 0 0 0 1 −2 2 0 2 −4 −9 0 1 0 0 0 1 6 1 −7 ~ 0 −9 0 4 1 −3 ~ 0 1 0 0 1 0 0 0 1 0 1 −2 .5 0 0 a .5a 0 .5 0 b = .5b 0 0 .5 c .5c −2 −1 1 0 −2 −1 2 5 ~ 0 1 2 5 1 0 0 0 5 10 3 1 2 3 x = 1 , unique solution 2 −3 1 4 2 −4 −15 6 1 −7 ~ 0 −27 0 −3 1 0 2 −4 1 −5 −5 −3 x = −3 , unique solution 1 1 Copyright © 2016 Pearson Education, Ltd. 6 −7 1 1-56 CHAPTER 1 1 5. [ A b ] = −3 • Linear Equations in Linear Algebra −5 7 −7 5 −2 1 ~ −2 0 −5 1 −7 2 −2 1 ~ 1 0 0 1 3 2 3 1 3 Note that a solution is not . To avoid this common error, write the equations: 1 x1 x2 x1= 3 − 3 x3 = 3 and solve for the basic variables: x2 = 1 − 2 x3 = 1 x is free 3 + 3 x3 + 2 x3 x1 3 − 3 x3 3 −3 General solution x = x2 =1 − 2 x3 = 1 + x3 −2 . For a particular solution, one might choose x3 x3 0 1 3 x3 = 0 and x = 1 . 0 1 3 6. [ A b ] = 0 −3 x1 x2 −2 −4 1 5 + 3 x3 + x3 1 5 1 −4 1 1 9 0 ~ 3 0 −6 0 −2 2 1 −1 1 2 1 −1 1 1 6 0 ~ 3 0 −3 0 −2 1 0 0 1 1 0 0 1 1 3 0 ~ 0 0 0 0 0 1 0 0 3 1 0 0 7 3 0 0 x1= 7 − 3 x3 = 7 . x2 = 3 − x3 = 3 x is free 3 x1 7 − 3 x3 7 −3 7 General solution: x = x2 = 3 − x3 = 3 + x3 −1 , one choice: 3 . x3 x3 0 1 0 7. a = 5; the domain of T is 5 , because a 6×5 matrix has 5 columns and for Ax to be defined, x must be in 5 . b = 6; the codomain of T is 6 , because Ax is a linear combination of the columns of A, and each column of A is in 6 . 8. A must have 5 rows and 4 columns. For the domain of T to be 4 , A must have four columns so that Ax is defined for x in 4 . For the codomain of T to be 5 , the columns of A must have five entries (in which case A must have five rows), because Ax is a linear combination of the columns of A. 1 9. Solve Ax = 0. 0 2 −4 1 −6 7 −4 6 −5 3 −4 0 1 0 ~ 0 0 0 −4 1 2 7 −4 −8 −5 3 6 0 1 0 ~ 0 0 0 Copyright © 2016 Pearson Education, Ltd. −4 1 0 7 −4 0 −5 3 0 0 0 0 1.9 1 ~ 0 0 0 1 0 x1 x 2 x== x3 x4 −9 −4 0 3 1 0 0 9 2 0 0 3 0 1 3 2 3 3 −18 + 3 x3 + 2 x3 x1 x2 x1 x2 − 9 x3 − 4 x3 x 9 x3 − 7 x4 = + 7 x4 = 0 1 x 4 x3 − 3 x4 = + 3 x4 = 0, 2 x is free 0 = 0 3 x4 is free 9 x3 − 7 x4 9 −7 4 x − 3x 4 −3 4 3 = x3 + x4 1 0 x3 0 1 x4 1 1 10. Solve Ax = 0. 0 −2 1 0 ~ 0 0 0 0 0 7 3 0 • Solutions x4 9 3 2 0 0 1 0 0 ~ 0 0 0 0 2 −4 3 5 0 1 0 0 ~ 0 0 0 0 3 1 0 0 9 2 0 0 0 0 1 0 3 −3 1 9 9 −6 2 18 2 −6 3 9 0 1 0 0 ~ 0 0 0 0 x1 = −3 x3 = 0 x = −2 x 2 3 = 0 = x x is free 3 = 0 x4 = 0 0 1 0 0 0 1 0 0 ~ 0 0 0 0 3 1 −3 9 3 2 0 0 0 0 0 0 0 0 1 0 9 2 −6 18 2 3 −6 9 −3 x3 −3 −2 x −2 3 = x3 x3 1 0 0 11. Is the system represented by [A b] consistent? Yes, as the following calculation shows. 1 0 2 −4 1 −6 7 −4 6 −5 3 −4 −1 1 1 ~ 0 0 0 −4 1 2 7 −4 −8 −5 3 6 −1 1 1 ~ 0 2 0 −4 1 0 7 −4 0 −5 3 0 −1 1 0 The system is consistent, so b is in the range of the transformation x Ax . 12. Is the system represented by [A b] consistent? 1 1 0 −2 3 0 1 3 9 3 2 0 2 −4 3 5 −1 1 3 0 ~ −1 0 4 0 3 −3 1 9 9 −6 2 18 2 −6 3 9 −1 1 4 0 ~ −1 0 2 0 3 1 −3 9 9 2 −6 18 2 3 −6 9 −1 −1 4 2 2 −1 1 3 9 2 −1 1 3 9 0 1 2 3 −1 0 1 2 3 −1 ~ ~ 0 0 0 3 1 0 0 0 3 1 0 0 0 −18 11 0 0 0 0 17 The system is inconsistent, so b is not in the range of the transformation x Ax . Copyright © 2016 Pearson Education, Ltd. 0 0 0 0 1-57 1-58 CHAPTER 1 • Linear Equations in Linear Algebra 13. 14. A reflection through the origin. A contraction by the factor .5. The transformation in Exercise 13 may also be described as a rotation of π radians about the origin or a rotation of –π radians about the origin. 15. 16. A projection onto the x2-axis A reflection through the line x2 = x1. −1 −2 2 6 17. T(3u) = 3T(u) = 3 = , T(2v) = 2T(v) = 2 = , and 3 6 1 3 6 −2 4 T(3u + 2v) = 3T(u) + 2T(v) = + = . 3 6 9 18. Draw a line through w parallel to v, and draw a line through w parallel to u. See the left part of the figure below. From this, estimate that w = u + 2v. Since T is linear, T(w) = T(u) + 2T(v). Locate T(u) and 2T(v) as in the right part of the figure and form the associated parallelogram to locate T(w). 19. All we know are the images of e1 and e2 and the fact that T is linear. The key idea is to write 5 1 0 x = =5 − 3 =5 e1 − 3 e 2 . Then, from the linearity of T, write −3 0 1 2 −1 13 T(x) = T(5e1 – 3e2) = 5T(e1) – 3T(e2) = 5y1 – 3y2 = 5 − 3 = . 5 6 7 Copyright © 2016 Pearson Education, Ltd. 1.9 • Solutions 1-59 x1 x 1 0 To find the image of 1 , observe that x = x1 + x2 = x1e1 + x2e 2 . Then = 0 1 x2 x2 2 −1 2 x1 − x2 T(x) = T(x1e1 + x2e2) = x1T(e1) + x2T(e2) = x1 + x2 = 5 6 5 x1 + 6 x2 20. Use the basic definition of Ax to construct A. Write x −2 T (x) =x1 v1 + x2 v 2 =[ v1 v 2 ] 1 = x2 5 7 −2 x, A = −3 5 7 −3 21. a. True. Functions from n to m are defined before Fig. 2. A linear transformation is a function with certain properties. b. False. The domain is 5 . See the paragraph before Example 1. c. False. The range is the set of all linear combinations of the columns of A. See the paragraph before Example 1. d. False. See the paragraph after the definition of a linear transformation. e. True. See the paragraph following the box that contains equation (4). 22. a. True. See the paragraph following the definition of a linear transformation. b. False. If A is an m×n matrix, the codomain is m . See the paragraph before Example 1. c. False. The question is an existence question. See the remark about Example 1(d), following the solution of Example 1. d. True. See the discussion following the definition of a linear transformation. e. True. See the paragraph following equation (5). 23. 24. Given any x in n , there are constants c1, …, cp such that x = c1v1 + ∙∙∙ cpvp, because v1, …, vp span n n. Then, from property (5) of a linear transformation, T(x) = c1T(v1) + ∙∙∙ + cpT(vp) = c10 + ∙∙ + cp0 = 0 25. Any point x on the line through p in the direction of v satisfies the parametric equation x = p + tv for some value of t. By linearity, the image T(x) satisfies the parametric equation T(x) = T(p + tv) = T(p) + tT(v). If T(v) = 0, then T(x) = T(p) for all values of t, and the image of the original line is just a single point. Otherwise, T(p) + tT(v) is the parametric equation of a line through T(p) in the direction of T(v). 26. Any point x on the plane P satisfies the parametric equation x = su + tv for some values of s and t. By linearity, the image T(x) satisfies the parametric equation T(x) = sT(u) + tT(v), s, t in . The set Copyright © 2016 Pearson Education, Ltd. 1-60 CHAPTER 1 • Linear Equations in Linear Algebra of images is just Span{T(u), T(v)}. If T(u) and T(v) are linearly independent, Span{T(u), T(v)} is a plane through T(u), T(v), and 0. If T(u) and T(v) are linearly dependent and not both zero, then Span{T(u), T(v)} is a line through 0. If T(u) = T(v) = 0, then Span{T(u), T(v)} is {0}. 27. a. From the figure in exercises 21 and 22 for Section 1.5, the line through p and q is in the direction of q – p, and so the equation of the line is x = p + t(q – p) = p + tq – tp = (1 – t)p + tq. b. Consider x = (1 – t)p + tq for t such that 0 < t < 1. Then, by linearity of T, T(x) = T((1 – t)p + tq) = (1 – t)T(p) + tT(q), for 0 < t < 1. If T(p) and T(q) are distinct, then (1 – t)T(p) + tT(q) is the equation for the line segment between T(p) and T(q), as shown in part (a) Otherwise, the set of images is just the single point T(p), because (1 – t)T(p) + tT(q) =(1 – t)T(p) + tT(p) = T(p) 28. Consider a point x in the parallelogram determined by u and v, say x = au + bv for 0 < a < 1, 0 < b < 1. By linearity of T, the image of x is T(x) = T(au + bv) = aT(u) + bT(v), for 0 < a < 1, and 0 < b < 1. This image point lies in the parallelogram determined by T(u) and T(v). Special “degenerate” cases arise when T(u) and T(v) are linearly dependent. If one of the images is not zero, then the “parallelogram” is actually the line segment from 0 to T(u) + T(v). If both T(u) and T(v) are zero, then the parallelogram is just {0}. Another possibility is that even u and v are linearly dependent, in which case the original parallelogram is degenerate (either a line segment or the zero vector). In this case, the set of images must be degenerate, too. 29. a. When b = 0, f (x) = mx. In this case, for all x,y in and all scalars c and d, f (cx + dy) = m(cx + dy) = mcx + mdy = c(mx) + d(my) = c·f (x) + d∙f (y) This shows that f is linear. b. When f (x) = mx + b, with b nonzero, f(0) = m(0) + b = b ≠ 0. This shows that f is not linear, because every linear transformation maps the zero vector in its domain into the zero vector in the codomain. (In this case, both zero vectors are just the number 0.) Another argument, for instance, would be to calculate f (2x) = m(2x) + b and 2f (x) = 2mx + 2b. If b is nonzero, then f (2x) is not equal to 2f (x) and so f is not a linear transformation. c. In calculus, f is called a “linear function” because the graph of f is a line. 30. Let T(x) = Ax + b for x in n . If b is not zero, T(0) = A0 + b = b ≠ 0. Actually, T fails both properties of a linear transformation. For instance, T(2x) = A(2x) + b = 2Ax + b, which is not the same as 2T(x) = 2(Ax + b) = 2Ax + 2b. Also, T(x + y) = A(x + y) + b = Ax + Ay + b which is not the same as T(x) + T(y) = Ax + b + Ay + b. 31. (The Study Guide has a more detailed discussion of the proof.) Suppose that {v1, v2, v3} is linearly dependent. Then there exist scalars c1, c2, c3, not all zero, such that c1v1 + c2v2 + c3v3 = 0. Then T(c1v1 + c2v2 + c3v3) = T(0) = 0. Since T is linear,c1T(v1) + c2T(v2) + c3T(v3) = 0. Since not all the weights are zero, {T(v1), T(v2), T(v3)} is a linearly dependent set. 32. Take any vector (x1, x2) with x2 ≠ 0, and use a negative scalar. For instance, T(0, 1) = (–2, 3), but T(–1∙(0, 1)) = T(0, –1) = (2, 3) ≠ (–1)∙T(0, 1). 33. One possibility is to show that T does not map the zero vector into the zero vector, something that every linear transformation does do. T(0, 0) = (0, 4, 0). 34. Suppose that {u, v} is a linearly independent set in n and yet T(u) and T(v) are linearly dependent. Then there exist weights c1, c2, not both zero, such that c1T(u) + c2T(v) = 0. Because T is linear, Copyright © 2016 Pearson Education, Ltd. 1.9 • Solutions 1-61 T(c1u + c2v) = 0. That is, the vector x = c1u + c2v satisfies T(x) = 0. Furthermore, x cannot be the zero vector, since that would mean that a nontrivial linear combination of u and v is zero, which is impossible because u and v are linearly independent. Thus, the equation T(x) = 0 has a nontrivial solution. 35. Take u and v in 3 and let c and d be scalars. Then cu + dv = (cu1 + dv1, cu2 + dv2, cu3 + dv3). The transformation T is linear because T(cu + dv) = (cu1 + dv1, cu2 + dv2, – (cu3 + dv3)) = (cu1 + dv1, cu2 + dv2, − cu3 − dv3) = (cu1, cu2, − cu3) + (dv1, dv2, − dv3) = c(u1, u2, − u3) + d(v1, v2, − v3) = cT(u) + dT(v) 36. Take u and v in 3 and let c and d be scalars. Then cu + dv = (cu1 + dv1, cu2 + dv2, cu3 + dv3). The transformation T is linear because T(cu + dv) = (cu1 + dv1, 0, cu3 + dv3) = (cu1, 0, cu3) + (dv1, 0, dv3) = c(u1, 0, u3) + d(v1, 0, v3) = cT(u) + dT(v) 4 −9 37. [M] −6 5 −2 7 4 −3 5 −8 5 8 −5 0 3 −4 0 1 0 0 ~ 0 0 0 0 0 1 0 0 0 0 1 0 −7 / 2 −9 / 2 0 0 −9 5 38. [M] 7 9 −4 −8 11 −7 −9 −7 16 −4 4 6 −9 5 0 1 0 0 ~ 0 0 0 0 0 1 0 0 0 0 1 0 3/ 4 5/ 4 −7 / 4 0 0 0 , 0 0 0 0 , 0 0 x1 = (7 / 2) x4 x = (9 / 2) x 2 4 x 0 = 3 x4 is free x1 = −(3/ 4) x4 x = −(5 / 4) x 2 4 x (7 / 4) x = 4 3 x is free 4 7 / 2 9 / 2 x = x4 0 1 −3/ 4 −5 / 4 x = x4 7/4 1 5 −5 7 1 0 0 −7 / 2 4 4 −2 −9 7 −8 0 5 0 1 0 −9 / 2 7 39. [M] , yes, b is in the range of the ~ −6 4 5 3 9 0 0 1 0 1 8 −4 7 0 0 0 0 0 5 −3 transformation, because the augmented matrix shows a consistent system. In fact, x1= 4 + (7 / 2) x4 4 x = 7 + (9 / 2) x 7 4 ; when x4 = 0 a solution is x = . the general solution is 2 1 x3 = 1 0 x4 is free 4 − 7 1 0 0 3/ 4 −5 / 4 −9 −4 −9 5 −8 −7 6 −7 0 1 0 5 / 4 −11/ 4 40. [M] , yes, b is in the range of the ~ 7 11 16 −9 13 0 0 1 −7 / 4 13/ 4 5 −5 0 0 0 0 0 9 −7 −4 transformation, because the augmented matrix shows a consistent system. In fact, Copyright © 2016 Pearson Education, Ltd. 1-62 CHAPTER 1 • Linear Equations in Linear Algebra −5 / 4 − (3/ 4) x4 x1 = −2 x = −4 −11/ 4 − (5 / 4) x4 the general solution is 2 ; when x4 = 1 a solution is x = . 5 = x3 13/ 4 + (7 / 4) x4 x4 is free 1 Notes: At the end of Section 1.8, the Study Guide provides a list of equations, figures, examples, and connections with concepts that will strengthen a student’s understanding of linear transformations. I encourage my students to continue the construction of review sheets similar to those for “span” and “linear independence,” but I refrain from collecting these sheets. At some point the students have to assume the responsibility for mastering this material. If your students are using MATLAB or another matrix program, you might insert the definition of matrix multiplication after this section, and then assign a project that uses random matrices to explore properties of matrix multiplication. See Exercises 34–36 in Section 2.1. Meanwhile, in class you can continue with your plans for finishing Chapter 1. When you get to Section 2.1, you won’t have much to do. The Study Guide’s MATLAB note for Section 2.1 contains the matrix notation students will need for a project on matrix multiplication. The appendices in the Study Guide have the corresponding material for Mathematica, Maple, and the T-83+/86/89 and HP-48G graphic calculators. 1.9 SOLUTIONS Notes: This section is optional if you plan to treat linear transformations only lightly, but many instructors will want to cover at least Theorem 10 and a few geometric examples. Exercises 15 and 16 illustrate a fast way to solve Exercises 17–22 without explicitly computing the images of the standard basis. The purpose of introducing one-to-one and onto is to prepare for the term isomorphism (in Section 4.4) and to acquaint math majors with these terms. Mastery of these concepts would require a substantial digression, and some instructors prefer to omit these topics (and Exercises 25–40). In this case, you can use the result of Exercise 31 in Section 1.8 to show that the coordinate mapping from a vector space onto n (in Section 4.4) preserves linear independence and dependence of sets of vectors. (See Example 6 in Section 4.4.) The notions of one-to-one and onto appear in the Invertible Matrix Theorem (Section 2.3), but can be omitted there if desired Exercises 25–28 and 31–36 offer fairly easy writing practice. Exercises 31, 32, and 35 provide important links to earlier material. −5 2 . 0 0 4 1 2. A = [T(e1) T(e2) T(e3)] = 3 −7 3 1 1. A = [T(e1) T(e2)] = 3 1 3. T(e1) = –e2, T(e2) = e1. A = [ −e 2 −5 4 0 1 e1 ] = −1 0 1/ 2 1/ 2 1/ 2 4. T(e1) = , A = , T(e2) = −1/ 2 1/ 2 −1/ 2 1/ 2 1/ 2 Copyright © 2016 Pearson Education, Ltd. 1.9 1 5. T(e1) = e1 – 2e2 = , T(e2) = e2, A = −2 3 6. T(e1) = e1, T(e2) = e2 + 3e1 = , A = 1 1 −2 1 0 • Solutions 1-63 0 1 3 1 7. Follow what happens to e1 and e2. Since e1 is on the unit circle in the plane, it rotates through –3π/4 radians into a point on the unit circle that lies in the third quadrant and on the line x2 = x1 (that is, y = x in more familiar notation). The point (–1,–1) is on the line x2 = x1 , but its distance from the origin is 2. So the rotational image of e1 is (–1/ 2, –1/ 2) . Then this image reflects in the horizontal axis to (–1/ 2,1/ 2) . Similarly, e2 rotates into a point on the unit circle that lies in the fourth quadrant and on the line x2 = − x1 , namely, (1 / 2, –1 / 2) . Then this image reflects in the horizontal axis to (1 / 2,1 / 2) . When the two calculations described above are written in vertical vector notation, the transformation’s standard matrix [T(e1) T(e2)] is easily seen: −1/ 2 −1/ 2 −1/ 2 1/ 2 1/ 2 e1 → → , e2 → → , A= −1/ 2 1/ 2 1/ 2 −1/ 2 1/ 2 8. e1 → e1 → e 2 and e 2 → −e 2 → −e1 , so= A [e 2 0 e1 ] −= 1 1/ 2 1/ 2 −1 0 9. The horizontal shear maps e1 onto e1, and then the reflection in the line x2 = –x1 maps e1 into –e2. (See Table 1.) The horizontal shear maps e2 into e2 into e2 – 2e1. To find the image of e2 – 2e1 when it is reflected in the line x2 = –x1, use the fact that such a reflection is a linear transformation. So, the image of e2 – 2e1 is the same linear combination of the images of e2 and e1, namely, –e1 – 2(–e2) = – e1 + 2e2. To summarize, 0 −1 . e1 → e1 → −e 2 and e 2 → e 2 − 2e1 → −e1 + 2e 2 , so A = 2 −1 To find the image of e2 – 2e1 when it is reflected through the vertical axis use the fact that such a reflection is a linear transformation. So, the image of e2 – 2e1 is the same linear combination of the images of e2 and e1, namely, e2 + 2e1. 0 −1 10. e1 → −e1 → −e 2 and e 2 → e 2 → −e1 , so A = −1 0 11. The transformation T described maps e1 → e1 → −e1 and maps e 2 → −e 2 → −e 2 . A rotation through π radians also maps e1 into –e1 and maps e2 into –e2. Since a linear transformation is completely Copyright © 2016 Pearson Education, Ltd. 1-64 CHAPTER 1 • Linear Equations in Linear Algebra determined by what it does to the columns of the identity matrix, the rotation transformation has the same effect as T on every vector in . 2 12. The transformation T in Exercise 8 maps e1 → e1 → e 2 and maps e 2 → −e 2 → −e1 . A rotation about the origin through π / 2 radians also maps e1 into e2 and maps e2 into –e1. Since a linear transformation is completely determined by what it does to the columns of the identity matrix, the rotation transformation has the same effect as T on every vector in . 2 13. Since (2, 1) = 2e1 + e2, the image of (2, 1) under T is 2T(e1) + T(e2), by linearity of T. On the figure in the exercise, locate 2T(e1) and use it with T(e2) to form the parallelogram shown below. 14. Since T(x) = Ax = [a1 a2]x = x1a1 + x2a2 = –a1 + 3a2, when x = (–1, 3), the image of x is located by forming the parallelogram shown below. 3 15. By inspection, 4 1 0 0 −1 1 16. By inspection, −2 1 −2 x1 3 x1 − 2 x3 0 x2 = 4 x1 1 x3 x1 − x2 + x3 −1 x1 − x2 x1 1 =− 2 x1 + x2 x 0 2 x1 17. To express T(x) as Ax , write T(x) and x as column vectors, and then fill in the entries in A by inspection, as done in Exercises 15 and 16. Note that since T(x) and x have four entries, A must be a 4×4 matrix. 0 x + x T(x) = 1 2 = x2 + x3 x3 + x4 x1 x 2 = A x3 x4 0 1 0 0 0 1 1 0 0 0 1 1 0 x1 0 x2 0 x3 1 x4 Copyright © 2016 Pearson Education, Ltd. 1.9 • Solutions 1-65 18. As in Exercise 17, write T(x) and x as column vectors. Since x has 2 entries, A has 2 columns. Since T(x) has 4 entries, A has 4 rows. 2 x2 − 3 x1 x − 4x 1 2 = 0 x2 x = A 1 x2 −3 1 0 0 2 −4 x1 0 x2 1 19. Since T(x) has 2 entries, A has 2 rows. Since x has 3 entries, A has 3 columns. x1 x1 − 5 x2 + 4 x3 1 A x2 = = x 0 x2 − 6 x3 3 −5 1 x1 4 x2 −6 x3 20. Since T(x) has 1 entry, A has 1 row. Since x has 4 entries, A has 4 columns. x1 x1 x 2 [2 0 3 −4] x2 x A [2 x1 + 3 x= − 4 ] [ = ] 3 4 x3 x3 x4 x4 3 x1 + x2 x 1 1 x1 21. T(x) = = . To solve T(x) = , row reduce the augmented = A 1 8 x2 4 5 x2 4 x1 + 5 x2 3 1 0 7 1 1 3 1 1 7 matrix: ~ ~ , x= . 4 5 8 0 1 −4 0 1 −4 −4 x1 − 2 x2 1 x1 22. T(x) = − x1 + 3 x2 = A = −1 x 3 x1 − 2 x2 2 3 augmented matrix: 1 −1 3 −2 3 −2 −1 1 4 ~ 0 9 0 −2 1 4 −1 1 3 ~ 0 12 0 −2 x 3 1 . To solve T(x) = x −2 2 −2 1 0 −1 1 3 ~ 0 0 0 0 1 0 −1 4 , row reduce the 9 5 5 3 , x = . 3 0 23. a. True. See Theorem 10. b. True. See Example 3. c. False. See the paragraph before Table 1. d. False. See the definition of onto. Any function from n to m maps each vector onto another vector. e. False. See Example 5. 24. a. False. See the paragraph preceding Example 2. b. True. See Theorem 10. c. True. See Table 1. Copyright © 2016 Pearson Education, Ltd. 1-66 CHAPTER 1 • Linear Equations in Linear Algebra d. False. See the definition of one-to-one. Any function from n to m maps a vector onto a single (unique) vector. e. True. See the solution of Example 5. 25. Three row interchanges on the standard matrix A of the transformation T in Exercise 17 produce 1 1 0 0 0 1 1 0 . This matrix shows that A has only three pivot positions, so the equation Ax = 0 has 0 0 1 1 0 0 0 0 a nontrivial solution. By Theorem 11, the transformation T is not one-to-one. Also, since A does not have a pivot in each row, the columns of A do not span 4 . By Theorem 12, T does not map 4 onto 4 . 26. The standard matrix A of the transformation T in Exercise 2 is 2×3. Its columns are linearly dependent because A has more columns than rows. So T is not one-to-one, by Theorem 12. Also, A is 4 −5 1 , which shows that the rows of A span 2 . By Theorem 12, T row equivalent to 0 −19 19 maps 3 onto 2 . 4 1 −5 27. The standard matrix A of the transformation T in Exercise 19 is . The columns of A 1 −6 0 are linearly dependent because A has more columns than rows. So T is not one-to-one, by Theorem 12. Also, A has a pivot in each row, so the rows of A span 2 . By Theorem 12, T maps 3 onto 2 . 28. The standard matrix A of the transformation T in Exercise 14 has linearly independent columns, because the figure in that exercise shows that a1 and a2 are not multiples. So T is one-to-one, by Theorem 12. Also, A must have a pivot in each column because the equation Ax = 0 has no free * variables. Thus, the echelon form of A is . Since A has a pivot in each row, the columns of A 0 span 2 . So T maps 2 onto 2 . An alternate argument for the second part is to observe directly from the figure in Exercise 14 that a1 and a2 span 2 . This is more or less evident, based on experience with grids such as those in Figure 8 and Exercise 7 of Section 1.3. 29. By Theorem 12, the columns of the standard matrix A must be linearly independent and hence the equation Ax = 0 has no free variables. So each column of A must be a pivot column: * * 0 * . Note that T cannot be onto because of the shape of A. A~ 0 0 0 0 0 Copyright © 2016 Pearson Education, Ltd. 1.9 • Solutions 1-67 30. By Theorem 12, the columns of the standard matrix A must span 3 . By Theorem 4, the matrix must have a pivot in each row. There are four possibilities for the echelon form: * * * * * * * * * 0 * * 0 * * , 0 * * , 0 0 * , 0 0 * 0 0 * 0 0 0 0 0 0 0 0 0 Note that T cannot be one-to-one because of the shape of A. 31. “T is one-to-one if and only if A has n pivot columns.” By Theorem 12(b), T is one-to-one if and only if the columns of A are linearly independent. And from the statement in Exercise 30 in Section 1.7, the columns of A are linearly independent if and only if A has n pivot columns. 32. The transformation T maps n onto m if and only if the columns of A span m , by Theorem 12. This happens if and only if A has a pivot position in each row, by Theorem 4 in Section 1.4. Since A has m rows, this happens if and only if A has m pivot columns. Thus, “T maps n onto m if and only A has m pivot columns.” 33. Define T : → by T(x) = Bx for some m×n matrix B, and let A be the standard matrix for T. By definition, A = [T(e1) ⋅ ⋅ ⋅ T(en)], where ej is the jth column of In. However, by matrix-vector multiplication, T(ej) = Bej = bj, the jth column of B. So A = [b1 ⋅ ⋅ ⋅ bn] = B. n m 34. The transformation T maps n onto m if and only if for each y in m there exists an x in n such that y = T(x). 35. If T : → maps n onto m , then its standard matrix A has a pivot in each row, by Theorem 12 and by Theorem 4 in Section 1.4. So A must have at least as many columns as rows. That is, m < n. When T is one-to-one, A must have a pivot in each column, by Theorem 12, so m > n. n m 36. Take u and v in p and let c and d be scalars. Then T(S(cu + dv)) = T(c⋅S(u) + d⋅S(v)) because S is linear = c⋅T(S(u)) + d⋅T(S(v)) because T is linear This calculation shows that the mapping x → T(S(x)) is linear. See equation (4) in Section 1.8 −5 8 37. [M] 4 −3 −5 −4 5 5 4 1 0 0 44 / 35 1 0 0 1.2571 0 1 0 79 / 35 0 1 0 2.2571 7 ~ . There is no pivot in ~ ⋅⋅⋅ ~ 0 0 1 86 / 35 0 0 1 2.4571 −3 4 0 0 0 0 0 0 0 0 the fourth column of the standard matrix A, so the equation Ax = 0 has a nontrivial solution. By Theorem 11, the transformation T is not one-to-one. (For a shorter argument, use the result of Exercise 31.) 10 3 −9 −2 5 4 −9 0 7 0 7 1 10 6 16 −4 0 1 −9 0 . No. There is no pivot in the third column of 38. [M] ~ ⋅⋅⋅ ~ 12 0 8 12 7 0 0 1 5 0 0 0 −8 −6 −2 0 the standard matrix A, so the equation Ax = 0 has a nontrivial solution. By Theorem 11, the transformation T is not one-to-one. (For a shorter argument, use the result of Exercise 31.) Copyright © 2016 Pearson Education, Ltd. 1-68 CHAPTER 1 • Linear Equations in Linear Algebra 3 7 5 0 0 5 0 4 −7 1 6 −8 5 12 −8 1 0 1 0 0 39. [M] −7 10 −8 −9 14 ~ ⋅⋅⋅ ~ 0 0 1 −2 0 . There is not a pivot in every row, 4 2 −6 0 0 0 1 3 −5 0 −5 0 6 −6 −7 3 0 0 0 0 so the columns of the standard matrix do not span 5 . By Theorem 12, the transformation T does not map 5 onto 5 . −1 0 0 0 5 1 4 1 0 0 −4 0 −9 ~ ⋅⋅⋅ ~ 0 0 1 0 0 . There is not a pivot in every row, 8 0 0 1 1 0 11 0 0 0 0 0 so the columns of the standard matrix do not span 5 . By Theorem 12, the transformation T does not map 5 onto 5 9 14 40. [M] −8 −5 13 1.10 13 15 −9 −6 14 5 −7 12 −8 15 6 −6 −5 9 2 SOLUTIONS 1. a. If x1 is the number of servings of Cheerios and x2 is the number of servings of 100% Natural Cereal, then x1 and x2 should satisfy nutrients nutrients quantities x1 per serving + x2 per serving of = of nutrients of Cheerios 100% Natural required 110 130 295 4 3 9 . That is, x1 +x = 20 2 18 48 2 5 8 Copyright © 2016 Pearson Education, Ltd. 1.10 110 4 b. The equivalent matrix equation is 20 2 2.5 −7 −16 −145 4 1 −7 0 ~ −16 0 −145 0 2.5 1 0 0 1-69 130 295 3 x1 9 . To solve this, row reduce the = 18 x2 48 5 8 110 4 augmented matrix for this equation. 20 2 1 0 ~ 0 0 • Solutions 130 3 18 5 4 1 1 0 ~ 0 0 0 0 0 1 0 0 295 2 9 4 ~ 48 20 8 110 1.5 1 0 0 5 3 18 130 8 1 2.5 9 4 3 ~ 48 10 9 295 110 130 4 9 24 295 The desired nutrients are provided by 1.5 servings of Cheerios together with 1 serving of 100% Natural Cereal. 2. Set up nutrient vectors for one serving of Shredded Wheat (SW) and Kellogg's Crispix (Crp): Nutrients: SW Crp calories 160 5 6 1 110 2 . .1 .4 protein fiber fat 160 110 5 2 3 a. Let B [SW = = = Crp ] , u . 6 .1 2 .4 1 Then Bu lists the amounts of calories, protein, carbohydrate, and fat in a mixture of three servings of Shredded Wheat and two servings of Crispix. b. [M] Let u1 and u2 be the number of servings of Shredded Wheat and Crispix, respectively. Can 120 u1 3.2 ? To find out, row reduce the augmented these numbers satisfy the equation B = u2 2.46 .64 matrix .4 .64 1 .4 .64 1 .4 .64 1 0 .4 160 110 130 1 5 2 3.2 0 0 0 0 46 27.6 0 1 .6 0 1 .6 ~ ~ ~ ~ 6 .1 2.46 0 −2.3 −1.38 0 −2.3 −1.38 0 0 0 0 0 0 .4 .64 0 46 27.6 0 0 0 0 0 0 0 0 0 1 Since the system is consistent, it is possible for a mixture of the two creals to provide the desired nutrients. The mixture is .4 servings of Shredded Wheat and .6 servings of Crispix. Copyright © 2016 Pearson Education, Ltd. 1-70 3. CHAPTER 1 • Linear Equations in Linear Algebra a. [M] Let x1, x2, and x3 be the number of servings of Annies’s Mac and Cheese, broccoli, and chicken, respectively, needed for the lunch. The values of x1, x2, and x3 should satisfy nutrients nutrients nutrients quantities x1 per serving + x2 per serving + x3 per serving = of nutrients of Mac and Cheese of broccoli of chicken required From the given data, 270 51 70 400 30 x1 10 + x2 5.4 + x3 15 = 2 5.2 0 10 To solve, row reduce the corresponding augmented matrix: 270 10 2 51 5.4 5.2 70 15 0 400 1 30 ~ ... ~ 0 0 10 0 1 0 0 .99 .99 servings of Mac and Cheese 0 1.54 ; x ≈ 1.54 = servings of broccoli 1 .79 .79 servings of chicken b. [M] Changing from Annie’s Mac and Cheese to Annie’s Whole Wheat Shells and White Cheddar 260 51 70 400 30 9 + x 5.4 + x 15 = changes the vector equation to x1 2 3 5 5.2 0 10 To solve, row reduce the corresponding augmented matrix: 260 9 5 51 5.4 5.2 70 15 0 400 1 30 ~ ... ~ 0 0 10 0 1 0 0 1.09 1.09 servings of Shells 0 .88 ; x ≈ .88 = servings of broccoli 1 1.03 1.03 servings of chicken Notice that the number of servings of broccoli has decreased as was desired. 4. Here are the data, assembled from Table 1 and Exercise 4: Nutrient Mg of Nutrients/Unit soy soy milk flour whey prot. protein carboh. fat calcium 36 52 0 1.26 51 34 7 .19 13 74 1.1 .8 80 0 3.4 .18 Nutrients Required (milligrams) 33 45 3 .8 a. Let x1, x2, x3, x4 represent the number of units of nonfat milk, soy flour, whey, and isolated soy protein, respectively. These amounts must satisfy the following matrix equation 36 52 0 1.26 51 34 7 .19 13 74 1.1 .8 80 x1 33 0 x2 45 = 3.4 x3 3 .18 x4 .8 Copyright © 2016 Pearson Education, Ltd. 1.10 36 52 b. [M] 0 1.26 51 34 7 .19 13 74 1.1 .8 80 0 3.4 .18 1 33 45 0 ~ ⋅⋅⋅ ~ 3 0 .8 0 0 1 0 0 0 0 1 0 0 0 0 1 • Solutions 1-71 .64 .54 −.09 −.21 The “solution” is x1 = .64, x2 = .54, x3 = –.09, x4 = –.21. This solution is not feasible, because the mixture cannot include negative amounts of whey and isolated soy protein. Although the coefficients of these two ingredients are fairly small, they cannot be ignored. The mixture of .64 units of nonfat milk and .54 units of soy flour provide 50.6 g of protein, 51.6 g of carbohydrate, 3.8 g of fat, and .9 g of calcium. Some of these nutrients are nowhere close to the desired amounts. 5. Loop 1: The resistance vector is 11 −5 r1 = 0 0 Total of RI voltage drops for current I 1 Voltage drop for I 2 is negative; I 2 flows in opposite direction Current I 3 does not flow in loop 1 Current I4 does not flow in loop 1 Loop 2: The resistance vector is −5 10 r2 = −1 0 Voltage drop for I1 is negative; I1 flows in opposite direction Total of RI voltage drops for current I 2 Voltage drop for I 3 is negative; I 3 flows in opposite direction Current I 4 does not flow in loop 2 0 −1 Also, r3 = , r4 = 9 −2 0 0 , and R = [r1 r2 r3 r4] = −2 10 11 −5 0 0 −5 10 −1 0 0 −1 9 −2 0 0 . −2 10 Notice that each off-diagonal entry of R is negative (or zero). This happens because the loop current directions are all chosen in the same direction on the figure. (For each loop j, this choice forces the currents in other loops adjacent to loop j to flow in the direction opposite to current Ij.) 50 −40 Next, set v = . The voltages in loops 2 and 4 are negative because the battery orientation in 30 −30 each loop is opposite to the direction chosen for positive current flow. Thus, the equation Ri = v becomes 11 −5 0 0 −5 10 −1 0 0 −1 9 −2 0 I1 50 0 I 2 −40 . = −2 I 3 30 10 I 4 −30 I1 3.68 I −1.90 [M]: The solution is i = 2 = . I 3 2.57 I 4 −2.49 Copyright © 2016 Pearson Education, Ltd. 1-72 CHAPTER 1 • Linear Equations in Linear Algebra 6. Loop 1: The resistance vector is 6 −1 r1 = 0 0 Total of RI voltage drops for current I1 Voltage drop for I 2 is negative; I 2 flows in opposite direction Current I 3 does not flow in loop 1 Current I 4 does not flow in loop 1 Loop 2: The resistance vector is −1 9 r2 = −4 0 Voltage drop for I 1 is negative; I 1 flows in opposite direction Total of RI voltage drops for current I 2 Voltage drop for I 3 is negative; I 3 flows in opposite direction Current I 4 does not flow in loop 2 0 0 −4 0 Also, r3 = , r4 = , and R = [r1 r2 r3 r4]= −2 7 7 −2 v becomes 6 −1 0 0 −1 9 −4 0 0 −4 7 −2 6 −1 0 0 −1 9 −4 0 0 −4 7 −2 0 0 . Set v = −2 7 30 20 . Then Ri = 40 10 0 I1 30 I1 6.36 I 8.14 0 I 2 20 . . [M]: The solution is i = 2 = = I 3 11.73 −2 I 3 40 7 I 4 10 I 4 4.78 7. Loop 1: The resistance vector is 12 −7 r1 = 0 −4 Total of RI voltage drops for current I 1 Voltage drop for I 2 is negative; I 2 flows in opposite direction Current I 3 does not flow in loop 1 Voltage drop for I 4 is negative; I 4 flows in opposite direction Loop 2: The resistance vector is −7 15 r2 = −6 0 Voltage drop for I1 is negative; I1 flows in opposite direction Total of RI voltage drops for current I 2 Voltage drop for I 3 is negative; I 3 flows in opposite direction Current I 4 does not flow in loop 2 0 −6 Also, r3 = , r4 = 14 −5 −4 0 , and R = [r1 r2 r3 r4] = −5 13 12 −7 0 −4 −7 15 −6 0 0 −6 14 −5 −4 0 . −5 13 Notice that each off-diagonal entry of R is negative (or zero). This happens because the loop current directions are all chosen in the same direction on the figure. (For each loop j, this choice forces the currents in other loops adjacent to loop j to flow in the direction opposite to current Ij.) Copyright © 2016 Pearson Education, Ltd. 1.10 • Solutions 1-73 40 30 . Note the negative voltage in loop 4. The current direction chosen in loop 4 is Next, set v = 20 −10 opposed by the orientation of the voltage source in that loop. Thus Ri = v becomes 12 −7 0 −4 −7 15 −6 0 0 −6 14 −5 −4 I1 40 I1 11.43 I 10.55 0 I 2 30 . . [M]: The solution is i = 2 = = I 3 8.04 −5 I 3 20 13 I 4 −10 I 4 5.84 8. Loop 1: The resistance vector is 9 −1 r1 = 0 −1 −4 Total of RI voltage drops for current I1 Voltage drop for I 2 is negative; I 2 flows in opposite direction Current I 3 does not flow in loop 1 Voltage drop for I 4 is negative; I 4 flows in opposite direction Voltage drop for I 5 is negative; I 5 flows in opposite direction Loop 2: The resistance vector is −1 7 r2 = −2 0 −3 Voltage drop for I1 is negative; I1 flows in opposite direction Total of RI voltage drops for current I 2 Voltage drop for I 3 is negative; I 3 flows in opposite direction Current I4 does not flow in loop 2 Voltage drop for I5 is negative; I5 flows in opposite direction 0 −1 −4 0 −1 −4 9 −1 50 −2 0 −3 −1 7 −2 0 −3 −30 Also, r3 = 10 , r4 = −3 , r5 = −3 , and R = 0 −2 10 −3 −3 . Set v = 20 . Note the 0 −3 7 −2 7 −2 −1 −40 −3 −2 12 −4 −3 −3 −2 12 −3 0 negative voltages for loops where the chosen current direction is opposed by the orientation of the voltage source in that loop. Thus Ri = v becomes: 0 −1 −4 I1 50 I1 4.00 9 −1 −1 7 −2 0 −3 I 2 −30 I 2 −4.38 0 −2 10 −3 −3 I 3 = 20 . [M] The solution is I 3 = −.90 . 0 −3 7 −2 I 4 −40 I 4 −5.80 −1 I 5 −.96 −4 −3 −3 −2 12 I 5 0 9. The population movement problems in this section assume that the total population is constant, with no migration or immigration. The statement that “about 7% of the city’s population moves to the suburbs” means also that the rest of the city’s population (93%) remain in the city. This determines the entries in the first column of the migration matrix (which concerns movement from the city). Copyright © 2016 Pearson Education, Ltd. 1-74 CHAPTER 1 • Linear Equations in Linear Algebra From: City Suburbs .93 .07 To: City Suburbs Likewise, if 5% of the suburban population moves to the city, then the other 95% remain in the .93 .05 suburbs. This determines the second column of the migration matrix: M = . The .07 .95 800,000 difference equation is xk+1 = Mxk for k = 0, 1, 2, …. Also, x0 = 500,000 10. .93 The population in 2016 (when k = 1) is x1 = Mx0 = .07 .05 800,000 769,000 = .95 500,000 531,000 .93 The population in 2017 (when k = 2) is x2 = Mx1 = .07 .05 769,000 741,720 = .95 531,000 558, 280 The data in the first sentence implies that the migration matrix has the form: From: City Suburbs .06 .04 To: City Suburbs The remaining entries are determined by the fact that the numbers in each column must sum to 1. (For instance, if 6% of the city people move to the suburbs, then the rest, or 94%, remain in the city.) .94 .04 10,000,000 . The initial population is x0 = So the migration matrix is M = . .06 .96 800,000 .94 The population in 2016 (when k = 1) is x1 = Mx0 = .06 .04 10,000,000 9, 432,000 = .96 800,000 1,368,000 .94 The population in 2017 (when k = 2) is x2 = Mx1 = .06 .04 9, 432,000 8,920,800 = .96 1,368,000 1,879, 200 11. The problem concerns two groups of people–those living in California and those living outside California (and in the United States). It is reasonable, but not essential, to consider the people living inside California first. That is, the first entry in a column or row of a vector will concern the people living in California. With this choice, the migration matrix has the form: From: Calif. Outside To: Calif. Outside a. For the first column of the migration matrix M, compute persons {Calif. who moved } = {Total Calif. pop.} 748,252 = .01967 38,041,430 Copyright © 2016 Pearson Education, Ltd. 1.10 • Solutions 1-75 The other entry in the first column is 1 – .01967 = .98033. Whatever number of decimal places is used, it is important that the two entries sum to 1. For the second column of M, compute persons {outside who moved } = {Total outside pop.} 493,641 = .00179 . The other 275,872,610 entry is 1 – .00179 = .99821. Thus, the migration matrix is From: Calif. Outside To: .98033 .01967 .00179 .99821 Calif. Outside b. [M] The initial vector is x0 = (38.041, 275.872), with data in millions of persons. Since x0 describes the population in 2012, and x1 describes the population in 2013, the vector x10 describes the projected population for the year 2022, assuming that the migration rates remain constant and there are no deaths, births, or migration. Here are the vectors x0 through x10 with the first 5 figures displayed. Numbers are in millions of persons: 38.041 37.787 37.538 37.294 37.056 36.822 x0 = , , , , , = x5 275.87 276.13 276.62 276.62 276.86 277.09 36.594 36.371 36.152 35.938 35.729 x6 = , , , , = x10. 277.32 277.54 277.76 277.98 278.18 .97 12. Set M = .00 .03 .97 x2 = .00 .03 .05 .90 .05 .10 .97 .05 .10 295 304 295 .05 and x0 = 55 . Then x1 = .00 .90 .05 55 ≈ 57 , and 150 .03 .05 .85 150 139 .85 .10 304 312 .05 57 ≈ 58 . The entries in x2 give the approximate distribution of cars on .85 139 130 .05 .90 .05 Wednesday, two days after Monday. 13. [M] The order of entries in a column of a migration matrix must match the order of the columns. For instance, if the first column concerns the population in the city, then the first entry in each column must be the fraction of the population that moves to (or remains in) the city. In this case, the data in .95 .03 600,000 and x0 = the exercise leads to M = .05 .97 400,000 a. Some of the population vectors are 523, 293 472,737 439, 417 417, 456 x5 = = , x10 = , x15 = , x 20 476,707 527, 263 560,583 582,544 The data here shows that the city population is declining and the suburban population is increasing, but the changes in population each year seem to grow smaller. Copyright © 2016 Pearson Education, Ltd. 1-76 CHAPTER 1 • Linear Equations in Linear Algebra 350,000 b. When x0 = , the situation is different. Now 650,000 358,523 364,140 367,843 370, 283 = x5 = , x10 = , x15 = , x 20 641, 477 635,860 632,157 629,717 The city population is increasing slowly and the suburban population is decreasing. No other conclusions are expected. (This example will be analyzed in greater detail later in the text.) 14. Here are Figs. (a) and (b) for Exercise 14, followed by the figure for Exercise 33 in Section 1.1: 4T1 =0 + 20 + T2 + T4 4T2 = T1 + 20 + 0 + T3 For Fig. (a), the equations are 4T3 = T4 + T2 + 0 + 20 4T4 =0 + T1 + T3 + 20 To solve the system, rearrange the equations and row reduce the augmented matrix. Interchanging rows 1 and 4 speeds up the calculations. The first five steps are shown in detail. 4 −1 0 −1 1 0 ~ 0 0 −1 0 −1 4 −1 0 −1 4 −1 0 −1 4 0 1 −4 1 0 −1 0 4 −2 0 −4 14 20 1 20 −1 ~ 20 0 20 4 −20 1 0 0 ~ 20 0 100 0 0 1 −4 4 −1 0 −1 4 −1 −1 0 −1 0 1 −4 1 0 −1 0 4 −2 0 0 12 −20 1 20 0 ~ 20 0 20 0 0 1 −4 4 0 −4 −1 4 −1 −1 −4 15 1 0 0 ~ 20 0 100 0 0 0 0 1 0 0 10 0 1 0 10 0 0 1 10 −20 1 0 0 ~ ⋅⋅⋅ ~ 20 0 0 120 −20 4T1 = 10 + 0 + T2 + T4 For Fig (b), the equations are 4T2 = T1 + 0 + 40 + T3 4T3 = T4 + T2 + 40 + 10 4T4 = 10 + T1 + T3 + 10 Copyright © 2016 Pearson Education, Ltd. 10 −20 0 1 −4 1 0 −1 0 −1 4 −1 20 −1 −4 15 100 1.10 • Solutions 1-77 Rearrange the equations and row reduce the augmented matrix: 4 −1 0 −1 −1 0 −1 4 −1 0 −1 4 −1 0 −1 4 10 1 0 40 ~ ⋅⋅⋅ ~ 50 0 20 0 0 0 0 1 0 0 0 1 0 0 0 1 17.5 20 12.5 10 a. Here are the solution temperatures for the three problems studied: Fig. (a) in Exercise 14 of Section 1.10: (10, 10, 10, 10) Fig. (b) in Exercise 14 of Section 1.10: (10, 17.5, 20, 12.5) Figure for Exercises 33 in Section 1.1 (20, 27.5, 30, 22.5) When the solutions are arranged this way, it is evident that the third solution is the sum of the first two solutions. What might not be so evident is that the list of boundary temperatures of the third problem is the sum of the lists of boundary temperatures of the first two problems. (The temperatures are listed clockwise, starting at the left of T1.) Fig. (a): ( 0, 20, 20, 0, 0, 20, 20, 0) Fig. (b): (10, 0, 0, 40, 40, 10, 10, 10) Fig. from Section 1.1: (10, 20, 20, 40, 40, 30, 30, 10) b. When the boundary temperatures in Fig. (a) are multiplied by 3, the new interior temperatures are also multiplied by 3. c. The correspondence from the list of eight boundary temperatures to the list of four interior temperatures is a linear transformation. A verification of this statement is not expected. However, it can be shown that the solutions of the steady-state temperature problem here satisfy a superposition principle. The system of equations that approximate the interior temperatures can be written in the form Ax = b, where A is determined by the arrangement of the four interior points on the plate and b is a vector in 4 determined by the boundary temperatures. Note: The MATLAB box in the Study Guide for Section 1.10 discusses scientific notation and shows how to generate a matrix whose columns list the vectors x0, x1, x2, …, determined by an equation xk+1 = Mxk for k = 0 , 1, …. Chapter 1 SUPPLEMENTARY EXERCISES 1. a. False. (The word “reduced” is missing.) Counterexample: 2 1 2 1 1 2 = A = , B = , C 3 4 0 −2 0 1 The matrix A is row equivalent to matrices B and C, both in echelon form. b. False. Counterexample: Let A be any n×n matrix with fewer than n pivot columns. Then the equation Ax = 0 has infinitely many solutions. (Theorem 2 in Section 1.2 says that a system has either zero, one, or infinitely many solutions, but it does not say that a system with infinitely many solutions exists. Some counterexample is needed.) c. True. If a linear system has more than one solution, it is a consistent system and has a free variable. By the Existence and Uniqueness Theorem in Section 1.2, the system has infinitely many solutions. Copyright © 2016 Pearson Education, Ltd. 1-78 CHAPTER 1 • Linear Equations in Linear Algebra d. False. Counterexample: The following system has no free variables and no solution: x1 + x2 = 1 x1 x2 + x2 = 5 = 2 e. True. See the box after the definition of elementary row operations, in Section 1.1. If [A b] is transformed into [C d] by elementary row operations, then the two augmented matrices are row equivalent, and thus have the same solution set. f. True. Theorem 6 in Section 1.5 essentially says that when Ax = b is consistent, the solution sets of the nonhomogeneous equation and the homogeneous equation are translates of each other. In this case, the two equations have the same number of solutions. g. False. For the columns of A to span m , the equation Ax = b must be consistent for all b in m , not for just one vector b in m . h. False. Any matrix can be transformed by elementary row operations into reduced echelon form, but not every matrix equation Ax = b is consistent. i. True. If A is row equivalent to B, then A can be transformed by elementary row operations first into B and then further transformed into the reduced echelon form U of B. Since the reduced echelon form of A is unique, it must be U. j. False. Every equation Ax = 0 has the trivial solution whether or not some variables are free. k. True, by Theorem 4 in Section 1.4. If the equation Ax = b is consistent for every b in m , then A must have a pivot position in every one of its m rows. If A has m pivot positions, then A has m pivot columns, each containing one pivot position. l. False. The word “unique” should be deleted. Let A be any matrix with m pivot columns but more than m columns altogether. Then the equation Ax = b is consistent and has m basic variables and at least one free variable. Thus the equation does not have a unique solution. m. True. If A has n pivot positions, it has a pivot in each of its n columns and in each of its n rows. The reduced echelon form has a 1 in each pivot position, so the reduced echelon form is the n×n identity matrix. n. True. Both matrices A and B can be row reduced to the 3×3 identity matrix, as discussed in the previous question. Since the row operations that transform B into I3 are reversible, A can be transformed first into I3 and then into B. o. True. The reason is essentially the same as that given for question f. p. True. If the columns of A span m , then the reduced echelon form of A is a matrix U with a pivot in each row, by Theorem 4 in Section 1.4. Since B is row equivalent to A, B can be transformed by row operations first into A and then further transformed into U. Since U has a pivot in each row, so does B. By Theorem 4, the columns of B span m . q. False. r. True. Any set of three vectors in 2 would have to be linearly dependent. s. False. If a set {v1, v2, v3, v4} were to span 5 , then the matrix A = [v1 v2 v3 v4] would have a pivot position in each of its five rows, which is impossible since A has only four columns. t. True. The vector –u is a linear combination of u and v, namely, –u = (–1)u + 0v. u. False. If u and v are multiples, then Span{u, v} is a line, and w need not be on that line. v. False. Let u and v be any linearly independent pair of vectors and let w = 2v. Then w = 0u + 2v, so w is a linear combination of u and v. However, u cannot be a linear combination of v and w Copyright © 2016 Pearson Education, Ltd. Chapter 1 • Supplementary Exercises 1-79 because if it were, u would be a multiple of v. That is not possible since {u, v} is linearly independent. w. False. The statement would be true if the condition v1 is not zero were present. See Theorem 7 in Section 1.7. However, if v1 = 0, then {v1, v2, v3} is linearly dependent, no matter what else might be true about v2 and v3. x. True. “Function” is another word used for “transformation” (as mentioned in the definition of “transformation” in Section 1.8), and a linear transformation is a special type of transformation. y. True. For the transformation x Ax to map 5 onto 6 , the matrix A would have to have a pivot in every row and hence have six pivot columns. This is impossible because A has only five columns. z. False. For the transformation x Ax to be one-to-one, A must have a pivot in each column. Since A has n columns and m pivots, m might be less than n. 2. If a ≠ 0, then x = b/a; the solution is unique. If a = 0, and b ≠ 0, the solution set is empty, because 0x = 0 ≠ b. If a = 0 and b = 0, the equation 0x = 0 has infinitely many solutions. 3. a. Any consistent linear system whose echelon form is * * 0 * 0 0 0 * * * * 0 * * * or 0 0 * or 0 0 * 0 0 0 0 0 0 0 0 0 b. Any consistent linear system whose coefficient matrix has reduced echelon form I3. c. Any inconsistent linear system of three equations in three variables. 4. If the coefficient matrix has a pivot position in every column, the number of rows of the coefficient matrix is greater or equal than the number of columns. We distinguish both situations: If the number of rows equals the number of columns, the coefficient matrix has a pivot position in each row. Hence there is a pivot position in the bottom row. As a consequence, there is no room for a pivot in the augmented column. So the system is consistent, by Theorem 2, and the solution is unique as there are no free variables in this case. If the number of rows is greater than the number of columns, an echelon form of the augmented matrix may have a row of the form [0· · ·0 b] with b = 6 0, in which case the system is inconsistent. If there are no rows of this form, then the system is consistent and the solution is unique. 3 k 1 3 k 1 5. a. ~ . If h = 12 and k ¹ 2, the second row of the augmented matrix 4 h 8 0 h − 12 8 − 4k indicates an inconsistent system of the form 0x2 = b, with b nonzero. If h = 12, and k = 2, there is only one nonzero equation, and the system has infinitely many solutions. Finally, if h ¹ 12, the coefficient matrix has two pivots and the system has a unique solution. h 1 −2 h 1 −2 b. ~ . If k + 3h = 0, the system is inconsistent. Otherwise, the 6 k −2 0 k + 3h 1 coefficient matrix has two pivots and the system has a unique solution. Copyright © 2016 Pearson Education, Ltd. 1-80 CHAPTER 1 • Linear Equations in Linear Algebra 4 −2 7 −5 6. a. = Set v1 = , v2 = , v 3 , and b = . “Determine if b is a linear combination of v1, 8 −3 10 −3 v2, v3.” Or, “Determine if b is in Span{v1, v2, v3}.” To do this, compute 7 −5 4 −2 7 −5 4 −2 . The system is consistent, so b is in Span{v1, v2, v3}. 8 −3 10 −3 ~ 0 1 −4 7 7 4 −2 −5 b. Set A = , b = . “Determine if b is a linear combination of the columns of A.” 8 −3 10 −3 c. Define T(x) = Ax. “Determine if b is in the range of T.” b1 2 −4 −2 3 7. a. Set v1 = −5 , v 2 = 1 , v 3 = 1 and b = b2 . “Determine if v1, v2, v3 span .” To do this, b3 7 −5 −3 row reduce [v1 v2 v3]: 2 −4 −2 2 −4 −2 2 −4 −2 −5 1 1 ~ 0 −9 −4 ~ 0 −9 −4 . The matrix does not have a pivot in each row, 7 −5 −3 0 9 4 0 0 0 so its columns do not span 3 , by Theorem 4 in Section 1.4. 2 b. Set A = −5 7 −4 1 −5 −2 1 . “Determine if the columns of A span 3 .” −3 c. Define T(x) = Ax. “Determine if T maps 3 onto 3 .” ∗ 8. a. 0 0 0 ∗ ∗ ∗ b. 0 0 ∗ ∗ 0 ∗ ∗, 0 ∗ 0 ∗ 0 ∗ ∗ 0 ∗ ∗ , 0 0 ∗ 0 0 ∗ 0 ∗ 0 ∗ , 0 0 ∗ 0 0 0 ∗ ∗ 1 2 9. The first line is the line spanned by . The second line is spanned by . So the problem is to 2 1 5 1 2 write as the sum of a multiple of and a multiple of . That is, find x1 and x2 such that 6 2 1 2 1 5 x1 + x2 = . Reduce the augmented matrix for this equation: 1 2 6 2 1 1 2 5 1 ~ 6 2 5 Thus, = 6 2 1 6 1 ~ 5 0 2 −3 6 1 ~ −7 0 2 1 6 1 ~ 7 / 3 0 0 1 4 2 7 1 + 3 1 3 2 5 8 / 3 7 / 3 or= 6 4 / 3 + 14 / 3 . Copyright © 2016 Pearson Education, Ltd. 4 / 3 7 / 3 Chapter 1 • Supplementary Exercises 1-81 10. The line through a1 and the origin and the line through a2 and the origin determine a “grid” on the x1x2-plane as shown below. Every point in 2 can be described uniquely in terms of this grid. Thus, b can be reached from the origin by traveling a certain number of units in the a1-direction and a certain number of units in the a2-direction. 11. A solution set is a line when the system has one free variable. If the coefficient matrix is 2×3, then 1 2 * two of the columns should be pivot columns. For instance, take . Put anything in column 0 3 * 3. The resulting matrix will be in echelon form. Make one row replacement operation on the second 1 2 1 1 2 1 row to create a matrix not in echelon form, such as ~ 0 3 1 1 5 2 12. A solution set is a plane where there are two free variables. If the coefficient matrix is 2×3, then only one column can be a pivot column. The echelon form will have all zeros in the second row. Use a 1 2 3 row replacement to create a matrix not in echelon form. For instance, let A = . 1 2 3 1 13. The reduced echelon form of A looks like E = 0 0 0 1 0 * * . Since E is row equivalent to A, the 0 1 equation Ex = 0 has the same solutions as Ax = 0. Thus 0 0 1 By inspection, E = 0 0 0 1 0 0 1 0 * 3 0 0 . * −2 = 0 1 0 −3 2 . 0 1 a+2 0 14. Row reduce the augmented matrix for x1 + x2 = (∗). a a+6 0 1 a+2 0 1 a+2 0 1 a+2 0 ∼ ∼ a a+6 0 0 a + 6 − a(a + 2) 0 0 −a2 − a + 6 0 The equation (∗) has a nontrivial solution only when −a2 − a + 6 = (3 + a)(2 − a) = 0. So the vectors are linearly independent for all a except a = −3 and a = 2. Copyright © 2016 Pearson Education, Ltd. 1-82 CHAPTER 1 • Linear Equations in Linear Algebra 15. a. If the three vectors are linearly independent, then a, c, and f must all be nonzero. (The converse is true, too.) Let A be the matrix whose columns are the three linearly independent vectors. Then A must have three pivot columns. (See Exercise 30 in Section 1.7, or realize that the equation Ax = 0 has only the trivial solution and so there can be no free variables in the system of equations.) Since A is 3×3, the pivot positions are exactly where a, c, and f are located. b. The numbers a, …, f can have any values. Here's why. Denote the columns by v1, v2, and v3. Observe that v1 is not the zero vector. Next, v2 is not a multiple of v1 because the third entry of v2 is nonzero. Finally, v3 is not a linear combination of v1 and v2 because the fourth entry of v3 is nonzero. By Theorem 7 in Section 1.7, {v1, v2, v3} is linearly independent. 16. Denote the columns from right to left by v1, …, v4. The “first” vector v1 is nonzero, v2 is not a multiple of v1 (because the third entry of v2 is nonzero), and v3 is not a linear combination of v1 and v2 (because the second entry of v3 is nonzero). Finally, by looking at first entries in the vectors, v4 cannot be a linear combination of v1, v2, and v3. By Theorem 7 in Section 1.7, the columns are linearly independent. 17. Here are two arguments. The first is a “direct” proof. The second is called a “proof by contradiction.” i. Since {v1, v2, v3} is a linearly independent set, v1 ≠ 0. Also, Theorem 7 shows that v2 cannot be a multiple of v1, and v3 cannot be a linear combination of v1 and v2. By hypothesis, v4 is not a linear combination of v1, v2, and v3. Thus, by Theorem 7, {v1, v2, v3, v4} cannot be a linearly dependent set and so must be linearly independent. ii. Suppose that {v1, v2, v3, v4} is linearly dependent. Then by Theorem 7, one of the vectors in the set is a linear combination of the preceding vectors. This vector cannot be v4 because v4 is not in Span{v1, v2, v3}. Also, none of the vectors in {v1, v2, v3} is a linear combinations of the preceding vectors, by Theorem 7. So the linear dependence of {v1, v2, v3, v4} is impossible. Thus {v1, v2, v3, v4} is linearly independent. 18. Suppose that c1 and c2 are constants such that (∗) c1 (v1 + v2 ) + c2 (v1 − v2 ) = 0 Then (c1 + c2 )v1 + (c1 − c2 )v2 = 0. Since v1 and v2 are linearly independent, both c1 + c2 = 0 and c1 − c2 = 0. If follows that both c1 and c2 in (∗) must be zero, which shows that {v1 + v2 , v1 − v2 } is linearly independent. 19. Let M be the line through the origin that is parallel to the line through v1, v2, and v3. Then v2 – v1 and v3 – v1 are both on M. So one of these two vectors is a multiple of the other, say v2 – v1 = k(v3 – v1). This equation produces a linear dependence relation (k – 1)v1 + v2 – kv3 = 0. A second solution: A parametric equation of the line is x = v1 + t(v2 – v1). Since v3 is on the line, there is some t0 such that v3 = v1 + t0(v2 – v1) = (1 – t0)v1 + t0v2. So v3 is a linear combination of v1 and v2, and {v1, v2, v3} is linearly dependent. 20. If T(u) = v, then since T is linear, T(–u) = T((–1)u) = (–1)T(u) = –v. 21. Either compute T (e1), T (e2) and T (e3) to make the columns of A, or write the vectors vertically in the definition of T and fill in the entries of A by inspection: ? Ax = ? ? ? ? ? ? x1 −x2 ? x2 = x1 , ? x3 x3 0 −1 A= 1 0 0 0 0 0 1 Copyright © 2016 Pearson Education, Ltd. Chapter 1 • Supplementary Exercises 1-83 22. By Theorem 12 in Section 1.9, the columns of A are linearly independent. Hence, the equation Ax = 0 has only the trivial solution and each of the three columns of A is a pivot column. Since A has also three rows, A has a pivot position in each row. By Theorem 4 in Section 1.4, the columns of A span R3 . By Theorem 12 in Section 1.9, the transforma-tion x 7→ Ax maps R3 onto R3 .
a −b 10 26 10a − 24b = 26 = implies that . Solve: 23. b a 24 0 24a + 10b = 0 10 −24 26 5 −12 13 5 −12 13 ∼ ∼ 24 10 0 12 5 0 0 169/5 −156/5 5 −12 13 5 0 25/13 1 0 ∼ ∼ ∼ 0 1 −12/13 0 1 −12/13 0 1 Thus a = 5/13 and b = −12/13. 5/13 −12/13
24. The matrix equation displayed gives the information 3a − 4b = 5 and 4a + 3b = 0. Solve for a and b: 3 −4 5 3 −4 5 3 −4 5 3 0 ∼ ∼ ∼ 4 3 0 0 25/3 −20/3 0 1 −4/5 0 1 1 0 3/5 ∼ 0 1 −4/5 So a = 3/5 and b = −4/5. 9/5 −4/5
25. a. The vector lists the number of three-, two-, and one-bedroom apartments provided when x1 floors of plan A are constructed. 3 4 5 b. x1 7 + x2 4 + x3 3 8 8 9 3 4 5 66 74 c. [M] Solve x1 7 + x2 4 + x3 3 = 8 8 9 136 3 7 8 4 4 8 5 3 9 66 1 74 ~ ⋅⋅⋅ 0 136 0 0 1 0 −1/ 2 13/ 8 0 2 15 0 x1 x2 − + (1/ 2) x3 (13/ 8) x3 0 = 2 = 15 0 = The general solution is x1 x = x2 = x3 2 + (1/ 2) x3 15 − (13/ 8) x = 3 x3 2 1/ 2 15 + x −13/ 8 3 0 1 However, the only feasible solutions must have whole numbers of floors for each plan. Thus, x3 must be a multiple of 8, to avoid fractions. One solution, for x3 = 0, is to use 2 floors of plan A and 15 floors of plan B. Another solution, for x3 = 8, is to use 6 floors of plan A , 2 floors of plan B, and 8 floors of plan C. These are the only feasible solutions. A larger positive multiple of 8 for x3 makes x2 negative. A negative value for x3, of course, is not feasible either. Copyright © 2016 Pearson Education, Ltd. 2.1 SOLUTIONS Notes: The definition here of a matrix product AB gives the proper view of AB for nearly all matrix calculations. (The dual fact about the rows of A and the rows of AB is seldom needed, mainly because vectors here are usually written as columns.) I assign Exercise 13 and most of Exercises 17–22 to reinforce the definition of AB. Exercises 23 and 24 are used in the proof of the Invertible Matrix Theorem, in Section 2.3. Exercises 23–25 are mentioned in a footnote in Section 2.2. A class discussion of the solutions of Exercises 23–25 can provide a transition to Section 2.2. Or, these exercises could be assigned after starting Section 2.2. Exercises 27 and 28 are optional, but they are mentioned in Example 4 of Section 2.4. Outer products also appear in Exercises 31–34 of Section 4.6 and in the spectral decomposition of a symmetric matrix, in Section 7.1. Exercises 29–33 provide good training for mathematics majors. 0 −1 −4 0 2 2 −2 A = (−2) = . Next, use B – 2A = B + (–2A): 2 −8 10 −4 4 −5 1 −4 0 2 3 −5 3 7 −5 = B − 2A + = . 6 −7 1 −4 −3 −8 10 −4 −7 1. The product AC is not defined because the number of columns of A does not match the number of rows 1 ⋅ 5 + 2 ⋅ 4 1 13 1 2 3 5 1 ⋅ 3 + 2(−1) = = of C. CD = . For mental computation, the −2 1 −1 4 −2 ⋅ 3 + 1(−1) −2 ⋅ 5 + 1 ⋅ 4 −7 −6 row-column rule is probably easier to use than the definition. 0 −1 1 2 + 14 0 − 10 −1 + 2 16 −10 1 2 7 −5 + 2 = = A + 2B 2.= 2 2 − 6 6 −13 −4 4 −5 1 −4 −3 4 + 2 −5 − 8 The expression 3C – E is not defined because 3C has 2 columns and –E has only 1 column. 1 1 ⋅ 7 + 2 ⋅ 1 1(−5) + 2(−4) 1 ⋅ 1 + 2(−3) 9 −13 −5 1 2 7 −5 = CB = = 6 −5 −2 1 1 −4 −3 −2 ⋅ 7 + 1 ⋅ 1 −2(−5) + 1(−4) −2 ⋅ 1 + 1(−3) −13 The product EB is not defined because the number of columns of E does not match the number of rows of B. 2-1 Copyright © 2016 Pearson Education, Ltd. 2-2 CHAPTER 2 • Matrix Algebra 3 −A 3. 3I 2 = 0 0 4 − 3 5 −1 3 − 4 = −2 0 − 5 4 −1 12 (3= I 2 ) A 3(= I 2 A) 3 = 5 −2 15 0 − (−1) −1 = 3 − (−2) −5 1 5 −3 , or −6 3 0 4 −1 3 ⋅ 4 + 0 3(−1) + 0 12 = (3I 2 ) A = = 0 3 5 −2 0 + 3 ⋅ 5 0 + 3(−2) 15 9 −1 −8 7 4. A − 5 I 3 = −4 1 3 5 −6 − 0 8 0 9 (5 I 3 ) A =5( I 3 A) =5 A =5 −8 −4 5 0 0 9 (5 I 3 )= A 0 5 0 −8 0 0 5 −4 15 45 −5 = −45 35 −30 20 5 40 − −1 7 1 0 5 0 −1 7 1 0 4 −1 −8 0 = 2 5 −4 1 3 45 −6 = −40 8 −20 3 5 ⋅ 9 + 0 + 0 −6 = 0 + 5( −8) + 0 8 0 + 0 + 5( −4) 2 −1 −7 3 5 4 = 7 , Ab 2 = 5. a. Ab 1 = −2 2 −3 12 AB = −1 b. 5 2 [ Ab1 −7 Ab 2 ] 7 = 12 2 3 4 −2 −3 −5 35 5 −3 −6 3 −6 3 15 −30 , or 40 5( −1) + 0 + 0 0 + 5⋅ 7 + 0 0 + 0 + 5 ⋅1 2 −1 = −2 5 4 1 2 −3 4 −6 −7 4 −6 −7 −1 ⋅ 3 + 2(−2) −2 5 ⋅ 3 + 4(−2) = 1 2 ⋅ 3 − 3(−2) −1(−2) + 2 ⋅ 1 5(−2) + 4 ⋅ 1= 2(−2) − 3 ⋅ 1 −7 7 12 4 −6 −7 4 −2 0 4 −2 14 1 3 −3 , Ab 2 = −9 0 = 0 = 6. a. Ab1 = −3 −3 −1 2 3 13 3 4 5 5 AB = [ Ab1 0 −3 Ab 2 ] = 13 14 −9 4 Copyright © 2016 Pearson Education, Ltd. 5 ⋅ 3 + 0 + 0 0 + 5( −6) + 0 0 + 0 + 5 ⋅ 8 2.1 4 b. −3 3 −2 1 0 2 5 4 ⋅1 − 2 ⋅ 2 3 = −3 ⋅ 1 + 0 ⋅ 2 −1 3 ⋅ 1 + 5 ⋅ 2 4 ⋅ 3 − 2(−1) −3 ⋅ 3 + 0(−1) = 3 ⋅ 3 + 5(−1) 0 −3 13 • Solutions 2-3 14 −9 4 7. Since A has 3 columns, B must match with 3 rows. Otherwise, AB is undefined. Since AB has 7 columns, so does B. Thus, B is 3×7. 8. The number of rows of B matches the number of rows of BC, so B has 3 rows. 15 2 5 4 −5 23 −10 + 5k 4 −5 2 5 23 = = , while BA = 9. AB = . k −9 15 + k k −3 1 6 − 3k 15 + k −3 1 3 3 Then AB = BA if and only if –10 + 5k = 15 and –9 = 6 – 3k, which happens if and only if k = 5. 2 −3 8 4 1 −7 2 −3 5 −2 1 = = , AC = 10. AB = 6 5 5 −2 14 6 3 1 −2 −4 −4 1 1 1 2 0 0 1 2 3 0 3 0 = 11. AD = 1 4 5 0 0 5 2 2 2 3 6 12 5 15 25 2 0 0 1 1 1 DA = 0 3 0 1 2 3 = 0 0 5 1 4 5 2 3 5 2 6 20 2 9 25 −7 14 Right-multiplication (that is, multiplication on the right) by the diagonal matrix D multiplies each column of A by the corresponding diagonal entry of D. Left-multiplication by D multiplies each row of A by the corresponding diagonal entry of D. To make AB = BA, one can take B to be a multiple of I3. For instance, if B = 4I3, then AB and BA are both the same as 4A. 12. Consider B = [b1 b2]. To make AB = 0, one needs Ab1 = 0 and Ab2 = 0. By inspection of A, a suitable 2 2 2 6 b1 is , or any multiple of . Example: B = . 1 3 1 1 13. Use the definition of AB written in reverse order: [Ab1 ⋅ ⋅ ⋅ Abp] = A[b1 ⋅ ⋅ ⋅ bp]. Thus [Qr1 ⋅ ⋅ ⋅ Qrp] = QR, when R = [r1 ⋅ ⋅ ⋅ rp]. 14. By definition, UQ = U[q1 ⋅ ⋅ ⋅ q4] = [Uq1 ⋅ ⋅ ⋅ Uq4]. From Example 6 of Section 1.8, the vector Uq1 lists the total costs (material, labor, and overhead) corresponding to the amounts of products B and C specified in the vector q1. That is, the first column of UQ lists the total costs for materials, labor, and overhead used to manufacture products B and C during the first quarter of the year. Columns 2, 3, and 4 of UQ list the total amounts spent to manufacture B and C during the 2nd, 3rd, and 4th quarters, respectively. 15. a. False. See the definition of AB. b. False. The roles of A and B should be reversed in the second half of the statement. See the box after Example 3. c. True. See Theorem 2(b), read right to left. d. True. See Theorem 3(b), read right to left. e. False. The phrase “in the same order” should be “in the reverse order.” See the box after Theorem 3. Copyright © 2016 Pearson Education, Ltd. 2-4 CHAPTER 2 • Matrix Algebra 16. a. False. AB must be a 3×3 matrix, but the formula for AB implies that it is 3×1. The plus signs should be just spaces (between columns). This is a common mistake. b. True. See the box after Example 6. c. False. The left-to-right order of B and C cannot be changed, in general. d. False. See Theorem 3(d). e. True. This general statement follows from Theorem 3(b). 2 −1 −1 = AB = [ Ab1 Ab 2 Ab3 ] , the first column of B satisfies the equation 17. Since 3 6 −9 −1 1 −2 −1 1 0 7 7 Ax = . Row reduction: [ A Ab1 ] ~ ~ . So b1 = . Similarly, 5 6 0 1 4 6 −2 4 2 1 0 − 8 1 −2 −8 [ A Ab 2 ] ~ −2 5 −9 ~ 0 1 −5 and b2 = −5 . Note: An alternative solution of Exercise 17 is to row reduce [A Ab1 Ab2] with one sequence of row operations. This observation can prepare the way for the inversion algorithm in Section 2.2. 18. The first two columns of AB are Ab1 and Ab2. They are equal since b1 and b2 are equal. 19. (A solution is in the text). Write B = [b1 b2 b3]. By definition, the third column of AB is Ab3. By hypothesis, b3 = b1 + b2. So Ab3 = A(b1 + b2) = Ab1 + Ab2, by a property of matrix-vector multiplication. Thus, the third column of AB is the sum of the first two columns of AB. 20. The second column of AB is also all zeros because Ab2 = A0 = 0. 21. Let bp be the last column of B. By hypothesis, the last column of AB is zero. Thus, Abp = 0. However, bp is not the zero vector, because B has no column of zeros. Thus, the equation Abp = 0 is a linear dependence relation among the columns of A, and so the columns of A are linearly dependent. Note: The text answer for Exercise 21 is, “The columns of A are linearly dependent. Why?” The Study Guide supplies the argument above in case a student needs help. 22. If the columns of B are linearly dependent, then there exists a nonzero vector x such that Bx = 0. From this, A(Bx) = A0 and (AB)x = 0 (by associativity). Since x is nonzero, the columns of AB must be linearly dependent. 23. If x satisfies Ax = 0, then CAx = C0 = 0 and so Inx = 0 and x = 0. This shows that the equation Ax = 0 has no free variables. So every variable is a basic variable and every column of A is a pivot column. (A variation of this argument could be made using linear independence and Exercise 30 in Section 1.7.) Since each pivot is in a different row, A must have at least as many rows as columns. 24. Take any b in m . By hypothesis, ADb = Imb = b. Rewrite this equation as A(Db) = b. Thus, the vector x = Db satisfies Ax = b. This proves that the equation Ax = b has a solution for each b in m . By Theorem 4 in Section 1.4, A has a pivot position in each row. Since each pivot is in a different column, A must have at least as many columns as rows. Copyright © 2016 Pearson Education, Ltd. 2.1 • Solutions 2-5 25. By Exercise 23, the equation CA = In implies that (number of rows in A) > (number of columns), that is, m > n. By Exercise 24, the equation AD = Im implies that (number of rows in A) < (number of columns), that is, m < n. Thus m = n. To prove the second statement, observe that DAC = (DA)C = InC = C, and also DAC = D(AC) = DIm = D. Thus C = D. A shorter calculation is C = InC = (DA)C = D(AC) = DIn = D 26. Write I3 =[e1 e2 e3] and D = [d1 d2 d3]. By definition of AD, the equation AD = I3 is equivalent |to the three equations Ad1 = e1, Ad2 = e2, and Ad3 = e3. Each of these equations has at least one solution because the columns of A span 3 . (See Theorem 4 in Section 1.4.) Select one solution of each equation and use them for the columns of D. Then AD = I3. 27. The product uTv is a 1×1 matrix, which usually is identified with a real number and is written without the matrix brackets. u v= [ −2 T 3 a −4] b = −2a + 3b − 4c , vT u = [a c −2 = = uv 3 [ a b c ] −4 T a vu = b [ −2 c T −2a 3a −4a −2b 3b −4b −2a −4] = −2b −2c 3 3a 3b 3c b −2 c ] 3 = −2a + 3b − 4c −4 −2c 3c −4c −4a −4b −4c 28. Since the inner product uTv is a real number, it equals its transpose. That is, uTv = (uTv)T = vT (uT)T = vTu, by Theorem 3(d) regarding the transpose of a product of matrices and by Theorem 3(a). The outer product uvT is an n×n matrix. By Theorem 3, (uvT)T = (vT)TuT = vuT. 29. The (i, j)-entry of A(B + C) equals the (i, j)-entry of AB + AC, because n n n ∑ aik (bkj + c= ∑ aik bkj + ∑ aik ckj kj ) k 1 = k 1= k 1 = The (i, j)-entry of (B + C)A equals the (i, j)-entry of BA + CA, because n n n ∑ (bik + cik )akj = ∑ bik akj + ∑ cik akj k 1 = k 1= k 1 = n 30. The (i, j))-entries of r(AB), (rA)B, and A(rB) are all equal, because = r ∑ aik bkj n (raik )bkj ∑= k 1= k 1 = n ∑ aik (rbkj ) . k 1 = 31. Use the definition of the product ImA and the fact that Imx = x for x in m . ImA = Im[a1 ⋅ ⋅ ⋅ an] = [Ima1 ⋅ ⋅ ⋅ Iman] = [a1 ⋅ ⋅ ⋅ an] = A 32. Let ej and aj denote the jth columns of In and A, respectively. By definition, the jth column of AIn is Aej, which is simply aj because ej has 1 in the jth position and zeros elsewhere. Thus corresponding columns of AIn and A are equal. Hence AIn = A. Copyright © 2016 Pearson Education, Ltd. 2-6 CHAPTER 2 • Matrix Algebra 33. The (i, j)-entry of (AB)T is the ( j, i)-entry of AB, which is a j1b1i + ⋅⋅⋅ + a jn bni The entries in row i of BT are b1i, … , bni, because they come from column i of B. Likewise, the entries in column j of AT are aj1, …, ajn, because they come from row j of A. Thus the (i, j)-entry in BTAT is a j1b1i + + a jnbni , as above. 34. Use Theorem 3(d), treating x as an n×1 matrix: (ABx)T = xT(AB)T = xTBTAT. 35. [M] The answer here depends on the choice of matrix program. For MATLAB, use the help command to read about zeros, ones, eye, and diag. For other programs see the appendices in the Study Guide. (The TI calculators have fewer single commands that produce special matrices.) 36. [M] The answer depends on the choice of matrix program. In MATLAB, the command rand(6,4) creates a 6×4 matrix with random entries uniformly distributed between 0 and 1. The command round(19*(rand(6,4)–.5))creates a random 6×4 matrix with integer entries between –9 and 9. The same result is produced by the command randomint in the Laydata Toolbox on text website. For other matrix programs see the appendices in the Study Guide. 37. [M] (A + I)(A – I) – (A2 – I) = 0 for all 4×4 matrices. However, (A + B)(A – B) – A2 – B2 is the zero matrix only in the special cases when AB = BA. In general,(A + B)(A – B) = A(A – B) + B(A – B) = AA – AB + BA – BB. 38. [M] The equality (AB)T = ATBT is very likely to be false for 4×4 matrices selected at random. 39. [M] The matrix S “shifts” the entries in a vector (a, b, c, d, e) to yield (b, c, d, e, 0). The entries in S2 result from applying S to the columns of S, and similarly for S 3 , and so on. This explains the patterns of entries in the powers of S: S2 0 0 1 0 0 0 0 0 1 0 = 0 0 0 0 1 , S 3 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 = 0 0 0 0 0 , S 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 S 5 is the 5×5 zero matrix. S 6 is also the 5×5 zero matrix. 5 40. [M] A .3318 .3346 .3336 .3346 .3323 .3331 , A10 = .3336 .3331 .3333 .333337 .333330 .333333 .333330 .333336 .333334 .333333 .333334 .333333 The entries in A20 all agree with .3333333333 to 9 or 10 decimal places. The entries in A30 all agree with .33333333333333 to at least 14 decimal places. The matrices appear to approach the matrix 1/ 3 1/ 3 1/ 3 1/ 3 1/ 3 1/ 3 . Further exploration of this behavior appears in Sections 4.9 and 5.2. 1/ 3 1/ 3 1/ 3 Note: The MATLAB box in the Study Guide introduces basic matrix notation and operations, including the commands that create special matrices needed in Exercises 35, 36 and elsewhere. The Study Guide appendices treat the corresponding information for the other matrix programs. Copyright © 2016 Pearson Education, Ltd. 2.2 2.2 • Solutions 2-7 SOLUTIONS Notes: The text includes the matrix inversion algorithm at the end of the section because this topic is popular. Students like it because it is a simple mechanical procedure. The final subsection is independent of the inversion algorithm and is needed for Exercises 35 and 36. Key Exercises: 8, 11–24, 35. (Actually, Exercise 8 is only helpful for some exercises in this section. Section 2.3 has a stronger result.) Exercises 23 and 24 are used in the proof of the Invertible Matrix Theorem (IMT) in Section 2.3, along with Exercises 23 and 24 in Section 2.1. I recommend letting students work on two or more of these four exercises before proceeding to Section 2.3. In this way students participate in the proof of the IMT rather than simply watch an instructor carry out the proof. Also, this activity will help students understand why the theorem is true. −1 8 6 1 4 −6 2 = = 1. 8 −5 / 2 32 − 30 −5 5 4 −1 3 2 1 4 −2 −2 = = 2. 3 7 / 2 12 − 14 −7 7 4 8 3. −7 5 −5 −1 = −5 1 −40 − (−35) 7 −1 −3 4 1 −3/ 2 −5 1 −5 = − 8 5 7 3 −4 −8 4 1 −8 1 = = 4. −24 − ( −28) −7 3 4 −7 7 −8 4 −2 or 3 −7 / 4 8 5. The system is equivalent to Ax = b, where A = 5 2 x = A–1b = −5 / 2 −5 1 or 8 −1.4 1 −1.6 1 3/ 4 6 2 and b = , and the solution is 4 −1 −3 2 7 = . Thus x1 = 7 and x2 = –9. 4 −1 −9 5 8 −9 = and b , and the solution is x = A–1b. To 6. The system is equivalent to Ax = b, where A = −7 −5 11 compute this by hand, the arithmetic is simplified by keeping the fraction 1/det(A) in front of the matrix for A–1. (The Study Guide comments on this in its discussion of Exercise 7.) From Exercise 3, 1 −5 x = A–1b = − 5 7 −5 −9 1 −10 2 = − = . Thus x1 = 2 and x2 = –5. 8 11 5 25 −5 Copyright © 2016 Pearson Education, Ltd. 2-8 CHAPTER 2 • Matrix Algebra −1 1 2 12 −2 1 12 1 = = 7. a. 1 2 −5 1 ⋅ 12 − 2 ⋅ 5 −5 5 12 x = A–1b1 = 1 12 2 −5 −2 6 or 1 −2.5 −1 .5 −2 −1 1 −18 −9 = = . Similar calculations give 1 3 2 8 4 11 −1 6 −1 13 = A−1b 2 = , A b3 = , A b4 . −5 −2 −5 −1 3 1 2 b. [A b1 b2 b3 b4] = 5 12 1 ~ 0 2 2 −1 8 1 −10 1 ~ 0 0 1 −9 4 11 −5 2 −4 6 −2 −9 The solutions are , 4 1 −5 3 1 ~ −10 0 2 6 2 1 3 5 −1 4 1 −5 2 −2 3 −5 13 −5 11 −5 , 6 13 −2 , and −5 , the same as in part (a). Note: The Study Guide also discusses the number of arithmetic calculations for this Exercise 7, stating that when A is large, the method used in (b) is much faster than using A–1. 8. Left-multiply each side of the equation AD = I by A–1 to obtain A–1AD = A–1I, ID = A–1, and D = A–1. Parentheses are routinely suppressed because of the associative property of matrix multiplication. 9. a. True, by definition of invertible. b. False. See Theorem 6(b). 1 1 c. False. If A = , then ab – cd = 1 – 0 ≠ 0, but Theorem 4 shows that this matrix is not invertible, 0 0 because ad – bc = 0. d. True. This follows from Theorem 5, which also says that the solution of Ax = b is unique, for each b. e. True, by the box just before Example 6. 10. a. False. The product matrix is invertible, but the product of inverses should be in the reverse order. See Theorem 6(b). b. True, by Theorem 6(a). c. True, by Theorem 4. d. True, by Theorem 7. e. False. The last part of Theorem 7 is misstated here. 11. (The proof can be modeled after the proof of Theorem 5.) The n×p matrix B is given (but is arbitrary). Since A is invertible, the matrix A–1B satisfies AX = B, because A(A–1B) = A A–1B = IB = B. To show this solution is unique, let X be any solution of AX = B. Then, left-multiplication of each side by A–1 shows that X must be A–1B: Thus A–1 (AX) = A–1B, so IX = A–1B, and thus X = A–1B. Copyright © 2016 Pearson Education, Ltd. 2.2 • Solutions 2-9 12. If you assign this exercise, consider giving the following Hint: Use elementary matrices and imitate the proof of Theorem 7. The solution in the Instructor’s Edition follows this hint. Here is another solution, based on the idea at the end of Section 2.2. Write B = [b1 ⋅ ⋅ ⋅ bp] and X = [u1 ⋅ ⋅ ⋅ up]. By definition of matrix multiplication, AX = [Au1 ⋅ ⋅ ⋅ Aup]. Thus, the equation AX = B is equivalent to the p systems: Au1 = b1, … Aup = bp Since A is the coefficient matrix in each system, these systems may be solved simultaneously, placing the augmented columns of these systems next to A to form [A b1 ⋅ ⋅ ⋅ bp] = [A B]. Since A is invertible, the solutions u1, …, up are uniquely determined, and [A b1 ⋅ ⋅ ⋅ bp] must row reduce to [I u1 ⋅ ⋅ ⋅ up] = [I X]. By Exercise 11, X is the unique solution A–1B of AX = B. 13. Left-multiply each side of the equation AB = AC by A–1 to obtain A–1AB = A–1AC, so IB = IC, and B = C. This conclusion does not always follow when A is singular. Exercise 10 of Section 2.1 provides a counterexample. 14. Right-multiply each side of the equation (B – C)D = 0 by D–1 to obtain(B – C)DD–1 = 0D–1, so (B – C)I = 0, thus B – C = 0, and B = C. 15. The box following Theorem 6 suggests what the inverse of ABC should be, namely, C–1B–1A–1. To verify that this is correct, compute: (ABC) C–1B–1A–1 = ABCC–1B–1A–1 = ABIB–1A–1 = ABB–1A–1 = AIA–1 = AA–1 = I and C–1B–1A–1 (ABC) = C–1B–1A–1ABC = C–1B–1IBC = C–1B–1BC = C–1IC = C–1C = I 16. Let C = AB. Then CB–1 = ABB–1, so CB–1 = AI = A. This shows that A is the product of invertible matrices and hence is invertible, by Theorem 6. Note: The Study Guide warns against using the formula (AB) –1 = B–1A–1 here, because this formula can be used only when both A and B are already known to be invertible. 17. Right-multiply each side of AB = BC by B–1, thus ABB–1 = BCB–1, so AI = BCB–1, and A = BCB–1. 18. Left-multiply each side of A = PBP–1 by P–1: thus P–1A = P–1PBP–1, so P–1A = IBP–1, and P–1A = BP–1 Then right-multiply each side of the result by P: thus P–1AP = BP–1P, so P–1AP = BI, and P–1AP = B 19. Unlike Exercise 17, this exercise asks two things, “Does a solution exist and what is it?” First, find what the solution must be, if it exists. That is, suppose X satisfies the equation C–1(A + X)B–1 = I. Left-multiply each side by C, and then right-multiply each side by B: thus CC–1(A + X)B–1 = CI, so I(A + X)B–1 = C, thus (A + X)B–1B = CB, and (A + X)I = CB Expand the left side and then subtract A from both sides: thus AI + XI = CB, so A + X = CB, and X = CB – A If a solution exists, it must be CB – A. To show that CB – A really is a solution, substitute it for X: C–1[A + (CB – A)]B–1 = C–1[CB]B–1 = C–1CBB–1 = II = I. Note: The Study Guide suggests that students ask their instructor about how many details to include in their proofs. After some practice with algebra, an expression such as CC–1(A + X)B–1 could be simplified directly to (A + X)B–1 without first replacing CC–1 by I. However, you may wish this detail to be included in the homework for this section. Copyright © 2016 Pearson Education, Ltd. 2-10 CHAPTER 2 • Matrix Algebra 20. a. Left-multiply both sides of (A – AX)–1 = X–1B by X to see that B is invertible because it is the product of invertible matrices. b. Invert both sides of the original equation and use Theorem 6 about the inverse of a product (which applies because X–1 and B are invertible): A – AX = (X–1B)–1 = B–1(X–1)–1 = B–1X Then A = AX + B–1X = (A + B–1)X. The product (A + B–1)X is invertible because A is invertible. Since X is known to be invertible, so is the other factor, A + B–1, by Exercise 16 or by an argument similar to part (a). Finally, (A + B–1)–1A = (A + B–1)–1(A + B–1)X = X Note: This exercise is difficult. The algebra is not trivial, and at this point in the course, most students will not recognize the need to verify that a matrix is invertible. 21. Suppose A is invertible. By Theorem 5, the equation Ax = 0 has only one solution, namely, the zero solution. This means that the columns of A are linearly independent, by a remark in Section 1.7. 22. Suppose A is invertible. By Theorem 5, the equation Ax = b has a solution (in fact, a unique solution) for each b. By Theorem 4 in Section 1.4, the columns of A span n . 23. Suppose A is n×n and the equation Ax = 0 has only the trivial solution. Then there are no free variables in this equation, and so A has n pivot columns. Since A is square and the n pivot positions must be in different rows, the pivots in an echelon form of A must be on the main diagonal. Hence A is row equivalent to the n×n identity matrix. 24. If the equation Ax = b has a solution for each b in n , then A has a pivot position in each row, by Theorem 4 in Section 1.4. Since A is square, the pivots must be on the diagonal of A. It follows that A is row equivalent to In. By Theorem 7, A is invertible. 0 0 x1 0 a b 25. Suppose A = and ad – bc = 0. If a = b = 0, then examine = This has the c d c d x2 0 d solution x1 = . This solution is nonzero, except when a = b = c = d. In that case, however, A is the −c −b zero matrix, and Ax = 0 for every vector x. Finally, if a and b are not both zero, set x2 = . Then a a b −b − ab + ba 0 = Ax 2 = = , because –cb + da = 0. Thus, x2 is a nontrivial solution of Ax = 0. c d a − cb + da 0 So, in all cases, the equation Ax = 0 has more than one solution. This is impossible when A is invertible (by Theorem 5), so A is not invertible. d 26. −c a c −b a a c b d d −c b da − bc = d 0 0 . Divide both sides by ad – bc to get CA = I. −cb + ad −b ad − bc = a 0 side is AC, because 0 . Divide both sides by ad – bc. The right side is I. The left −cb + da 1 a ad − bc c b d d −c −b a = a c b 1 d d ad − bc −c Copyright © 2016 Pearson Education, Ltd. −b = AC. a 2.2 • Solutions 2-11 27. a. Interchange A and B in equation (1) after Example 6 in Section 2.1: rowi (BA) = rowi (B)⋅A. Then replace B by the identity matrix: rowi (A) = rowi (IA) = rowi (I)⋅A. b. Using part (a), when rows 1 and 2 of A are interchanged, write the result as row 2 ( A) row (= 1 A) row 3 ( A) row 2 ( I ) ⋅ A row ( I )= ⋅ A 1 row 3 ( I ) ⋅ A row 2 ( I ) row ( I = EA 1 ) A row 3 ( I ) (*) Here, E is obtained by interchanging rows 1 and 2 of I. The second equality in (*) is a consequence of the fact that rowi (EA) = rowi (E)×A. c. Using part (a), when row 3 of A is multiplied by 5, write the result as row1 ( A) row ( A ) 2 = 5 ⋅ row 3 ( A) row1 ( I ) ⋅ A row ( I ) ⋅= A 2 5 ⋅ row 3 ( I ) ⋅ A row1 ( I ) row ( I )= A EA 2 5 ⋅ row 3 ( I ) Here, E is obtained by multiplying row 3 of I by 5. 28. When row 3 of A is replaced by row3(A) – 4⋅row1(A), write the result as row1 ( I ) ⋅ A row 2 ( I ) ⋅ A row 3 ( I ) ⋅ A − 4 ⋅ row1 ( I ) ⋅ A row1 ( A) = row 2 ( A) row 3 ( A) − 4 ⋅ row1 ( A) row1 ( I ) ⋅ A = row 2 ( I ) ⋅ A = [row 3 ( I ) − 4 ⋅ row1 ( I )] ⋅ A row1 ( I ) A EA = row 2 ( I ) row 3 ( I ) − 4 ⋅ row1 ( I ) Here, E is obtained by replacing row3(I) by row3(I) – 4⋅row1(I). 1 29. [ A I ] = 4 2 7 −7 A–1 = 4 30. [ A 1 ~ 0 2 −1 1 −4 0 1 ~ 1 0 2 1 1 4 0 1 ~ −1 0 1/ 5 0 0 1 ~ 1 0 2 −1 1/ 5 −4 / 5 2 −1 5 I] = 4 2 1 0 1 ~ 1 0 1 0 10 7 1/ 5 4/5 1 0 0 1 ~ 1 4 0 1 ~ −1 0 0 1 2 7 −7 / 5 4/5 2 −7 / 5 . A−1 = −1 4/5 0 1 2 −1 Copyright © 2016 Pearson Education, Ltd. 0 1 −7 4 2 −1 2-12 CHAPTER 2 31. [ A • Matrix Algebra 0 −2 1 0 0 1 0 1 I] = 3 1 4 0 1 0 ~ 0 1 − 4 0 0 1 0 −3 2 −3 1 ~ 0 0 0 1 0 −2 −2 2 1 3 7 1 ~ 0 0 0 1 0 0 0 1 8 10 7/2 32. [ A 1 ~ 0 0 −2 −7 6 1 I= ] 4 −2 −2 1 0 0 1 3 1 −1 0 3 4 3/ 2 1 3 −4 1 −4 10 0 1 0 ~ 0 1 0 0 1 −2 0 1 0 0 0 2 8 10 7 1 8 −1 1 . A = 10 1/ 2 7 / 2 1 0 0 0 1 0 0 1 0 ~ 0 1 0 −2 1 2 −2 −2 8 1 3 −2 3 4 3 1 1 1 3 4 3/ 2 1 −1 −2 0 0 1 0 1 0 1 1 1/ 2 1 −4 2 0 1 0 0 0 1 0 0 . The matrix A is not invertible. 1 0 0 0 1 −1 1 0 0 , and for j = 1, …, n, let aj, bj, and ej denote the jth columns of A, B, 33. Let B = 0 −1 1 0 0 −1 1 and I, respectively. Note that for j = 1, …, n – 1, aj – aj+1 = ej (because aj and aj+1 have the same entries except for the jth row), bj = ej – ej+1 and an = bn = en. To show that AB = I, it suffices to show that Abj = ej for each j. For j = 1, …, n – 1, Abj = A(ej – ej+1) = Aej – Aej+1 = aj – aj+1 = ej and Abn = Aen = an = en. Next, observe that aj = ej + ⋅ ⋅ ⋅ + en for each j. Thus, Baj = B(ej + ⋅ ⋅ ⋅ + en) = bj + ⋅ ⋅ ⋅ + bn = (ej – ej+1) + (ej+1 – ej+2) + ⋅ ⋅ ⋅ + (en–1 – en) + en = ej This proves that BA = I. Combined with the first part, this proves that B = A–1. Note: Students who do this problem and then do the corresponding exercise in Section 2.4 will appreciate the Invertible Matrix Theorem, partitioned matrix notation, and the power of a proof by induction. 1 1 34. Let A = 1 1 0 2 2 0 0 3 2 3 0 0 0 33 1 0 1/ 2 0 −1/ 2 0 , and B = 0 −1/ 3 1/ 3 3 n 0 −1/ n 0 0 1/ n and for j = 1, …, n, let aj, bj, and ej denote the jth columns of A, B, and I, respectively. Note that for 1 1 1 j = 1, …, n–1, aj = j(ej + ⋅ ⋅ ⋅ + en), bj = e j − e j +1 , and b n = e n . j j +1 n Copyright © 2016 Pearson Education, Ltd. 2.2 • Solutions 2-13 To show that AB = I, it suffices to show that Abj = ej for each j. For j = 1, …, n–1, 1 1 1 1 e j +1 = a j − Abj = A e j − a j +1 = (ej + ⋅ ⋅ ⋅ + en) – (ej+1 + ⋅ ⋅ ⋅ + en) = ej j j +1 j +1 j 1 1 e = a e n . Finally, for j = 1, …, n, the sum bj + ⋅ ⋅ ⋅ + bn is a “telescoping sum” Also, Abn = A = n n n n whose value is 1 1 e j . Thus, Baj = j(Bej + ⋅ ⋅ ⋅ + Ben) = j(bj + ⋅ ⋅ ⋅ + bn) = j e j = e j j j which proves that BA = I. Combined with the first part, this proves that B = A–1. Note: If you assign Exercise 34, you may wish to supply a hint using the notation from Exercise 33: Express each column of A in terms of the columns e1, …, en of the identity matrix. Do the same for B. 35. Row reduce [A e3]: −2 2 1 −7 5 3 −9 6 4 0 1 0 ~ 2 1 −2 3 5 −7 4 6 −9 1 1 0 ~ 0 0 0 1 ~ 0 0 3 −1 0 0 0 1 −15 1 6 ~ 0 4 0 3 1 0 0 0 1 −15 1 −6 ~ 0 4 0 3 −1 −1 0 1 0 4 −2 −1 0 0 1 1 1 −2 ~ 0 2 0 3 −1 0 4 −2 1 1 −2 4 3 −6 . 4 3 Answer: The third column of A is −6 . 4 36. [M] Write B = [A F], where F consists of the last two columns of I3, and row reduce: 3/ 2 −9 / 2 −25 −9 −27 0 0 1 0 0 B = 546 180 537 1 0 ~ 0 1 0 −433/ 6 439 / 2 154 50 149 0 1 0 0 1 68 / 3 −69 –1 1.5000 The last two columns of A are −72.1667 22.6667 –1 −4.5000 219.5000 −69.0000 1 1 −1 37. There are many possibilities for C, but C = is the only one whose entries are 1, –1, and 0. 0 −1 1 With only three possibilities for each entry, the construction of C can be done by trial and error. This is probably faster than setting up a system of 4 equations in 6 unknowns. The fact that A cannot be invertible follows from Exercise 25 in Section 2.1, because A is not square. 1 0 0 0 . 38. Write AD = A[d1 d2] = [Ad1 Ad2]. The structure of A shows that D = 0 0 0 1 [There are 25 possibilities for D if entries of D are allowed to be 1, –1, and 0.] There is no 4×2 matrix C such that CA = I4. If this were true, then CAx would equal x for all x in 4 . This cannot happen because Copyright © 2016 Pearson Education, Ltd. 2-14 CHAPTER 2 • Matrix Algebra the columns of A are linearly dependent and so Ax = 0 for some nonzero vector x. For such an x, CAx = C(0) = 0. An alternate justification would be to cite Exercise 23 or 25 in Section 2.1. .005 .002 39. y = Df = .002 .004 .001 .002 and 3, respectively. .001 30 .27 .002 50 = .30 . The deflections are .27 in., .30 in., and .23 in. at points 1, 2, .005 20 .23 2 40. [M] The stiffness matrix is D . Use an “inverse” command to produce D = 125 −1 0 –1 –1 −1 3 −1 0 −1 2 To find the forces (in pounds) required to produce a deflection of .04 cm at point 3, most students will use technology to solve Df = (0, 0, .04) and obtain (0, –5, 10). Here is another method, based on the idea suggested in Exercise 42. The first column of D–1 lists the forces required to produce a deflection of 1 in. at point 1 (with zero deflection at the other points). Since the transformation y D–1y is linear, the forces required to produce a deflection of .04 cm at point 3 is given by .04 times the third column of D–1, namely (.04)(125) times (0, –1, 2), or (0, –5, 10) pounds. 41. To determine the forces that produce a deflections of .08, .12, .16, and .12 cm at the four points on the beam, use technology to solve Df = y, where y = (.08, .12, .16, .12). The forces at the four points are 12, 1.5, 21.5, and 12 newtons, respectively. 42. [M] To determine the forces that produce a deflection of .24 cm at the second point on the beam, use technology to solve Df = y, where y = (0, .24, 0, 0). The forces at the four points are –104, 167, –113, and 56.0 newtons, respectively. These forces are .24 times the entries in the second column of D–1. Reason: The transformation y D −1y is linear, so the forces required to produce a deflection of .24 cm at the second point are .24 times the forces required to produce a deflection of 1 cm at the second point. These forces are listed in the second column of D–1. Another possible discussion: The solution of Dx = (0, 1, 0, 0) is the second column of D–1. Multiply both sides of this equation by .24 to obtain D(.24x) = (0, .24, 0, 0). So .24x is the solution of Df = (0, .24, 0, 0). (The argument uses linearity, but students may not mention this.) Note: The Study Guide suggests using gauss, swap, bgauss, and scale to reduce [A I], because I prefer to postpone the use of ref (or rref) until later. If you wish to introduce ref now, see the Study Guide’s technology notes for Sections 2.8 or 4.3. (Recall that Sections 2.8 and 2.9 are only covered when an instructor plans to skip Chapter 4 and get quickly to eigenvalues.) 2.3 SOLUTIONS Notes: This section ties together most of the concepts studied thus far. With strong encouragement from an instructor, most students can use this opportunity to review and reflect upon what they have learned, and form a solid foundation for future work. Students who fail to do this now usually struggle throughout the rest of the course. Section 2.3 can be used in at least three different ways. (1) Stop after Example 1 and assign exercises only from among the Practice Problems and Exercises 1 to 28. I do this when teaching “Course 3” described in the text's “Notes to the Instructor. ” If you did not cover Theorem 12 in Section 1.9, omit statements (f) and (i) from the Invertible Matrix Theorem. Copyright © 2016 Pearson Education, Ltd. 2.3 • Solutions 2-15 (2) Include the subsection “Invertible Linear Transformations” in Section 2.3, if you covered Section 1.9. I do this when teaching “Course 1” because our mathematics and computer science majors take this class. Exercises 29–40 support this material. (3) Skip the linear transformation material here, but discuss the condition number and the Numerical Notes. Assign exercises from among 1–28 and 41–45, and perhaps add a computer project on the condition number. (See the projects on our web site.) I do this when teaching “Course 2” for our engineers. The abbreviation IMT (here and in the Study Guide) denotes the Invertible Matrix Theorem (Theorem 8). 7 5 1. The columns of the matrix are not multiples, so they are linearly independent. By (e) in the −3 −6 IMT, the matrix is invertible. Also, the matrix is invertible by Theorem 4 in Section 2.2 because the determinant is nonzero. 6 −4 2. The fact that the columns of are multiples is not so obvious. The fastest check in this case 6 −9 may be the determinant, which is easily seen to be zero. By Theorem 4 in Section 2.2, the matrix is not invertible. 3. Row reduction to echelon form is trivial because there is really no need for arithmetic calculations: 0 0 5 0 0 5 0 0 5 −3 −7 0 ~ 0 −7 0 ~ 0 −7 0 The 3×3 matrix has 3 pivot positions and hence is 8 5 −1 0 5 −1 0 0 −1 invertible, by (c) of the IMT. [Another explanation could be given using the transposed matrix. But see the note below that follows the solution of Exercise 14.] 4 −7 0 4. The matrix 3 0 −1 obviously has linearly dependent columns (because one column is zero), and 2 0 9 so the matrix is not invertible (or singular) by (e) in the IMT. 0 5. 1 −4 3 0 −9 −5 1 2 ~ 0 7 −4 0 3 −9 2 1 −5 ~ 0 7 0 0 3 −9 2 1 −5 ~ 0 15 0 0 3 0 2 −5 0 The matrix is not invertible because it is not row equivalent to the identity matrix. 1 6. 0 −3 −5 3 6 −4 1 4 ~ 0 0 0 −5 3 −9 −4 1 4 ~ 0 −12 0 −5 3 0 −4 4 0 The matrix is not invertible because it is not row equivalent to the identity matrix. −1 3 7. −2 0 −3 5 −6 −1 0 8 3 2 1 −1 −3 0 ~ 2 0 1 0 −3 −4 0 −1 0 8 3 2 1 −1 0 0 ~ 0 0 1 0 −3 −4 0 0 0 8 3 0 1 0 0 1 The 4×4 matrix has four pivot positions and so is invertible by (c) of the IMT. Copyright © 2016 Pearson Education, Ltd. 2-16 CHAPTER 2 • Matrix Algebra 1 0 8. The 4×4 matrix 0 0 4 −6 9. [M] 7 −1 0 1 −5 2 −1 0 ~ 0 0 2 8 0 0 −1 0 ~ 0 0 2 8 0 0 3 5 0 0 7 9 2 0 4 6 is invertible because it has four pivot positions, by (c) of the IMT. 8 10 −7 11 10 3 −7 −1 2 3 −1 −1 2 3 9 −6 1 11 9 0 −11 − 7 ~ ~ 19 7 −5 10 19 0 9 31 0 −7 −7 0 8 5 −1 4 3 − 1 −1 2 3 −1 −1 5 0 8 5 −11 −11 0 ~ ~ 25.375 24.375 0 0 25.375 24.375 0 0 1 1 0 −.1250 −.1250 0 3 −1 5 −11 1 1 0 −1 2 3 −1 −1 −1 0 8 5 −11 15 ~ 9 31 12 12 0 15 −11 0 −11 − 7 2 3 −1 8 5 −11 0 1 1 0 25.375 24.375 The 4×4 matrix is invertible because it has four pivot positions, by (c) of the IMT. 7 7 9 5 3 1 7 9 5 3 1 5 3 1 6 4 2 0 .4 .8 −.4 8 −8 0 .4 .8 −.4 −18.8 10. [M] 7 5 3 10 1 −3.6 ~ 0 0 0 9 ~ 0 .8 1.6 .2 .6 2.2 −21.6 −21.2 0 0 1 −21 9 6 4 −9 −5 0 8 5 2 11 0 4 0 .2 .4 −.2 −10.4 0 0 0 5 0 ~ 0 0 0 3 .4 0 0 0 1 .8 1 0 0 7 −.4 −21 1 0 9 −18.8 34 7 −1 9 −18.8 7 34 −1 The 5×5 matrix is invertible because it has five pivot positions, by (c) of the IMT. 11. a. b. c. d. True, by the IMT. If statement (d) of the IMT is true, then so is statement (b). True. If statement (h) of the IMT is true, then so is statement (e). False. Statement (g) of the IMT is true only for invertible matrices. True, by the IMT. If the equation Ax = 0 has a nontrivial solution, then statement (d) of the IMT is false. In this case, all the lettered statements in the IMT are false, including statement (c), which means that A must have fewer than n pivot positions. e. True, by the IMT. If AT is not invertible, then statement (1) of the IMT is false, and hence statement (a) must also be false. 12. a. True. If statement (k) of the IMT is true, then so is statement ( j). b. True. If statement (e) of the IMT is true, then so is statement (h). c. True. See the remark immediately following the proof of the IMT. Copyright © 2016 Pearson Education, Ltd. 2.3 • Solutions 2-17 d. False. The first part of the statement is not part (i) of the IMT. In fact, if A is any n×n matrix, the linear transformation x Ax maps n into n , yet not every such matrix has n pivot positions. e. True, by the IMT. If there is a b in n such that the equation Ax = b is inconsistent, then statement (g) of the IMT is false, and hence statement (f) is also false. That is, the transformation x Ax cannot be one-to-one. Note: The solutions below for Exercises 13–30 refer mostly to the IMT. In many cases, however, part or all of an acceptable solution could also be based on various results that were used to establish the IMT. 13. If a square upper triangular n×n matrix has nonzero diagonal entries, then because it is already in echelon form, the matrix is row equivalent to In and hence is invertible, by the IMT. Conversely, if the matrix is invertible, it has n pivots on the diagonal and hence the diagonal entries are nonzero. 14. If A is lower triangular with nonzero entries on the diagonal, then these n diagonal entries can be used as pivots to produce zeros below the diagonal. Thus A has n pivots and so is invertible, by the IMT. If one of the diagonal entries in A is zero, A will have fewer than n pivots and hence be singular. Notes: For Exercise 14, another correct analysis of the case when A has nonzero diagonal entries is to apply the IMT (or Exercise 13) to AT. Then use Theorem 6 in Section 2.2 to conclude that since AT is invertible so is its transpose, A. You might mention this idea in class, but I recommend that you not spend much time discussing AT and problems related to it, in order to keep from making this section too lengthy. (The transpose is treated infrequently in the text until Chapter 6.) If you do plan to ask a test question that involves AT and the IMT, then you should give the students some extra homework that develops skill using AT. For instance, in Exercise 14 replace “columns” by “rows.” Also, you could ask students to explain why an n×n matrix with linearly independent columns must also have linearly independent rows. 15. If A has two identical columns then its columns are linearly dependent. Part (e) of the IMT shows that A cannot be invertible. 16. Part (h) of the IMT shows that a 5×5 matrix cannot be invertible when its columns do not span . 5 17. If A is invertible, so is A–1, by Theorem 6 in Section 2.2. By (e) of the IMT applied to A–1, the columns of A–1 are linearly independent. 18. By (g) of the IMT, C is invertible. Hence, each equation Cx = v has a unique solution, by Theorem 5 in Section 2.2. This fact was pointed out in the paragraph following the proof of the IMT. 19. By (e) of the IMT, D is invertible. Thus the equation Dx = b has a solution for each b in , by (g) of the IMT. Even better, the equation Dx = b has a unique solution for each b in 7 , by Theorem 5 in Section 2.2. (See the paragraph following the proof of the IMT.) 7 20. By the box following the IMT, E and F are invertible and are inverses. So FE = I = EF, and so E and F commute. 21. The matrix G cannot be invertible, by Theorem 5 in Section 2.2 or by the box following the IMT. So (g), and hence (h), of the IMT are false and the columns of G do not span n . 22. Statement (g) of the IMT is false for H, so statement (d) is false, too. That is, the equation Hx = 0 has a nontrivial solution. Copyright © 2016 Pearson Education, Ltd. 2-18 CHAPTER 2 • Matrix Algebra 23. Statement (b) of the IMT is false for K, so statements (e) and (h) are also false. That is, the columns of K are linearly dependent and the columns do not span n . 24. No conclusion about the columns of L may be drawn, because no information about L has been given. The equation Lx = 0 always has the trivial solution. 25. Suppose that A is square and AB = I. Then A is invertible, by the (k) of the IMT. Left-multiplying each side of the equation AB = I by A–1, one has A–1AB = A–1I, IB = A–1, and B = A–1. By Theorem 6 in Section 2.2, the matrix B (which is A–1) is invertible, and its inverse is (A–1)–1, which is A. 26. If the columns of A are linearly independent, then since A is square, A is invertible, by the IMT. So A2, n which is the product of invertible matrices, is invertible. By the IMT, the columns of A2 span . 27. Let W be the inverse of AB. Then ABW = I and A(BW) = I. Since A is square, A is invertible, by (k) of the IMT. Note: The Study Guide for Exercise 27 emphasizes here that the equation A(BW) = I, by itself, does not show that A is invertible. Students are referred to Exercise 38 in Section 2.2 for a counterexample. Although there is an overall assumption that matrices in this section are square, I insist that my students mention this fact when using the IMT. Even so, at the end of the course, I still sometimes find a student who thinks that an equation AB = I implies that A is invertible. 28. Let W be the inverse of AB. Then WAB = I and (WA)B = I. By (j) of the IMT applied to B in place of A, the matrix B is invertible. 29. Since the transformation x Ax is not one-to-one, statement (f) of the IMT is false. Then (i) is also false and the transformation x Ax does not map n onto n . Also, A is not invertible, which implies that the transformation x Ax is not invertible, by Theorem 9. 30. Since the transformation x Ax is one-to-one, statement (f) of the IMT is true. Then (i) is also true and the transformation x Ax maps n onto n . Also, A is invertible, which implies that the transformation x Ax is invertible, by Theorem 9. 31. Since the equation Ax = b has a solution for each b, the matrix A has a pivot in each row (Theorem 4 in Section 1.4). Since A is square, A has a pivot in each column, and so there are no free variables in the equation Ax = b, which shows that the solution is unique. Note: The preceding argument shows that the (square) shape of A plays a crucial role. A less revealing proof is to use the “pivot in each row” and the IMT to conclude that A is invertible. Then Theorem 5 in Section 2.2 shows that the solution of Ax = b is unique. 32. If Ax = 0 has only the trivial solution, then A must have a pivot in each of its n columns. Since A is square (and this is the key point), there must be a pivot in each row of A. By Theorem 4 in Section 1.4, the equation Ax = b has a solution for each b in n . Another argument: Statement (d) of the IMT is true, so A is invertible. By Theorem 5 in Section 2.2, the equation Ax = b has a (unique) solution for each b in n . Copyright © 2016 Pearson Education, Ltd. 2.3 • Solutions 2-19 9 −5 33. (Solution in Study Guide) The standard matrix of T is A = , which is invertible because 4 −7 det A ≠ 0. By Theorem 9, the transformation T is invertible and the standard matrix of T–1 is A–1. From 7 9 the formula for a 2×2 inverse, A−1 = . So 4 5 7 9 x1 T −1 ( x1 , x2 ) = ( 7 x1 + 9 x2 , 4 x1 + 5 x2 ) 4 5 x = 2 6 34. The standard matrix of T is A = −5 −8 , which is invertible because det A = 2 ≠ 0. By Theorem 9, T 7 −1 B A= is invertible, and T −1 (x) = Bx, where= 1 7 T −1 ( x1 , x2 ) = 2 5 1 7 2 5 8 . Thus 6 8 x1 7 5 x1 + 4 x2 , x1 + 3 x2 = 6 x2 2 2 35. (Solution in Study Guide) To show that T is one-to-one, suppose that T(u) = T(v) for some vectors u and v in n . Then S(T(u)) = S(T(v)), where S is the inverse of T. By Equation (1), u = S(T(u)) and S(T(v)) = v, so u = v. Thus T is one-to-one. To show that T is onto, suppose y represents an arbitrary vector in n n n and define x = S(y). Then, using Equation (2), T(x) = T(S(y)) = y, which shows that T maps onto Second proof: By Theorem 9, the standard matrix A of T is invertible. By the IMT, the columns of A are n n n linearly independent and span . By Theorem 12 in Section 1.9, T is one-to-one and maps onto 36. If T maps n onto n , then the columns of its standard matrix A span n by Theorem 12 in Section 1.9. By the IMT, A is invertible. Hence, by Theorem 9 in Section 2.3, T is invertible, and A–1 is the standard matrix of T–1. Since A–1 is also invertible, by the IMT, its columns are linearly independent and span n . Applying Theorem 12 in Section 1.9 to the transformation T–1, we conclude that T–1 is a one-to-one mapping of n onto n . 37. Let A and B be the standard matrices of T and U, respectively. Then AB is the standard matrix of the mapping x T (U (x)) , because of the way matrix multiplication is defined (in Section 2.1). By hypothesis, this mapping is the identity mapping, so AB = I. Since A and B are square, they are invertible, by the IMT, and B = A–1. Thus, BA = I. This means that the mapping x U (T (x)) is the identity mapping, i.e., U(T(x)) = x for all x in n . 38. Let A be the standard matrix of T. By hypothesis, T is not a one-to-one mapping. So, by Theorem 12 in Section 1.9, the standard matrix A of T has linearly dependent columns. Since A is square, the columns of A do not span n . By Theorem 12, again, T cannot map n onto n 39. Given any v in n , we may write v = T(x) for some x, because T is an onto mapping. Then, the assumed properties of S and U show that S(v) = S(T(x)) = x and U(v) = U(T(x)) = x. So S(v) and U(v) are equal for each v. That is, S and U are the same function from n into n Copyright © 2016 Pearson Education, Ltd. 2-20 CHAPTER 2 • Matrix Algebra 40. Given u, v in n let x = S(u) and y = S(v). Then T(x)=T(S(u)) = u and T(y) = T(S(v)) = v, by equation (2). Hence S ( u += v ) S (T ( x ) + T ( y )) = S (T ( x + y )) Because T islinear = x+y By equation (1) = S ( u) + S ( v ) So, S preserves sums. For any scalar r, = S ( r u) S= ( rT ( x )) S (T ( r x )) Because T islinear = rx By equation (1) = rS ( u) So S preserves scalar multiples. Thus S ia a linear transformation. 41. [M] a. The exact solution of (3) is x1 = 3.94 and x2 = .49. The exact solution of (4) is x1 = 2.90 and x2 = 2.00. b. When the solution of (4) is used as an approximation for the solution in (3) , the error in using the value of 2.90 for x1 is about 26%, and the error in using 2.0 for x2 is about 308%. c. The condition number of the coefficient matrix is 3363. The percentage change in the solution from (3) to (4) is about 7700 times the percentage change in the right side of the equation. This is the same order of magnitude as the condition number. The condition number gives a rough measure of how sensitive the solution of Ax = b can be to changes in b. Further information about the condition number is given at the end of Chapter 6 and in Chapter 7. Note: See the Study Guide’s MATLAB box, or a technology appendix, for information on condition number. Only the TI-83+ and TI-89 lack a command for this. 42. [M] MATLAB gives cond(A) = 23683, which is approximately 104. If you make several trials with MATLAB, which records 16 digits accurately, you should find that x and x1 agree to at least 12 or 13 significant digits. So about 4 significant digits are lost. Here is the result of one experiment. The vectors were all computed to the maximum 16 decimal places but are here displayed with only four decimal places: .9501 .21311 , b = Ax = = = x rand(4,1) .6068 .4860 −3.8493 5.5795 . The MATLAB solution is x1 = A\b = 20.7973 .8467 .9501 .2311 . .6068 .4860 .0171 .4858 ×10–12. The computed solution x1 is accurate to about 12 decimal places. However, x – x1 = −.2360 .2456 Copyright © 2016 Pearson Education, Ltd. 2.5 • Solutions 2-21 43. [M] MATLAB gives cond(A) = 68,622. Since this has magnitude between 104 and 105, the estimated accuracy of a solution of Ax = b should be to about four or five decimal places less than the 16 decimal places that MATLAB usually computes accurately. That is, one should expect the solution to be accurate to only about 11 or 12 decimal places. Here is the result of one experiment. The vectors were all computed to the maximum 16 decimal places but are here displayed with only four decimal places: .2190 .0470 x = rand(5,1) = .6789 , b = Ax = .6793 .9347 15.0821 .8165 19.0097 . The MATLAB solution is x1 = A\b = −5.8188 14.5557 .2190 .0470 .6789 . .6793 .9347 .3165 −.6743 However, x – x1 = .3343 × 10−11 . The computed solution x1 is accurate to about 11 decimal places. .0158 −.0005 44. [M] Solve Ax = (0, 0, 0, 0, 1). MATLAB shows that cond( A) ≈ 4.8 × 105. Since MATLAB computes numbers accurately to 16 decimal places, the entries in the computed value of x should be accurate to at least 11 digits. The exact solution is (630, –12600, 56700, –88200, 44100). 45. [M] Some versions of MATLAB issue a warning when asked to invert a Hilbert matrix of order 12 or larger using floating-point arithmetic. The product AA–1 should have several off-diagonal entries that are far from being zero. If not, try a larger matrix. Note: All matrix programs supported by the Study Guide have data for Exercise 45, but only MATLAB and Maple have a single command to create a Hilbert matrix. The HP-48G data for Exercise 45 contain a program that can be edited to create other Hilbert matrices. Notes: The Study Guide for Section 2.3 organizes the statements of the Invertible Matrix Theorem in a table that imbeds these ideas in a broader discussion of rectangular matrices. The statements are arranged in three columns: statements that are logically equivalent for any m×n matrix and are related to existence concepts, those that are equivalent only for any n×n matrix, and those that are equivalent for any n×p matrix and are related to uniqueness concepts. Four statements are included that are not in the text’s official list of statements, to give more symmetry to the three columns. You may or may not wish to comment on them. I believe that students cannot fully understand the concepts in the IMT if they do not know the correct wording of each statement. (Of course, this knowledge is not sufficient for understanding.) The Study Guide’s Section 2.3 has an example of the type of question I often put on an exam at this point in the course. The section concludes with a discussion of reviewing and reflecting, as important steps to a mastery of linear algebra. 2.4 SOLUTIONS Notes: Partitioned matrices arise in theoretical discussions in essentially every field that makes use of matrices. The Study Guide mentions some examples (with references). Every student should be exposed to some of the ideas in this section. If time is short, you might omit Example 4 and Theorem 10, and replace Example 5 by a problem similar to one in Exercises 1–10. (A sample Copyright © 2016 Pearson Education, Ltd. 2-22 CHAPTER 2 • Matrix Algebra replacement is given at the end of these solutions.) Then select homework from Exercises 1–13, 15, and 21– 24. The exercises just mentioned provide a good environment for practicing matrix manipulation. Also, students will be reminded that an equation of the form AB = I does not by itself make A or B invertible. (The matrices must be square and the IMT is required.) 1. Apply the row-column rule as if the matrix entries were numbers, but for each product always write the B I 0 A B IA + 0C IB + 0 D A = entry of the left block-matrix on the left. E I C D = EA + IC EB + ID EA + C EB + D 2. Apply the row-column rule as if the matrix entries were numbers, but for each product always write the E 0 A B EA + 0C EB + 0 D EA EB = entry of the left block-matrix on the left. 0 F C D = 0 A + FC 0 B + FD FC FD 3. Apply the row-column rule as if the matrix entries were numbers, but for each product always write the 0 I W X 0W + IY 0 X + IZ Y Z = entry of the left block-matrix on the left. I 0 Y Z = IW + 0Y IX + 0 Z W X 4. Apply the row-column rule as if the matrix entries were numbers, but for each product always write the entry of the left block-matrix on the left. 0 A B IA + 0C IB + 0 D A B I = − X I C D = − XA + IC − XB + ID − XA + C − XB + D A B I 0 AI + BX A0 + BY 5. Compute the left side of the equation: = C 0 X Y CI + 0 X C 0 + 0Y Set this equal to the right side of the equation: A + BX = 0 BY = I A + BX BY 0 I = so that C 0 Z 0 = C Z= 0 0 Since the (2, 1) blocks are equal, Z = C. Since the (1, 2) blocks are equal, BY = I. To proceed further, assume that B and Y are square. Then the equation BY =I implies that B is invertible, by the IMT, and Y = B–1. (See the boxed remark that follows the IMT.) Finally, from the equality of the (1, 1) blocks, BX = –A, B–1BX = B–1(–A), and X = –B–1A. The order of the factors for X is crucial. Note: For simplicity, statements (j) and (k) in the Invertible Matrix Theorem involve square matrices C and D. Actually, if A is n×n and if C is any matrix such that AC is the n×n identity matrix, then C must be n×n, too. (For AC to be defined, C must have n rows, and the equation AC = I implies that C has n columns.) Similarly, DA = I implies that D is n×n. Rather than discuss this in class, I expect that in Exercises 5–8, when students see an equation such as BY = I, they will decide that both B and Y should be square in order to use the IMT. X 0 A 0 XA + 0 B X 0 + 0C XA = 6. Compute the left side of the equation: Y Z B C = YA + ZB Y 0 + ZC YA + ZB XA Set this equal to the right side of the equation: YA + ZB 0 I = ZC 0 Copyright © 2016 Pearson Education, Ltd. 0 ZC 0 = XA I= 0 0 so that I YA + ZB = 0 ZC = I 2.5 • Solutions 2-23 To use the equality of the (1, 1) blocks, assume that A and X are square. By the IMT, the equation XA =I implies that A is invertible and X = A–1. (See the boxed remark that follows the IMT.) Similarly, if C and Z are assumed to be square, then the equation ZC = I implies that C is invertible, by the IMT, and Z = C–1. Finally, use the (2, 1) blocks and right-multiplication by A–1 to get YA = –ZB = –C–1B, then YAA–1 = (–C–1B)A–1, and Y = –C–1BA–1. The order of the factors for Y is crucial. X 7. Compute the left side of the equation: Y 0 0 A 0 0 I B Z XA + 0 + 0 B 0 = YA + 0 + IB I XZ + 0 + 0 I YZ + 0 + II Set this equal to the right side of the equation: XZ I 0 = XA I= XZ 0 XA YA + B YZ + I = 0 I so that YA = +I I + B 0 YZ= To use the equality of the (1, 1) blocks, assume that A and X are square. By the IMT, the equation XA =I implies that A is invertible and X = A–1. (See the boxed remark that follows the IMT) Also, X is invertible. Since XZ = 0, X – 1 XZ = X – 1 0 = 0, so Z must be 0. Finally, from the equality of the (2, 1) blocks, YA = –B. Right-multiplication by A–1 shows that YAA–1 = –BA–1 and Y = –BA–1. The order of the factors for Y is crucial. A 8. Compute the left side of the equation: 0 B X I 0 Y 0 Z AX + B 0 = I 0 X + I 0 AY + B 0 0Y + I 0 AZ + BI 0 Z + II AX AY AZ + B I 0 0 = Set this equal to the right side of the equation: 0 I 0 0 I 0 To use the equality of the (1, 1) blocks, assume that A and X are square. By the IMT, the equation XA =I implies that A is invertible and X = A–1. (See the boxed remark that follows the IMT. Since AY = 0, from the equality of the (1, 2) blocks, left-multiplication by A–1 gives A–1AY = A–10 = 0, so Y = 0. Finally, from the (1, 3) blocks, AZ = –B. Left-multiplication by A–1 gives A–1AZ = A–1(–B), and Z = – A–1B. The order of the factors for Z is crucial. Note: The Study Guide tells students, “Problems such as 5–10 make good exam questions. Remember to mention the IMT when appropriate, and remember that matrix multiplication is generally not commutative.” When a problem statement includes a condition that a matrix is square, I expect my students to mention this fact when they apply the IMT. 9. Compute the left side of the equation: I 0 0 A11 A12 IA11 + 0 A21 + 0 A31 X I 0 A 21 A22 = XA11 + IA21 + 0 A31 Y 0 I A31 A32 YA11 + 0 A21 + IA31 IA12 + 0 A22 + 0 A32 XA12 + IA22 + 0 A32 YA12 + 0 A22 + IA32 A11 Set this equal to the right side of the equation: XA11 + A21 YA11 + A31 A12 B11 B12 0 XA12 + A22 = B22 YA12 + A32 0 B32 = A11 B= A12 B12 11 + A21 0 XA12 = + A22 B22 . so that XA11 = YA11 = + A31 0 YA12 = + A32 B32 Copyright © 2016 Pearson Education, Ltd. 2-24 CHAPTER 2 • Matrix Algebra Since the (2,1) blocks are equal, XA11 + A21 = 0 and XA11 = − A21. Since A11 is invertible, right multiplication by A11−1 gives X = − A21 A11−1. Likewise since the (3,1) blocks are equal, −1 −1 YA11 + A31 = 0 and YA11 = − A31. Since A11 is invertible, right multiplication by A11 gives Y = − A31 A11 . B22 . Since X = − A21 A11−1 , B22 = A22 − A21 A11−1 A12 . Finally, from the (2,2) entries, XA12 + A22 = I 10. Since the two matrices are inverses, C A 0 I B 0 I 0 Z I X Compute the left side of the equation: I 0 0 I 0 0 II + 0 Z + 0 X C I 0 Z I 0 = CI + IZ + 0 X A B I X Y I AI + BZ + I X 0 I Y I 0 + 0 I + 0Y C 0 + II + 0Y A0 + BI + IY I Set this equal to the right side of the equation: C + Z A + BZ + X = I I C+Z = 0 so that A + BZ + X = 0 0 I 0 = 0 I 0 0 I 0 0 0 I I 0 + 00 + 0 I C 0 + I 0 + 0 I A0 + B0 + II 0 I B +Y 0 I 0 0 0 = 0 I 0 I 0 0 I = 0 0= 0 0 = I I = 0 0 .. B += Y 0 = I I 0 and Z = −C . Likewise since the (3, 2) blocks are equal, Since the (2,1) blocks are equal, C + Z = B +Y = 0 and Y = − B. Finally, from the (3,1) entries, A + BZ + X =0 and X =− A − BZ . Since Z =−C , X =− A − B (−C ) =− A + BC . 11. a. True. See the subsection Addition and Scalar Multiplication. b. False. See the paragraph before Example 3. 12. a. True. See the paragraph before Example 4. b. False. See the paragraph before Example 3. 13. You are asked to establish an if and only if statement. First, supose that A is invertible, D E B 0 D E BD BE I 0 = . Then and let A−1 = = F G 0 C F G CF CG 0 I Since B is square, the equation BD = I implies that B is invertible, by the IMT. Similarly, CG = I implies that C is invertible. Also, the equation BE = 0 imples that E = B −1 0 = 0. Similarly F = 0. Thus −1 −1 B 0 D E B = = = A 0 C E G 0 −1 0 C −1 (*) This proves that A is invertible only if B and C are invertible. For the “if ” part of the statement, suppose that B and C are invertible. Then (*) provides a likely candidate for A−1 which can be used to show that −1 0 BB −1 0 I 0 B 0 B A is invertible. Compute: = = . C −1 0 CC −1 0 I 0 C 0 Copyright © 2016 Pearson Education, Ltd. 2.5 • Solutions 2-25 Since A is square, this calculation and the IMT imply that A is invertible. (Don’t forget this final sentence. Without it, the argument is incomplete.) Instead of that sentence, you could add the equation: B −1 0 B 0 B −1 B 0 I = = −1 0 − C 0 C C 1C 0 0 0 I 14. You are asked to establish an if and only if statement. First suppose that A is invertible. Example 5 shows that A11 and A22 are invertible. This proves that A is invertible only if A11 and A22 are invertible. For the if part of this statement, suppose that A11 and A22 are invertible. Then the formula in Example 5 provides a likely candidate for A−1 which can be used to show that A is invertible . Compute: A11 0 −1 A12 A 11 A22 0 −1 −1 −1 + A12 0 A11 A 11 − A 11 A12 A 22 = −1 −1 A 22 0 A 11 + A22 0 I = 0 I = 0 −1 −1 −1 + A12 A 22 A11 (− A 11 ) A 12 A 22 −1 0(− A 11 ) A12 A −221 + A 22 A −221 −1 −1 −1 −( A11 A 11 + A12 A 22 ) A12 A 22 I −1 −1 I − A12 A 22 + A 12 A 22 = I 0 0 I Since A is square, this calculation and the IMT imply that A is invertible. 15. Compute the right side of the equation: A11Y 0 I Y A11 I 0 A11 0 I Y A11 = = X I 0 S 0 I X A 11 S 0 I X A11 X A11Y + S Set this equal to the left side of the equation: A 11 = A 11 A 11Y A 11 A 12 A 11Y = A 12 A 11 so that = X A X A11 = A 21 X A 11 Y + S = X A 11Y + S A 21 A 22 A 22 11 −1 Since the (1, 2) blocks are equal, A 11Y = A 12. Since A11 is invertible, left multiplication by A 11 gives −1 A 12. Likewise since the (2,1) blocks are equal, X A11 = A21. Since A11 is invertible, right Y = A 11 −1 gives that X = A21 A11−1. One can check that the matrix S as given in the exercise multiplication by A11 satisfies the equation X A11Y + S = A22 with the calculated values of X and Y given above. 0 I 0 I 0 I 16. Suppose that A and A11 are invertible. First note that = and X I − X I 0 I I Y I −Y I 0 I 0 I Y = and . Since the matrices 0 I 0 are square, they are both I 0 I X I 0 I I invertible by the IMT. Equation (7) may be left multipled by X I 0 −1 Y A to find 11 I 0 0 I = S X −1 0 I −1 −1 and right multipled by 0 I Y A . I 0 I 0 A is invertible as the product of invertible matrices. Finally, Thus by Theorem 6, the matrix 11 S 0 Exercise 13 above may be used to show that S is invertible. Copyright © 2016 Pearson Education, Ltd. 2-26 CHAPTER 2 • Matrix Algebra 17. The column-row expansions of Gk and Gk+1 are: Gk = X k X kT = col1 ( X k ) row1 ( X kT ) + ... + colk ( X k ) row k ( X kT ) and Gk +1 = X k +1 X kT+1 = col1 ( X k +1 ) row1 ( X kT+1 ) + ... + colk ( X k +1 ) row k ( X kT+1 ) + colk +1 ( X k +1 ) row k +1 ( X kT+1 ) = col1 ( X k ) row1 ( X kT ) + ... + colk ( X k ) row k ( X kT ) + colk +1 ( X k +1 ) row k +1 ( X kT ) = Gk + colk +1 ( X k +1 ) row k +1 ( X kT ) since the first k columns of Xk+1 are identical to the first k columns of Xk. Thus to update Gk to produce Gk+1, the number colk+1 (Xk+1) rowk+1 ( X kT ) should be added to Gk. X T 18. Since W = [ X x0 ] , the product W W = = [ X x0 ] T x0 from Exercise 15, S may be computed: = S xT0 x0 − xT0 X ( X T X ) −1 X T x0 T X T X T x0 X X T x0 . By applying the formula for S xT0 x0 = xT0 ( I m − X ( X T X ) −1 X T )x0 = xT0 Mx0 19. The matrix equation (8) in the text is equivalent to ( A − sI n )= x + Bu 0 and = Cx + u y Rewrite the first equation as ( A − sI n )x = − Bu. When A − sI n is invertible, ( A − sI n ) −1 (− Bu) = x= −( A − sI n ) −1 Bu Substitute this formula for x into the second equation above: −1 C (−( A − sI= y, so that I m u − C ( A= − sI n ) −1 Bu y n ) Bu ) + u Thus y =( I m − C ( A − sI n ) −1 B )u. If W ( s ) =I m − C ( A − sI n ) −1 B, then y = W ( s )u. The matrix W(s) is the Schur complement of the matrix A − sI n in the system matrix in equation (8) A − BC − sI n 20. The matrix in question is −C be computed: S= I m − (−C )( A − BC − sI m ) −1 B B . By applying the formula for S from Exercise 15, S may I m =I m + C ( A − BC − sI m ) −1 B 0 + 0 1 0 1 0 1 + 0 1 21. a. A2 = = = 2 3 −1 3 −1 3 − 3 0 + (−1) 0 0 1 0 A 0 A2 + 0 0 + 0 I A b. M 2 = = = 2 I − A I − A A − A 0 + (− A) 0 0 I Copyright © 2016 Pearson Education, Ltd. 2.5 • Solutions 2-27 22. Let C be any nonzero 2×3 matrix. Following the pattern in 21(a) with block matrices instead of numbers, 0 + 0 I3 0 0 0 I3 0 I3 + 0 I I3 set M = 3 = and verify M 2 = = . 2 C − I 2 C − I 2 CI 3 − I 2C 0 + ( − I 2 ) 0 I 2 C −I2 23. The product of two 1×1 “lower triangular” matrices is “lower triangular.” Suppose that for n = k, the product of two k×k lower triangular matrices is lower triangular, and consider any (k+1)× (k+1) matrices a 0T b 0T as A1 = A1 and B1. Partition these matrices = , B1 v A w B where A and B are k×k matrices, v and w are in k , and a and b are scalars. Since A1 and B1 are lower triangular, so are A and B. Then a 0T b 0T ab + 0T w a 0T + 0T B ab 0T = A1B1 = = v0T + AB bv + Aw AB v A w B vb + Aw Since A and B are k×k, AB is lower triangular. The form of A1B1 shows that it, too, is lower triangular. Thus the statement about lower triangular matrices is true for n = k +1 if it is true for n = k. By the principle of induction, the statement is true for all n > 1. Note: Exercise 23 is good for mathematics and computer science students. The solution of Exercise 23 in the Study Guide shows students how to use the principle of induction. The Study Guide also has an appendix on “The Principle of Induction,” at the end of Section 2.4. The text presents more applications of induction in Section 3.2 and in the Supplementary Exercises for Chapter 3. 1 0 0 0 1 1 0 0 24. Let An = = 1 1 1 0 , Bn 1 1 1 1 1 −1 0 0 0 1 −1 0 0 1 −1 0 0 0 . 1 By direct computation A2B2 = I2. Assume that for n = k, the matrix AkBk is Ik, and write 1 0T 1 Ak +1 = = and Bk +1 v Ak w 0T Bk where v and w are in k , vT = [1 1 ⋅ ⋅ ⋅ 1], and wT = [–1 0 ⋅ ⋅ ⋅ 0]. Then 1 + 0T w 0T + 0T Bk 1 0T = = I k +1 T v + Ak w v0 + Ak Bk 0 I k The (2,1)-entry is 0 because v equals the first column of Ak., and Akw is –1 times the first column of Ak. By the principle of induction, AnBn = In for all n > 2. Since An and Bn are square, the IMT shows that 1 = Ak +1 Bk +1 v 0T 1 0T = Ak w Bk these matrices are invertible, and Bn = An−1. Copyright © 2016 Pearson Education, Ltd. 2-28 CHAPTER 2 • Matrix Algebra Note: An induction proof can also be given using partitions with the form shown below. The details are slightly more complicated. Ak 0 Bk = Ak +1 = and Bk +1 T T 1 v w 0 1 0 Bk 0 Ak Bk + 0wT Ak 0 + 0 I k 0 = = = I k +1 T 1 wT 1 vT Bk + wT 1 vT 0 + 1 0 The (2,1)-entry is 0T because vT times a column of Bk equals the sum of the entries in the column, and all of such sums are zero except the last, which is 1. So vTBk is the negative of wT. By the principle of induction, AnBn = In for all n > 2. Since An and Bn are square, the IMT shows that these matrices are Ak = A T k +1 Bk +1 v invertible, and Bn = An−1. 25. First, visualize a partition of A as a 2×2 block–diagonal matrix, as below, and then visualize the (2,2)-block itself as a block-diagonal matrix. That is, 1 2 0 0 0 3 5 0 0 0 A11 = A = 0 0 2 0 0 0 0 0 0 7 8 0 0 0 5 6 2 0 0 0 2 0 7 8 ,= where A22 = A22 0 0 5 6 3 Observe that B is invertible and B–1 = −2.5 0 .5 .5 −1 3 −4 0 = = invertible, and A22 0 0 2.5 3.5 − −4 . By Exercise 13, the block diagonal matrix A22 is 3.5 0 3 −2.5 0 −4 3.5 −5 Next, observe that A11 is also invertible, with inverse 3 and its inverse is block diagonal: −5 2 3 −1 −1 0 A 11 = A−1 = −1 A22 0 0 0 0 = −4 3 −2.5 3.5 0 .5 0 0 0 B −5 3 0 0 0 2 . By Exercise 13, A itself is invertible, −1 2 −1 0 0 0 0 0 .5 0 0 0 0 0 3 −2.5 0 0 0 −4 3.5 26. [M] This exercise and the next, which involve large matrices, are more appropriate for MATLAB, Maple, and Mathematica, than for the graphic calculators. a. Display the submatrix of A obtained from rows 15 to 20 and columns 5 to 10. MATLAB: A(15:20, 5:10) Maple: submatrix(A, 15..20, 5..10) Mathematica: Take[ A, {15,20}, {5,10} ] Copyright © 2016 Pearson Education, Ltd. 2.5 • Solutions 2-29 b. Insert a 5×10 matrix B into rows 10 to 14 and columns 20 to 29 of matrix A: MATLAB: A(10:14, 20:29) = B ; The semicolon suppresses output display. Maple: copyinto(B, A, 10, 20): The colon suppresses output display. Mathematica: For [ i=10, i 0. Then H contains a basis S consisting of n vectors. But applying the Basis Theorem to V, S is also a basis for V. Thus H = V = SpanS. 27. Suppose that dim = k < ∞. Now n is a subspace of for all n, and dim k −1 = k, so dim k −1 = dim . This would imply that k −1 = , which is clearly untrue: for example p(t ) = t k is in but not in k −1 . Thus the dimension of cannot be finite. 28. The space C( ) contains as a subspace. If C( ) were finite-dimensional, then would also be finite-dimensional by Theorem 11. But is infinite-dimensional by Exercise 27, so C( ) must also be infinite-dimensional. 29. a. True. Apply the Spanning Set Theorem to the set {v1 ,…, v p } and produce a basis for V. This basis will not have more than p elements in it, so dimV ≤ p. b. True. By Theorem 11, {v1 ,…, v p } can be expanded to find a basis for V. This basis will have at least p elements in it, so dimV ≥ p. c. True. Take any basis (which will contain p vectors) for V and adjoin the zero vector to it. 30. a. False. For a counterexample, let v be a non-zero vector in 3 , and consider the set {v, 2v}. This is a linearly dependent set in 3 , but dim 3= 3 > 2 . b. True. If dimV ≤ p, there is a basis for V with p or fewer vectors. This basis would be a spanning set for V with p or fewer vectors, which contradicts the assumption. c. False. For a counterexample, let v be a non-zero vector in 3 , and consider the set {v, 2v}. This is a linearly dependent set in 3 with 3 – 1 = 2 vectors, and dim 3 = 3 . 31. Since H is a nonzero subspace of a finite-dimensional vector space V, H is finite-dimensional and has a basis. Let {u1 ,…, u p } be a basis for H. We show that the set {T (u1 ),…, T (u p )} spans T(H). Let y Copyright © 2016 Pearson Education, Ltd. 4-34 CHAPTER 4 • Vector Spaces be in T(H). Then there is a vector x in H with T(x) = y. Since x is in H and {u1 ,…, u p } is a basis for H, x may be written as= x c1u1 + …+ c p u p for some scalars c1 ,…, c p . Since the transformation T is linear,= y T= c1T (u1 ) + …+ c pT (u p ) . Thus y is a linear combination of (x) T (c1u1 + …+ c p u= p) T (u1 ),…, T (u p ) , and {T (u1 ),…, T (u p )} spans T(H). By the Spanning Set Theorem, this set contains a basis for T(H). This basis then has not more than p vectors, and dimT(H) ≤ p = dim H. 32. Since H is a nonzero subspace of a finite-dimensional vector space V, H is finite-dimensional and has a basis. Let {u1 ,…u p } be a basis for H. In Exercise 31 above it was shown that {T (u1 ),…, T (u p )} spans T(H). In Exercise 32 in Section 4.3, it was shown that {T (u1 ),…, T (u p )} is linearly independent. Thus {T (u1 ),…, T (u p )} is a basis for T(H), and dimT(H) = p = dim H. 33. [M] a. To find a basis for 5 which contains the given vectors, we row reduce −9 −7 8 −5 7 9 4 1 6 −7 6 7 −8 5 −7 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 0 0 ∼ 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 −1/ 3 0 −1/ 3 0 0 0 0 0 1 0 0 0 0 0 1 1 1 0 3 −9 3/ 7 5 / 7 −3/ 7 . 22 / 7 −53/ 7 The first, second, third, fifth, and sixth columns are pivot columns, so these columns of the original matrix ( {v1 , v 2 , v 3 , e 2 , e3 } ) form a basis for 5 : b. The original vectors are the first k columns of A. Since the set of original vectors is assumed to be linearly independent, these columns of A will be pivot columns and the original set of vectors will be included in the basis. Since the columns of A include all the columns of the identity matrix, Col A = n . 34. [M] a. The -coordinate vectors of the vectors in are the columns of the matrix 1 0 0 P = 0 0 0 0 0 1 0 0 0 0 0 −1 0 2 0 0 0 0 0 −3 0 4 0 0 0 1 0 −8 0 8 0 0 0 5 0 −20 0 16 0 −1 0 18 0 . −48 0 32 The matrix P is invertible because it is triangular with nonzero entries along its main diagonal. Thus its columns are linearly independent. Since the coordinate mapping is an isomorphism, this shows that the vectors in are linearly independent. b. We know that dim H = 7 because is a basis for H. Now is a linearly independent set, and the vectors in lie in H by the trigonometric identities. Thus by the Basis Theorem, is a basis for H. Copyright © 2016 Pearson Education, Ltd. 4.6 4.6 • Solutions 4-35 SOLUTIONS Notes: This section puts together most of the ideas from Chapter 4. The Rank Theorem is the main result in this section. Many students have difficulty with the difference in finding bases for the row space and the column space of a matrix. The first process uses the nonzero rows of an echelon form of the matrix. The second process uses the pivots columns of the original matrix, which are usually found through row reduction. Students may also have problems with the varied effects of row operations on the linear dependence relations among the rows and columns of a matrix. Problems of the type found in Exercises 19–26 make excellent test questions. Figure 1 and Example 4 prepare the way for Theorem 3 in Section 6.1; Exercises 27–29 anticipate Example 6 in Section 7.4. 1. The matrix B is in echelon form. There are two pivot columns, so the dimension of Col A is 2. There are two pivot rows, so the dimension of Row A is 2. There are two columns without pivots, so the equation Ax = 0 has two free variables. Thus the dimension of Nul A is 2. A basis for Col A is the 1 −4 pivot columns of A: −1 , 2 . A basis for Row A consists of the pivot rows of B: 5 −6 {(1,0, −1,5),(0, −2,5, −6)}. To find a basis for Nul A row reduce to reduced echelon form: 1 A∼ 0 0 1 −1 −5 / 2 5 . The solution to Ax = 0 in terms of free variables is x= x3 − 5 x4 , 1 3 = x2 (5 / 2) x3 − 3 x4 with x3 and x4 free. Thus a basis for Nul A is 1 −5 5 / 2 −3 , 1 0 0 1 . 2. The matrix B is in echelon form. There are three pivot columns, so the dimension of Col A is 3. There are three pivot rows, so the dimension of Row A is 3. There are two columns without pivots, so the equation Ax = 0 has two free variables. Thus the dimension of Nul A is 2. A basis for Col A is 1 4 9 −6 −10 −2 the pivot columns of A: , , . A basis for Row A is the pivot rows of B: −3 −6 −3 3 4 0 {(1, −3,0,5, −7),(0,0, 2, −3,8),(0,0,0,0,5)}. To find a basis for Nul A row reduce to reduced echelon 1 0 form: A ∼ 0 0 −3 0 0 0 0 1 0 0 5 −3/ 2 0 0 0 0 . The solution to Ax = 0 in terms of free variables is 1 0 Copyright © 2016 Pearson Education, Ltd. 4-36 CHAPTER 4 • Vector Spaces = x1 3 x2 − 5 x4 , x3 = (3/ 2) x4 , x5 = 0 , with x2 and x4 free. Thus a basis for Nul A is 3 −5 1 0 0 , 3/ 2 0 1 0 0 . 3. The matrix B is in echelon form. There are three pivot columns, so the dimension of Col A is 3. There are three pivot rows, so the dimension of Row A is 3. There are two columns without pivots, so the equation Ax = 0 has two free variables. Thus the dimension of Nul A is 2. A basis for Col A is 2 6 2 −3 −3 −2 the pivot columns of A: , , . A basis for Row A is the pivot rows of B: 4 9 5 −2 3 −4 {(2, −3,6, 2,5),(0,0,3, −1,1),(0,0,0,1,3)}. To find a basis for Nul A row reduce to reduced echelon −9 / 2 4 / 3 . The solution to Ax = 0 in terms of free variables is 3 0 = x1 (3/ 2) x2 + (9 / 2) x5 , x3 = −(4 / 3) x5 , x4 = −3 x5 , with x2 and x5 free. Thus a basis for Nul A is 1 0 form: A ∼ 0 0 −3/ 2 0 0 0 3/ 2 9 / 2 1 0 0 , −4 / 3 0 −3 0 1 0 1 0 0 0 0 1 0 . 4. The matrix B is in echelon form. There are three pivot columns, so the dimension of Col A is 3. There are three pivot rows, so the dimension of Row A is 3. There are three columns without pivots, so the equation Ax = 0 has three free variables. Thus the dimension of Nul A is 3. A basis for Col A 1 1 7 1 2 10 is the pivot columns of A: 1 , −1 , 1 . A basis for Row A is the pivot rows of B: 1 −3 −5 1 −2 0 {(1,1, − 3, 7, 9, − 9), (0,1, − 1, 3, 4, − 3), (0, 0, 0,1, − 1, − 2)}. To find a basis for Nul A row reduce to 1 0 reduced echelon form: A ∼ 0 0 0 0 1 0 0 0 −2 −1 0 0 0 0 0 1 0 0 9 7 −1 0 0 2 3 −2 . The solution to Ax = 0 in terms of free 0 0 Copyright © 2016 Pearson Education, Ltd. 4.6 • Solutions 4-37 variables is x1 = 2 x3 − 9 x5 − 2 x6 , x2 =x3 − 7 x5 − 3 x6 , x= x5 + 2 x6 , with x3 , x5 , and x6 free. Thus a 4 basis for Nul A is 2 −9 −2 1 −7 −3 1 0 0 , , 0 1 2 0 1 0 0 0 1 . 5. By the Rank Theorem, dim Nul A = 8 − rank A = 8 − 3 = 5. Since = rank AT dim = Col AT dim Row A, rank AT = 3. = dim Row A rank = A,dim Row A 3. Since 6. By the Rank Theorem, dim Nul A = 3 − rank A = 3 − 3 = 0. Since = rank AT dim = Col AT dim Row = A, rank AT 3. = dim Row A rank = A, dim Row A 3. Since 7. Yes, Col A = 4 . Since A has four pivot columns, dim Col A = 4. Thus Col A is a four-dimensional subspace of 4 , and Col A = 4 . No, Nul A ≠ 3 . It is true that dim Nul A = 3 , but Nul A is a subspace of 7 . 8. Since A has four pivot columns, rank A = 4, and dim Nul A = 6 − rank A = 6 − 4 = 2. No. Col A ≠ 4 . It is true that dim = Col A rank = A 4, but Col A is a subspace of 5 . 9. Since dim Nul A = 4, rank A = 6 − dim Nul A = 6 − 4 = 2. So dim = Col A rank = A 2. 10. Since dim Nul A = 5, rank A = 6 − dim Nul A = 6 − 5 = 1. So dim = Col A rank = A 1. 11. Since dim Nul A = 2, rank A = 5 − dim Nul A = 5 − 2 = 3. So dim Row = A dim = Col A rank = A 3. 12. Since dim Nul A = 4, rank A = 6 − dim Nul A = 6 − 4 = 2. So dim Row = A dim = Col A rank = A 2. 13. The rank of a matrix A equals the number of pivot positions which the matrix has. If A is either a 7 × 5 matrix or a 5 × 7 matrix, the largest number of pivot positions that A could have is 5. Thus the largest possible value for rank A is 5. 14. The dimension of the row space of a matrix A is equal to rank A, which equals the number of pivot positions which the matrix has. If A is either a 4 × 3 matrix or a 3 × 4 matrix, the largest number of pivot positions that A could have is 3. Thus the largest possible value for dim Row A is 3. 15. Since the rank of A equals the number of pivot positions which the matrix has, and A could have at most 6 pivot positions, rank A ≤ 6. Thus dim Nul A = 8 − rank A ≥ 8 − 6 = 2. 16. Since the rank of A equals the number of pivot positions which the matrix has, and A could have at most 4 pivot positions, rank A ≤ 4. Thus dim Nul A = 4 − rank A ≥ 4 − 4 = 0. 17. a. True. The rows of A are identified with the columns of AT . See the paragraph before Example 1. b. False. See the warning after Example 2. Copyright © 2016 Pearson Education, Ltd. 4-38 CHAPTER 4 • Vector Spaces c. True. See the Rank Theorem. d. False. See the Rank Theorem. e. True. See the Numerical Note before the Practice Problem. 18. a. b. c. d. False. Review the warning after Theorem 6 in Section 4.3. False. See the warning after Example 2. True. See the remark in the proof of the Rank Theorem. True. This fact was noted in the paragraph before Example 4. It also follows from the fact that the rows of AT are the columns of ( AT )T = A. e. True. See Theorem 13. 19. Yes. Consider the system as Ax = 0, where A is a 5 × 6 matrix. The problem states that dim Nul A = 1 . By the Rank Theorem, rank A = 6 − dim Nul A = 5. Thus dim = Col A rank = A 5, and since Col A is a subspace of 5 , Col A = 5 . So every vector b in 5 is also in Col A, and Ax = b, has a solution for all b. 20. No. Consider the system as Ax = b, where A is a 6 × 8 matrix. The problem states that dim Nul A = 2. By the Rank Theorem, rank A = 8 − dim Nul A = 6. Thus dim = Col A rank = A 6, and since Col A is a subspace of 6 , Col A = 6 So every vector b in 6 is also in Col A, and Ax = b has a solution for all b. Thus it is impossible to change the entries in b to make Ax = b into an inconsistent system. 21. No. Consider the system as Ax = b, where A is a 9 × 10 matrix. Since the system has a solution for all b in 9 , A must have a pivot in each row, and so rankA = 9. By the Rank Theorem, dimNulA = 10 − 9 = 1. Thus it is impossible to find two linearly independent vectors in Nul A. 22. No. Consider the system as Ax = 0, where A is a 10 × 12 matrix. Since A has at most 10 pivot 12 rankA ≥ 2. Thus it is impossible to find positions, rankA ≤ 10. By the Rank Theorem, dimNulA =− a single vector in Nul A which spans Nul A. 23. Yes, six equations are sufficient. Consider the system as Ax = 0, where A is a 12 × 8 matrix. The problem states that dimNul A = 2. By the Rank Theorem, rank A = 8 − dimNul A = 6. Thus dimCol = A rank = A 6. So the system Ax = 0 is equivalent to the system Bx = 0, where B is an echelon form of A with 6 nonzero rows. So the six equations in this system are sufficient to describe the solution set of Ax = 0. 24. Yes, No. Consider the system as Ax = b, where A is a 7 × 6 matrix. Since A has at most 6 pivot positions, rank A ≤ 6. By the Rank Theorem, dim Nul A = 6 − rank A ≥ 0. If dimNul A = 0, then the system Ax = b will have no free variables. The solution to Ax = b, if it exists, would thus have to be unique. Since rank A ≤ 6, Col A will be a proper subspace of 7 . Thus there exists a b in 7 for which the system Ax = b is inconsistent, and the system Ax = b cannot have a unique solution for all b. Copyright © 2016 Pearson Education, Ltd. 4.6 • Solutions 4-39 25. No. Consider the system as Ax = b, where A is a 10 × 12 matrix. The problem states that dim Nul A = 3. By the Rank Theorem, dimCol A = rank A = 12 − dimNul A = 9. Thus Col A will be a proper subspace of 10 . Thus there exists a b in 10 for which the system Ax = b is inconsistent, and the system Ax = b cannot have a solution for all b. 26. Consider the system Ax = 0, where A is a m × n matrix with m > n. Since the rank of A is the number of pivot positions that A has and A is assumed to have full rank, rank A = n. By the Rank Theorem, dim Nul A = n − rank A = 0. So Nul A = {0}, and the system Ax = 0 has only the trivial solution. This happens if and only if the columns of A are linearly independent. 27. Since A is an m × n matrix, Row A is a subspace of n , Col A is a subspace of m , and Nul A is a subspace of n . Likewise since AT is an n × m matrix, Row AT is a subspace of m , Col AT is a subspace of n , and Nul AT is a subspace of m . Since Row A = Col AT and Col A = Row AT , there are four dinstict subspaces in the list: Row A, Col A, Nul A, and Nul AT . 28. a. Since A is an m × n matrix and dim Row A = rank A, dim Row A + dim Nul A = rank A + dim Nul A = n. = = dim Col A dim Row A dim Col AT rank AT , b. Since AT is an n × m matrix and = m. dim Col A + dim Nul AT = rank AT + dim Nul AT = 29. Let A be an m × n matrix. The system Ax = b will have a solution for all b in m if and only if A has a pivot position in each row, which happens if and only if dim Col A = m. By Exercise 28 b., dim Col A = m if and only if dim Nul AT = m − m = 0 , or Nul AT = {0}. Finally, Nul AT = {0} if and only if the equation AT x = 0 has only the trivial solution. 30. The equation Ax = b is consistent if and only if rank [ A equal if and only if b is not a pivot column of [ A Section 1.2. 2 −3 a 31. Compute that uv = [ 5 T b 2a −3a c] = 5a 2b −3b 5b b ] = rank A because the two ranks will be b ]. The result then follows from Theorem 2 in 2c −3c . Each column of uvT is a multiple of u, 5c so dimCol uvT = 1 , unless a = b = c = 0, in which case uvT is the 3 × 3 zero matrix and dimCol uvT = 0. rank uvT dimCol uvT ≤ 1 In any case,= 32. Note that the second row of the matrix is twice the first row. Thus if v = (1, –3, 4), which is the first row of the matrix, 1 uvT = [1 2 −3 1 4]= 2 −3 −6 4 . 8 Copyright © 2016 Pearson Education, Ltd. 4-40 CHAPTER 4 33. Let A = [u1 • Vector Spaces u3 ] , and assume that rank A = 1. Suppose that u1 ≠ 0 . Then {u1} is basis for u2 Col A, since Col A is assumed to be one-dimensional. Thus there are scalars x and y with u 2 = xu1 1 T and u3 = yu1 , and A = u1 v , where v = x . If u1 = 0 but u 2 ≠ 0 , then similarly {u 2 } is basis for y Col A, since Col A is assumed to be one-dimensional. Thus there is a scalar x with u3 = xu 2 , and 0 0 T A = u 2 v , where v = 1 . If u= u= 0 but u3 ≠ 0, then A = u3 v , where v = 0 . 1 2 x 1 T 34. Let A be an m × n matrix with of rank r > 0, and let U be an echelon form of A. Since A can be reduced to U by row operations, there exist invertible elementary matrices E1 , …, E p with ( E p ⋅⋅⋅ E1 ) A = U . Thus = A ( E p ⋅⋅⋅ E1 ) −1U , since the product of invertible matrices is invertible. Let = E ( E p ⋅⋅⋅ E1 ) −1 ; then A = EU. Let the columns of E be denoted by c1 , … , c m . Since the rank of A is r, U has r nonzero rows, which can be denoted d1T ,…, dTr . By the column-row expansion of A (Theorem 10 in Section 2.4): A EU = = [c1 … d1T dT r cm ] = c1d1T + …+ c r dTr , which is the sum of r rank 1 matrices. 0 0 35. [M] a. Begin by reducing A to reduced echelon form: 1 0 A ∼ 0 0 0 0 13/ 2 1 11/ 2 0 0 0 0 0 0 0 0 1 0 0 5 1/ 2 −11/ 2 0 0 A basis for Col A is the pivot columns of A, so matrix C contains these columns: 7 −4 C= 5 −3 6 −9 6 −7 5 −8 5 −2 5 −1 4 −3 −5 2 . −4 9 Copyright © 2016 Pearson Education, Ltd. 0 0 0 1 0 −3 2 7 . 1 0 4.6 • Solutions 4-41 A basis for Row A is the pivot rows of the reduced echelon form of A, so matrix R contains 5 0 −3 1 0 13/ 2 0 0 1 11/ 2 0 1/ 2 0 2 these rows: R = . 0 0 0 1 −11/ 2 0 7 0 0 0 1 1 0 0 To find a basis for Nul A row reduce to reduced echelon form, note that the solution to Ax = 0 in terms of free variables is x1 = −(13/ 2) x3 − 5 x5 + 3 x7 , x2 = −(11/ 2) x3 − (1/ 2) x5 − 2 x7 , = x4 (11/ 2) x5 − 7 x7 , x6 = − x7 , with x3 , x5 , and x7 free. Thus matrix N is −13/ 2 −11/ 2 1 N = 0 0 0 0 −5 −1/ 2 0 11/ 2 1 0 0 3 −2 0 −7 . 0 −1 1 1 0 0 T T b. The reduced echelon form of A is A ∼ 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 −2 /11 −41/11 0 28 /11 , so the solution to 0 0 0 AT x = 0 in terms of free variables is x1 = (2 /11) x5 , x2 = (41/11) x5 , x3 = 0, x4 = −(28 /11) x5 , 2 /11 41/11 with x5 free. Thus matrix M is M = 0 . The matrix S = RT N is 7 × 7 because the −28 /11 1 columns of RT and N are in 7 and dimRow A + dimNul A = 7. The matrix T = [C M ] is 5 5. Both S and T are × 5 because the columns of C and M are in 5 and dimCol A + dimNul AT = invertible because their columns are linearly independent. This fact will be proven in general in Theorem 3 of Section 6.1. 36.[M] Answers will vary, but in most cases C will be 6 × 4, and will be constructed from the first 4 columns of A. In most cases R will be 4 × 7, N will be 7 × 3, and M will be 6 × 2. 37. [M] The C and R from Exercise 35 work here, and A = CR. = 38. [M] If A is nonzero, then A = CR. Note that CR [Cr1 Cr2 … Crn ] , where r1 , …, rn are the columns of R. The columns of R are either pivot columns of R or are not pivot columns of R. Copyright © 2016 Pearson Education, Ltd. 4-42 CHAPTER 4 • Vector Spaces Consider first the pivot columns of R. The i th pivot column of R is ei , the i th column in the identity matrix, so Cei is the i th pivot column of A. Since A and R have pivot columns in the same locations, when C multiplies a pivot column of R, the result is the corresponding pivot column of A in its proper location. Suppose r j is a nonpivot column of R. Then r j contains the weights needed to construct the j th column of A from the pivot columns of A, as is discussed in Example 9 of Section 4.3 and in the paragraph preceding that example. Thus r j contains the weights needed to construct the j th column of A from the columns of C, and Cr j = a j . 4.7 SOLUTIONS Notes: This section depends heavily on the coordinate systems introduced in Section 4.4. The row reduction algorithm that produces P can also be deduced from Exercise 12 in Section 2.2, by row ← reducing P P . to I P −1P . The change-of-coordinates matrix here is interpreted in Section 5.4 as the matrix of the identity transformation relative to two bases. 9 6 6 1. a. Since b 9c1 − 4c 2 , [b1 ] = , [b 2 ] = , and P = = 6c1 − 2c 2 and b= 1 2 ← −4 −2 −2 −3 = [x ] b. Since x = −3b1 + 2b 2 , [x ] = and 2 6 = P [ x ] ← −2 9 −3 0 = −4 2 −2 5 −1 −1 2. a. Since b1 = −c1 + 4c 2 and b= 5c1 − 3c 2 , [b1 ] = , [b 2 ] = , and P = 2 ← −3 4 4 5 [x ] = b. Since= x 5b1 + 3b 2 , [x ] = and 3 9 . −4 5 . −3 5 5 10 −1 P [x ] = = ← 4 −3 3 11 3. Equation (ii) is satisfied by P for all x in V. 4. Equation (i) is satisfied by P for all x in V. 4 −1 5. a. Since = a1 4b1 − b 2 , a 2 = −b1 + b 2 + b3 , and a= b 2 − 2b3 , [a1 ] = −1 , [a 2 ] = 1 , 3 0 1 0 0 4 −1 [a3 ] = 1 , and P = −1 1 1 . ← − 1 −2 0 2 Copyright © 2016 Pearson Education, Ltd. 4.7 • Solutions 0 3 8 3 4 −1 2 . P = 1 1 4 = −1 b. Since x =3a1 + 4a 2 + a3 , [x] A = 4 and [x ] = ← 1 1 −2 1 2 0 0 2 6.a. Since f1 = 2d1 − d 2 + d3 , = f 2 3d 2 + d3 , and f3 = −3d1 + 2d3 , [f1 ] = −1 , [f2 ] = 3 , 1 1 2 0 −3 −3 [f3 ] = 0 , and P = −1 3 0 . ← 2 1 1 2 1 2 b. Since x =f1 − 2f 2 + 2f3 , [x ] = −2 and [x ] = P [x ] = −1 ← 2 1 7. To find P , row reduce the matrix [c1 ← [c1 c2 b1 1 b2 ] ∼ 0 0 1 −3 −5 ← c2 b1 1 b2 ] ∼ 0 0 1 3 −4 ← c2 b1 1 b2 ] ∼ 0 0 1 9 −4 [c1 c2 b1 1 b2 ] ∼ 0 0 1 8 −5 b1 c2 b1 c2 b1 1 , and= P ← 2 −2 −1 P = −5 ← 1 . 3 b 2 ] : so −2 , and= P ← 3 3 −1 = P 4 ← 2 . 3 −2 , and= P ← 1 1 −1 = P 4 ← 2 . 9 b2 ] : −2 9 . Thus P = ← 1 −4 10. To find P , row reduce the matrix [c1 ← c2 −3 1 −4 0 −2 = −7 . 2 2 3 b2 ] : −2 3 . Thus P = ← 3 −4 9. To find P , row reduce the matrix [c1 [c1 b1 1 −3 . Thus P = ← 2 −5 8. To find P , row reduce the matrix [c1 [c1 c2 0 3 1 b2 ] : 3 8 . Thus P = 8 ← −2 −5 3 2 −1 = , and= P P −5 ←88 ← −2 3 . −8 11. a. False. See Theorem 15. b. True. See the first paragraph in the subsection “Change of Basis in n .” 12. a. True. The columns of P are coordinate vectors of the linearly independent set B. See the ← second paragraph after Theorem 15. b. False. The row reduction is discussed after Example 2. The matrix P obtained there satisfies [x ] = P[x ] Copyright © 2016 Pearson Education, Ltd. 4-43 4-44 CHAPTER 4 • Vector Spaces c1 , c 2 , c 3} {1, t , t 2 }. The 13. Let = {b1 , b 2 , b3} = {1 − 2t + t 2, 3 − 5t + 4t 2, 2t + 3t 2 } = and let {= 1 3 0 2 . So −2 , [b ] = −5 , [b ] = -coordinate vectors of b1 , b 2 , and b3 are [b1 ] = 2 3 1 4 3 3 0 1 P =−2 −5 2 . Let x = –1 + 2t. Then the coordinate vector [x ] satisfies ← 4 3 1 −1 P [x= ] [x= ] 2 . This system may be solved by row reducing its augmented matrix: ← 0 3 0 −1 1 0 0 5 1 5 −2 −5 2 −2 . 2 ∼ 0 1 0 −2 , so [x ] = 4 3 0 0 0 1 1 1 1 c1 , c 2 , c 3} {1, t , t 2 }. The -coordinate 14. Let = {b1 , b 2 , b3} = {1 − 3t 2 , 2 + t − 5t 2 , 1 + 2t} = and let {= 1 0 , [b ] vectors of b1 , b 2 , and b3 = are [b1 ] = 2 − 3 2 = 1 , [b ] 3 − 5 2 1 1 1 2 . So P = 0 1 2 . ← − 0 3 −5 0 0 2 ] [x= ] 0 . This system may be Let x = t . Then the coordinate vector [x ] satisfies P [x= ← 1 1 solved by row reducing its augmented matrix: 0 − 3 2 1 −5 1 2 0 0 1 0 ∼ 0 1 0 and t 2 = 3(1 − 3t 2 ) − 2(2 + t − 5t 2 ) + (1 + 2t ). 15. (a) (b) (c) (d) is a basis for V the coordinate mapping is a linear transformation the product of a matrix and a vector the coordinate vector of v relative to 1 0 Qe1 16. (a) [= b1 ] Q[= b1 ] Q= 0 (b) [b k ] [b k ] Q= [b k ] Qe k (c) = Copyright © 2016 Pearson Education, Ltd. 0 1 0 0 0 1 3 3 −2 , so [x ] = −2 1 1 4.7 • Solutions 17. [M] a. Since we found P in Exercise 34 of Section 4.5, we can calculate that 32 0 0 1 −1 P = 0 32 0 0 0 0 32 0 0 0 0 0 16 0 16 0 0 0 0 0 24 0 8 0 0 0 12 0 16 0 4 0 0 0 20 0 10 0 2 0 10 0 15 0 . 6 0 1 b. Since P is the change-of-coordinates matrix from to , P −1 will be the change-ofcoordinates matrix from to . By Theorem 15, the columns of P −1 will be the coordinate vectors of the basis vectors in . Thus 1 (1 + cos 2t ) 2 1 = cos3t (3cos t + cos 3t ) 4 1 cos 4t = (3 + 4cos 2t + cos 4t ) 8 1 5 cos= t (10cos t + 5cos 3t + cos 5t ) 16 1 cos6t = (10 + 15cos 2t + 6cos 4t + cos 6t ) 32 2 cos= t 18. [M] The -coordinate vector of the integrand is (0, 0, 0, 5, –6, 5, –12). Using P −1 from the previous exercise, the - coordinate vector of the integrand will be P −1 (0, 0, 0, 5, − 6, 5, − 12) = (−6, 55 / 8, − 69 / 8, 45 /16, − 3, 5 /16, − 3/ 8) Thus the integral may be rewritten as 55 69 45 5 3 cos t − cos 2t + cos 3t − 3cos 4t + cos 5t − cos 6t dt , 8 8 16 16 8 which equals 55 69 15 3 1 1 −6t + sin t − sin 2t + sin 3t − sin 4t + sin 5t − sin 6t + C. 8 16 16 4 16 16 ∫ −6 + Copyright © 2016 Pearson Education, Ltd. 4-45 4-46 CHAPTER 4 • Vector Spaces 19. [M] a. If is the basis {v1 , v 2 , v 3 }, then the columns of P are [u1 ] , [u 2 ] , and [u3 ] . So v 3 ][u j ] , and [u1 u j = [ v1 v2 [u1 −2 u3 ] = 2 3 u2 −8 5 2 b. Analogously to part a., [ v1 [ v1 −7 1 2 −3 6 4 v2 u3 ] = [ v1 u2 v2 −1 −6 −5 0 = 1 21 2 −5 6 v 3 ] = [ w1 v 3 ] P. In the current exercise, −6 −9 32 −5 0 . 3 w 3 ] P, so [ w1 w2 w3 ] = w2 v 3 ] P . In the current exercise, −1 v2 −1 2 −1 −2 −8 −7 1 5 2 −3 −5 0 w 2 w3 ] 2 [ w1 = 3 2 6 4 6 1 8 5 28 38 21 −2 −8 −7 5 5 2 −3 −5 −3 = −9 −13 −7 . = 2 3 2 6 −2 −2 −1 −3 2 3 20. a. P = P ← P ← ← Let x be any vector in the two-dimensional vector space. Since P is the change-of-coordinates ← matrix from to and P is the change-of-coordinates matrix from to , ← [x ] = P [x ] and [x ] = P [x ] = ← ← P P [x ] . But since P is the change-of-coordinates ← ← ← matrix from to , [x ] = P [x ] . Thus P [x ] = P ← ← P [x ] ← ← for any vector [x ] in 2 , and P = P P . ← ← ← 1 −2 −1 1 7 −3 b. [M] For example, let 7 = , , = , , and = , −5 2 8 −5 5 −1 we can calculate the change-of-coordinates matrices: 1 −2 7 −3 1 0 −3 1 −3 1 ∼ ⇒ P = −5 2 5 −1 0 1 −5 2 ←7 −5 2 1 1 −2 1 0 0 −8 / 3 −8 / 3 −1 0 ∼ ⇒ P = 8 −5 −5 ← 2 0 1 1 −14 / 3 8 1 −14 / 3 1 7 −3 1 0 40 / 3 −16 / 3 −1 40 / 3 −16 / 3 = 8 −5 5 −1 ∼ 0 1 61 / 3 −25 / 3 ⇒ P ←7 61 / 3 −25 / 3 40 / 3 −16 / 3 0 = One confirms= easily that P ← 61 / 3 −25 / 3 1 −8 / 3 −3 1 = −14 / 3 −5 2 Copyright © 2016 Pearson Education, Ltd. P P ←88 ← . Then 4.8 4.8 • Solutions 4-47 SOLUTIONS Notes: This is an important section for engineering students and worth extra class time. To spend only one lecture on this section, you could cover through Example 5, but assign the somewhat lengthy Example 3 for reading. Finding a spanning set for the solution space of a difference equation uses the Basis Theorem (Section 4.5) and Theorem 17 in this section, and demonstrates the power of the theory of Chapter 4 in helping to solve applied problems. This section anticipates Section 5.7 on differential equations. The reduction of an n th order difference equation to a linear system of first order difference equations was introduced in Section 1.10, and is revisited in Sections 4.9 and 5.6. Example 3 is the background for Exercise 26 in Section 6.5. 1. Let yk = 2k . Then k yk + 2 + 2 yk +1 − 8 yk = 2k + 2 + 2(2k +1 ) − 8(2k = ) 2k (22 + 22 = − 8) 2= (0) 0 for all k Since the difference equation holds for all k, 2k is a solution. Let yk = (−4) k . Then (−4) k (0) = 0 for all k yk + 2 + 2 yk +1 − 8 yk =(−4) k + 2 + 2(−4) k +1 − 8(−4) k =(−4) k ((−4) 2 + 2(−4) − 8) = Since the difference equation holds for all k, (−4) k is a solution. 2. Let yk = 3k . Then k k = ) 3k (32 = − 9) 3= (0) 0 for all k yk + 2 − 9 yk = 3k + 2 − 9(3 Since the difference equation holds for all k, 3k is a solution. Let yk = (−3) k . Then yk + 2 − 9 yk =(−3) k + 2 − 9(−3) k = (−3) k ((−3) 2 − 9) = (−3) k (0) = 0 for all k Since the difference equation holds for all k, (−3) k is a solution. 3. The signals 2k and (−4) k are linearly independent because neither is a multiple of the other; that is, there is no scalar c such that 2k= c(−4) k for all k. By Theorem 17, the solution set H of the difference equation yk + 2 + 2 yk +1 − 8 yk = 0 is two-dimensional. By the Basis Theorem, the two linearly independent signals 2k and (−4) k form a basis for H. 4. The signals 3k and (−3) k are linearly independent because neither is a multiple of the other; that is, there is no scalar c such that 3k= c(−3) k for all k. By Theorem 17, the solution set H of the difference equation yk + 2 − 9 yk = 0 is two-dimensional. By the Basis Theorem, the two linearly independent signals k 3k and ( −3) form a basis for H. 5. Let yk = (−3) k . Then yk + 2 + 6 yk +1 + 9 yk =− ( 3) k + 2 + 6(−3) k +1 + 9(−3) k =(−3) k ((−3) 2 + 6(−3) + 9) = (−3) k (0) = 0 for all k Since the difference equation holds for all k, (−3) k is in the solution set H. Copyright © 2016 Pearson Education, Ltd. 4-48 CHAPTER 4 • Vector Spaces Let yk= k (−3) k . Then yk + 2 + 6 yk +1 + 9 yk = (k + 2)(−3) k + 2 + 6(k + 1)(−3) k +1 + 9k (−3) k =(−3) k ((k + 2)(−3) 2 + 6(k + 1)(−3) + 9k ) =− ( 3) k (9k + 18 − 18k − 18 + 9k ) = (−3) k (0) = 0 for all k Since the difference equation holds for all k, k (−3) k is in the solution set H. The signals (−3) k and k (−3) k are linearly independent because neither is a multiple of the other; that is, there is no scalar c such that (−3) k = ck (−3) k for all k and there is no scalar c such that c(−3) k =k (−3) k for all k . By Theorem 17, dim H = 2, so the two linearly independent signals 3k and (−3) k form a basis for H by the Basis Theorem. 6. Let yk = 5k cos k2π . Then kπ (k + 2)π (k + 2)π kπ + 25 5k cos 5k 52 cos = + 25 cos 2 2 2 2 kπ kπ = 25 ⋅ 5k cos + π + cos = 25 ⋅ 5k (0) = 0 for all k 2 2 yk = 5k + 2 cos + 2 + 25 yk since cos(t + π) = –cos t for all t. Since the difference equation holds for all k, 5k cos k2π is in the solution set H. Let yk = 5k sin k2π . Then (k + 2)π kπ (k + 2)π kπ k 2 = + 25 sin + 25 5k sin 5 5 sin 2 2 2 2 kπ kπ = 25 ⋅ 5k sin + π + sin 25 ⋅ 5k (0) = 0 for all k = 2 2 5k + 2 sin yk += 2 + 25 yk since sin(t + π) = –sin t for all t. Since the difference equation holds for all k, 5k sin k2π is in the solution set H. The signals 5k cos k2π and 5k sin k2π are linearly independent because neither is a multiple of the other. By Theorem 17, dim H = 2, so the two linearly independent signals 5k cos k2π and 5k sin k2π form a basis for H by the Basis Theorem. 7. Compute and row reduce the Casorati matrix for the signals 1k , 2k , and (−2) k , setting k = 0 for convenience: 10 1 1 2 1 20 1 2 22 (−2)0 1 (−2)1 ∼ 0 (−2) 2 0 0 1 0 0 0 1 This Casorati matrix is row equivalent to the identity matrix, thus is invertible by the IMT. Hence the set of signals {1k , 2k ,(−2) k } is linearly independent in . The exercise states that these signals are in the solution set H of a third-order difference equation. By Theorem 17, dim H = 3, so the three linearly independent signals 1k , 2k , (−2) k form a basis for H by the Basis Theorem. Copyright © 2016 Pearson Education, Ltd. 4.8 • Solutions 4-49 8. Compute and row reduce the Casorati matrix for the signals 2k , 4k , and (−5) k , setting k = 0 for 20 convenience: 21 2 2 40 1 4 42 (−5)0 1 (−5)1 ∼ 0 (−5) 2 0 0 1 0 0 0 . This Casorati matrix is row equivalent to the identity 1 matrix, thus is invertible by the IMT. Hence the set of signals {2k , 4k ,(−5) k } is linearly independent in . The exercise states that these signals are in the solution set H of a third-order difference equation. By Theorem 17, dim H = 3, so the three linearly independent signals 2k , 4k , (−5) k form a basis for H by the Basis Theorem. 9. Compute and row reduce the Casorati matrix for the signals 1k , 3k cos k2π , and 3k sin k2π , setting k = 0 for 10 30 cos 0 30 sin 0 1 0 0 convenience: 11 31 cos π2 31 sin π2 ∼ 0 1 0 . This Casorati matrix is row equivalent to the 2 0 0 1 2 2 1 3 cos 3 sin π π identity matrix, thus is invertible by the IMT. Hence the set of signals {1k ,3k cos k2π ,3k sin k2π } is linearly independent in . The exercise states that these signals are in the solution set H of a third-order difference equation. By Theorem 17, dim H = 3, so the three linearly independent signals 1k , 3k cos k2π , and 3k sin k2π , form a basis for H by the Basis Theorem. 10. Compute and row reduce the Casorati matrix for the signals (−1) k , k (−1) k , and 5k , setting k = 0 for (−1)0 convenience: (−1)1 2 (−1) 0(−1)0 1(−1) 1 2(−1) 2 50 1 51 ∼ 0 52 0 0 1 0 0 0 . This Casorati matrix is row equivalent to the 1 identity matrix, thus is invertible by the IMT. Hence the set of signals {(−1) k , k (−1) k , 5k } is linearly independent in . The exercise states that these signals are in the solution set H of a third-order difference equation. By Theorem 17, dim H = 3, so the three linearly independent signals (−1) k , k (−1) k , and 5k form a basis for H by the Basis Theorem. 11. The solution set H of this third-order difference equation has dim H = 3 by Theorem 17. The two signals (−1) k and 3k cannot possibly span a three-dimensional space, and so cannot be a basis for H. 12. The solution set H of this fourth-order difference equation has dim H = 4 by Theorem 17. The two signals 1k and (−1) k cannot possibly span a four-dimensional space, and so cannot be a basis for H. 13. The auxiliary equation for this difference equation is r 2 − r + 2 / 9 = 0. By the quadratic formula (or factoring), r = 2/3 or r = 1/3, so two solutions of the difference equation are (2 / 3) k and (1/ 3) k . The signals (2 / 3) k and (1/ 3) k are linearly independent because neither is a multiple of the other. By Theorem 17, the solution space is two-dimensional, so the two linearly independent signals (2 / 3) k and (1/ 3) k form a basis for the solution space by the Basis Theorem. Copyright © 2016 Pearson Education, Ltd. 4-50 CHAPTER 4 • Vector Spaces 14. The auxiliary equation for this difference equation is r 2 − 7 r + 12 = 0. By the quadratic formula (or factoring), r = 3 or r = 4, so two solutions of the difference equation are 3k and 4k . The signals 3k and 4k are linearly independent because neither is a multiple of the other. By Theorem 17, the solution space is two-dimensional, so the two linearly independent signals 3k and 4k form a basis for the solution space by the Basis Theorem. 15. The auxiliary equation for this difference equation is r 2 − 25 = 0. By the quadratic formula (or factoring), r = 5 or r = –5, so two solutions of the difference equation are 5k and (−5) k . The signals 5k and (−5) k are linearly independent because neither is a multiple of the other. By Theorem 17, the solution space is two-dimensional, so the two linearly independent signals 5k and (−5) k form a basis for the solution space by the Basis Theorem. 16. The auxiliary equation for this difference equation is 16r 2 + 8r − 3 = 0. By the quadratic formula (or factoring), r = 1/4 or r = –3/4, so two solutions of the difference equation are (1/ 4) k and (−3/ 4) k . The signals (1/ 4) k and (−3/ 4) k are linearly independent because neither is a multiple of the other. By Theorem 17, the solution space is two-dimensional, so the two linearly independent signals (1/ 4) k and (−3/ 4) k form a basis for the solution space by the Basis Theorem. 17. Letting a = .9 and b = 4/9 gives the difference equation Yk + 2 − 1.3Yk +1 + .4Yk = 1. First we find a particular solution Yk = T of this equation, where T is a constant. The solution of the equation T – 1.3T + .4T = 1 is T = 10, so 10 is a particular solution to Yk + 2 − 1.3Yk +1 + .4Yk = 1 . Next we solve the homogeneous difference equation Yk + 2 − 1.3Yk +1 + .4Yk = 0. The auxiliary equation for this difference equation is r 2 − 1.3r + .4 = 0. By the quadratic formula (or factoring), r = .8 or r = .5, so two solutions of the homogeneous difference equation are .8k and .5k . The signals (.8) k and (.5) k are linearly independent because neither is a multiple of the other. By Theorem 17, the solution space is two-dimensional, so the two linearly independent signals (.8) k and (.5) k form a basis for the solution space of the homogeneous difference equation by the Basis Theorem. Translating the solution space of the homogeneous difference equation by the particular solution 10 of the nonhomogeneous difference equation gives us the general k k solution of Yk + 2 − 1.3Yk +1 + .4Yk = 1 : Yk = c1 (.8) + c2 (.5) + 10. As k increases the first two terms in the solution approach 0, so Yk approaches 10. 18. Letting a = .9 and b = .5 gives the difference equation Yk + 2 − 1.35Yk +1 + .45Yk = 1. First we find a particular solution Yk = T of this equation, where T is a constant. The solution of the equation T – 1.35T + .45T = 1 is T = 10, so 10 is a particular solution to Yk + 2 − 1.35Yk +1 + .45Yk = 1 . Next we solve the homogeneous difference equation Yk + 2 − 1.35Yk +1 + .45Yk = 0. The auxiliary equation for this difference equation is r 2 − 1.35r + .45 = 0. By the quadratic formula (or factoring), r = .6 or r = .75, so two solutions of the homogeneous difference equation are .6k and .75k . The signals (.6) k and (.75) k are linearly independent because neither is a multiple of the other. By Theorem 17, the solution space is twodimensional, so the two linearly independent signals (.6) k and (.75) k form a basis for the solution space of the homogeneous difference equation by the Basis Theorem. Translating the solution space of the homogeneous difference equation by the particular solution 10 of the nonhomogeneous difference k k equation gives us the general solution of Yk + 2 − 1.35Yk +1 + .45Yk = 1 : Yk =c1 (.6) + c2 (.75) + 10. Copyright © 2016 Pearson Education, Ltd. 4.8 • Solutions 4-51 19. The auxiliary equation for this difference equation is r 2 + 4r + 1 = 0. By the quadratic formula, r =−2 + 3 or r =−2 − 3, so two solutions of the difference equation are (−2 + 3) k and (−2 − 3) k . The signals (−2 + 3) k and (−2 − 3) k are linearly independent because neither is a multiple of the other. By Theorem 17, the solution space is two-dimensional, so the two linearly independent signals (−2 + 3) k and (−2 − 3) k form a basis for the solution space by the Basis Theorem. Thus a general solution to this difference equation is yk = c1 (−2 + 3) k + c2 (−2 − 3) k . 20. Let a =−2 + 3 and b =−2 − 3 . Using the solution from the previous exercise, we find that N N y1 = c1a + c2b = 5000 and y N = c1a + c2b = 0. This is a system of linear equations with variables a c1 and c2 whose augmented matrix may be row reduced: N a b bN 1 5000 ∼ 0 0 0 1 5000b N bN a − aN b 5000a N bN a − aN b 5000b N 5000a N = c , . (Alternatively, Cramer’s Rule may be applied to get the same 2 bN a − aN b bN a − aN b solution). Thus = So c1 = yk c1a k + c2b k = 5000(a k b N − a N b k ) bN a − aN b 21. The smoothed signal zk has the following values: z1 = (9 + 5 + 7) / 3 = 7, z2 = (5 + 7 + 3) / 3 = 5, z3 = (7 + 3 + 2) / 3 = 4, z4 = (3 + 2 + 4) / 3 = 3, z5 = (2 + 4 + 6) / 3 = 4, z6 = (4 + 6 + 5) / 3 = 5, z7 = (6 + 5 + 7) / 3 = 6, z8 = (5 + 7 + 6) / 3 = 6, z9 = (7 + 6 + 8) / 3 = 7, z10 = (6 + 8 + 10) / 3 = 8, z11 = (8 + 10 + 9) / 3 = 9, z12 = (10 + 9 + 5) / 3= 8, z13 = (9 + 5 + 7) / 3 = 7. 22. a. The smoothed signal zk has the following values: z0 = .35 y2 + .5 y1 + .35 y0 = .35(0) + .5(.7) + .35(3) = 1.4, z1 = .35 y3 + .5 y2 + .35 y1 = .35(−.7) + .5(0) + .35(.7) = 0, z2 = .35 y4 + .5 y3 + .35 y2 = .35(−.3) + .5(−.7) + .35(0) = −1.4, z3 = .35 y5 + .5 y4 + .35 y3 = .35(−.7) + .5(−.3) + .35(−.7) = −2, z4 = .35 y6 + .5 y5 + .35 y4 = .35(0) + .5(−.7) + .35(−.3) = −1.4, z= .35 y7 + .5 y6 + .35 y= .35(.7) + .5(0) + .35(−.7)= 0, 5 5 z6 = .35 y8 + .5 y7 + .35 y6 = .35(3) + .5(.7) + .35(0) = 1.4, z7 = .35 y9 + .5 y8 + .35 y7 = .35(.7) + .5(3) + .35(.7) = 2, Copyright © 2016 Pearson Education, Ltd. 4-52 CHAPTER 4 • Vector Spaces z8 =.35 y10 + .5 y9 + .35 y8 =.35(0) + .5(.7) + .35(3) =1.4,… b. This signal is two times the signal output by the filter when the input (in Example 3) was y = cos(π t/4). This is expected because the filter is linear. The output from the input 2cos(π t/4) + cos(3π t/4) should be two times the output from cos(π t/4) plus the output from cos(3π t/4) (which is zero). 23. a. yk +1 − 1.01 yk = −450, y0 = 10,000. b. [M] MATLAB code to create the table: pay = 450, y = 10000, m = 0, table = [0;y] while y>450 y = 1.01*y-pay m = m+1 table = [table [m;y]] end m,y Mathematica code to create the table: pay = 450; y = 10000; m = 0; balancetable = {{0, y}}; While[y > 450, {y = 1.01*y - pay; m = m + 1, AppendTo[balancetable, {m, y}]}]; m y c. [M] At month 26, the last payment is $114.88. The total paid by the borrower is $11,364.88. 24. a. yk +1 − 1.005 yk = 200, y0 = 1,000. b. [M] MATLAB code to create the table: pay = 200, y = 1000, m = 0, table = [0;y] for m = 1: 60 y = 1.005*y+pay table = [table [m;y]] end interest = y-60*pay-1000 Mathematica code to create the table: pay = 200; y = 1000; amounttable = {{0, y}}; Do[{y = 1.005*y + pay; AppendTo[amounttable, {m, y}]},{m,1,60}]; interest = y-60*pay-1000 c. [M] The total is $6213.55 at k = 24, $12,090.06 at k = 48, and $15,302.86 at k = 60. When k = 60, the interest earned is $2302.86. (k + 1) 2 , 25. To show that yk = k 2 is a solution of yk + 2 + 3 yk +1 − 4 yk = 10k + 7, substitute yk = k 2 , yk += 1 (k + 2) 2 : and yk += 2 Copyright © 2016 Pearson Education, Ltd. 4.8 • Solutions 4-53 yk + 2 + 3 yk +1 − 4 yk =( k + 2)2 + 3( k + 1)2 − 4k 2 = (k 2 + 4k + 4) + 3(k 2 + 2k + 1) − 4k 2 = k 2 + 4k + 4 + 3k 2 + 6k + 3 − 4= k 2 10k + 7 for all k The auxiliary equation for the homogeneous difference equation yk + 2 + 3 yk +1 − 4 yk = 0. 0 is r 2 + 3r − 4 = By the quadratic formula (or factoring), r = –4 or r = 1, so two solutions of the difference equation are (−4) k and 1k . The signals (−4) k and 1k are linearly independent because neither is a multiple of the other. By Theorem 17, the solution space is two-dimensional, so the two linearly independent signals (−4) k and 1k form a basis for the solution space of the homogeneous difference equation by the Basis Theorem. The general solution to the homogeneous difference equation is thus c1 (−4) k + c2 ⋅ 1k =c1 (−4) k + c2 . Adding the particular solution k 2 of the nonhomogeneous difference equation, we find that the general solution of the difference equation yk + 2 + 3 yk +1 − 4 yk =10k + 7 is yk = k 2 + c1 (−4) k + c2 . 26. To show that yk = 1 + k is a solution of yk + 2 − 8 yk +1 + 15 yk = 8k + 2, substitute yk = 1 + k , yk +1 =1 + (k + 1) =2 + k , and yk + 2 =1 + (k + 2) =3 + k : yk + 2 − 8 yk +1 + 15 yk = (3 + k ) − 8(2 + k ) + 15(1 + k ) = 3 + k − 16 − 8k + 15 + 15k= 8k + 2 for all k The auxiliary equation for the homogeneous difference equation yk + 2 − 8 yk +1 + 15 yk = 0 is r 2 − 8r + 15 = 0. By the quadratic formula (or factoring), r = 5 or r = 3, so two solutions of the difference equation are 5k and 3k . The signals 5k and 3k are linearly independent because neither is a multiple of the other. By Theorem 17, the solution space is two-dimensional, so the two linearly independent signals 5k and 3k form a basis for the solution space of the homogeneous difference equation by the Basis Theorem. The general solution to the homogeneous difference equation is thus c1 ⋅ 5k + c2 ⋅ 3k . Adding the particular solution 1+ k of the nonhomogeneous difference equation, we find that the general solution k k of the difference equation yk + 2 − 8 yk +1 + 15 yk = 8k + 2 is yk =1 + k + c1 ⋅ 5 + c2 ⋅ 3 . 27. To show that yk = 2 − 2k is a solution of yk + 2 − (9 / 2) yk +1 + 2 yk =3k + 2 , substitute yk = 2 − 2k , yk +1 =− 2 2(k + 1) = −2k , and yk + 2 =2 − 2(k + 2) =−2 − 2k : yk + 2 − (9 / 2) yk +1 + 2 yk = (−2 − 2k ) − (9 / 2)(−2k ) + 2(2 − 2k ) =−2 − 2k + 9k + 4 − 4k= 3k + 2 for all k The auxiliary equation for the homogeneous difference equation yk + 2 − (9 / 2) yk +1 + 2 yk = 0 is r 2 − (9 / 2)r + 2 = 0. By the quadratic formula (or factoring), r = 4 or r = 1/2, so two solutions of the difference equation are 4k and (1/ 2) k . The signals 4k and (1/ 2) k are linearly independent because neither is a multiple of the other. By Theorem 17, the solution space is two-dimensional, so the two linearly independent signals 4k and (1/ 2) k form a basis for the solution space of the homogeneous difference equation by the Basis Theorem. The general solution to the homogeneous difference equation is thus c1 ⋅ 4k + c2 ⋅ (1/ 2) k = c1 ⋅ 4k + c2 ⋅ 2− k . Adding the particular solution 2 – 2k of the nonhomogeneous difference equation, we find that the general solution of the difference equation k −k yk + 2 − (9 / 2) yk +1 + 2 yk =3k + 2 is yk = 2 − 2k + c1 ⋅ 4 + c2 ⋅ 2 . 28. To show that y= 2k − 4 is a solution of yk + 2 + (3/ 2) yk +1 − yk =+ 1 3k , substitute y= 2k − 4 , k k and : yk +1 = 2(k + 1) − 4 = 2k − 2, yk + 2= 2(k + 2) − 4= 2k yk + 2 + (3/ 2) yk +1 − yk = 2k + (3/ 2)(2k − 2) − (2k − 4) = 2k + 3k − 3 − 2k + 4 = 1 + 3k for all k Copyright © 2016 Pearson Education, Ltd. 4-54 CHAPTER 4 • Vector Spaces The auxiliary equation for the homogeneous difference equation yk + 2 + (3/ 2) yk +1 − yk = 0 is r 2 + (3/ 2)r − 1 =0. By the quadratic formula (or factoring), r = –2 or r = 1/2, so two solutions of the difference equation are (−2) k and (1/ 2) k . The signals (−2) k and (1/ 2) k are linearly independent because neither is a multiple of the other. By Theorem 17, the solution space is two-dimensional, so the two linearly independent signals (−2) k and (1/ 2) k form a basis for the solution space of the homogeneous difference equation by the Basis Theorem. The general solution to the homogeneous difference equation is thus c1 ⋅ (−2) k + c2 ⋅ (1/ 2) k = c1 ⋅ (−2) k + c2 ⋅ 2− k . Adding the particular solution 2k – 4 of the nonhomogeneous difference equation, we find that the general solution of the difference k −k equation yk + 2 + (3/ 2) yk +1 − yk =+ 1 3k is yk= 2k − 4 + c1 ⋅ (−2) + c2 ⋅ 2 . yk y 29. Let x k = k +1 . Then= x k +1 yk + 2 yk +3 yk +1 y k +2 = yk + 3 yk + 4 0 0 0 9 yk x k +1 30. Let x k = yk +1 . Then= yk + 2 yk +1 = yk + 2 yk +3 0 0 −1/16 1 0 0 −6 0 1 0 −8 1 0 0 0 yk 0 yk +1 = Ax k . 1 yk + 2 6 yk +3 0 yk 1 = yk +1 Ax k . 3/ 4 yk + 2 31. The difference equation is of order 2. Since the equation yk +3 + 5 yk + 2 + 6 yk +1 = 0 holds for all k, it holds if k is replaced by k − 1. Performing this replacement transforms the equation into yk + 2 + 5 yk +1 + 6 yk = 0, which is also true for all k. The transformed equation has order 2. 32. The order of the difference equation depends on the values of a1 , a2 , and a3 . If a3 ≠ 0, then the order is 3. If a3 = 0 and a2 ≠ 0, then the order is 2. If a= a= 0 and a1 ≠ 0, then the order is 1. 3 2 If a= a= a= 0, then the order is 0, and the equation has only the zero signal for a solution. 3 2 1 zk k 2 yk 33. The Casorati matrix C(k) is C (k ) = = 2 yk +1 zk +1 (k + 1) . In particular, 2(k + 1) | k + 1| 2k | k | 0 0 1 −2 4 −8 C (0) , C (−1) , and= = = C (−2) , none of which are invertible. In fact, C(k) is 0 1 2 0 1 −2 not invertible for all k, since det C (k= ) 2k 2 (k + 1) | k + 1| −2(k + 1) 2 k | k=| 2k (k + 1) ( k | k + 1| − (k + 1) | k |) If k = 0 or k = –1, det C(k) = 0. If k > 0, then k + 1 > 0 and k| k + 1 | – (k + 1)| k | = k(k + 1) – (k + 1)k = 0, so det C(k) = 0. If k < –1, then k + 1 < 0 and k| k + 1 | – (k + 1)| k | = –k(k + 1) + (k + 1)k = 0, so det C(k) = 0. Thus detC(k)=0 for all k, and C(k) is not invertible for all k. Since C(k) is not invertible for all k, it provides no information about whether the signals { yk } and {zk } are linearly dependent or linearly independent. In fact, neither signal is a multiple of the other, so the signals { yk } and {zk } are linearly independent. 34. No, the signals could be linearly dependent, since the vector space V of functions considered on the entire real line is not the vector space of signals. For example, consider the functions f (t) = sinπt, g(t) = sin 2πt, and h(t) = sin 3πt. The functions f, g, and h are linearly independent in V since they have Copyright © 2016 Pearson Education, Ltd. 4.9 • Solutions 4-55 different periods and thus no function could be a linear combination of the other two. However, sampling the functions at any integer n gives f (n) = g(n) = h(n) = 0, so the signals are linearly dependent in . 35. Let { yk } and {zk } be in , and let r be any scalar. The k th term of { yk } + {zk } is yk + zk , while the k th term of r{ yk } is ryk . Thus T ({ yk } + {zk }) = T { yk + zk } = ( yk + 2 + zk + 2 ) + a ( yk +1 + zk +1 ) + b( yk + zk ) = ( yk + 2 + ayk +1 + byk ) + ( zk + 2 + azk +1 += bzk ) T { yk } + T {zk },and T (r{ yk }) = T {r yk } = r yk + 2 + a (r yk +1 ) + b(r yk ) = r ( yk + 2 + ayk +1 + byk ) = rT { yk } so T has the two properties that define a linear transformation. 36. Let z be in V, and suppose that x p in V satisfies T (x p ) = z. Let u be in the kernel of T; then T(u) = 0. Since T is a linear transformation, T (u + x p ) = T (u) + T (x p ) = 0 + z = z, so the vector x= u + x p satisfies the nonhomogeneous equation T(x) = z. 37. We compute that (TD)( y0 , y1= , y2 ,…) T ( D( y0 , y1 ,= y2 ,…)) T (0, y0 , y1= , y2 ,…) ( y0 , y1 , y2 ,…) while ( DT )( y0 ,= y1 , y2 ,…) D(T ( y0 , y= D( y1 , = y2 , y3 ,…) (0, y1 , y2 , y3 ,…) 1 , y2 ,…)) Thus TD = I (the identity transformation on 4.9 0), while DT ≠ I. SOLUTIONS Notes: This section builds on the population movement example in Section 1.10. The migration matrix is examined again in Section 5.2, where an eigenvector decomposition shows explicitly why the sequence of state vectors x k tends to a steady state vector. The discussion in Section 5.2 does not depend on prior knowledge of this section. 1. a. Let N stand for “News” and M stand for “Music.” Then the listeners’ behavior is given by the table From: N M To: .7 .6 N .3 .4 M .7 so the stochastic matrix is P = .3 .6 . .4 1 b. Since 100% of the listeners are listening to news at 8:15, the initial state vector is x0 = . 0 c. There are two breaks between 8:15 and 9:25, so we calculate x 2 : .7 = x1 P= x0 .3 .6 1 .7 = .4 0 .3 .7 .6 .7 .67 = x 2 P= x1 = .3 .4 .3 .33 Thus 33% of the listeners are listening to news at 9:25. Copyright © 2016 Pearson Education, Ltd. 4-56 CHAPTER 4 • Vector Spaces 2. a. Let the foods be labelled “1,” “2,” and “3.” Then the animals’ behavior is given by the table From: 1 2 3 To: .5 .25 .25 1 .25 .5 .25 2 .25 .25 .5 3 .5 so the stochastic matrix is P = .25 .25 .25 .25 . .5 .25 .5 .25 1 b. There are two trials after the initial trial, so we calculate x 2 . The initial state vector is 0 . 0 .5 = x1 P= x0 .25 .25 .25 .5 .25 .25 1 .25 = 0 .5 0 .5 = x 2 P= x1 .25 .25 .25 .5 .25 .25 .5 .25 .25 = .5 .25 .5 .25 .25 .375 .3125 .3125 Thus the probability that the animal will choose food #2 is .3125. 3. a. Let H stand for “Healthy” and I stand for “Ill.” Then the students’ conditions are given by the table From: H I To: .95 .45 H .05 .55 I .95 so the stochastic matrix is P = .05 .45 . .55 .8 b. Since 20% of the students are ill on Monday, the initial state vector is x0 = . For Tuesday’s .2 percentages, we calculate x1 ; for Wednesday’s percentages, we calculate x 2 : .95 = x1 P= x0 .05 .45 .8 .85 = .55 .2 .15 .95 .45 .85 .875 = x 2 P= x1 = .05 .55 .15 .125 Thus 15% of the students are ill on Tuesday, and 12.5% are ill on Wednesday. Copyright © 2016 Pearson Education, Ltd. 4.9 • Solutions 4-57 1 c. Since the student is well today, the initial state vector is x0 = . We calculate x 2 : 0 .95 = x1 P= x0 .05 .45 1 .95 = .55 0 .05 .95 .45 .95 .925 = x 2 P= x1 = .05 .55 .05 .075 Thus the probability that the student is well two days from now is .925. 4. a. Let G stand for good weather, I for indifferent weather, and B for bad weather. Then the change in the weather is given by the table From: G I B To: .6 .4 .4 G .3 .3 .5 I .1 .3 .1 B .6 so the stochastic matrix is P = .3 .1 .4 .3 .3 .4 .5 . .1 .5 b. The initial state vector is .5 . We calculate x1 : 0 .6 .4 = x1 P= x0 .3 .3 .1 .3 .4 .5 .5 = .5 .1 0 .5 .3 .2 Thus the chance of bad weather tomorrow is 20%. 0 c. The initial state vector is x0 = .4 . We calculate x 2 : .6 .6 = x1 P= x0 .3 .1 .4 .3 .3 .4 0 .5 = .4 .1 .6 .6 = x 2 P= x1 .3 .1 .4 .3 .3 .4 .4 .5 .42 = .1 .18 .4 .42 .18 .48 .336 .184 Thus the chance of good weather on Wednesday is 48%. Copyright © 2016 Pearson Education, Ltd. 4-58 CHAPTER 4 • Vector Spaces .6 −.9 5. We solve Px = x by rewriting the equation as (P – I )x = 0, where P − I = .9 −.6 . Row reducing the augmented matrix for the homogeneous system (P – I )x = 0 gives .6 0 1 −2 / 3 0 x1 −.9 2 / 3 2 x2 = , and one solution is . Since x = . Thus .9 −.6 0 ∼ 0 0 0 1 3 x2 2 2 / 5 .4 = q = the entries in sum to 5, multiply by 1/5 to obtain the steady-state vector . 3/ 5 .6 3 .5 −.2 6. We solve Px = x by rewriting the equation as (P – I )x = 0, where P − I = .2 −.5 . Row reducing the augmented matrix for the homogeneous system (P – I )x = 0 gives .5 0 1 −5 / 2 0 x1 5 / 2 −.2 5 x2 = , and one solution is . Since x = . Thus .2 −.5 0 ∼ 0 0 0 1 2 x2 5 / 7 .714 5 = q the entries in sum to 7, multiply by 1/7 to obtain the steady-state vector ≈ . 2 / 7 .286 2 .1 .1 −.3 .2 . Row − I .2 −.2 7. We solve Px = x by rewriting the equation as (P – I )x = 0, where P= .1 .1 −.3 reducing the augmented matrix for the homogeneous system (P – I )x = 0 gives .1 .1 0 1 0 −1 0 x1 1 1 −.3 .2 −.2 .2 0 ∼ 0 1 −2 0 . Thus = x2 x3 2 , and one solution is 2 . Since x = x3 1 1 .1 .1 −.3 0 0 0 0 0 1 = q the entries in 2 sum to 4, multiply by 1/4 to obtain the steady-state vector 1 −.3 − I 0 8. We solve Px = x by rewriting the equation as (P – I )x = 0, where P= .3 1/ 4 = 1/ 2 1/ 4 .2 −.8 .6 .25 .5 . .25 .2 .4 . Row −.6 reducing the augmented matrix for the homogeneous system (P – I )x = 0 gives .2 .2 0 1 0 −1 0 x1 1 −.3 0 −.8 = x2 x3 1/ 2 , and one solution is .4 0 ∼ 0 1 −1/ 2 0 . Thus x = x3 1 .3 .6 −.6 0 0 0 0 0 2 Since the entries in 1 2 .84 9. Since P 2 = .16 2 1 . 2 2 / 5 .4 = 1/ 5 .2 . q = sum to 5, multiply by 1/5 to obtain the steady-state vector 2 / 5 .4 .2 has all positive entries, P is a regular stochastic matrix. .8 Copyright © 2016 Pearson Education, Ltd. 4.9 • Solutions 4-59 1 1 − .8k 10. Since P k = will have a zero as its (2,1) entry for all k, so P is not a regular .8k 0 stochastic matrix. .6 .6 −.3 , so P − I = . Solving (P – I )x = 0 by row reducing the .4 .3 −.6 .6 0 1 −2 0 x1 −.3 2 x2 , and one solution ∼ = x = augmented matrix gives . Thus 0 0 .3 −.6 0 0 1 x2 .7 11. From Exercise 1, P = .3 2 is . Since the entries in 1 2 / 3 .667 = q ≈ . 1/ 3 .333 .5 12. From Exercise 2, P = .25 .25 2 1 sum to 3, multiply by 1/3 to obtain the steady-state vector .25 .5 .25 .25 .25 , so P= −I .5 −.5 .25 .25 .25 .25 −.5 .25 −.5 .25 .25 .25 . Solving (P – I )x = 0 by row −.5 0 1 −.5 .25 0 ∼ 0 reducing the augmented matrix gives .25 −.5 .25 .25 0 0 x1 1 1 1 = x2 x3 1 , and one solution is 1 . Since the entries in 1 x = x3 1 1 1 0 1 0 −1 −1 0 0 0 . Thus 0 sum to 3, multiply by 1/3 to 1/ 3 .333 = q 1/ 3 ≈ .333 . Thus in the long run each food will be preferred obtain the steady-state vector 1/ 3 .333 equally. .45 .45 −.05 , so P − I = . Solving (P – I )x = 0 by row .55 .05 −.45 .45 0 1 −9 0 −.05 reducing the augmented matrix gives . Thus ∼ 0 0 .05 −.45 0 0 x1 9 9 9 x2 , and one solution is . Since the entries in sum to 10, multiply by 1/10 = x = 1 1 1 x2 .95 13. a. From Exercise 3, P = .05 9 /10 .9 = q = to obtain the steady-state vector . 1/10 .1 b. After many days, a specific student is ill with probability .1, and it does not matter whether that student is ill today or not. Copyright © 2016 Pearson Education, Ltd. 4-60 CHAPTER 4 • Vector Spaces .6 .4 14. From Exercise 4, P = .3 .3 .1 .3 .4 .4 −.4 .3 −.7 .5 . Solving (P – I )x = 0 by row .1 .3 −.9 .4 .4 0 1 0 −3 0 .5 0 ∼ 0 1 −2 0 −.7 .3 −.9 0 0 0 0 0 .4 .5 , so P= −I .1 −.4 reducing the augmented matrix gives .3 .1 x1 3 3 3 = x2 x3 2 , and one solution is 2 . Since the entries in 2 sum to 6, multiply by 1/6 x = Thus x3 1 1 1 1/ 2 .5 = q 1/ 3 ≈ .333 . Thus in the long run the chance that a day has to obtain the steady-state vector 1/ 6 .167 good weather is 50%. .0027 .0027 −.0129 , so P − I = . Solving (P – I )x = 0 by row reducing .9973 .0129 −.0027 .0027 0 1 −.209302 0 −.0129 the augmented matrix gives . Thus ∼ 0 0 .0129 −.0027 0 0 .9871 15. [M] Let P = .0129 x1 .209302 .209302 .209302 x = = x2 , and one solution is . Since the entries in sum to 1. 1 1 1 x2 209302 , multiply by .173077 1/1. 209302 to obtain the steady-state vector q = . Thus about 17.3% of the total U.S. .826923 population would eventually live in California. .01 .09 −.10 .90 .01 .09 = − I .01 −.10 .01 . Solving (P – I )x = 0 by row 16. [M] Let P = .01 .90 .01 , so P .09 .09 .09 .90 .09 −.1 .01 .09 0 1 0 −.919192 0 −.10 reducing the augmented matrix gives .01 −.10 .01 0 ∼ 0 1 −.191919 0 .09 .09 −.1 0 0 0 0 0 x1 .919192 = x2 x3 .191919 , and one solution is x = Thus x3 1 .919192 .191919 . Since the entries in 1 .919192 .191919 sum 1 .435407 to 2.111111, multiply by 1/2.111111 to obtain the steady-state vector q = .090909 . Thus on a .473684 Copyright © 2016 Pearson Education, Ltd. 4.9 • Solutions 4-61 typical day, about (.090909)(2000) = 182 cars will be rented or available from the downtown location. 17. a. The entries in each column of P sum to 1. Each column in the matrix P – I has the same entries as in P except one of the entries is decreased by 1. Thus the entries in each column of P – I sum to 0, and adding all of the other rows of P – I to its bottom row produces a row of zeros. b. By part a., the bottom row of P – I is the negative of the sum of the other rows, so the rows of P – I are linearly dependent. c. By part b. and the Spanning Set Theorem, the bottom row of P – I can be removed and the remaining (n – 1) rows will still span the row space of P – I. Thus the dimension of the row space of P – I is less than n. Alternatively, let A be the matrix obtained from P – I by adding to the bottom row all the other rows. These row operations did not change the row space, so the row space of P – I is spanned by the nonzero rows of A. By part a., the bottom row of A is a zero row, so the row space of P – I is spanned by the first (n – 1) rows of A. d. By part c., the rank of P – I is less than n, so the Rank Theorem may be used to show that dimNul(P – I ) = n – rank(P – I ) > 0. Alternatively the Invertible Martix Theorem may be used since P – I is a square matrix. 1 0 1 0 . Notice that Px = x for any vector x in 2 , and that and are 18. If α = β = 0 then P = 0 1 0 1 two linearly independent steady-state vectors in this case. −α If α ≠ 0 or β ≠ 0, we solve (P – I )x = 0 where P − I = α −α β 0 α − β 0 matrix gives ∼ α −β 0 0 0 0 β . Row reducing the augmented − β x1 β = x = So α x1 = β x2 , and one possible solution is to let x1 = β , x2 = α . Thus . Since the x2 α β 1 β . entries in sum to α + β, multiply by 1/(α + β ) to obtain the steady-state vector q = α + β α α 19. a. The product Sx equals the sum of the entries in x. Thus x is a probability vector if and only if its entries are nonnegative and Sx = 1. = b. Let P [p1 SP = p2 [ Sp1 … p n ] , where p1 , p 2 , …, p n are probability vectors. By part a., Sp 2 … Sp n=] [1 1 … 1=] S c. By part b., S(Px) = (SP)x = Sx = 1. The entries in Px are nonnegative since P and x have only nonnegative entries. By part a., the condition S(Px) = 1 shows that Px is a probability vector. = 20. Let P [p1 p2 2 … p n ] , so P= PP = [ Pp1 Pp 2 … Pp n ]. By Exercise 19c., the columns of P 2 are probability vectors, so P 2 is a stochastic matrix. Alternatively, SP = S by Exercise 19b., since P is a stochastic matrix. Right multiplication by P gives SP 2 = SP, so SP = S implies that SP 2 = S . Since the entries in P are nonnegative, so are the entries in P 2 , and P 2 is stochastic matrix. Copyright © 2016 Pearson Education, Ltd. 4-62 CHAPTER 4 • Vector Spaces 21. [M] a. To four decimal places .2779 .2780 .2803 .2941 .2817 .2817 .3368 .3355 .3357 .3335 .3356 .3356 2 3 P ,P = .1847 .1861 .1833 .1697 .1817 .1817 .2005 .2004 .2007 .2027 .2010 .2010 .2816 .2816 .2816 .2816 .3355 .3355 .3355 .3355 4 5 . The columns of P= P= .1819 .1819 .1819 .1819 .2009 .2009 .2009 .2009 .2817 .3355 .1819 .2010 .2814 .3352 , .1825 .2009 P k are converging to a common .2816 .3355 , which is the vector to which vector as k increases. The steady state vector q for P is q = .1819 .2009 the columns of P k are converging. b. To four decimal places, 10 Q .8222 .4044 .5385 = 20 .0324 .3966 .1666 , Q .1453 .1990 .2949 .7477 .6815 .7105 = 40 Q = .0783 .1329 .1074 , Q .1740 .1856 .1821 30 Q 50 .6000 .2036 .1964 .6690 .1326 , .1984 .7401 .7140 .0843 .1057 .1756 .1802 .7257 .0960 , .1783 .7372 .7269 .7315 .0867 .0951 .0913 , Q 60 = .1761 .1780 .1772 .7356 .7340 .7347 = 80 Q = .0880 .0893 .0887 , Q .1764 .1767 .1766 70 .7353 Q= Q= .0882 .1765 116 .7674 .0637 .1688 117 .7353 .0882 .1765 .7360 .0876 .1763 .7320 .7338 .0909 .0894 , .1771 .1767 .7354 .7348 .0881 .0887 .1764 .1766 .7351 .0884 , .1765 .7353 .0882 .1765 Copyright © 2016 Pearson Education, Ltd. Chapter 4 • Supplementary Exercises 4-63 .7353 The steady state vector q for Q is q = .0882 Conjecture: the columns of P k , where P is a .1765 regular stochastic matrix, converge to the steady state vector for P as k increases. c. Let P be an n × n regular stochastic matrix, q the steady state vector of P, and e j the j th column of the n × n identity matrix. Consider the Markov chain {x k } where x k +1 = P x k and x0 = e j . By Theorem 18, x k = P k x0 converges to q as k → ∞. But P k x0 = P k e j , which is the j th column of P k . Thus the j th column of P k converges to q as k → ∞; that is, P k → [q q … q] . 22. [M] Answers will vary. MATLAB Student Version 4.0 code for Method (1): A=randstoc(32); flops(0); tic, x=nulbasis(A-eye(32)); q=x/sum(x); toc, flops MATLAB Student Version 4.0 code for Method (2): A=randstoc(32); flops(0); tic, B=A^100; q=B(: ,1); toc, flops Chapter 4 SUPPLEMENTARY EXERCISES 1. a. True. This set is Span{v1 , ... v p } , and every subspace is itself a vector space. b. True. Any linear combination of v1 , …, v p−1 is also a linear combination of v1 , …, v p−1 , v p using the zero weight on v p . c. False. Counterexample: Take v p = 2 v1 . Then {v1 , ... v p } is linearly dependent. d. False. Counterexample: Let {e1 , e 2 , e3 } be the standard basis for . Then {e1 , e 2 } is a linearly 3 independent set but is not a basis for . e. True. See the Spanning Set Theorem (Section 4.3). f. True. By the Basis Theorem, S is a basis for V because S spans V and has exactly p elements. So S must be linearly independent. g. False. The plane must pass through the origin to be a subspace. 3 2 h. False. Counterexample: 0 0 5 0 0 −2 7 0 0 3 . 0 i. True. This statement appears before Theorem 13 in Section 4.6. j. False. Row operations on A do not change the solutions of Ax = 0. 1 k. False. Counterexample: A = 3 2 ; A has two nonzero rows but the rank of A is 1. 6 Copyright © 2016 Pearson Education, Ltd. 4-64 CHAPTER 4 • Vector Spaces l. m. n. o. False. If U has k nonzero rows, then rank A = k and dimNul A = n – k by the Rank Theorem. True. Row equivalent matrices have the same number of pivot columns. False. The nonzero rows of A span Row A but they may not be linearly independent. True. The nonzero rows of the reduced echelon form E form a basis for the row space of each matrix that is row equivalent to E. p. True. If H is the zero subspace, let A be the 3 × 3 zero matrix. If dim H = 1, let {v} be a basis for H and set A = [ v v v ] . If dim H = 2, let {u,v} be a basis for H and set A = [u v v ] , for example. If dim H = 3, then H = , so A can be any 3 × 3 invertible matrix. Or, let 3 {u, v, w} be a basis for H and set A = [u v w] . 1 0 0 q. False. Counterexample: A = . If rank A = n (the number of columns in A), then the 0 1 0 transformation x Ax is one-to-one. r. True. If x Ax is onto, then Col A = and rank A = m. See Theorem 12(a) in Section 1.9. s. True. See the second paragraph after Theorem 15 in Section 4.7. m t. False. The j th column of P is b j . C C ←B 1 2 3 2 3 4 2. The set is SpanS, where S = , , 3 4 5 . Note that S is a linearly dependent 4 5 6 set, but each pair of vectors in S forms a linearly independent set. Thus any two of the three 1 2 3 2 3 4 vectors 3 , 4 , 5 will be a basis for SpanS. 4 5 6 3. The vector b will be in W = Span{u1 , u 2 } if and only if there exist constants c1 and c2 with c1u1 + c2u 2 = b. Row reducing the augmented matrix gives −2 4 −6 1 2 −5 b1 −2 b2 ∼ 0 b3 0 1 4 0 b1 2b1 + b2 b1 + 2b2 + b3 so W = Span{u1 , u 2 } is the set of all (b1 , b2 , b3 ) satisfying b1 + 2b2 + b3 = 0. 4. The vector g is not a scalar multiple of the vector f, and f is not a scalar multiple of g, so the set {f, g} is linearly independent. Even though the number g(t) is a scalar multiple of f(t) for each t , the scalar depends on t . Copyright © 2016 Pearson Education, Ltd. Chapter 4 • Supplementary Exercises 4-65 5. The vector p1 is not zero, and p 2 is not a multiple of p1. However, p3 is 2p1 + 2p 2 , so p3 is discarded. The vector p 4 cannot be a linear combination of p1 and p 2 since p 4 involves t 2 but p1 and p 2 do not involve t 2 . The vector p5 is (3/ 2)p1 − (1/ 2)p 2 + p 4 (which may not be so easy to see at first.) Thus p5 is a linear combination of p1 , p 2 , and p 4 , so p5 is discarded. So the resulting basis is {p1 , p 2 , p 4 }. 6. Find two polynomials from the set {p1 , . . . , p 4 } that are not multiples of one another. This is easy, because one compares only two polynomials at a time. Since these two polynomials form a linearly independent set in a two-dimensional space, they form a basis for H by the Basis Theorem. 7. You would have to know that the solution set of the homogeneous system is spanned by two solutions. In this case, the null space of the 23×25 coefficient matrix A is at most two-dimensional. By the Rank Theorem, dimColA = 25−dimNulA ≥ 25−2 = 23. Since ColA is a subspace of R23 , Col A = R23 . Thus Ax = b has a solution for every b in R23 . 8. If n = 0, then H and V are both the zero subspace, and H = V. If n > 0, then a basis for H consists of n linearly independent vectors u1 , . . . , u n . These vectors are also linearly independent as elements of V. But since dimV = n, any set of n linearly independent vectors in V must be a basis for V by the Basis span V, and H Span{ Theorem. So u1 , . . . , u n = = u1 , . . . , u n } V . 9. Let T : n → m be a linear transformation, and let A be the m × n standard matrix of T. a. If T is one-to-one, then the columns of A are linearly independent by Theoerm 12 in Section 1.9, so dimNul A = 0. By the Rank Theorem, dimCol A = n – 0 = n, which is the number of columns of A. As noted in Section 4.2, the range of T is Col A, so the dimension of the range of T is n. b. If T maps onto , then the columns of A span by Theoerm 12 in Section 1.9, so dimCol A = m. By the Rank Theorem, dimNul A = n – m. As noted in Section 4.2, the kernel of T is Nul A, so the dimension of the kernel of T is n – m. Note that n – m must be nonnegative in this case: since A must have a pivot in each row, n ≥ m. n m m 10. Let S = {v1 , ... , v p }. If S were linearly independent and not a basis for V, then S would not span V. In this case, there would be a vector v p+1 in V that is not in Span{v1 , ... , v p }. Let S ′ = {v1 , ... , v p , v p +1}. Then S ′ is linearly independent since none of the vectors in S ′ is a linear combination of vectors that precede it. Since S ′ has more elements than S, this would contradict the maximality of S. Hence S must be a basis for V. 11. If S is a finite spanning set for V, then a subset of S is a basis for V. Denote this subset of S by S ′. Since S ′ is a basis for V, S ′ must span V. Since S is a minimal spanning set, S ′ cannot be a proper subset of S. Thus S ′ = S, and S is a basis for V. 12. a. Let y be in Col AB. Then y = ABx for some x. But ABx = A(Bx), so y = A(Bx), and y is in Col A. Thus Col AB is a subspace of Col A, so rank AB = dimCol AB ≤ dimCol A = rank A by Theorem 11 in Section 4.5. T T T T b. By the Rank Theorem and part a.: rank AB= rank( AB ) = rank B A ≤ rank B = rank B Copyright © 2016 Pearson Education, Ltd. 4-66 CHAPTER 4 • Vector Spaces 13. By Exercise 12, rankAQ ≤ rankA, and rankA = rankA(QQ−1) = rank(AQ)Q−1 ≤ rankAQ, so rankAQ = rankA. 14. Note that (PA)T = AT PT . Since PT is invertible, we can use Exercise 13 to conclude that rank(PA)T = rankAT PT = rankAT . Since the ranks of a matrix and its transpose are equal (by the Rank Theorem), rankPA = rankA. 15. The equation AB = 0 shows that each column of B is in Nul A. Since Nul A is a subspace of n , all linear combinations of the columns of B are in Nul A. That is, Col B is a subspace of Nul A. By Theorem 11 in Section 4.5, rank B = dimCol B ≤ dimNul A. By this inequality and the Rank Theorem applied to A, n = rank A + dimNul A ≥ rank A + rank B 16. Suppose that rank A = r1 and rank B = r2 . Then there are rank factorizations A = C1 R1 and B = C2 R2 of A and B, where C1 is m × r1 with rank r1 , C2 is m × r2 with rank r2 , R1 is r1 × n with rank r1 , and R2 is r2 × n with rank r2 . Create an m × (r1 + r2 ) matrix C = [C1 C2 ] and an (r1 + r2 ) × n matrix R R C2 ] 1 = CR . R2 Since the matrix CR is a product, its rank cannot exceed the rank of either of its factors by Exercise 12. Since C has r1 + r2 columns, the rank of C cannot exceed r1 + r2 . Likewise R has r1 + r2 rows, so the rank of R cannot exceed r1 + r2 . Thus the rank of A + B cannot exceed r1 += r2 rank A + rank B, or rank (A + B) ≤ rank A + rank B. by stacking R1 over R2 . Then A + B= C1 R1 + C2 R2= [C1 17. Let A be an m × n matrix with rank r. (a) Let A1 consist of the r pivot columns of A. The columns of A1 are linearly independent, so A1 is an m × r matrix with rank r. (b) By the Rank Theorem applied to A1, the dimension of RowA1 is r, so A1 has r linearly independent rows. Let A2 consist of the r linearly independent rows of A1. Then A2 is an r × r matrix with linearly independent rows. By the Invertible Matrix Theorem, A2 is invertible. 18. Let A be a 4 × 4 matrix and B be a 4 × 2 matrix, and let u 0 , . . . , u3 be a sequence of input vectors in 2 . a. Use the equation x= Ax k + Bu k for k = 0, ... , 4, k = 0, . . . ,4, with x0 = 0. k +1 x1 = Ax 0 + Bu 0 = Bu 0 x 2 = Ax1 + Bu1 = ABu 0 + Bu1 x3 = Ax 2 + Bu 2 = A( ABu 0 + Bu1 ) + Bu 2 = A2 Bu 0 + ABu1 + Bu 2 x 4 = Ax3 + Bu3 = A( A2 Bu 0 + ABu1 + Bu 2 ) + Bu3 = A3 Bu 0 + A2 Bu1 + ABu 2 + Bu3 u3 u 2 3 2 = B AB A B A B u M u 1 u 0 Copyright © 2016 Pearson Education, Ltd. Chapter 4 • Supplementary Exercises 4-67 Note that M has 4 rows because B does, and that M has 8 columns because B and each of the 8 k matrices A B have 2 columns. The vector u in the final equation is in , because each u k is in 2 . b. If (A, B) is controllable, then the controlability matrix has rank 4, with a pivot in each row, and the columns of M span 4 . Therefore, for any vector v in 4 , there is a vector u in 8 such that v = Mu. However, from part a. we know that x 4 = M u when u is partitioned into a control sequence u 0 ,…, u3 . This particular control sequence makes x 4 = v. 19. To determine if the matrix pair (A, B) is controllable, we compute the rank of the matrix B AB A2B . To find the rank, we row reduce. a . Since
1 2 B AB A B = 1 0 1 0 1 0 0 0 3 ∼ 0 1 0, 3 5 0 0 1 the rank of the matrix B AB A2B is 3, and the pair (A, B) is controllable. b . Since 1 −1 0 1 2 0 1 ∼ 0 B AB A B = −1 0 1 −1 0 0 −1 1 −1 , 0 0 the rank of the matrix B AB A2B is 2, and the pair (A, B) is not controllable. 20. a . We have b1 M = B AB A2 B = b2 b3 a11 b1 a22 b2 a33 b3 a211 b1 a222 b2 2 a33 b3 b . When bj = 0 for some j, the entries in one of the rows of M are all equal to zero. Hence rankM < 3 and the matrix pair (A, B) is not controllable. c . If two of the diagonal entries of A are equal, then two rows of M are multiples of one another. Hence rankM < 3 and the matrix pair (A, B) is not controllable. d . If bj 6= 0 for all j, elementary row operations give us b1 b 2 b3 a11 b1 a22 b2 a33 b3 1 a211 b1 a222 b2 ∼ 1 1 a233 b3 a11 a22 a33 a211 a222 2 a33 Using Exercise 9 in the Supplementary Exercises for Chapter 3 we obtain 1 a11 1 a det 22 1 a33 a211 a222 = (a22 − a11 )(a33 − a11 )(a33 − a22 ) a233 If the three diagonal entries of A are all distinct, this determinant is not equal to zero. Hence rankM = 3 and the matrix pair (A, B) is controllable. Copyright © 2016 Pearson Education, Ltd. 4-68 CHAPTER 4 • Vector Spaces 21. [M] To determine if the matrix pair (A, B) is controllable, we compute the rank of the matrix B AB A2 B A3 B . To find the rank, we row reduce: 0 0 −1 1 0 0 −1 1 0 0 1.6 0 1 0 −1.6 −1 2 3 . = ∼ B AB A B A B 0 −1 1.6 −.96 0 0 1 −1.6 0 −1 1.6 −.96 −.024 0 0 0 The rank of the matrix is 3, and the pair (A, B) is not controllable. 22. [M] To determine if the matrix pair (A, B) is controllable, we compute the rank of the matrix B AB A2 B A3 B . To find the rank, we row reduce: 0 0 −1 1 0 0 0 1 0 0 .5 0 1 0 0 −1 B AB A2 B A3 B . = ∼ 0 −1 .5 11.45 0 0 1 0 −1 .5 11.45 −10.275 0 0 0 1 The rank of the matrix is 4, and the pair (A, B) is controllable. Copyright © 2016 Pearson Education, Ltd. 5.1 SOLUTIONS Notes: Exercises 1–6 reinforce the definitions of eigenvalues and eigenvectors. The subsection on eigenvectors and difference equations, along with Exercises 33 and 34, refers to the chapter introductory example and anticipates discussions of dynamical systems in Sections 5.2 and 5.6. 1. The number 2 is an eigenvalue of A if and only if the equation Ax = 2x has a nontrivial solution. 3 2 2 0 1 2 This equation is equivalent to ( A − 2 I )x = 0. Compute A −= 2I − = . The 3 8 0 2 3 6 columns of A are obviously linearly dependent, so ( A − 2 I )x = 0 has a nontrivial solution, and so 2 is an eigenvalue of A. 2. The number −2 is an eigenvalue of A if and only if the equation Ax = −2x has a nontrivial solution. 3 2 0 9 3 7 This equation is equivalent to ( A + 2 I )x = = 0. Compute A + 2I + = . The 3 −1 0 2 3 1 columns of A are obviously linearly dependent, so ( A + 2 I )x = 0 has a nontrivial solution, and so −2 is an eigenvalue of A. −3 3. Is Ax a multiple of x? Compute −3 1 1 1 1 1 = ≠ λ . So is not an eigenvector of A. 8 4 29 4 4 2 4. Is Ax a multiple of x? Compute 1 1 −1 + 2 −1 + 2 2 The second entries of x and Ax = 4 1 3 + 2 shows that if Ax is a multiple of x, then that multiple must be 3 + 2. Check 3 + 2 times the first entry of x: 2 (3 + 2)(−1 + 2) =−3 + 2 + 2 2 =−1 + 2 2 Copyright © 2016 Pearson Education, Ltd. 5-1 5-2 CHAPTER 5 • Eigenvalues and Eigenvectors −1 + 2 This matches the first entry of Ax, so is an eigenvector of A, and the corresponding 1 eigenvalue is 3 + 2. 3 5. Is Ax a multiple of x? Compute −4 2 eigenvalue 0. 3 6. Is Ax a multiple of x? Compute 3 5 A for the eigenvalue −2. 7 −5 4 6 3 6 9 4 0 4 1 −3 = 0 . So −3 is an eigenvector of A for the 1 4 1 0 7 1 −2 1 1 7 −2 = 4 =− ( 2) −2 So −2 is an eigenvector of 1 1 5 1 −2 7. To determine if 4 is an eigenvalue of A, decide if the matrix A − 4 I is invertible. 0 −1 3 0 −1 4 0 0 −1 A − 4= I 2 3 1 − 0 4 0= 1 . Invertibility can be checked in several 2 −1 −3 4 5 0 0 4 −3 4 1 ways, but since an eigenvector is needed in the event that one exists, the best strategy is to row reduce the augmented matrix for ( A − 4 I )x = 0: 0 −1 0 −1 0 −1 0 1 0 1 0 −1 2 −1 1 0 � 0 −1 −1 0 � 0 −1 −1 0 . The equation ( A − 4 I )x = 0 has a −3 4 1 0 0 4 4 0 0 0 0 0 nontrivial solution, so 4 is an eigenvalue. Any nonzero solution of ( A − 4 I )x = 0 is a corresponding eigenvector. The entries in a solution satisfy x1 + x3 = 0, with x3 free. The general 0 and − x2 − x3 = solution is not requested, so to save time, simply take any nonzero value for x3 to produce an eigenvector. If x3 = 1, then x = (−1, − 1, 1). Note: The answer in the text is (1, 1, − 1), written in this form to make the students wonder whether the more common answer given above is also correct. This may initiate a class discussion of what answers are “correct.” 8. To determine if 3 is an eigenvalue of A, decide if the matrix A − 3I is invertible. 2 2 3 0 0 −2 2 2 1 A − 3I= 3 −2 1 − 0 3 0 = 3 −5 1 . Row reducing the augmented matrix 0 1 1 0 0 3 0 1 −2 2 2 0 1 −1 −1 0 1 0 −3 0 −2 1 0 � 0 1 −2 0 � 0 1 −2 0 . The equation [(A − 3I ) 0] yields: 3 −5 0 1 −2 0 0 −2 4 0 0 0 0 0 0 ( A − 3I ) x = 0 has a nontrivial solution, so 3 is an eigenvalue. Any nonzero solution of ( A − 3I )x = is a corresponding eigenvector. The entries in a solution satisfy x1 − 3 x3 = 0 and x2 − 2 x3 = 0, with Copyright © 2016 Pearson Education, Ltd. 5.1 • Solutions 5-3 x3 free. The general solution is not requested, so to save time, simply take any nonzero value for x3 3 to produce an eigenvector. If x3 = 1, then x = 2 . 1 5 0 1 0 4 0 9. For λ =1 : A − 1I = 0 is − = . The augmented matrix for ( A − I )x = 2 1 0 1 2 0 4 2 0 0 0 0 is x2e 2 , where . Thus x1 = 0 and x2 is free. The general solution of ( A − I )x = 0 0 e 2 = , and so e 2 is a basis for the eigenspace corresponding to the eigenvalue 1. 1 5 For λ =5 : A − 5 I = 2 0 5 − 1 0 0 0 = 5 2 0 . The equation ( A − 5 I )x = 0 leads to −4 x 1 2 x1 − 4 x2 = 0, so that x1 = 2 x2 and x2 is free. The general solution is = x2 2x 2 = x2 2 2 x2 . So 1 1 is a basis for the eigenspace. 10 −9 4 0 6 −9 10. For λ =4 : A − 4 I = 0 is − = . The augmented matrix for ( A − 4 I )x = 4 −2 0 4 4 −6 6 −9 0 1 −9 / 6 0 . Thus x1= (3/ 2) x2 and x2 is free. The general solution is 4 −6 0 � 0 0 0 x (3/ 2) x 3/ 2 3/ 2 2 1 = x2 . A basis for the eigenspace corresponding to 4 is . Another choice = x2 x2 1 1 3 is . 2 4 −2 10 0 −6 −2 11. A −= . The augmented matrix for ( A − 10 I )x = 0 is 10 I − = 9 0 10 −3 −1 −3 −6 −2 0 1 1/ 3 0 −3 −1 0 � 0 0 0 . Thus x1 = (−1/ 3) x2 and x2 is free. The general solution is x1 −(1/ 3) x2 −1/ 3 −1/ 3 = x2 . A basis for the eigenspace corresponding to 10 is x = . Another x2 1 1 2 −1 choice is . 3 Copyright © 2016 Pearson Education, Ltd. 5-4 CHAPTER 5 • Eigenvalues and Eigenvectors 4 1 0 6 4 7 12. For λ =1 : A − I = 0 is − = . The augmented matrix for ( A − I )x = −3 −1 0 1 −3 −2 4 0 1 2 / 3 0 6 . Thus x1 = (−2 / 3) x2 and x2 is free. A basis for the eigenspace −3 −2 0 � 0 0 0 −2 −2 / 3 corresponding to 1 is . Another choice is . 3 1 7 For λ =5 : A − 5 I = −3 4 0 1 2 2 −3 −6 0 � 0 0 x 1 = x2 −2 x 2 = x2 4 5 − −1 0 0 2 = 5 −3 4 0 is . The augmented matrix for ( A − 5 I )x = −6 0 . Thus x1 = −2 x2 and x2 is free. The general solution is 0 −2 −2 x2 . A basis for the eigenspace is . 1 1 4 13. For λ = 1 : A − 1I = −2 − 2 0 1 0 1 1 0 − 0 1 0 0 1 0 0 3 0 = −2 1 − 2 0 0 0 1 0 . The equations for ( A − I )x = 0 are 0 0 3 x1 + x3 = easy to solve: . Row operations hardly seem necessary. Obviously x1 is zero, and = 0 −2 x1 hence x3 is also zero. There are three-variables, so x2 is free. The general solution of ( A − I )x = 0 is 0 x2e 2 , where e 2 = 1 and so e 2 provides a basis for the eigenspace. 0 4 0 1 2 0 0 2 2 For λ = 2: A − 2 I =−2 1 0 − 0 2 0 =− − 2 0 1 0 0 2 − 2 0 1 0 2 0 1 0 2 [( A − 2 I ) 0] =− 0 0 � 0 −1 1 0 � 2 −1 0 −1 0 0 0 0 0 −2 0 −1 0 1 0 0 1 0 . −1 0 1 0 1/ 2 −1 0 0 0 . So 0 −1/ 2 0 is x3 1 . A nice basis x1 =−(1/ 2) x3 , x2 =x3 , with x3 free. The general solution of ( A − 2 I )x = 1 −1 vector for the eigenspace is 2 . 2 Copyright © 2016 Pearson Education, Ltd. 5.1 4 2 For λ = 3: A − 3I =− −2 0 1 0 1 [( A − 3I ) 0] = −2 −2 1 0 −2 0 −2 0 1 3 0 − 0 1 0 0 0 � 0 0 1 2 0 =− 3 −2 0 3 0 1 0 0 0 −2 0 1 2 0 0 −2 0 1 0 0 0 0 � 0 • Solutions 5-5 1 0 . −2 0 1 0 1 −1 0 0 0 . So x1 =− x3 , x2 =x3 , 0 −1 with x3 free. A basis vector for the eigenspace is 1 . 1 1 14. For λ =−2 : A − (−2 I ) =A + 2 I = 1 4 matrix for [ A − (−2) I ]x =0, or 0 −1 3 0] 1 −1 [( A + 2 I ) = 0 4 −13 3 −1 2 0 + 0 1 0 0 −3 −13 0 2 0 0 3 0 = 1 2 4 0 −1 −13 −1 0 . The augmented 3 ( A + 2 I )x = 0, is 0 1 0 � 0 0 0 0 1 −13 −1/ 3 −1/ 3 13/ 3 0 1 0 � 0 0 0 0 1 0 −1/ 3 −1/ 3 0 0 0 . Thus 0 1/ 3 0 is x3 1/ 3 . A basis for x1 =(1/ 3) x3 , x2 =(1/ 3) x3 , with x3 free. The general solution of ( A + 2 I )x = 1 1/ 3 the eigenspace corresponding to −2 is 1/ 3 ; another is 1 1 −1 15. For λ =: 3 [( A − 3I ) 0] = 2 2 −2 4 3 −3 6 0 1 0 � 0 0 0 2 0 0 and x3 free. The general solution of ( A − 3I )x = 0, is x = for the eigenspace is: −2 −3 1 , 0 0 1 1 1 . 3 3 0 0 0 0 . Thus x1 + 2 x2 + 3 x3 = 0, with x2 0 −2 x − 3 x 2 3 x2 x 3 −2 −3 = x2 1 + x3 0 . A basis 0 1 . Note: For simplicity, the text answer omits the set brackets. I permit my students to list a basis without the set brackets. Some instructors may prefer to include brackets. Copyright © 2016 Pearson Education, Ltd. 5-6 CHAPTER 5 • Eigenvalues and Eigenvectors 16. For 3 1 λ =4 : A − 4 I = 0 0 0 3 1 0 0 4 0 0 − 0 0 4 0 2 1 1 0 0 −1 1 −1 [( A − 4 I ) 0] = 0 1 0 0 2 1 −3 0 0 0 0 0 0 4 0 0 0 1 0 0 � 0 0 0 0 0 −1 0 1 = 0 0 4 0 0 0 4 0 0 1 0 0 −2 −3 0 0 0 0 0 0 0 −1 1 0 2 0 3 0 , 1 0 0 1 0 0 . 0 0 0 0 . So x1 = 2 x3 , x2 = 3 x3 , with x3 and 0 0 0 is x x4 free variables. The general solution of ( A − 4 I )x = basis for the eigenspace is: 2 1 −3 0 x 1 x = 2 x 3 x 4 2 x 3 3 x3 = x 3 x 4 2 0 3 0 =x3 + x4 . A 1 0 0 1 . Note: I urge my students always to include the extra column of zeros when solving a homogeneous system. Exercise 16 provides a situation in which failing to add the column is likely to create problems for a student, because the matrix A − 4 I itself has a column of zeros. 0 17. The eigenvalues of 0 0 0 2 0 0 5 are 0, 2, and −1, on the main diagonal, by Theorem 1. −1 4 0 1 3 3 3 0 0 0 0 0 are 4, 0, and −3, on the main diagonal, by Theorem 1. −3 18. The eigenvalues of 1 19. The matrix 1 1 2 2 2 is not invertible because its columns are linearly dependent. So the number 0 is an eigenvalue of the matrix. See the discussion following Example 5. 5 5 5 20. The matrix A = 5 5 5 is not invertible because its columns are linearly dependent. So the 5 5 5 number 0 is an eigenvalue of A. Eigenvectors for the eigenvalue 0 are solutions of Ax = 0 and therefore have entries that produce a linear dependence relation among the columns of A. Any nonzero vector (in R 3 ) whose entries sum to 0 will work. Find any two such vectors that are not multiples; for instance, (1, 1, − 2) and (1, − 1, 0). Copyright © 2016 Pearson Education, Ltd. 5.1 • Solutions 5-7 21. a. False. The equation Ax = λx must have a nontrivial solution. b. True. See the paragraph after Example 5. c. True. See the discussion of equation (3). d. True. See Example 2 and the paragraph preceding it. Also, see the Numerical Note. e. False. See the warning after Example 3. 22. a. False. The vector x in Ax = λx must be nonzero. b. False. See Example 4 for a two-dimensional eigenspace, which contains two linearly independent eigenvectors corresponding to the same eigenvalue. The statement given is not at all the same as Theorem 2. In fact, it is the converse of Theorem 2 (for the case r = 2 ). c. True. See the paragraph after Example 1. d. False. Theorem 1 concerns a triangular matrix. See Examples 3 and 4 for counterexamples. e. True. See the paragraph following Example 3. The eigenspace of A corresponding to λ is the null space of the matrix A − λI . 23. If a 2 × 2 matrix A were to have three distinct eigenvalues, then by Theorem 2 there would correspond three linearly independent eigenvectors (one for each eigenvalue). This is impossible because the vectors all belong to a two-dimensional vector space, in which any set of three vectors is linearly dependent. See Theorem 8 in Section 1.7. In general, if an n × n matrix has p distinct eigenvalues, then by Theorem 2 there would be a linearly independent set of p eigenvectors (one for each eigenvalue). Since these vectors belong to an n-dimensional vector space, p cannot exceed n. 24. A simple example of a 2 × 2 matrix with only one distinct eigenvalue is a triangular matrix with the same number on the diagonal. By experimentation, one finds that if such a matrix is actually a diagonal matrix then the eigenspace is two dimensional, and otherwise the eigenspace is only one dimensional. 4 Examples: 0 1 4 and 4 0 5 . 4 25. If λ is an eigenvalue of A, then there is a nonzero vector x such that Ax = λx. Since A is invertible, A−1 Ax = A−1 (λx), and so x = λ( A−1x). Since x ≠ 0 (and since A is invertible), λ cannot be zero. Then λ −1x = A−1x, which shows that λ −1 is an eigenvalue of A−1. Note: The Study Guide points out here that the relation between the eigenvalues of A and A−1 is important in the so-called inverse power method for estimating an eigenvalue of a matrix. See Section 5.8. 26. Suppose that A2 is the zero matrix. If Ax = λx for some x ≠ 0, then 2 A= x A( A= x) A(λ= x) λ= Ax λ 2 x. Since x is nonzero, λ must be zero. Thus each eigenvalue of A is zero. 27. Use the Hint in the text to write, for any λ, ( A − λI )T = AT − (λI )T = AT − λI . Since ( A − λI )T is invertible if and only if A − λI is invertible (by Theorem 6(c) in Section 2.2), it follows that AT − λI is not invertible if and only if A − λI is not invertible. That is, λ is an eigenvalue of AT if and only if λ is an eigenvalue of A. Copyright © 2016 Pearson Education, Ltd. 5-8 CHAPTER 5 • Eigenvalues and Eigenvectors Note: If you discuss Exercise 27, you might ask students on a test to show that A and AT have the same characteristic polynomial (discussed in Section 5.2). Since det A = det AT , for any square matrix A, T I ) det( A − λI= )T det( AT − (λI )= ) det( AT − λI ). det( A − λ= 28. If A is lower triangular, then AT is upper triangular and has the same diagonal entries as A. Hence, by the part of Theorem 1 already proved in the text, these diagonal entries are eigenvalues of AT. By Exercise 27, they are also eigenvalues of A. 29. Let v be the vector in n whose entries are all ones. Then Av = sv. 30. Suppose the column sums of an n × n matrix A all equal the same number s. By Exercise 29 applied to AT in place of A, the number s is an eigenvalue of AT. By Exercise 27, s is an eigenvalue of A. 31. Suppose T reflects points across (or through) a line that passes through the origin. That line consists of all multiples of some nonzero vector v. The points on this line do not move under the action of A. So T ( v ) = v. If A is the standard matrix of T, then Av = v. Thus v is an eigenvector of A corresponding to the eigenvalue 1. The eigenspace is Span {v}. Another eigenspace is generated by any nonzero vector u that is perpendicular to the given line. (Perpendicularity in R 2 should be a familiar concept even though orthogonality in R n has not been discussed yet.) Each vector x on the line through u is transformed into the vector −x. The eigenvalue is −1. 33. (The solution is given in the text.) a. Replace k by k + 1 in the definition of x k , and obtain= x k +1 c1λ k +1u + c2 µ k +1v. b. = Ax k A(c1l k u + c2 µ k v ) = c1l k Au + c2 µ k Av by linearity k = c1ll u + c2 µ k µ v since u and v are eigenvectors = x k +1 34. You could try to write x0 as linear combination of eigenvectors, v1 , …, v p . If λ1 , …,λ p are corresponding eigenvalues, and if x0 = c1 v1 + + c p v p , then you could define x= c1λ1k v1 + + c p λkp v p k In this case, for k = 0, 1, 2, …, = Ax k A(c1λ1k v1 + + c p λkp v p ) = c1λ1k Av1 + + c p λkp Av p c1 1k +1 v1 λ = + + c p kp +1 v p λ Linearity The v i are eigenvectors. = x k +1 35. Using the figure in the exercise, plot T (u) as 2u, because u is an eigenvector for the eigenvalue 2 of the standard matrix A. Likewise, plot T ( v ) as 3v, because v is an eigenvector for the eigenvalue 3. Since T is linear, the image of w is T (w ) = T (u + v ) = T (u) + T ( v ). Copyright © 2016 Pearson Education, Ltd. 5.1 • Solutions 5-9 36. As in Exercise 35, T (u) = −u and T ( v ) = 3v because u and v are eigenvectors for the eigenvalues −1 and 3, respectively, of the standard matrix A. Since T is linear, the image of w is T (w ) = T (u + v ) = T (u) + T ( v ). Note: The matrix programs supported by this text all have an eigenvalue command. In some cases, such as MATLAB, the command can be structured so it provides eigenvectors as well as a list of the eigenvalues. At this point in the course, students should not use the extra power that produces eigenvectors. Students need to be reminded frequently that eigenvectors of A are null vectors of a translate of A. That is why the instructions for Exercises 35–38 tell students to use the method of Example 4. It is my experience that nearly all students need manual practice finding eigenvectors by the method of Example 4, at least in this section if not also in Sections 5.2 and 5.3. However, [M] exercises do create a burden if eigenvectors must be found manually. For this reason, the data files for the text include a special command, nulbasis for each matrix program (MATLAB, Maple, etc.). The output of nulbasis (A) is a matrix whose columns provide a basis for the null space of A, and these columns are identical to the ones a student would find by row reducing the augmented matrix [ A 0]. With nulbasis, student answers will be the same (up to multiples) as those in the text. I encourage my students to use technology to speed up all numerical homework here, not just the [M ] exercises, 37. [M] Let A be the given matrix. Use the MATLAB commands eig and nulbasis (or equivalent commands). The command ev = eig(A) computes the three eigenvalues of A and stores them in a vector ev. In this exercise, ev = (3, 13, 13). The eigenspace for the eigenvalue 3 is the null space of A − 3I . Use nulbasis to produce a basis for each null space. If the format is set for rational 5/ 9 display, the result is nulbasis(A -ev(1)*eye(3))= −2 / 9 . For simplicity, scale the entries 1 5 by 9. A basis for the eigenspace for λ = 3 : −2 9 −2 −1 For the next eigenvalue, 13, compute nulbasis(A -ev(2)*eye(3))= 1 0 . Basis for 0 1 −2 −1 eigenspace for λ = 13 : 1 , 0 . There is no need to use ev(3) because it is the same as 0 1 ev(2). 38. [M] ev = eig(A) = (13, − 12, − 12, 13). For λ = 13 : −1/ 2 0 nulbasis (A -ev(1)*eye(4))= 1 0 1/ 3 −4 / 3 . Basis for eigenspace : 0 1 Copyright © 2016 Pearson Education, Ltd. −1 1 0 , −4 2 0 0 3 . 5-10 CHAPTER 5 • Eigenvalues and Eigenvectors 2/ 7 1 For λ = −12 : nulbasis(A -ev(2)*eye(4))= 1 0 39. [M] For λ = 5, basis: 2 −1 2 −1 1 0 1 , 0 , 0 0 1 0 0 0 1 0 2 0 −1 7 −1 . Basis: , 0 7 0 0 1 1 −2 3 7 7 . For λ = −2, basis: −5 , −5 5 0 0 5 . 40. [M] ev = eig(A)= (21.68984106239549, −16.68984106239549, 3, 2, 2). The first two eigenvalues are the roots of λ 2 − 5λ − 362 = 0. −0.33333333333333 −0.33333333333333 2.39082008853296 −0.80748675519962 Basis for = λ ev(1) : 0.33333333333333 , for = λ ev(2) : 0.33333333333333 . 0.58333333333333 0.58333333333333 1.000000000000000 1.00000000000000 −2 −.5 0 −2 1 .5 For the eigenvalues 3 and 2, the eigenbases are 0 , and 0 , 0 , respectively. 1 0 1 0 0 1 Note: Since so many eigenvalues in text problems are small integers, it is easy for students to form a habit of entering a value for λ in nulbasis (A - λI) based on a visual examination of the eigenvalues produced by eig(A)when only a few decimal places for λ are displayed. Exercise 40 may help your students discover the dangers of this approach. 5.2 SOLUTIONS Notes: Exercises 9–14 can be omitted, unless you want your students to have some facility with determinants of 3 × 3 matrices. In later sections, the text will provide eigenvalues when they are needed for matrices larger than 2 × 2. If you discussed partitioned matrices in Section 2.4, you might wish to bring in Supplementary Exercises 12–14 in Chapter 5. (Also, see Exercise 14 of Section 2.4.) Exercises 25 and 27 support the subsection on dynamical systems. The calculations in these exercises and Example 5 prepare for the discussion in Section 5.6 about eigenvector decompositions. Copyright © 2016 Pearson Education, Ltd. 5.2 • Solutions 5-11 7 2 7 2 7 λ 0 2 − λ 1.= A λI , A −= − = . The characteristic polynomial 2 − λ 7 2 7 2 0 λ 7 is det( A − λI ) = (2 − λ) 2 − 7 2 = 4 − 4λ + λ 2 − 49 = λ 2 − 4λ − 45 . In factored form, the characteristic equation is (λ − 9)(λ + 5) =0, so the eigenvalues of A are 9 and −5. 3 5 3 5 − λ 2.= λI A , A −= . The characteristic polynomial is 5 − λ 3 5 3 det( A − λ I ) = (5 − λ )(5 − λ ) − 3 ⋅ (3) = λ 2 − 10λ + 16 . Since λ 2 − 10λ + 16 = (λ − 8)(λ − 2), the eigenvalues of A are 8 and 2. −2 3 −2 3 − λ 3. A = , A= − λI . The characteristic polynomial is −1 − λ 1 −1 1 det( A − λI ) = (3 − λ)(−1 − λ) − (−2)(1) = λ 2 − 2λ − 1 . Use the quadratic formula to solve the characteristic equation and find the eigenvalues: λ = −b ± b 2 − 4ac 2 ± 4 + 4 = = 1± 2 . 2a 2 −3 5 −3 5 − λ 4. A = , A= − λI . The characteristic polynomial of A is 3 3 − λ −4 −4 det( A − λI ) = (5 − λ)(3 − λ) − (−3)(−4) = λ2 − 8λ + 3 . Use the quadratic formula to solve the characteristic equation and find the eigenvalues: λ= 8 ± 64 − 4(3) 8 ± 2 13 = = 4 ± 13 2 2 1 2 1 2 − λ 5. A = ,= A − λI . The characteristic polynomial of A is 4 − λ −1 4 −1 det( A − λI ) = (2 − λ)(4 − λ) − (1)(−1) = λ 2 − 6λ + 9 = (λ − 3) 2 . Thus, A has only one eigenvalue 3, with multiplicity 2. −4 3 −4 3 − λ 6. A = , A= − λI . The characteristic polynomial is 8 8 − λ 4 4 det( A − λI ) = (3 − λ)(8 − λ) − (−4)(4) = λ2 − 11λ + 40 . Use the quadratic formula to solve −11 ± 121 − 4(40) −11 ± −39 . These values are complex numbers, not = 2 2 real numbers, so A has no real eigenvalues. There is no nonzero vector x in 2 such that Ax = λ x, because a real vector Ax cannot equal a complex multiple of x. det= (A − λI ) = 0: λ 3 5 3 5 − λ 7. A = , A= − λI . The characteristic polynomial is 4 − λ −4 4 −4 det( A − λI ) = (5 − λ)(4 − λ) − (3)(−4) = λ2 − 9λ + 32 . Use the quadratic formula to solve Copyright © 2016 Pearson Education, Ltd. 5-12 CHAPTER 5 • Eigenvalues and Eigenvectors 9 ± 81 − 4(32) 9 ± −47 . These values are complex numbers, not real = 2 2 numbers, so A has no real eigenvalues. There is no nonzero vector x in 2 such that Ax = λ x, because a real vector Ax cannot equal a complex multiple of x. det ( A= − λI ) = 0: λ −2 7 −2 7 − λ 8. A = , A= − λI . The characteristic polynomial is 3 3 − λ 2 2 det( A − λI ) = (7 − λ)(3 − λ) − (−2)(2) = λ2 − 10λ + 25 . Since λ 2 − 10λ + 25 = (λ − 5) 2 , the only eigenvalue is 5, with multiplicity 2. −1 0 1 − λ 9. det(= A − λ I ) det 2 −1 . From the special formula for 3 × 3 determinants, the 3−λ 0 6 0 − λ characteristic polynomial is det( A − λI ) = (1 − λ)(3 − λ)(−λ) + 0 + (−1)(2)(6) − 0 − (6)(−1)(1 − λ) − 0 = (λ2 − 4λ + 3)(−λ) − 12 + 6(1 − λ) =−λ3 + 4λ2 − 3λ − 12 + 6 − 6λ = −λ3 + 4λ2 − 9λ − 6 (This polynomial has one irrational zero and two imaginary zeros.) Another way to evaluate the determinant is to interchange rows 1 and 2 (which reverses the sign of the determinant) and then make one row replacement: −1 −1 0 3−λ 1 − λ 2 −1 = − det 1 − λ −1 det 2 3−λ 0 0 0 6 0 − λ 6 0 − λ −1 3−λ 2 = − det 0 0 + (.5λ − .5)(3 − λ ) −1 + (.5λ − .5)(−1) . Next, expand by cofactors down the first 0 6 0−λ column. The quantity above equals (.5λ − .5)(3 − λ) −.5 − .5λ −2det = −2[(.5λ − .5)(3 − λ)(−λ) − (−.5 − .5λ)(6)] −λ 6 = −λ3 + 4λ2 − 9λ − 6 (1 − λ)(3 − λ)(−λ) − (1 + λ)(6) = (λ2 − 4λ + 3)(−λ) − 6 − 6λ = 3 0 − λ 10. det(= 0−λ A − λ I ) det 3 1 2 characteristic polynomial is 1 2 . From the special formula for 3 × 3 determinants, the 0 − λ det( A − λ I ) = (−λ )(−λ )(−λ ) + 3 ⋅ 2 ⋅ 1 + 1 ⋅ 3 ⋅ 2 − 1 ⋅ (−λ ) ⋅ 1 − 2 ⋅ 2 ⋅ (−λ ) − (−λ ) ⋅ 3 ⋅ 3 =−λ 3 + 6 + 6 + λ + 4λ + 9λ =−λ 3 + 14λ + 12 11. The special arrangements of zeros in A makes a cofactor expansion along the first row highly effective. Copyright © 2016 Pearson Education, Ltd. 5.2 4 − λ det( A − λI ) =det 5 −2 0 3−λ 0 0 3 − λ 2 =(4 − λ) det 0 2 − λ • Solutions 2 2 − λ = (4 − λ)(3 − λ )(2 − λ)= (4 − λ)(λ2 − 5λ + 6)= −λ3 + 9λ2 − 26λ + 24 If only the eigenvalues were required, there would be no need here to write the characteristic polynomial in expanded form. 12. Make a cofactor expansion along the third row: −1 − λ det( A − λI ) = det −3 0 1 −1 − λ 1 = (2 − λ) ⋅ det −3 2 − λ 0 4−λ 0 0 4 − λ =(2 − λ)(−1 − λ)(4 − λ) =−λ3 + 5λ2 − 2λ − 8 13. Make a cofactor expansion down the third column: 6 − λ det( A − λ I ) = det −2 5 −2 9−λ 8 0 6 − λ 0 = (3 − λ ) ⋅ det −2 3 − λ −2 9 − λ = (3 − λ )[(6 − λ )(9 − λ ) − (−2)(−2)] = (3 − λ )(λ 2 − 15λ + 50) = −λ 3 + 18λ 2 − 95λ + 150 or (3 − λ )(λ − 5)(λ − 10) 14. Make a cofactor expansion along the second row: 5 − λ det( A − λI ) = det 0 6 −2 1− λ 7 3 5 − λ 0 = (1 − λ) ⋅ det 6 −2 − λ 3 −2 − λ = (1 − λ) ⋅ [(5 − λ)(−2 − λ) − 3 ⋅ 6] = (1 − λ)(λ2 − 3λ − 28) = −λ3 + 4λ2 + 25λ − 28 or (1 − λ)(λ − 7)(λ + 4) 15. Use the fact that the determinant of a triangular matrix is the product of the diagonal entries: 4 − λ 0 det( A − λI ) =det 0 0 −7 3−λ 0 0 0 −4 3−λ 0 2 6 =(4 − λ)(3 − λ) 2 (1 − λ) −8 1 − λ The eigenvalues are 4, 3, 3, and 1. 16. The determinant of a triangular matrix is the product of its diagonal entries: 5 − λ 8 det( A − λI ) = det 0 1 0 −4 − λ 7 −5 0 0 1− λ 2 0 0 = (5 − λ)(−4 − λ)(1 − λ) 2 0 1 − λ The eigenvalues are 5, 1, 1, and −4. Copyright © 2016 Pearson Education, Ltd. 5-13 5-14 CHAPTER 5 • Eigenvalues and Eigenvectors 17. The determinant of a triangular matrix is the product of its diagonal entries: 3 − λ −5 3 0 −4 0 1− λ 8 −7 1 0 0 0−λ 2 9 0 0 0 1− λ −2 0 0 (3 − λ ) 2 (1 − λ ) 2 (−λ ) 0 = 0 3 − λ The eigenvalues are 3, 3, 1, 1, and 0. 18. Row reduce the augmented matrix for the equation ( A − 5 I )x = 0: 0 0 0 0 −2 −2 0 0 6 h 0 0 −1 0 4 −4 −2 0 0 0 0 0 0 0 � 0 0 0 0 6 h−6 0 0 −1 1 4 4 0 0 0 0 � 0 0 0 0 1 0 0 0 −3 h−6 0 0 0 0 1 0 0 0 0 0 For a two-dimensional eigenspace, the system above needs two free variables. This happens if and only if h = 6. 19. Since the equation det( A − λI ) = (λ1 − λ)(λ 2 − λ) (λ n − λ) holds for all λ , set λ =0 and conclude that det A = λ1λ 2 λ n . 20. det( AT = − λI ) det( AT − λ I T ) = det( A − λ I )T Transpose property = det( A − λ I ) Theorem 3(c) 21. a. b. c. d. False. See Example 1. False. See Theorem 3. True. See Theorem 3. False. See the solution of Example 4. 22. a. b. c. d. False. See the paragraph before Theorem 3. The absolute value of det A equals the volume. False. See Theorem 3. True. See the paragraph before Example 4. False. See the warning after Theorem 4. −1 23. If A = QR, with Q invertible, and if A1 = RQ, then write = A1 Q= QRQ Q −1 AQ, which shows that A1 is similar to A. 24. First, observe that if P is invertible, then Theorem 3(b) shows that −1 = 1 det = I det( PP= ) (det P )(det P −1 ) . Use Theorem 3(b) again when A = PBP −1 , = = = P )(det B )(det P −1 ) (det = B )(det P )(det P −1 ) det B . det A det( PBP −1 ) (det 25. Example 5 of Section 4.9 showed that Av1 = v1 , which means that v1 is an eigenvector of A corresponding to the eigenvalue 1. Copyright © 2016 Pearson Education, Ltd. 5.2 • Solutions 5-15 a. Since A is a 2 × 2 matrix, the eigenvalues are easy to find, and factoring the characteristic polynomial is easy when one of the two factors is known. .3 .6 − λ 2 det = (.6 − λ)(.7 − λ) − (.3)(.4) = λ − 1.3λ + .3 = (λ − 1)(λ − .3) . The eigenvalues . . − 4 7 λ are 1 and .3. .6 − .3 For the eigenvalue .3, solve ( A − .3I )x = 0 : .4 .3 .7 − .3 0 .3 = 0 .4 .3 .4 0 1 � 0 0 1 0 0 . 0 Here x1 + x2 = 0, with x2 free. The general solution is not needed. Set x2 = 1 to find an −1 eigenvector v 2 = . A suitable basis for 2 is {v1 , v 2 }. 1 1/ 2 3/ 7 −1 b. Write x= v1 + cv 2 : = + c . By inspection, c is −1/14. (The value of c depends 0 1/ 2 4 / 7 1 on how v 2 is scaled.) c. For k =1, 2, …, define x k = Ak x0 . Then x1 = A( v1 + cv 2 ) = Av1 + cAv 2 = v1 + c(.3) v 2 , because v1 and v 2 are eigenvectors. Again x 2 = Ax1 = A( v1 + c(.3) v 2 ) = Av1 + c(.3) Av 2 = v1 + c(.3)(.3) v 2 . Continuing, the general pattern is x k = v1 + c(.3) k v 2 . As k increases, the second term tends to 0 and so x k tends to v1. b a b a 26. If a ≠ 0, then A = = � U , and det A = (a )(d − ca −1b) = ad − bc. If a = 0, −1 c d 0 d − ca b 0 b c d then A = (−1)1 (cb) = 0 − bc = ad − bc. U (with one interchange), so det A = = � c d 0 b 27. a. Av1 = v1 , Av 2 = .5 v 2 , Av 3 = .2 v 3 . b. The set {v1 , v 2 , v 3 } is linearly independent because the eigenvectors correspond to different eigenvalues (Theorem 2). Since there are three vectors in the set, the set is a basis for 3 . So there exist unique constants such that x0 =c1 v1 + c2 v 2 + c3 v 3 , and wT x0 =c1wT v1 + c2 wT v 2 + c3 wT v 3 . Since x0 and v1 are probability vectors and since the entries in v 2 and v 3 sum to 0, the above equation shows that c1 = 1. c. By (b), x0 =c1 v1 + c2 v 2 + c3 v 3 . Using (a), x k= Ak x0= c1 Ak v1 + c2 Ak v 2 + c3 Ak v 3= v1 + c2 (.5) k v 2 + c3 (.2) k v 3 → v1 as k → ∞ 28. [M] Answers will vary, but should show that the eigenvectors of A are not the same as the eigenvectors of AT , unless, of course, AT = A. 29. [M] Answers will vary. The product of the eigenvalues of A should equal det A. Copyright © 2016 Pearson Education, Ltd. 5-16 CHAPTER 5 30. [M] • Eigenvalues and Eigenvectors The characteristic polynomials and the eigenvalues for the various values of a are given in the following table: a Characteristic Polynomial 31.8 −.4 − 2.6t + 4t − t 31.9 .8 − 3.8t + 4t 2 − t 3 2.7042, 1, .2958 32.0 2 − 5t + 4t 2 − t 3 2, 1, 1 32.1 3.2 − 6.2t + 4t 2 − t 3 1.5 ± .9747i, 1 32.2 4.4 − 7.4t + 4t 2 − t 3 1.5 ± 1.4663i, 1 2 3 Eigenvalues 3.1279, 1, − .1279 The graphs of the characteristic polynomials are: Notes: An appendix in Section 5.3 of the Study Guide gives an example of factoring a cubic polynomial with integer coefficients, in case you want your students to find integer eigenvalues of simple 3 × 3 or perhaps 4 × 4 matrices. The MATLAB box for Section 5.3 introduces the command poly (A), which lists the coefficients of the characteristic polynomial of the matrix A, and it gives MATLAB code that will produce a graph of the characteristic polynomial. (This is needed for Exercise 30.) The Maple and Mathematica appendices have corresponding information. The appendices for the TI and HP calculators contain only the commands that list the coefficients of the characteristic polynomial. 5.3 SOLUTIONS 5 7 2 0 1.= P = ,D = , A PDP −1 , and 2 3 0 1 3 −7 4 16 0 4 = P −1 = = , D 0 1 , and A − 2 5 A4 = PD 4 P −1. We compute 5 7 16 0 3 −7 226 = 5 90 2 3 0 1 −2 Copyright © 2016 Pearson Education, Ltd. −525 . −209 5.3 • Solutions 2 −3 1 0 4 4 −1 −1 2. P = = ,D = , A PDP , and A = PD P . We compute − / 3 5 0 1 2 5 3 1 0 0 5 3 1 151 2 −3 1 4 −1 4 P= 3 2= , D= 0 1/16 , and A = 5 0 1/16 3 2 16 −225 −3 1 k −1 = = P 3. Ak PD 3 0 a k 1 0 0 1 0 = b k −3 1 ak 3a k − 3b k 3 k −1 = = P 4. Ak PD 1 4 2k 1 0 0 −1 4 4 − 3⋅ 2k = 1k 1 −3 1 − 2k 5-17 90 . −134 0 . bk 12 ⋅ 2k −12 . 4 ⋅ 2k −3 5. By the Diagonalization Theorem, eigenvectors form the columns of the left factor, and they correspond respectively to the eigenvalues on the diagonal of the middle factor. 1 1 2 λ = 5 : 1 ; λ = 1 : 0 , −1 . −1 0 1 −1 −2 0 6. As in Exercise 5, inspection of the factorization gives: λ = 4 : 2 ; λ = 5 : 0 , 1 . 0 1 0 7. Since A is triangular, its eigenvalues are obviously ±1. 0 For λ = 1: A − 1I = 6 0 0 amounts to 6 x1 − 2 x2 = 0, so x1= (1/ 3) x2 . The equation ( A − 1I )x = −2 1/ 3 1 with x2 free. The general solution is x2 , and a nice basis vector for the eigenspace is v1 = . 1 3 2 For λ = −1: A + 1I = 6 0 0 amounts to 2 x1 = 0, so x1 = 0 with x2 . The equation ( A + 1I )x = 0 0 0 free. The general solution is x2 , and a basis vector for the eigenspace is v 2 = . 1 1 From v1 and v 2 construct = P = v v2 1 1 3 0 1 . Then set D = 1 0 in D correspond to v1 and v 2 respectively. Copyright © 2016 Pearson Education, Ltd. 0 , where the eigenvalues −1 5-18 CHAPTER 5 • Eigenvalues and Eigenvectors 8. Since A is triangular, its only eigenvalue is obviously 5. 0 1 For λ = 5: A − 5 I = 0 amounts to x2 = 0, so x2 = 0 with x1 free. 0 0 . The equation ( A − 5 I )x = 1 The general solution is x1 . Since we cannot generate an eigenvector basis for 2 , A is not 0 diagonalizable. 9. To find the eigenvalues of A, compute its characteristic −1 3 − λ 2 2 polynomial: det( A − λI ) = det = (3 − λ)(5 − λ) − (−1)(1) = λ − 8λ + 16 = (λ − 4) . Thus 1 5 − λ the only eigenvalue of A is 4. −1 For λ = 4: A − 4 I = 1 −1 0 amounts to x1 + x2 = 0, so x1 = − x2 with . The equation ( A − 4 I )x = 1 −1 x2 free. The general solution is x2 . Since we cannot generate an eigenvector basis for 2 , A is 1 not diagonalizable. 10. To find the eigenvalues of A, compute its characteristic polynomial: 3 2 − λ det( A − λI ) = det = (2 − λ)(1 − λ) − (3)(4) = λ 2 − 3λ − 10 = (λ − 5)(λ + 2) . 1 − λ 4 eigenvalues of A are 5 and −2 . −3 For λ = 5: A − 5 I = 4 Thus the 3 0 amounts to x1 − x2 = . The equation ( A − 5 I )x = 0, so x1 = x2 with −4 1 1 x2 free. The general solution is x2 , and a basis vector for the eigenspace is v1 = . 1 1 4 3 For λ = −2: A + 2 I = 0 amounts to 4 x1 + 3 x2 = 0, so 4 3 . The equation ( A + 1I )x = −3/ 4 , and a nice basis vector for the x1 = (−3/ 4) x2 with x2 free. The general solution is x2 1 −3 eigenspace is v 2 = . 4 From v1 and v 2 construct P = = v v2 1 1 1 −3 5 . Then set D = 4 0 in D correspond to v1 and v 2 respectively. Copyright © 2016 Pearson Education, Ltd. 0 , where the eigenvalues −2 5.3 • Solutions 5-19 11. The eigenvalues of A are given to be 1, 2, and 3. −4 3 For λ = 3: A − 3I =− −3 4 1 1 −2 0 , and row reducing [ A − 3I 0 1 0] yields 0 0 0 1 0 −1/ 4 −3/ 4 0 0 0 . The 0 −2 / 3 −1 0 0 0 . The 0 1/ 4 1 general solution is x3 3/ 4 , and a nice basis vector for the eigenspace is v1 = 3 . 1 4 −3 3 For λ = 2: A − 2 I =− −3 4 2 1 −2 0 , and row reducing [ A − 2 I 1 1 0] yields 0 0 0 1 0 2 / 3 2 general solution is x3 1 , and a nice basis vector for the eigenspace is v 2 = 3 . 1 3 −2 4 −2 1 For λ = 1: A − I = −3 3 0 , and row reducing [ A − 1I 0] yields 0 −3 1 0 2 1 1 general solution is x3 1 , and a basis vector for the eigenspace is v 3 = 1 . 1 1 −1 −1 0 0 1 0 1 2 1 3 3 1 . Then set D = From v1 , v 2 and v 3= construct P 4 3 1 the eigenvalues in D correspond to v1 , v 2 and v 3 respectively. = v v2 v3 1 3 0 0 0 0 . The 0 0 0 , where 1 0 2 0 12. The eigenvalues of A are given to be 2 and 8. 2 2 1 −4 For λ = 8: A − 8= 2 , and row reducing [ A − 8 I 0] yields 0 I 2 −4 0 2 2 −4 1 1 general solution is x3 1 , and a basis vector for the eigenspace is v1 = 1 . 1 1 0 1 0 −1 −1 0 2 2 For λ = 2: A − 2 I = 2 2 2 2 1 0 0 0 0 . The general 0 2 2 , and row reducing [ A − 2 I 2 1 0] yields 0 0 −1 −1 solution is x2 1 + x3 0 , and a basis for the eigenspace is {v 2 , v 3= } 0 1 Copyright © 2016 Pearson Education, Ltd. 1 0 0 −1 −1 1 , 0 0 1 . 0 0 . The 0 5-20 CHAPTER 5 • Eigenvalues and Eigenvectors 1 From v1 , v 2 and v 3= construct P = v1 v2 v3 1 1 −1 1 0 −1 8 0 . Then set D = 0 0 1 0 0 , where 2 0 2 0 the eigenvalues in D correspond to v1 , v 2 and v 3 respectively. 13. The eigenvalues of A are given to be 5 and 1. −3 For λ = 5: A − 5= I 1 −1 2 −2 −2 −1 −1 , and row reducing [ A − 5 I −3 1 0] yields 0 0 −1 general solution is x3 −1 , and a basis for the eigenspace is v1 = 1 From v1 , v 2 and v 3 construct P = v 1 v2 −2 1 0 0 0 . The 0 −1 0 0 0 0 . The 0 −1 −1 . 1 2 −1 1 1 For λ = 1:= 2 −1 , and row reducing [ A − I 0] yields 0 A − 1I 1 0 −1 −2 1 −2 1 general solution is x2 1 + x3 0 , and a basis for the eigenspace is {v 2 , v 3= } 0 1 −1 v3 = −1 1 1 1 0 0 1 0 2 0 0 −2 1 1 , 0 0 1 1 5 0 . Then set D = 0 1 0 0 1 0 . 0 0 , where 1 the eigenvalues in D correspond to v1 , v 2 and v 3 respectively. 14. The eigenvalues of A are given to be 5 and 4. −1 2 For λ = 5: A − 5 I = 0 0 0 0 −2 4 , and row reducing [ A − 5 I 0 1 0] yields 0 0 0 −2 general solution is x2 1 + x3 0 , and a basis for the eigenspace is {v1 , v 2= } 0 1 0 0 2 1 For λ = 4: A − 4 I = 0 0 −2 4 , and row reducing [ A − 4 I 1 0 0 0 1 1/ 2 0 0] yields 0 0 0 2 0 0 −2 0 0 , 1 1 0 0 1 0 −1 −1/ 2 general solution is x3 1 , and a nice basis vector for the eigenspace is v 3 = 2 . 0 0 Copyright © 2016 Pearson Education, Ltd. 0 0 . The 0 . 0 0 . The 0 5.3 • Solutions −2 0 −1 5 From v1 , v 2 and v 3= construct P = v1 v2 v3 0 1 2 . Then set D = 0 1 0 0 0 the eigenvalues in D correspond to v1 , v 2 and v 3 respectively. 5-21 0 0 , where 4 0 5 0 15. The eigenvalues of A are given to be 3 and 1. 4 2 For λ = 3: A − 3I = −2 4 2 −2 1 0] yields 0 0 16 8 , and row reducing [ A − 3I −8 1 0 0 −1 −4 general solution is x2 1 + x3 0 , and a basis for the eigenspace is {v1 , v 2= } 0 1 6 2 For λ = 1: A − I = −2 4 4 −2 −2 general solution is x3 −1 , 1 16 8 , and row reducing [ A − I −6 1 0] yields 0 0 −2 and a basis for the eigenspace is v 3 = −1 . 1 −1 From v1 , v 2 and v 3 construct = P v1 = v2 v3 1 0 −4 0 1 0 1 0 4 0 0 0 0 . The 0 −1 −4 1 , 0 0 1 0 0 . The 0 2 1 0 −2 3 −1 . Then set D = 0 1 0 0 3 0 0 0 , 1 where the eigenvalues in D correspond to v1 , v 2 and v 3 respectively. 16. The eigenvalues of A are given to be 2 and 1. −2 1 For λ = 2: A − 2 I =− 1 −4 −2 2 −6 −3 , and row reducing [ A − 2 I 3 1 0] yields 0 0 2 0 0 −2 −3 general solution is x2 1 + x3 0 , and a basis for the eigenspace is {v1 , v 2= } 0 1 −1 For λ = 1: A − I = −1 1 −4 −1 2 −6 −3 , and row reducing [ A − I 4 1 0] yields 0 0 −2 general solution is x3 −1 , and a basis for the eigenspace is v 3 = 1 −2 −1 . 1 Copyright © 2016 Pearson Education, Ltd. 0 1 0 3 0 0 0 0 . The 0 −2 −3 1 , 0 0 1 2 1 0 . 0 0 . The 0 5-22 CHAPTER 5 • Eigenvalues and Eigenvectors 2 0 0 −2 −3 −2 From v1 , v 2 and v 3 construct P v1 = v2 v3 1 = 0 −1 . Then set D = 0 2 0 , 0 0 0 1 1 1 where the eigenvalues in D correspond to v1 , v 2 and v 3 respectively. 17. Since A is triangular, its eigenvalues are obviously 4 and 5. 1 0 0 0 0 0 0 For λ = 4: A − 4 I = 1 0 0 , and row reducing [ A − 4 I 0] yields 0 0 1 0 . The general 0 0 1 0 0 0 0 0 0 solution is x2 1 , and a basis for the eigenspace is v1 = 1 . 0 0 Since λ = 5 must have only a one-dimensional eigenspace, we can find at most 2 linearly independent eigenvectors for A, so A is not diagonalizable. −2 18. An eigenvalue of A is given to be 5; an eigenvector v1 = 1 is also given. To find the eigenvalue 2 4 −2 6 −7 −16 3 = corresponding to v1 , compute Av1 = 6 −3v1. Thus the eigenvalue in 13 −2 1 =− 12 16 1 2 − 6 question is −3. 4 −12 −16 For λ = 5:= 8 −2 , and row reducing [ A − 5 I 0] yields A − 5I 6 12 16 −4 −4 / 3 1/ 3 1 4 / 3 −1/ 3 0 0 0 0 0 . The general solution is x2 1 + x3 0 , and a nice basis for the 0 1 0 0 0 0 eigenspace is { v 2 , v 3= } −4 1 3 , 0 0 3 . −2 From v1 , v 2 and v 3 construct = P = v1 v 2 v 3 1 2 −4 3 0 1 −3 0 . Then set D = 0 0 3 0 5 0 0 0 , where 5 the eigenvalues in D correspond to v1 , v 2 and v 3 respectively. Note that this answer differs from the text. There, P = v 2 v 3 v1 and the entries in D are rearranged to match the new order of the eigenvectors. According to the Diagonalization Theorem, both answers are correct. Copyright © 2016 Pearson Education, Ltd. 5.3 • Solutions 5-23 19. Since A is triangular, its eigenvalues are obviously 2, 3, and 5. 3 −3 0 1 For λ = 2: A − 2 I = 0 0 0 0 9 1 0 −2 , and row reducing [ A − 2 I 0] yields 0 0 0 0 −1 −1 −1 2 The general solution is x3 + x4 , and a nice basis for the eigenspace is 1 0 0 1 {v1 , v 2= } −1 −1 −1 2 , 1 0 0 1 −3/ 2 0 0 0 0 1 0 0 0 0 1 0 0 1 0 0 1 1 0 0 1 −2 0 0 0 0 . 0 0 . 2 −3 0 0 For λ = 3: A − 3I = 0 0 0 0 1 0 0 0 0 1 0 0 0 0 . 0 0 9 −2 , and row reducing [ A − 3I 0] yields 0 −1 3/ 2 1 The general solution is x2 , and a nice basis for the eigenspace is 0 0 0 1 −1 0 3 2 v3 = . 0 0 0 −3 0 −2 For λ = 5: A − 5 I = 0 0 0 0 1 0 The general solution is x1 , 0 0 9 −2 , and row reducing [ A − 5 I 0] yields 0 −3 1 0 and a basis for the eigenspace is v 4 = . 0 0 0 1 −3 0 Copyright © 2016 Pearson Education, Ltd. 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 . 0 0 5-24 CHAPTER 5 • Eigenvalues and Eigenvectors −1 −1 From v1 , v 2 , v 3 and v 4 = construct P = v1 v 2 v 3 v 4 1 0 2 0 D= 0 0 −1 2 0 1 1 0 . Then set 0 0 3 2 0 0 0 0 , where the eigenvalues in D correspond to v1 , v 2 , v 3 and v 4 respectively. Note 0 5 that this answer differs from the text. There, P = [ v 4 v 3 v1 v 2 ] and the entries in D are rearranged to match the new order of the eigenvectors. According to the Diagonalization Theorem, both answers are correct. 0 2 0 0 0 0 3 0 20. Since A is triangular, its eigenvalues are obviously 4 and 2. 0 0 For λ = 4: A − 4 I = 0 1 0 0 0 0 0 0 −2 0 0 1 0 0 , and row reducing [ A − 4 I 0] yields 0 0 −2 0 0 0 0 0 0 2 1 0 The general solution is x2 + x4 , and a basis for the eigenspace is { v1 , v 2 }= 0 0 0 1 2 0 For λ = 2: A − 2 I = 0 1 0 2 0 0 0 0 0 0 0 1 0 0 , and row reducing [ A − 2 I 0] yields 0 0 0 0 0 0 0 0 general solution is x3 + x4 , and a basis for the eigenspace is {v 3 , v 4 }= 1 0 0 1 0 1 construct P = From v1 , v 2 , v 3 and v 4 = v1 v 2 v 3 v 4 0 0 4 0 D= 0 0 0 4 0 0 0 0 2 0 2 0 0 1 0 0 1 0 0 1 0 0 −2 0 0 0 0 1 0 0 0 0 0 0 0 2 1 0 , 0 0 0 1 0 0 0 0 0 0 0 0 , 1 0 0 1 . 0 0 . Then set 0 1 0 0 , where the eigenvalues in D correspond to v1 , v 2 and v 3 respectively. 0 2 Copyright © 2016 Pearson Education, Ltd. 0 0 . 0 0 . 0 0 . The 0 0 5.3 • Solutions 5-25 21. a. False. The symbol D does not automatically denote a diagonal matrix. b. True. See the remark after the statement of the Diagonalization Theorem. c. False. The 3 × 3 matrix in Example 4 has 3 eigenvalues, counting multiplicities, but it is not diagonalizable. d. False. Invertibility depends on 0 not being an eigenvalue. (See the Invertible Matrix Theorem.) A diagonalizable matrix may or may not have 0 as an eigenvalue. See Examples 3 and 5 for both possibilities. 22. a. False. The n eigenvectors must be linearly independent. See the Diagonalization Theorem. b. False. The matrix in Example 3 is diagonalizable, but it has only 2 distinct eigenvalues. (The statement given is the converse of Theorem 6.) c. True. This follows from AP = PD and formulas (1) and (2) in the proof of the Diagonalization Theorem. d. False. See Example 4. The matrix there is invertible because 0 is not an eigenvalue, but the matrix is not diagonalizable. 23. A is diagonalizable because you know that five linearly independent eigenvectors exist: three in the three-dimensional eigenspace and two in the two-dimensional eigenspace. Theorem 7 guarantees that the set of all five eigenvectors is linearly independent. 24. No, by Theorem 7(b). Here is an explanation that does not appeal to Theorem 7: Let v1 and v 2 be eigenvectors that span the two one-dimensional eigenspaces. If v is any other eigenvector, then it belongs to one of the eigenspaces and hence is a multiple of either v1 or v 2 . So there cannot exist three linearly independent eigenvectors. By the Diagonalization Theorem, A cannot be diagonalizable. 25. Let {v1} be a basis for the one-dimensional eigenspace, let v 2 and v 3 form a basis for the twodimensional eigenspace, and let v 4 be any eigenvector in the remaining eigenspace. By Theorem 7, {v1 , v 2 , v 3 , v 4 } is linearly independent. Since A is 4 × 4, the Diagonalization Theorem shows that A is diagonalizable. 26. Yes, if the third eigenspace is only one-dimensional. In this case, the sum of the dimensions of the eigenspaces will be six, whereas the matrix is 7 × 7. See Theorem 7(b). An argument similar to that for Exercise 24 can also be given. 27. If A is diagonalizable, then A = PDP −1 for some invertible P and diagonal D. Since A is invertible, 0 is not an eigenvalue of A. So the diagonal entries in D (which are eigenvalues of A) are not zero, and D is invertible. By the theorem on the inverse of a product, −1 −1 −1 = A−1 (= PDP −1 ) −1 ( P −1 )= D P PD −1 P −1 . Since D −1 is obviously diagonal, A−1 is diagonalizable. 28. If A has n linearly independent eigenvectors, then by the Diagonalization Theorem, A = PDP −1 for some invertible P and diagonal D. Using properties of transposes, T −1 = AT (= PDP −1 )T ( P −1= )T D T PT ( P= ) DPT QDQ −1 , where Q = ( PT ) −1. Thus AT is diagonalizable. By the Diagonalization Theorem, the columns of Q are n linearly independent eigenvectors of AT. Copyright © 2016 Pearson Education, Ltd. 5-26 CHAPTER 5 • Eigenvalues and Eigenvectors 29. The diagonal entries in D1 are reversed from those in D. So interchange the (eigenvector) columns of 1 1 P to make them correspond properly to the eigenvalues in D1. In this case, P1 = and −2 −1 3 0 D1 = . Although the first column of P must be an eigenvector corresponding to the 0 5 −3 1 eigenvalue 3, there is nothing to prevent us from selecting some multiple of , say , and 6 −2 1 −3 letting P2 = . We now have three different factorizations or “diagonalizations” of A: 6 −1 −1 = A PDP = P1= D1 P1−1 P2 D1 P2−1 30. A nonzero multiple of an eigenvector is another eigenvector. To produce P2 , simply multiply one or both columns of P by a nonzero scalar unequal to 1. 31. For a 2 × 2 matrix A to be invertible, its eigenvalues must be nonzero. A first attempt at a 2 3 construction might be something such as , whose eigenvalues are 2 and 4. Unfortunately, 0 4 a 2 × 2 matrix with two distinct eigenvalues is diagonalizable (Theorem 6). So, adjust the 2 3 a b construction to , which works. In fact, any matrix of the form has the desired 0 2 0 a properties when a and b are nonzero. The eigenspace for the eigenvalue a is one-dimensional, as a simple calculation shows, and there is no other eigenvalue to produce a second eigenvector. 32. Any 2 × 2 matrix with two distinct eigenvalues is diagonalizable, by Theorem 6. If one of those a b eigenvalues is zero, then the matrix will not be invertible. Any matrix of the form has the 0 0 desired properties when a and b are nonzero. The number a must be nonzero to make the matrix 0 0 diagonalizable; b must be nonzero to make the matrix not diagonal. Other solutions are and a b 0 0 a . b −6 −3 33. A = −1 −4 4 0 −2 4 0 1 1 0 9 6 , ev = eig(A)=(5,1,-2,-2). 0 7 1.0000 0.5000 , A basis for the eigenspace of λ =5 is nulbasis(A-ev(1)*eye(4)) = −0.5000 1.0000 Copyright © 2016 Pearson Education, Ltd. 2 1 . −1 2 5.3 • Solutions 5-27 2 1.0000 −0.5000 , A basis for the eigenspace of λ =1 is −1 . nulbasis(A-ev(2)*eye(4)) = −7 −3.5000 2 1.0000 1.0000 1.5000 1.0000 −0.7500 , , A basis for the eigenspace of nulbasis(A-ev(3)*eye(4)) = 1.0000 0 0 1.0000 1 6 1 −3 λ = −2 is , . 1 0 0 4 2 1 Thus we construct P = −1 2 0 4 34. A = 8 0 13 8 9 8 6 12 5 0 2 −1 −7 2 1 1 1 0 6 5 0 −3 and D = 0 0 4 0 0 1 0 0 0 0 −2 0 0 0 . 0 −2 4 4 , ev = eig(A) =(-4,24,1,-4). 8 −4 −2 −1 0 0 nulbasis(A-ev(1)*eye(4)) = , . 1 0 0 1 −2 −1 0 0 λ = −4 is , . 1 0 0 1 A basis for the eigenspace of 5.6000 5.6000 . nulbasis(A-ev(2)*eye(4)) = 7.2000 1.0000 28 28 A basis for the eigenspace of λ =24 is . 36 5 1.0000 1.0000 . nulbasis(A-ev(3)*eye(4)) = −2.0000 1.0000 1 1 A basis for the eigenspace of λ =1 is . −2 1 Copyright © 2016 Pearson Education, Ltd. 5-28 CHAPTER 5 • Eigenvalues and Eigenvectors −2 0 Thus we construct P = 1 0 11 −3 35. A = −8 1 8 −6 5 12 6 −18 4 −2 −3 −2 8 −10 4 12 3 −14 −1 0 0 1 28 28 36 5 1 −4 0 1 and D = 0 −2 1 0 0 −4 0 0 0 0 24 0 0 0 . 0 1 −4 1 4 , ev = eig(A) =(5,1,3,5,1). −1 −1 2.0000 1.0000 −0.3333 −0.3333 nulbasis(A-ev(1)*eye(5)) = −1.0000 , −1.0000 . 0 1.0000 0 1.0000 6 3 −1 −1 = λ 5 is −3 , −3 . 3 0 0 3 0.8000 0.6000 −0.6000 −0.2000 nulbasis(A-ev(2)*eye(5)) = −0.4000 , −0.8000 . 0 1.0000 0 1.0000 4 3 −3 −1 = λ 1 is −2 , −4 . 5 0 0 5 A basis for the eigenspace of A basis for the eigenspace of Copyright © 2016 Pearson Education, Ltd. 5.3 0.5000 −0.2500 nulbasis(A-ev(3)*eye(5)) = −1.0000 . −0.2500 1.0000 6 −1 Thus we construct P = −3 3 0 4 0 36. A = 6 9 15 4 1 12 20 28 2 −2 11 10 14 3 −2 2 10 5 3 −1 −3 0 3 −2 2 −4 , −6 −3 4 −3 −2 5 0 3 −1 −4 0 5 • Solutions 5-29 2 −1 A basis for the eigenspace of λ =3 is −4 . −1 4 2 5 0 −1 −4 and D = 0 −1 0 0 4 0 5 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 . 0 3 ev = eig(A) =(3,5,7,5,3). 2.0000 −1.0000 −1.5000 0.5000 nulbasis(A-ev(1)*eye(5)) = 0.5000 , 0.5000 . 0 1.0000 0 1.0000 4 −2 −3 1 = λ 3 is 1 , 1 . 2 0 0 2 0 −1.0000 −0.5000 1.0000 nulbasis(A-ev(2)*eye(5)) = 1.0000 , 0 . 0 −1.0000 0 1.0000 A basis for the eigenspace of A basis for the eigenspace of 0 −1 −1 1 = λ 5 is 2 , 0 . 0 −1 0 1 Copyright © 2016 Pearson Education, Ltd. 5-30 CHAPTER 5 • Eigenvalues and Eigenvectors 0.3333 0.0000 nulbasis(A-ev(3)*eye(5)) = 0.0000 . 1.0000 1.0000 4 −3 Thus we construct P = 1 2 0 −2 1 1 0 2 0 −1 2 0 0 −1 1 0 −1 1 1 0 A basis for the eigenspace of λ =7 is 0 . 3 3 1 3 0 0 0 and D = 0 3 0 0 3 0 3 0 0 0 0 0 5 0 0 0 0 0 5 0 0 0 0 . 0 7 . Notes: For your use, here is another matrix with five distinct real eigenvalues. To four decimal places, they are 11.0654, 9.8785, 3.8238, −3.7332, and −6.0345. 5 −3 0 6 −8 −7 3 −5 3 0 −3 −7 5 −3 5 . 1 −7 5 0 −4 −5 −3 −2 0 8 The MATLAB box in the Study Guide encourages students to use eig (A) and nulbasis to practice the diagonalization procedure in this section. It also remarks that in later work, a student may automate the process, using the command [P D] = eig (A). You may wish to permit students to use the full power of eig in some problems in Sections 5.5 and 5.7. 5.4 SOLUTIONS 3 −1 1. Since T ( b1 ) = 3d1 − 5d 2 , [T ( b1 )] = . Likewise T (b 2 ) = −d1 + 6d 2 implies that [T ( b 2 )] = −5 6 0 and T (b3 ) = 4d 2 implies that [T ( b3 )] = . Thus the matrix for T relative to and is 4 [T ( b )] [T ( b )] [T ( b )] 1 2 3 3 = −5 −1 6 0 . 4 2 2. Since T (d1 ) = 2b1 − 3b 2 , [T (d1 )] = . Likewise T (d 2 ) = −4b1 + 5b 2 implies that −3 −4 2 [T (d 2 )] = . Thus the matrix for T relative to and is [T (d1 )] [T (d 2 )] = 5 −3 3. a. T (e1 ) = 0b1 − 1b 2 + b3 , T (e 2 ) = −1b1 − 0b 2 − 1b3 , T (e3 ) = 1b1 − 1b 2 + 0b3 Copyright © 2016 Pearson Education, Ltd. −4 . 5 5.4 • Solutions 5-31 0 −1 1 b. [T (e1 )] = −1 , [T (e 2 )] = 0 , [T (e 3 )] = −1 1 − 0 1 c. The matrix for T relative to E and is [ [T (e1 )] [T (e 2 )] 0 −1 −1 [T (e 3 )] ] = 0 1 −1 1 −1 . 0 2 −4 4. Let E = {e1 , e 2 } be the standard basis for 2 . Since [T (b1 )]E = T (b1 ) = T (b 2 ) = 0 , [T (b 2 )]E = −1 , 5 and [T (b= (b3 ) , the matrix for T relative to and T= 3 )]E 3 2 E is [[T (b1 )]E [T (b 2 )]E [T (b3 )]E ] = 0 −4 −1 5 . 3 5. a. T (p) = (t + 5)(2 − t + t 2 ) = 10 − 3t + 4t 2 + t 3 b. Let p and q be polynomials in P2, and let c be any scalar. Then T (p(t ) + q(t )) = (t + 5)[p(t ) + q(t )] = (t + 5)p(t ) + (t + 5)q(t ) = T (p(t )) + T (q(t )) T (c ⋅ p(t )) = (t + 5)[c ⋅ p(t )] = c ⋅ (t + 5)p(t ) = c ⋅ T [p(t )] and T is a linear transformation. 5 1 c. Let = {1, t, t 2 } and = {1, t, t 2, t 3}. Since T ( b1 ) = T (1) = (t + 5)(1) = t + 5, [T ( b1 )] = . 0 0 0 5 Likewise since T ( b 2 ) = T (t ) = (t + 5)(t ) = t 2 + 5t, [T ( b 2 )] = , and since 1 0 0 0 2 2 3 2 T ( b3 ) =T (t ) =(t + 5)(t ) =t + 5t , [T ( b3 )] = . Thus the matrix for T relative to and 5 1 is [ [T ( b1 )] [T ( b 2 )] 5 1 [T ( b3 )] ] = 0 0 0 5 1 0 0 0 . 5 1 Copyright © 2016 Pearson Education, Ltd. 5-32 CHAPTER 5 • Eigenvalues and Eigenvectors 6. a. T (p) = (2 − t + t 2 ) + t 2 (2 − t + t 2 ) = 2 − t + 3t 2 − t 3 + t 4 b. Let p and q be polynomials in 2 , and let c be any scalar. Then T (p(t ) + q(t )) = [p(t ) + q(t )] + t 2 [p(t ) + q(t )] = [p(t ) + t 2p(t )] + [q(t ) + t 2q(t )] = T (p(t )) + T (q(t )) T (c ⋅ p(t )) = [c ⋅ p(t )] + t 2 [c ⋅ p(t )] = c ⋅ [p(t ) + t 2p(t )] = c ⋅ T [p(t )] and T is a linear transformation. 1 0 c. Let = {1, t, t 2 } and = {1, t, t 2, t 3, t 4 } . Since T ( b1 ) = T (1) = 1 + t 2 (1) = t 2 + 1, [T ( b1 )] = 1 . 0 0 0 1 Likewise since T ( b 2 ) = T (t ) = t + (t 2 )(t ) = t 3 + t, [T ( b 2 )] = 0, and 1 0 0 0 since T ( b3 ) = T (t 2 ) = t 2 + (t 2 )(t 2 ) = t 4 + t 2 , [T ( b3 )] = 1 . Thus the matrix for T relative to 0 1 1 0 0 0 1 0 and [ [T ( b1 )] [T ( b 2 )] [T ( b3 )] ] = 1 0 1 . 0 1 0 0 0 1 3 7. Since T ( b1 ) = T (1) = 3 + 5t, [T ( b1 )] = 5 . Likewise since 0 0 0 2 2 0 . Thus the T (t ) = t , [T ( b3 )] = T (b2 ) = T (t ) = −2t + 4t , [T ( b 2 )] =−2 , and since T ( b3 ) = 1 4 2 Copyright © 2016 Pearson Education, Ltd. 5.4 • Solutions 5-33 0 0 3 matrix representation of T relative to the basis is [T ( b1 )] [T ( b 2 )] [T ( b3 = )] 5 −2 0 . 4 1 0 Perhaps a faster way is to realize that the information given provides the general form of T (p) as shown in the figure below: T a0 + a1t + a2t 2 → 3a0 + (5a0 − 2a1 )t + (4a1 + a2 )t 2 coordinate mapping coordinate mapping a 0 a1 a 2 3a 0 → 5a0 −2a1 by[T ] 4a +a 1 2 multiplication The matrix that implements the multiplication along the bottom of the figure is easily filled in by inspection: ? ? ? ? ? ? ? a0 ? a1 = ? a2 3a 0 − 5 a 2 a 0 1 4a +a 1 2 3 implies that [T ] = 5 0 0 −2 4 0 0 1 0 3 8. Since [3b1 − 4b 2 ] = =[T ] [3b1 − 4b 2 ] =0 −4 , [T (3b1 − 4b 2 )]777 1 0 T (3b1 − 4b 2 ) = 24b1 − 20b 2 + 11b3 . −6 5 −2 1 3 24 −1 −4 = −20 and 7 0 11 5 + 3(−1) 2 5 + 3(0) = 9. a. T (p) = 5 5 + 3(1) 8 b. Let p and q be polynomials in 2 , and let c be any scalar. Then (p + q)(−1) T (p + q) = (p + q)(0) = (p + q)(1) p(−1) + q(−1) p(0) + q(0) = p(1) + q(1) p(−1) q(−1) p(0) + q(0) = T (p) + T (q) p(1) q(1) p(−1) (c ⋅ p)(−1) c ⋅ (p(−1)) T (c ⋅ p) = (c ⋅ p)(0) = c ⋅ (p(0)) =c ⋅ p(0) =c ⋅ T (p) p(1) (c ⋅ p)(1) c ⋅ (p(1)) and T is a linear transformation. c. Let = {1, t, t 2 } and = {e1, e 2 , e 3} be the standard basis for 3 . Since 1 −1 [T (b1 )]E = T (b1 ) == T (1) 1 , [T (b 2 )]E = T (b 2 ) == T (t ) 0 , and 1 1 Copyright © 2016 Pearson Education, Ltd. 5-34 CHAPTER 5 • Eigenvalues and Eigenvectors 1 [T (b= T= (b3 ) T= (t ) 0 , the matrix for T relative to and is 3 )]E 1 1 −1 1 [T (b )] 0 0 . 1 E [T (b 2 )]E [T (b 3 )]E = 1 1 1 1 2 10. a. Let p and q be polynomials in 3 , and let c be any scalar. Then (p + q)(−3) (p + q)(−1) = T (p + q) = (p + q)(1) (p + q)(3) p(−3) + q(−3) p(−3) q(−3) p(−1) + q(−1) p(−1) q(−1) = + = T (p) + T (q) p(1) + q(1) p(1) q(1) p(3) + q(3) p(3) q(3) (c ⋅ p)(−3) c ⋅ (p(−3)) p(−3) (c ⋅ p)(−1) c ⋅ (p(−1)) = =c ⋅ p(−1) =c ⋅ T (p) T (c ⋅ p) = (c ⋅ p)(1) c ⋅ (p(1)) p(1) (c ⋅ p)(3) c ⋅ (p(3)) p(3) and T is a linear transformation. b. Let = {1, t, t 2 , t 3} and = {e1, e 2 , e 3 , e 4 } be the standard basis for 4 . Since 1 −3 9 1 −1 1 , [T (b 2 )]E = , [T (b3 )]E = , T (b1 ) = T (1) = T (b 2 ) = T (t ) = T (b3 ) = T (t 2 ) = [T (b1 )]E = 1 1 1 1 3 9 −27 −1 3 , the matrix for T relative to and E is and [T (b= T b T t = = )] ( ) ( ) 4 E 4 1 27 1 −3 9 −27 1 −1 1 −1 [T (b )] . 1 E [T (b 2 )]E [T (b 3 )]E [T (b 4 )]E = 1 1 1 1 3 9 27 1 11. Following Example= 4, if P = P −1 AP b b2 = 1 2 −1 1 , then the -matrix is 2 4 2 1 1 1 2 −1 3 = 2 −1 −1 −1 2 0 5 1 5 1 Copyright © 2016 Pearson Education, Ltd. 5.4 12. Following Example= 4, if P = P −1 AP b = b2 1 3 2 • Solutions 5-35 −1 , then the -matrix is 1 1 1 1 −1 4 3 −1 1 = 1 −2 5 −2 3 −2 3 2 2 1 13. Start by diagonalizing A. The characteristic polynomial is λ 2 − 4λ + 3 = (λ − 1)(λ − 3), so the eigenvalues of A are 1 and 3. −1 For λ = 1: A − I = −3 1 . The equation ( A − I )x = 0 amounts to − x1 + x2 = 0, so x1 = x2 with x2 3 1 free. A basis vector for the eigenspace is thus v1 = . 1 −3 1 For λ = 3: A − 3I = 0 amounts to −3 x1 + x2 = 0, so x1= (1/ 3) x2 −3 1 . The equation ( A − 3I )x = 1 with x2 free. A nice basis vector for the eigenspace is thus v 2 = . 3 1 1 1 3 which diagonalizes A. By Theorem 8, the basis = {v1, v 2 } has the property that the -matrix of the transformation x Ax is a diagonal matrix. From v1 and v 2 we may construct P = = v2 v1 14. Start by diagonalizing A. The characteristic polynomial is λ 2 − 6λ − 16 = (λ − 8)(λ + 2), so the eigenvalues of A are 8 and −2. −3 For λ = 8: A − 8 I = −7 −3 0 amounts to x1 + x2 = 0, so x1 = − x2 with . The equation ( A − 8 I )x = −7 −1 x2 free. A basis vector for the eigenspace is thus v1 = . 1 7 For λ = − 2: A + 2 I = −7 −3 0 amounts to 7 x1 − 3 x2 = 0, so . The equation ( A + 2 I )x = 3 3 x1= (3/ 7) x2 with x2 free. A nice basis vector for the eigenspace is thus v 2 = . 7 −1 3 1 7 which diagonalizes A. By Theorem 8, the basis = {v1, v 2 } has the property that the -matrix of the transformation x Ax is a diagonal matrix. From v1 and v 2 we may construct = P = v v2 1 Copyright © 2016 Pearson Education, Ltd. 5-36 CHAPTER 5 • Eigenvalues and Eigenvectors 15. Start by diagonalizing A. The characteristic polynomial is λ 2 − 7λ + 10 = (λ − 5)(λ − 2), so the eigenvalues of A are 5 and 2. −1 For λ = 5: A − 5 I = −1 −2 . The equation ( A − 5 I )x = 0 amounts to x1 + 2 x2 = 0, so x1 = −2 x2 −2 −2 with x2 free. A basis vector for the eigenspace is thus v1 = . 1 2 For λ = 2: A − 2 I = −1 −2 0 amounts to x1 − x2 = 0, so x1 = x2 with . The equation ( A − 2 I )x = 1 1 x2 free. A basis vector for the eigenspace is thus v 2 = . 1 −2 1 1 1 which diagonalizes A. By Theorem 8, the basis = {v1, v 2 } has the property that the -matrix of the transformation x Ax is a diagonal matrix. From v1 and v 2 we may construct = P = v v2 1 16. Start by diagonalizing A. The characteristic polynomial is λ 2 − 5λ = λ(λ − 5), so the eigenvalues of A are 5 and 0. −3 For λ = 5: A − 5 I = −1 −6 0 amounts to x1 + 2 x2 = 0, so x1 = −2 x2 . The equation ( A − 5 I )x = −2 −2 with x2 free. A basis vector for the eigenspace is thus v1 = . 1 2 For λ = 0: A − 0 I = −1 −6 0 amounts to x1 − 3 x2 = 0, so x1 = 3 x2 . The equation ( A − 0 I )x = 3 3 with x2 free. A basis vector for the eigenspace is thus v 2 = . 1 −2 3 1 1 which diagonalizes A. By Theorem 8, the basis = {v1, v 2 } has the property that the -matrix of the transformation x Ax is a diagonal matrix. From v1 and v 2 we may construct = P = v2 v1 1 1 1 2 17. a. We compute that A= b1 = = 2b1 . so b1 is an eigenvector of A corresponding to −1 3 1 2 the eigenvalue 2. The characteristic polynomial of A is λ 2 − 4λ + 4 = (λ − 2) 2 , so 2 is the only −1 1 eigenvalue for A. Now A − 2 I = −1 1 , which implies that the eigenspace corresponding to the eigenvalue 2 is one-dimensional. Thus the matrix A is not diagonalizable. Copyright © 2016 Pearson Education, Ltd. 5.4 b. Following Example 4, if P = b1 −4 −1 P= AP 1 5 1 −1 −1 • Solutions 5-37 b2 , then the -matrix for T is 1 1 5 1 5 2 = = 3 1 4 1 4 0 −1 . 2 18. If there is a basis B such that [T ] is diagonal, then A is similar to a diagonal matrix, by the second paragraph following Example 3. For this to happen, A would have three linearly independent eigenvectors. However, this is not necessarily the case, because A has only two distinct eigenvalues. 19. If A is similar to B, then there exists an invertible matrix P such that P −1 AP = B. Thus B is invertible because it is the product of invertible matrices. By a theorem about inverses of products, −1 −1 = B −1 P= A ( P −1 ) −1 P −1 A−1 P, which shows that A−1 is similar to B −1. 20. If A = PBP −1 , then A2= ( PBP −1 )( PBP −1 )= PB ( P −1 P ) BP −1= PB ⋅ I ⋅ BP −1= PB 2 P −1. So A2 is similar to B 2. 21. By hypothesis, there exist invertible P and Q such that P −1 BP = A and Q −1CQ = A. Then P −1 BP = Q −1CQ. Left-multiply by Q and right-multiply by Q −1 to obtain −1 −1 QP −1 BPQ −1 = QQ −1CQQ= . So C QP = BPQ −1 ( PQ −1 ) −1 B ( PQ −1 ), which shows that B is similar to C. 22. If A is diagonalizable, then A = PDP −1 for some P. Also, if B is similar to A, then B = QAQ −1 for some Q. Then B Q= = ( PDP −1 )Q −1 (QP = ) D( P −1Q −1 ) (QP) D(QP) −1 . So B is diagonalizable. 23. If Ax = λx, x ≠ 0, then P −1 Ax = λP −1x. If B = P −1 AP, then −1 B (= P −1x) P −1 AP (= P −1x) P= Ax λP −1x by the first calculation. Note that P −1x ≠ 0, because x ≠ 0 and P −1 is invertible. Hence (*) shows that P −1x is an eigenvector of B corresponding to λ . (Of course, λ is an eigenvalue of both A and B because the matrices are similar, by Theorem 4 in Section 5.2.) 24. If A = PBP −1 , = then rank A rank P ( BP −1 ) rank BP −1 , by Supplementary Exercise 13 in Chapter 4. = Also, rank BP −1 = rank B, by Supplementary Exercise 14 in Chapter 4, since P −1 is invertible. Thus rank A = rank B. 25. If A = PBP −1 , then = tr( A) tr(( = PB ) P −1 ) tr( P −1 ( PB )) By the trace property −1 = tr( P = PB ) tr( = IB ) tr( B ) If B is diagonal, then the diagonal entries of B must be the eigenvalues of A, by the Diagonalization Theorem (Theorem 5 in Section 5.3). So tr= A tr= B {sum of the eigenvalues of A}. 26. If A = PDP −1 for some P, then the general trace property from Exercise 25 shows that = tr A tr= [( PD) P −1 ] tr [ P −1 PD] = tr D. (Or, one can use the result of Exercise 25 that since A is similar to D, tr A = tr D. ) Since the eigenvalues of A are on the main diagonal of D, tr D is the sum of the eigenvalues of A. Copyright © 2016 Pearson Education, Ltd. (*) 5-38 CHAPTER 5 • Eigenvalues and Eigenvectors 27. For each j, I (b j ) = b j . Since the standard coordinate vector of any vector in n is just the vector itself, [ I (b j )]ε = b j . Thus the matrix for I relative to and the standard basis E is simply b 1 b2 b n . This matrix is precisely the change-of-coordinates matrix P defined in Section 4.4. 28. For each j, I (b j ) = b j , and [ I ( b j )] = [b j ] . By formula (4), the matrix for I relative to the bases and is M = …[b1 ] … [b n ] . In Theorem 15 of Section 4.7, this matrix was denoted [b2 ] by P and was called the change-of-coordinates matrix from to . ← 29. If = {b1, … , b n }, then the -coordinate vector of b j is e j , the standard basis vector for n . For instance, b1 =1 ⋅ b1 + 0 ⋅ b 2 + … + 0 ⋅ b n . Thus [ I ( b= [b = e j , and j )] j ] = [ I ] [ I ( b )] 1 [ I= ( b n )] [e= I 1 en ] 30. [M] If P is the matrix whose columns come from then the -matrix of the transformation x Ax is D = P −1 AP. From the data in the text, −14 A = −33 11 2 D = −2 −1 4 9 −4 −1 1 0 −14 −31 , P =b1 11 1 −14 0 −33 −1 11 4 9 −4 b3 −1 = −2 1 −14 −1 −31 −2 11 1 −1 −1 1 b2 −1 −1 1 −1 −2 , 0 −1 8 −2 =0 0 0 3 1 0 −6 3 −3 31. [M] If P is the matrix whose columns come from , then the -matrix of the transformation x Ax is D = P −1 AP. From the data in the text, −7 = A 1 −3 −1 = D 1 0 −48 14 −45 −3 3 −1 −16 6 , P = −19 3 −3 −2 b 1 −1 , b= b3 1 2 1 −3 −3 0 3 −7 −1/ 3 −7 −48 −16 −3 −2 0 1 14 6 1 1 −= 1 0 0 0 −1/ 3 −3 −45 −19 −3 −3 −2 −4 0 Copyright © 2016 Pearson Education, Ltd. −6 −6 −1 5.4 15 0 32. [M] A = 1 2 −66 13 −15 −18 −44 21 −21 −22 • Solutions 5-39 −33 −15 , ev = eig(A) = (2, 4, 4, 5). 12 8 0.0000 −1.5000 . A basis for the eigenspace of λ = 2 is nulbasis(A-ev(1)*eye(4)) = 1.5000 1.0000 0 −3 b1 = . 3 2 −10.0000 13.0000 −2.3333 1.6667 , . nulbasis(A-ev(2)*eye(4)) = 1.0000 0 0 1.0000 −30 39 −7 5 , } λ = 4 is {b 2 , b= . 3 3 0 0 3 2.7500 −0.7500 . nulbasis(A-ev(4)*eye(4)) = 1.0000 1.0000 11 −3 b4 = . 4 4 A basis for the eigenspace of A basis for the eigenspace of λ = 5 is The basis = {b1, b 2 , b3 , b 4 } is a basis for 4 with the property that [T ] is diagonal. Note: The Study Guide comments on Exercise 25 and tells students that the trace of any square matrix A equals the sum of the eigenvalues of A, counted according to multiplicities. This provides a quick check on the accuracy of an eigenvalue calculation. You could also refer students to the property of the determinant described in Exercise 19 of Section 5.2. Copyright © 2016 Pearson Education, Ltd. 5-40 5.5 CHAPTER 5 • Eigenvalues and Eigenvectors SOLUTIONS 1 1. = A 1 −2 1 − λ , A− = λI 3 1 −2 . det( A − λI ) = (1 − λ)(3 − λ) − (−2) = λ 2 − 4λ + 5 . Use the 3− λ quadratic formula to find the eigenvalues: λ= 4 ± 16 − 20= 2 ± i. Example 2 gives a shortcut for 2 finding one eigenvector, and Example 5 shows how to write the other eigenvector with no effort. −1 − i −2 For λ = 2 + i: A − (2 + i ) I = . The equation ( A − λ I )x = 0 gives 1 1 − i (−1 − i ) x1 − 2 x2 =0 x1 + (1 − i ) x2 = 0 As in Example 2, the two equations are equivalent—each determines the same relation between x1 and x2 . So use the second equation to obtain x1 =−(1 − i ) x2 , with x2 free. The general solution is −1 + i −1 + i x2 , and the vector v1 = provides a basis for the eigenspace. 1 1 −1 − i For λ = 2 – i: Let v= v= 1 2 1 . The remark prior to Example 5 shows that v 2 is automatically an eigenvector for 2 + i. In fact, calculations similar to those above would show that {v 2 } is a basis for the eigenspace. (In general, for a real matrix A, it can be shown that the set of complex conjugates of the vectors in a basis of the eigenspace for λ is a basis of the eigenspace for λ .) 5 2. A = 1 −5 . The characteristic polynomial is λ 2 − 6λ + 10, so the eigenvalues of A are 1 λ= 6 ± 36 − 40 = 3 ± i. 2 −5 2 − i For λ = 3 + i: A − (3 + i ) I = 0 amounts to . The equation ( A − (3 + i ) I )x = 1 −2 − i (2 + i ) x2 with x2 free. A basis vector for the eigenspace is thus x1 + (−2 − i ) x2 =0, so x= 1 2 + i v1 = . 1 2 − i For λ = 3 – i: A basis vector for the eigenspace is v= v= 1 2 1 . 1 3. A = −2 5 . The characteristic polynomial is λ 2 − 4λ + 13, so the eigenvalues of A are 3 λ= 4 ± −36= 2 ± 3i. 2 Copyright © 2016 Pearson Education, Ltd. 5.5 • Solutions 5 −1 − 3i For λ = 2 + 3i: A − (2 + 3i ) I = 0 amounts to . The equation ( A − (2 + 3i ) I )x = −2 1 − 3i 1 − 3i x2 with x2 free. A nice basis vector for the eigenspace is thus −2 x1 + (1 − 3i ) x2 = 0, so x1 = 2 1 − 3i v1 = . 2 1 + 3i For λ = 2 – 3i: A basis vector for the eigenspace is v= v= 1 2 2 . 5 −2 2 4. A = . The characteristic polynomial is λ − 8λ + 17, so the eigenvalues of A are 1 3 λ= 8 ± −4= 4 ± i. 2 −2 1 − i For λ = 4 + i: A − (4 + i ) I = 0 amounts to . The equation ( A − (4 + i ) I )x = 1 −1 − i x1 + (−1 − i ) x2 =0, so x1= (1 + i ) x2 with x2 free. A basis vector for the eigenspace is thus 1 + i v1 = . 1 1 − i For λ = 4 – i: A basis vector for the eigenspace is v= v= 1 2 1 . 0 5. A = −8 1 . The characteristic polynomial is λ 2 − 4λ + 8, so the eigenvalues of A are 4 λ= 4 ± −16= 2 ± 2i. 2 1 −2 − 2i For λ = 2 + 2i: A − (2 + 2i ) I = 0 amounts to . The equation ( A − (2 + 2i ) I )x = −8 2 − 2i (−2 − 2i ) x1 + x2 =0, so x= (2 + 2i ) x1 with x1 free. A basis vector for the eigenspace is thus 2 1 v1 = . 2 + 2i 1 For λ = 2 – 2i: A basis vector for the eigenspace is v= v= 1 2 2 − 2i . 4 6. A = −3 3 . The characteristic polynomial is λ 2 − 8λ + 25, so the eigenvalues of A are 4 λ= 8 ± −36= 4 ± 3i. 2 Copyright © 2016 Pearson Education, Ltd. 5-41 5-42 CHAPTER 5 • Eigenvalues and Eigenvectors −3i For λ = 4 + 3i: A − (4 + 3i ) I = −3 3 0 amounts to x1 + ix2 = 0, . The equation ( A − (4 + 3i ) I )x = −3i −i so x1 = −ix2 with x2 free. A basis vector for the eigenspace is thus v1 = . 1 i For λ = 4 – 3i: A basis vector for the eigenspace is v= v= 1 2 1 . 7. A = 3 −1 . 3 1 From Example 6, the eigenvalues are 3 ± i. The scale factor for the transformation x Ax is r =| λ |= ( 3) 2 + 12 = 2. For the angle of rotation, plot the point (a, b) = ( 3,1) in the xy-plane and use trigonometry: arctan (1/ 3) = π/ 6 radians. ϕ = arctan (b/a) = Note: Your students will want to know whether you permit them on an exam to omit calculations for a a matrix of the form b −b and simply write the eigenvalues a ± bi. A similar question may arise about a 1 1 the corresponding eigenvectors, and , which are announced in the Practice Problem. Students −i i may have trouble keeping track of the correspondence between eigenvalues and eigenvectors. 8. A = 3 . 3 3 −3 From Example 6, the eigenvalues are 3 ± 3i. The scale factor for the transformation x Ax is r = | λ | = ( 3) 2 + 32 = 2 3. From trigonometry, the angle of rotation ϕ is arctan (b/a ) = arctan (−3/ 3) = −π/ 3 radians. − 3/ 2 9. A = −1/ 2 1/ 2 . From Example 6, the eigenvalues are − 3/ 2 ± (1/ 2)i. The scale factor for the − 3/ 2 transformation x Ax is r = | λ | = (− 3/ 2) 2 + (1/ 2) 2 = 1. From trigonometry, the angle of rotation arctan ((−1/ 2) / (− 3/ 2)) = −5π/ 6 radians. ϕ is arctan (b/a) = Copyright © 2016 Pearson Education, Ltd. 5.5 −5 10. A = 5 • Solutions 5-43 −5 . From Example 6, the eigenvalues are −5 ± 5i. The scale factor for the transformation −5 x Ax is r = | λ | = (−5) 2 + 52 = 5 2. From trigonometry, the angle of rotation ϕ is arctan(b/a ) = arctan(5/ (−5)) = 3π/ 4 radians. .1 11. A = −.1 .1 . From Example 6, the eigenvalues are .1 ± .1i. The scale factor for the transformation .1 x Ax is r = | λ | = (.1) 2 + (.1) 2 = 2 /10. From trigonometry, the angle of rotation ϕ is arctan (b/a ) = arctan (−.1/.1) = −π/ 4 radians. 0 12. A = −.3 .3 . From Example 6, the eigenvalues are 0 ± .3i. The scale factor for the transformation 0 x Ax is r =| λ |= 02 + (.3) 2 = .3. From trigonometry, the angle of rotation ϕ is arctan (b/a ) = arctan (−∞) = −π/ 2 radians. −1 − i 13. From Exercise 1, λ= 2 ± i, and the eigenvector v = corresponds to λ= 2 − i. Since 1 −1 −1 −1 −1 Re v = and Im v = , take P = . Then compute 0 1 1 0 1 1 −2 −1 −1 0 1 −3 −1 2 −1 0 −1 . Actually, Theorem 9 C P= AP = = = 3 1 0 −1 −1 2 −1 1 2 −1 −1 1 gives the formula for C. Note that the eigenvector v corresponds to a − bi instead of a + bi. If, for 2 1 instance, you use the eigenvector for 2 + i, your C will be . −1 2 Notes: The Study Guide points out that the matrix C is described in Theorem 9 and the first column of C is the real part of the eigenvector corresponding to a − bi, not a + bi, as one might expect. Since students may forget this, they are encouraged to compute C from the formula C = P −1 AP, as in the solution above. The Study Guide also comments that because there are two possibilities for C in the factorization of a 2 × 2 matrix as in Exercise 13, the measure of rotation of the angle associated with the transformation x Ax is determined only up to a change of sign. The “orientation” of the angle is determined by the change of variable x = Pu. See Figure 4 in the text. 5 −5 14. A = . From Exercise 2, the eigenvalues of A are λ= 3 ± i, and the eigenvector 1 1 2 −1 2 − i corresponds to λ= 3 − i. By Theorem 9, P [Re and v Im v ] = = v= 0 1 1 0 −1 = C P= AP −1 1 5 2 1 −5 2 −1 3 = 1 1 0 1 −1 3 Copyright © 2016 Pearson Education, Ltd. 5-44 CHAPTER 5 • Eigenvalues and Eigenvectors 1 5 15. A = . From Exercise 3, the eigenvalues of A are λ= 2 ± 3i, and the eigenvector −2 3 1 + 3i 1 3 corresponds to λ= 2 − 3i. By Theorem 9, P [Re v= = = v Im v ] and 2 2 0 1 0 −3 1 5 1 3 2 −3 − = C= P −1 AP = 1 −2 3 2 0 3 2 6 −2 5 −2 16. A = . From Exercise 4, the eigenvalues of A are λ= 4 ± i, and the eigenvector 3 1 1 −1 1 − i corresponds to λ= 4 − i. By Theorem 9, P [ Re and v Im v ] = = v= 0 1 1 0 1 5 −1 = C P= AP −1 1 1 −2 1 −1 4 = 3 1 0 1 −1 . 4 −.8 1 2 17. A = . The characteristic polynomial is λ + 1.2λ + 1, so the eigenvalues of A are 4 − 2 . 2 λ = −.6 ± .8i. To find an eigenvector corresponding to −.6 − .8i, we compute −.8 1.6 + .8i . The equation ( A − (−.6 − .8i ) I )x = 0 amounts to A − (−.6 − .8i ) I = −1.6 + .8i 4 4 x1 + (−1.6 + .8i ) x2 = 0, so x1 = ((2 − i ) / 5) x2 with x2 free. A nice eigenvector corresponding to 2 − i 2 9, P [ Re −.6 − .8i is thus v = . By Theorem = = v Im v ] 5 5 −.8 2 −1 −.6 −.8 1 0 1 1 −1 . = C P= AP = 0 .8 −.6 5 −5 2 4 −2.2 5 1 18. A = .4 −1 and 0 −1 . The characteristic polynomial is λ 2 − 1.6λ + 1, so the eigenvalues of A are λ = .8 ± .6i. .6 .2 + .6i To find an eigenvector corresponding to .8 − .6i, we compute A − (.8 − .6i ) I = .4 −1 . −.2 + .6i The equation ( A − (.8 − .6i ) I )x = 0 amounts to .4 x1 + (−.2 + .6i ) x2 = 0, so x1 = ((1 − 3i ) / 2) x2 with x2 1 − 3i free. A nice eigenvector corresponding to .8 − .6i is thus v = . By Theorem 9, 2 1 −3 1 0 3 1 −1 1 −3 .8 −.6 −1 and . v Im v ] = P [ Re = = = C P= AP .8 0 .6 0 6 −2 1 .4 .6 2 2 1.52 −.7 2 19. A = . The characteristic polynomial is λ − 1.92λ + 1, so the eigenvalues of A are . 56 . 4 λ = .96 ± .28i. To find an eigenvector corresponding to .96 − .28i, we compute Copyright © 2016 Pearson Education, Ltd. 5.5 • Solutions 5-45 −.7 .56 + .28i A − (.96 − .28i ) I = . The equation ( A − (.96 − .28i ) I )x = 0 amounts to . −. + . i 56 56 28 .56 x1 + (−.56 + .28i ) x2 = 0, so x1 = ((2 − i ) / 2) x2 with x2 free. A nice eigenvector corresponding to 2 − i 9, P .96 − .28i is thus v = = . By Theorem 2 1 0 1 1.52 −.7 2 −1 −1 = = C P= AP .4 2 0 2 −2 2 .56 v Im v ] = [ Re .96 .28 2 2 −1 and 0 −.28 . .96 −1.64 −2.4 2 20. A = . The characteristic polynomial is λ − .56λ + 1, so the eigenvalues of A are . . 1 92 2 2 λ = .28 ± .96i. To find an eigenvector corresponding to .28 − .96i, we compute −2.4 −1.92 + .96i . The equation ( A − (.28 − .96i ) I )x = 0 amounts to A − (.28 − .96i ) I = 1.92 + .96i 1.92 1.92 x1 + (1.92 + .96i ) x2 = 0, so x1 = ((−2 − i ) / 2) x2 with x2 free. A nice eigenvector corresponding to −2 −2 − i 9, P [ Re .28 − .96i is thus v = = = v Im v ] . By Theorem 2 2 1 −1.64 −2.4 −2 −1 .28 −.96 1 0 −1 . = C P= AP = 2.2 2 0 .96 .28 2 −2 −2 1.92 −1 and 0 21. The first equation in (2) is (−.3 + .6i ) x1 − .6 x2 = 0. We solve this for x2 to find that 2 x2= ((−.3 + .6i ) /.6) x1= ((−1 + 2i ) / 2) x1. Letting x1 = 2, we find that y = is an eigenvector −1 + 2i 2 −1 + 2i −2 − 4i −1 + 2i for the matrix A. = Since y = = v1 the vector y is a complex 5 5 5 −1 + 2i multiple of the vector v1 used in Example 2. 22. Since A(= ( Ax) µ= (λx) λ( µ x), µ x is an eigenvector of A. µ x) µ= 23. (a) properties of conjugates and the fact that xT = xT (b) Ax = Ax and A is real (c) xT Ax is a scalar and hence may be viewed as a 1 × 1 matrix (d) properties of transposes (e) AT = A and the definition of q 24. xT Ax = xT (λx) = λ ⋅ xT x because x is an eigenvector. It is easy to see that xT x is real (and positive) because zz is nonnegative for every complex number z. Since xT Ax is real, by Exercise 23, so is λ. Next, write x= u + iv, where u and v are real vectors. Then Ax = A( u + iv ) = Au + iAv and λ= x λu + iλv . The real part of Ax is Au because the entries in A, u, and v are all real. The real part of λx is λu because λ and the entries in u and v are real. Since Ax and λx are equal, their real parts are equal, too. (Apply the corresponding statement about complex numbers to each entry of Ax.) Thus Au = λu, which shows that the real part of x is an eigenvector of A. Copyright © 2016 Pearson Education, Ltd. 5-46 CHAPTER 5 • Eigenvalues and Eigenvectors 25. Write Ax A(Re x) + iA(Im x). Since A is real, so are A(Re x) and = = x Re x + i (Im x), so that A(Im x). Thus A(Re x) is the real part of Ax and A(Im x) is the imaginary part of Ax. 26. a. If λ= a − bi, then Av = λv = ( a − bi )(Re v + i Im v ) = ( a Re v + b Im v ) + i ( a Im v − b Re v ) . By )) )))) ( )) )))) ( Re Av Im Av Exercise 25, A(Re = v ) Re = Av a Re v + b Im v A(Im v ) = Im Av = −b Re v + a Im v a −b Im v ]. By (a), A(Re v ) = P , A(Im v ) = P . So b a a −b a −b AP = [ A(Re v ) = A(Im v )] P = P P= PC . b a a b b. Let P = [ Re v .7 −2.0 27. [M ] A = 0 1.0 1.1 −4.0 −.5 2.8 2.0 −8.6 −1.0 6.0 1.7 −7.4 . v = eig(A) =(.2+.5i,.2-.5i,.3+.1i,.3-.1i) −1.0 5.3 For λ = .2 − .5i, an eigenvector is 0.5000 - 0.5000i -2.0000 + 0.0000i , so that nulbasis(A-ev(2)*eye(4)) = 0.0000 - 0.0000i 1 .0000 .5 − .5i −2 . v1 = 0 1 For λ = .3 − .1i, an eigenvector is −0.5000 − 0.0000i −.5 0.0000 + 0.5000i . , so that v 2 = .5i nulbasis(A -ev(4)*eye(4))= −.75 − .25i −0.7500 − 0.2500i 1 1.0000 Copyright © 2016 Pearson Education, Ltd. 5.6 .5 −2 Hence by Theorem 9, P Re = = v1 Im v1 Re v2 Im v2 0 1 .2 .5 C= 0 0 −.5 .2 0 0 −1.4 −1.3 28. [M ] A = .3 2.0 0 0 .3 .1 −.5 0 0 0 −.5 0 −.75 1 • Solutions 0 .5 and −.25 0 0 0 . Other choices are possible, but C must equal P −1 AP. −.1 .3 −2.0 −.8 −1.9 3.3 −2.0 −.1 −1.6 2.3 −2.0 −.6 . ev = eig(A) =(-.4+i,-.4-i,-.2+.5i,-.2-.5i) −1.4 2.6 For λ = −.4 − i, an eigenvector is −1 − i -1.0000 - 1.0000i −1 + i -1.0000 + 1.0000i . , so that v1 = nulbasis(A-ev(2)*eye(4)) = 1− i 1.0000 - 1.0000i 1 1.0000 For λ = −.2 − .5i, an eigenvector is 0 0.0000 - 0.0000i −1 − i -0.5000 - 0.5000i , so that v 2 = nulbasis(A-ev(4)*eye(4)) = −1 + i -0.5000 + 0.5000i 2 1.0000 −1 −1 Hence by Theorem 9, P Re v1 Im v1 Re v2 Im v2 = = 1 1 −.4 1 C= 0 0 5.6 −1 −.4 0 0 5-47 0 0 −.2 .5 −1 1 −1 0 0 −1 −1 2 0 −1 and 1 0 0 0 . Other choices are possible, but C must equal P −1 AP. −.5 −.2 SOLUTIONS 1. The exercise does not specify the matrix A, but only lists the eigenvalues 3 and 1/3, and the −1 1 9 corresponding eigenvectors v1 = and v 2 = . Also, x0 = . 1 1 1 Copyright © 2016 Pearson Education, Ltd. 5-48 CHAPTER 5 • Eigenvalues and Eigenvectors a. To find the action of A on x0 , express x0 in terms of v1 and v 2 . That is, find c1 and c2 such that = x0 c1 v1 + c2 v 2 . This is certainly possible because the eigenvectors v1 and v 2 are linearly independent (by inspection and also because they correspond to distinct eigenvalues) and hence form a basis for R 2 . (Two linearly independent vectors in R 2 automatically span R 2 . ) The row 5 1 −1 9 1 0 reduction v1 v2 x0 = shows that = x0 5 v1 − 4 v 2 . Since v1 and � 1 1 0 1 −4 1 v 2 are eigenvectors (for the eigenvalues 3 and 1/3): 15 −4 / 3 49 / 3 x1 = Ax0 = 5 Av1 − 4 Av 2 = 5 ⋅ 3v1 − 4 ⋅ (1/ 3) v 2 = − = 15 4 / 3 41/ 3 b. Each time A acts on a linear combination of v1 and v 2 , the v1 term is multiplied by the eigenvalue 3 and the v 2 term is multiplied by the eigenvalue 1/3: x 2 = Ax1 = A[5 ⋅ 3v1 − 4(1/ 3) v 2 ] = 5(3) 2 v1 − 4(1/ 3) 2 v 2 In general, = x k 5(3) k v1 − 4(1/ 3) k v 2 , for k ≥ 0. 1 2 −3 3 are eigenvectors of a 3 × 3 matrix A, corresponding to 2. The vectors v1 = 0 , v 2 = 1 , v 3 =− −3 −5 7 −2 eigenvalues 3, 4/5, and 3/5, respectively. Also, x0 = −5 . To describe the solution of the equation 3 x k +1 = Ax k (k = 1, 2, …), first write x0 in terms of the eigenvectors. 2 −3 −2 1 0 0 2 1 0 1 −3 −5 � 0 1 0 1 ⇒ x 0 = 2 v1 + v 2 + 2 v 3 v 2 v3 = 3 5 7 3 0 0 1 2 − − Then, x1 = A(2 v1 + v 2 + 2 v 3 ) = 2 Av1 + Av 2 + 2 Av 3 = 2 ⋅ 3v1 + (4 / 5) v 2 + 2 ⋅ (3/ 5) v 3 . In general, v 1 x0 1 x k = 2 ⋅ 3 v1 + (4 / 5) v 2 + 2 ⋅ (3/ 5) v 3 . For all k sufficiently large, x k ≈ 2 ⋅ 3 v1 = 2 ⋅ 3 0 . −3 k k k k k .5 .4 3. A = , det( A − λ I ) = (.5 − λ )(1.1 − λ ) + .08 = λ 2 − 1.6λ + .63. This characteristic −.2 1.1 polynomial factors as (λ − .9)(λ − .7), so the eigenvalues are .9 and .7. If v1 and v 2 denote corresponding eigenvectors, and if = x0 c1 v1 + c2 v 2 , then x1 = A(c1 v1 + c2 v 2 ) = c1 Av1 + c2 Av 2 = c1 (.9) v1 + c2 (.7) v 2 , and for k ≥ 1, x k = c1 (.9) k v1 + c2 (.7) k v 2 . For any choices of c1 and c2 , both the owl and wood rat populations decline over time. .5 .4 2 4. A = , det( A − λ I ) = (.5 − λ )(1.1 − λ ) − (.4)( −.125) = λ − 1.6λ + .6. This characteristic −. 125 1 . 1 polynomial factors as (λ − 1)(λ − .6), so the eigenvalues are 1 and .6. For the eigenvalue 1, solve Copyright © 2016 Pearson Education, Ltd. 5.6 • Solutions 5-49 −.5 .4 0 −5 4 0 4 � ( A − I )x =: 0 . A basis for the eigenspace is v1 = . Let v 2 be an −.125 .1 0 0 0 0 5 eigenvector for the eigenvalue .6. (The entries in v 2 are not important for the long-term behavior of the system.). If = x0 c1 v1 + c2 v 2 , then x1 = c1 Av1 + c2 Av 2 = c1 v1 + c2 (.6) v 2 , and for k sufficiently 4 4 large, x= c1 + c2 (.6) k v 2 ≈ c1 . Provided that c1 ≠ 0, the owl and wood rat populations each k 5 5 stabilize in size, and eventually the populations are in the ratio of 4 owls for each 5 thousand rats. If some aspect of the model were to change slightly, the characteristic equation would change slightly and the perturbed matrix A might not have 1 as an eigenvalue. If the eigenvalue becomes slightly large than 1, the two populations will grow; if the eigenvalue becomes slightly less than 1, both populations will decline. .4 5. A = −.325 .3 , det( A − λ I= ) λ 2 − 1.6λ + .5775. The quadratic formula provides the roots of the 1.2 1.6 ± 1.62 − 4(.5775) 1.6 ± .25 characteristic equation: λ = 1.05 and .55 . = = 2 2 Because one eigenvalue is larger than one, both populations grow in size. Their relative sizes are determined eventually by the entries in the eigenvector corresponding to 1.05. Solve .3 0 −13 6 0 −.65 6 ( A − 1.05 I )x =0 : � . An eigenvector is v1 = 13 . −.325 .15 0 0 0 0 Eventually, there will be about 6 spotted owls for every 13 (thousand) flying squirrels. .3 .4 6. When p =.5, A = ) λ 2 − 1.6λ + .63= (λ − .9)(λ − .7). , and det( A − λ I= 5 1 2 −. . The eigenvalues of A are .9 and .7, both less than 1 in magnitude. The origin is an attractor for the dynamical system and each trajectory tends toward 0. So both populations of owls and squirrels eventually perish. The calculations in Exercise 4 (as well as those in Exercises 35 and 27 in Section 5.1) show that if the largest eigenvalue of A is 1, then in most cases the population vector x k will tend toward a multiple of the eigenvector corresponding to the eigenvalue 1. [If v1 and v 2 are eigenvectors, with v1 corresponding to λ = 1, and if = x0 c1 v1 + c2 v 2 , then x k tends toward c1 v1 , provided c1 is not zero.] So the problem here is to determine the value of the predation parameter p such that the largest eigenvalue of A is 1. Compute the characteristic polynomial: .3 .4 − λ 2 det = (.4 − λ)(1.2 − λ) + .3 p = λ − 1.6λ + (.48 + .3 p ) . By the quadratic − . − p 1 2 λ formula, λ = 1.6 ± 1.62 − 4(.48 + .3 p ) . The larger eigenvalue is 1 when 2 1.6 + 1.62 − 4(.48 + .3 p ) =2 and 2.56 − 1.92 − 1.2 p =.4 . In this case, .64 − 1.2 p =.16, and p = .4. 7. a. The matrix A in Exercise 1 has eigenvalues 3 and 1/3. Since | 3 | > 1 and | 1/ 3 | < 1, the origin is a saddle point. Copyright © 2016 Pearson Education, Ltd. 5-50 CHAPTER 5 • Eigenvalues and Eigenvectors −1 b. The direction of greatest attraction is determined by v 2 = , the eigenvector corresponding to 1 the eigenvalue with absolute value less than 1. The direction of greatest repulsion is determined 1 by v1 = , the eigenvector corresponding to the eigenvalue greater than 1. 1 c. The drawing below shows: (1) lines through the eigenvectors and the origin, (2) arrows toward the origin (showing attraction) on the line through v 2 and arrows away from the origin (showing repulsion) on the line through v1 , (3) several typical trajectories (with arrows) that show the general flow of points. No specific points other than v1 and v 2 were computed. This type of drawing is about all that one can make without using a computer to plot points. Note: If you wish your class to sketch trajectories for anything except saddle points, you will need to go beyond the discussion in the text. The following remarks from the Study Guide are relevant. Sketching trajectories for a dynamical system in which the origin is an attractor or a repellor is more difficult than the sketch in Exercise 7. There has been no discussion of the direction in which the trajectories “bend” as they move toward or away from the origin. For instance, if you rotate Figure 1 of Section 5.6 through a quarter-turn and relabel the axes so that x1 is on the horizontal axis, then the new figure corresponds to the matrix A with the diagonal entries .8 and .64 interchanged. In general, if A is a diagonal matrix, with positive diagonal entries a and d, unequal to 1, then the trajectories lie on the axes or on curves whose equations have the form x2 = r ( x1 ) s , where = s (ln d ) / (ln a ) and r depends on the initial point x0 . (See Encounters with Chaos, by Denny Gulick, New York: McGraw-Hill, 1992, pp. 147– 150.) 8. The matrix from Exercise 2 has eigenvalues 3, 4/5, and 3/5. Since one eigenvalue is greater than 1 and the others are less than one in magnitude, the origin is a saddle point. The direction of greatest repulsion is the line through the origin and the eigenvector (1, 0, −3) for the eigenvalue 3. The direction of greatest attraction is the line through the origin and the eigenvector (−3, −3, 7) for the smallest eigenvalue 3/5. 1.7 9. A= −1.2 = λ −.3 , det( A − λI= ) λ 2 − 2.5λ + = 1 0, .8 2.5 ± 2.52 − 4(1) 2.5 ± 2.25 2.5 ± 1.5 = = = 2 and .5 . The origin is a saddle point because one 2 2 2 Copyright © 2016 Pearson Education, Ltd. 5.6 • Solutions 5-51 eigenvalue is greater than 1 and the other eigenvalue is less than 1 in magnitude. The direction of greatest repulsion is through the origin and the eigenvector v1 found below. Solve −.3 0 1 1 0 −.3 −1 0: � ( A − 2 I )x = , so x1 = –x2, and x2 is free. Take v1 = . The −1.2 −1.2 0 0 0 0 1 direction of greatest attraction is through the origin and the eigenvector v 2 found below. Solve 1.2 ( A − .5 I ) x = 0 : −1.2 −.3 .3 0 � 0 1 0 −.25 0 0 1 , so x1 = −.25 x2 , and x2 is free. Take v 2 = . 0 4 .3 .4 10. A= = 0, ) λ 2 − 1.4λ + .45 , det( A − λ I= −. . 3 1 1 1.4 ± 1.42 − 4(.45) 1.4 ± .16 1.4 ± .4 = = = .5 and .9 . The origin is an attractor because both λ= 2 2 2 eigenvalues are less than 1 in magnitude. The direction of greatest attraction is through the origin and −.2 .4 0 1 −2 0 the eigenvector v1 found below. Solve ( A − .5I )x = 0 : , so x1 = 2 x2 , � 0 0 −.3 .6 0 0 2 and x2 is free. Take v1 = . 1 .4 11. A= −.4 .5 , det( A − λ I= ) λ 2 − 1.7λ + .72= 0 , 1.3 1.7 ± 1.7 2 − 4(.72) 1.7 ± .01 1.7 ± .1 λ= = = = .8 and .9 . The origin is an attractor because both 2 2 2 eigenvalues are less than 1 in magnitude. The direction of greatest attraction is through the origin and −.4 .5 0 1 −1.25 0 the eigenvector v1 found below. Solve ( A − .8 I )x = 0 : , so � 0 0 −.4 .5 0 0 5 x1 = 1.25 x2 , and x2 is free. Take v1 = . 4 .6 .5 12. A= , det( A − λ I= ) λ 2 − 1.9λ + .88= 0 . −.3 1.4 1.9 ± 1.92 − 4(.88) 1.9 ± .09 1.9 ± .3 λ= = = = .8 and 1.1 . The origin is a saddle point because one 2 2 2 eigenvalue is greater than 1 and the other eigenvalue is less than 1 in magnitude. The direction of greatest repulsion is through the origin and the eigenvector v1 found below. Solve 1 −.6 .6 0 1 −1 0 ( A − 1.1I )x = 0 : � , so x1 = x2 , and x2 is free. Take v1 = . The 0 0 1 −.3 .3 0 0 direction of greatest attraction is through the origin and the eigenvector v 2 found below. Solve −.3 ( A − .8 I ) x = 0 : −.3 .6 .6 0 � 0 1 0 −2 0 0 2 , so x1 = 2 x2 , and x2 is free. Take v 2 = . 0 1 Copyright © 2016 Pearson Education, Ltd. 5-52 CHAPTER 5 .8 13. A= −.4 • Eigenvalues and Eigenvectors .3 , det( A − λ I )= λ 2 − 2.3λ + 1.32= 0 , 1.5 2.3 ± 2.32 − 4(1.32) 2.3 ± .01 2.3 ± .1 λ= = = = 1.1 and 1.2 . The origin is a repellor because both 2 2 2 eigenvalues are greater than 1 in magnitude. The direction of greatest repulsion is through the origin −.4 .3 0 1 −.75 0 and the eigenvector v1 found below. Solve ( A − 1.2 I )x = 0 : , so � 0 0 −.4 .3 0 0 3 x1 = .75 x2 , and x2 is free. Take v1 = . 4 1.7 14. A= −.4 .6 , det( A − λ I )= λ 2 − 2.4λ + 1.43= 0 , .7 2.4 ± 2.42 − 4(1.43) 2.4 ± .04 2.4 ± .2 = = = λ= 1.1 and 1.3 . The origin is a repellor because both 2 2 2 eigenvalues are greater than 1 in magnitude. The direction of greatest repulsion is through the origin .6 0 1 1.5 0 .4 and the eigenvector v1 found below. Solve ( A − 1.3I )x = 0 : , so � 0 0 −.4 −.6 0 0 −3 x1 =−1.5 x2 , and x2 is free. Take v1 = . 2 .4 .3 15. A = .3 0 .8 .2 .2 .3 . Given eigenvector v1 = .5 .4 3 v1 , compute Av1 =. .3 −.1 For λ = .5 : .3 .3 .2 For λ =.2 : .3 .3 0 .3 .2 0 .6 .2 0 .8 .2 .2 .3 0 .2 .3 .3 .1 .6 and eigenvalues .5 and .2. To find the eigenvalue for .3 .2 .1 .1 6 =1 ⋅ v Thus v is an eigenvector for λ =. 1 .3 .6 =. 1 1 .5 .3 .3 0 0 � 0 0 0 � 0 1 0 0 1 0 0 −2 3 0 0 1 0 0 1 0 1 0 0 0 x1 = 2 x3 2 0 , x2 = −3 x3 . Set v 2 = −3 . 1 0 x3 is free 0 x1 = − x3 −1 0 , x2 =0 . Set v 3 = 0 1 0 x3 is free Given x0= (0, .3, .7), find weights such that x0 2 −1 0 1 .1 v 0 .3 � 0 v2 v3 x0 =.6 −3 1 1 1 .7 0 . 3 = c1 v1 + cv 2 + c3 v 3 . 0 1 0 0 0 1 1 .1 . .3 Copyright © 2016 Pearson Education, Ltd. 5.6 • Solutions 5-53 x0= v1 + .1v 2 + .3v 3 x1= Av1 + .1Av 2 + .3 Av 3= v1 + .1(.5) v 2 + .3(.2) v 3 , and xk= v1 +.1(.5)k v2 +.3(.2)k v3 . As k increases, xk approaches v1. 16. [M] .90 01 A =. .09 .01 .90 .09 .09 1.0000 0.8900 . To four decimal places, .01 ⋅ ev = eig(A)= .90 .8100 0.9192 v1 = nulbasis(A -ev(1)*eye(3))= 0.1919 . Exact : 1.0000 91 / 99 19 / 99 1 -1 v2 = nulbasis(A -ev(2)*eye(3))= 1 0 -1 v3 = nulbasis(A -ev(3)*eye(3))= 0 1 The general solution of the dynamical system is x k = c1 v1 + c2 (.89) k v 2 + c3 (.81) k v 3 . Note: When working with stochastic matrices and starting with a probability vector (having nonnegative entries whose sum is 1), it helps to scale v1 to make its entries sum to 1. If v1 = (91/ 209, 19 / 209, 99 / 209), or (.435, .091, .474) to three decimal places, then the weight c1 above turns out to be 1. See the text’s discussion of Exercise 27 in Section 5.2. 0 1.6 17. a. A = .8 .3 −λ b. det .3 1.6 = λ 2 − .8λ − .48 = 0. The eigenvalues of A are given by .8 − λ .8 ± (−.8) 2 − 4(−.48) .8 ± 2.56 .8 ± 1.6 = = = 1.2 and − .4 . The numbers of juveniles and 2 2 2 adults are increasing because the largest eigenvalue is greater than 1. The eventual growth rate of each age class is 1.2, which is 20% per year. To find the eventual relative population sizes, solve −1.2 1.6 0 1 −4 / 3 0 x1= (4 / 3) x2 4 � . . Set v1 = ( A − 1.2 I )x =0: 3 . Eventually, 0 0 x2 is free .3 −.4 0 0 there will be about 4 juveniles for every 3 adults. λ= c. [M] Suppose that the initial populations are given by x= (15, 10). The Study Guide describes 0 how to generate the trajectory for as many years as desired and then to plot the values for each population. Let x= ( jk , a k ). Then we need to plot the sequences {jk }, {a k }, {jk + a k }, and k {jk /a k }. Adjacent points in a sequence can be connected with a line segment. When a sequence is Copyright © 2016 Pearson Education, Ltd. 5-54 CHAPTER 5 • Eigenvalues and Eigenvectors plotted, the resulting graph can be captured on the screen and printed (if done on a computer) or copied by hand onto paper (if working with a graphics calculator). 0 18. a. A = .6 0 0 0 .75 .42 0 .95 0.0774 + 0.4063i b. ev = eig(A)= 0.0774 − 0.4063i . The long-term growth rate is 1.105, about 10.5 % per year. 1.1048 0.3801 = 0.2064 . For each 100 adults, there will be v = nulbasis(A -ev(3)*eye(3)) 1.0000 approximately 38 calves and 21 yearlings. Note: The MATLAB box in the Study Guide and the various technology appendices all give directions for generating the sequence of points in a trajectory of a dynamical system. Details for producing a graphical representation of a trajectory are also given, with several options available in MATLAB, Maple, and Mathematica. 5.7 SOLUTIONS 1. From the “eigendata” (eigenvalues and corresponding eigenvectors) given, the eigenfunctions for the differential equation x′ = Ax are v1e 4t and v 2 e 2t . The general solution of x′ = Ax has the form −3 −1 −6 c1 e 4t + c2 e 2t . The initial condition x(0) = determines c1 and c2 : 1 1 1 −3 −1 −6 1 0 −3 −1 −6 . Solving the system: � c1 e 4(0) + c2 e 2(0) = 1 1 0 1 1 1 1 1 and x(t ) Thus c1 =5/ 2, c2 =−3/ 2, = 5 / 2 . −3 / 2 5 −3 4t 3 −1 2t e − e . 2 1 2 1 2. From the eigendata given, the eigenfunctions for the differential equation x′ = Ax are v1e −3t and −1 1 v 2 e −1t . The general solution of x′ = Ax has the form c1 e −3t + c2 e −1t . The initial condition 1 1 2 −1 1 2 x(0) = determines c1 and c2 : c1 e −3(0) + c2 e −1(0) = 3 . Solving the system: 3 1 1 −1 1 1 1 2 3. A = −1 2 � 3 1 0 0 1 1/ 2 1 −1 −3t 5 1 −t . Thus c1 = 1/ 2, c2 = 5/ 2,= and x(t ) e + e . 5 / 2 2 1 2 1 3 , det( A − λ I ) = λ 2 − 1 = (λ − 1)(λ + 1) = 0. Eigenvalues: 1 and −1. −2 Copyright © 2016 Pearson Education, Ltd. 5.7 1 For λ = 1: −1 • Solutions 5-55 −3 1 3 0 0 0 0 , so x1 = −3 x2 with x2 free. Take x2 = 1 and v1 = 1 . 3 0 1 1 0 � , so x1 = − x2 with x2 free. Take x2 = 1 and −1 0 0 0 0 0 � 0 3 −3 3 For λ = –1: −1 −1 v2 = . 1 3 For the initial condition x(0) = , find c1 and c2 such that c1 v1 + c2 v 2 = x(0) : 2 −3 −1 3 1 0 −5/ 2 . Thus c1 =−5/ 2, c2 =9 / 2, and � v1 v2 x(0) = 1 2 0 1 9 / 2 1 −3 −1 x(t ) = − 5 et + 9 e − t . 2 1 2 1 Since one eigenvalue is positive and the other is negative, the origin is a saddle point of the dynamical system described by x′ = Ax. The direction of greatest attraction is the line through v 2 and the origin. The direction of greatest repulsion is the line through v1 and the origin. −2 4. A = 1 −5 , det( A − λ I ) = λ 2 − 2λ − 3 = (λ + 1)(λ − 3) = 0. Eigenvalues: −1 and 3. 4 −5 For λ = 3: 1 −5 1 −1 For λ = –1: 1 0 1 � 0 0 −5 5 0 � 0 1 0 1 0 5 0 0 −1 , so x1 = − x2 with x2 free. Take x2 = 1 and v1 = . 0 1 0 −5 , so x1 = −5 x2 with x2 free. Take x2 = 1 and v 2 = . 0 1 3 For the initial condition x(0) = , find c1 and c2 such that c1 v1 + c2 v 2 = x(0) : 2 −1 −5 3 1 0 13/ 4 . Thus c1 =13/ 4, c2 =−5/ 4, and � v1 v2 x(0) = 1 1 2 0 1 −5/ 4 13 −1 3t 5 −5 −t e − e . x(t ) = 4 1 4 1 Since one eigenvalue is positive and the other is negative, the origin is a saddle point of the dynamical system described by x′ = Ax. The direction of greatest attraction is the line through v 2 and the origin. The direction of greatest repulsion is the line through v1 and the origin. 7 5. A = 3 −1 , det ( A − λ I ) = λ 2 − 10λ + 24 = (λ − 4)(λ − 6) = 0. Eigenvalues: 4 and 6. 3 3 For λ = 4: 3 −1 −1 0 � 0 1 0 −1/ 3 0 0 1 , so x1= (1/ 3) x2 with x2 free. Take x2 = 3 and v1 = . 0 3 Copyright © 2016 Pearson Education, Ltd. 5-56 CHAPTER 5 • Eigenvalues and Eigenvectors 1 For λ = 6: 3 −1 −3 0 � 0 1 0 −1 0 0 1 , so x1 = x2 with x2 free. Take x2 = 1 and v 2 = . 0 1 3 For the initial condition x(0) = , find c1 and c2 such that c1 v1 + c2 v 2 = x(0) : 2 1 1 3 1 0 −1/ 2 . Thus c1 =−1/ 2, c2 =7 / 2, and v1 v2 x(0) = � 7 / 2 3 1 2 0 1 1 1 7 1 x(t ) = − e 4t + e6t . 2 3 2 1 Since both eigenvalues are positive, the origin is a repellor of the dynamical system described by x′ = Ax. The direction of greatest repulsion is the line through v 2 and the origin. 1 6. A = 3 −2 , det ( A − λ I ) = λ 2 + 3λ + 2 = (λ + 1)(λ + 2) = 0. Eigenvalues: −1 and −2. −4 3 For λ = –2: 3 −2 −2 0 � 0 1 0 −2 / 3 0 2 For λ = –1: 3 −2 −3 0 � 0 1 0 −1 0 0 2 , so x1= (2 / 3) x2 with x2 free. Take x2 = 3 and v1 = . 0 3 0 1 , so x1 = x2 with x2 free. Take x2 = 1 and v 2 = . 0 1 3 For the initial condition x(0) = , find c1 and c2 such that c1 v1 + c2 v 2 = x(0) : 2 2 1 2 1 3 1 0 −1 . Thus c1 =−1, c2 =5, and x(t ) = − e −2t + 5 e −t . [ v1 v2 x(0)] = � 5 3 1 3 1 2 0 1 Since both eigenvalues are negative, the origin is an attractor of the dynamical system described by x′ = Ax. The direction of greatest attraction is the line through v1 and the origin. −1 1 1 , with eigenvectors v1 = and v 2 = corresponding to 3 1 3 1 1 eigenvalues 4 and 6 respectively. To decouple the equation x′ = Ax,= set P [ v= 1 v2 ] and let 3 1 7 7. From Exercise 5, A = 3 4 0 −1 −1 D= , so that A = PDP and D = P AP. Substituting x(t ) = Py (t ) into x′ = Ax we have 0 6 d = ( P y ) A= ( P y ) PDP −1= ( P y ) PD y . Since P has constant entries, dtd ( P y ) = P ( dtd (y )), so that dt y ′ (t ) 4 0 y1 (t ) left-multiplying the equality P ( dtd (y )) = PD y by P −1 yields y′ = D y , or 1 = . y2′ (t ) 0 6 y2 (t ) Copyright © 2016 Pearson Education, Ltd. 5.7 1 8. From Exercise 6, A = 3 • Solutions 5-57 −2 1 2 , with eigenvectors v1 = and v 2 = corresponding to −4 1 3 eigenvalues −2 and −1 respectively. To decouple the equation x′ = Ax= , set P = v v2 1 2 3 1 1 0 −2 −1 −1 and let D = , so that A = PDP and D = P AP. Substituting x(t ) = Py (t ) into x′ = Ax − 0 1 d we have = ( Py ) A= ( Py ) PDP −1= ( Py ) PDy . Since P has constant entries, dtd ( Py ) = P ( dtd (y ) ) , so dt that left-multiplying the equality P ( dtd (y ) ) = PDy by P −1 yields y′ = Dy , or y1′ (t ) −2 y ′ (t ) = 0 2 0 y1 (t ) . −1 y2 (t ) 2 −3 1 − i 9. A = . An eigenvalue of A is −2 + i with corresponding eigenvector v = . The −1 −1 1 complex eigenfunctions veλt and veλt form a basis for the set of all complex solutions to x′ = Ax. 1 − i ( −2+i )t 1 + i ( −2−i )t , where c1 and c2 are arbitrary The general complex solution is c1 e + c2 e 1 1 complex numbers. To build the general real solution, rewrite ve( −2+i )t as: 1 − i −2t it 1 − i −2t = ve( −2+i )t = e e e (cos t + i sin t ) 1 1 cos t − i cos t + i sin t − i 2 sin t −2t = e cos t + i sin t cos t + sin t −2t sin t − cos t −2t = e +i sin t e cos t cos t + sin t −2t sin t − cos t −2t The general real solution has the form c1 e + c2 e , where c1 and c2 cos t sin t now are real numbers. The trajectories are spirals because the eigenvalues are complex. The spirals tend toward the origin because the real parts of the eigenvalues are negative. 3 10. A = −2 1 1 + i . An eigenvalue of A is 2 + i with corresponding eigenvector v = . The complex 1 −2 eigenfunctions ve λt and ve λt form a basis for the set of all complex solutions to x′ = Ax. The 1 + i (2+i )t 1 − i (2−i )t general complex solution is c1 , where c1 and c2 are arbitrary complex + c2 e e −2 −2 numbers. To build the general real solution, rewrite ve(2+i )t as: Copyright © 2016 Pearson Education, Ltd. 5-58 CHAPTER 5 • Eigenvalues and Eigenvectors 1 + i 2t it 1 + i 2t = ve(2+i )t = e e −2 e (cos t + i sin t ) −2 cos t + i cos t + i sin t + i 2 sin t 2t = e −2cos t − 2i sin t cos t − sin t 2t sin t + cos t 2t = e + i −2sin t e −2cos t cos t − sin t 2t sin t + cos t 2t The general real solution has the form c1 e + c2 e , where c1 and c2 −2cos t −2sin t now are real numbers. The trajectories are spirals because the eigenvalues are complex. The spirals tend away from the origin because the real parts of the eigenvalues are positive. −3 11. A = 2 −9 −3 + 3i . An eigenvalue of A is 3i with corresponding eigenvector v = . The 3 2 complex eigenfunctions ve λt and ve λt form a basis for the set of all complex solutions to x′ = Ax. −3 + 3i (3i )t −3 − 3i ( −3i )t The general complex solution is c1 , where c1 and c2 are arbitrary e + c2 e 2 2 complex numbers. To build the general real solution, rewrite ve(3i )t as: −3 + 3i ve(3i )t = (cos3t + i sin 3t ) 2 −3cos3t − 3sin 3t −3sin 3t + 3cos3t = +i 2cos3t 2sin 3t −3cos3t − 3sin 3t −3sin 3t + 3cos3t The general real solution has the form c1 + c2 , where c1 and 2cos3t 2sin 3t c2 now are real numbers. The trajectories are ellipses about the origin because the real parts of the eigenvalues are zero. −7 12. A = −4 10 3 − i . An eigenvalue of A is −1 + 2i with corresponding eigenvector v = . The 5 2 complex eigenfunctions ve λt and ve λt form a basis for the set of all complex solutions to x′ = Ax. 3 − i ( −1+ 2i ) t 3 + i ( −1−2i ) t The general complex solution is c1 , where c1 and c2 are arbitrary e + c2 e 2 2 complex numbers. To build the general real solution, rewrite ve( −1+ 2i )t as: 3 − i − t ve( −1+ 2i )t = e (cos 2t + i sin 2t ) 2 3cos 2t + sin 2t −t 3sin 2t − cos 2t −t = e + i e 2cos 2t 2sin 2t Copyright © 2016 Pearson Education, Ltd. 5.7 • Solutions 5-59 3cos 2t + sin 2t −t 3sin 2t − cos 2t −t The general real solution has the form c1 e + c2 e , where c1 and 2cos 2t 2sin 2t c2 now are real numbers. The trajectories are spirals because the eigenvalues are complex. The spirals tend toward the origin because the real parts of the eigenvalues are negative. 4 13. A = 6 −3 1 + i . An eigenvalue of A is 1 + 3i with corresponding eigenvector v = . The complex −2 2 eigenfunctions ve λt and ve λt form a basis for the set of all complex solutions to x′ = Ax. The 1 + i (1+3i ) t 1 − i (1−3i ) t general complex solution is c1 , where c1 and c2 are arbitrary complex + c2 e e 2 2 numbers. To build the general real solution, rewrite ve(1+3i )t as: 1 + i t ve(1+3i )t = e (cos3t + i sin 3t ) 2 cos3t − sin 3t t sin 3t + cos3t t = e + i 2sin 3t e 2cos3t cos3t − sin 3t t sin 3t + cos3t t The general real solution has the form c1 e + c2 e , where c1 and c2 2cos3t 2sin 3t now are real numbers. The trajectories are spirals because the eigenvalues are complex. The spirals tend away from the origin because the real parts of the eigenvalues are positive. −2 14. A = −8 1 1 − i . An eigenvalue of A is 2i with corresponding eigenvector v = . The complex 2 4 eigenfunctions ve λt and ve λt form a basis for the set of all complex solutions to x′ = Ax. The 1 − i (2i )t 1 + i ( −2i )t general complex solution is c1 , where c1 and c2 are arbitrary complex e + c2 e 4 4 numbers. To build the general real solution, rewrite ve(2i )t as: 1 − i = ve(2i )t (cos 2t + i sin 2t ) 4 cos 2t + sin 2t sin 2t − cos 2t = +i 4cos 2t 4sin 2t cos 2t + sin 2t sin 2t − cos 2t The general real solution has the form c1 + c2 , where c1 and c2 now 4cos 2t 4sin 2t are real numbers. The trajectories are ellipses about the origin because the real parts of the eigenvalues are zero. Copyright © 2016 Pearson Education, Ltd. 5-60 CHAPTER 5 −8 15. [M] A = 2 7 • Eigenvalues and Eigenvectors −12 1 12 1.0000 −1.0000 . −2.0000 −6 2 . The eigenvalues of A are: ev = eig(A)= 5 −1.0000 nulbasis(A-ev(1)*eye(3)) = 0.2500 1.0000 −1.2000 nulbasis(A-ev(2)*eye(3)) = 0.2000 1.0000 − 4 , so that v = 1 . 1 4 −6 , so that v = 1 . 2 5 −1.0000 −1 nulbasis (A-ev(3)*eye(3)) = 0.0000 , so that v 3 = 0 . −1.0000 1 −4 −6 −1 −t t Hence the general solution is x(t ) =c1 1 e + c2 1 e + c3 0 e −2t . The origin is a saddle point. 4 5 1 A solution with c1 = 0 is attracted to the origin while a solution with c= c= 0 is repelled. 2 3 −6 16. [M] A 2 = −4 −11 5 −5 16 −4 . The eigenvalues of A are: ev = eig(A)= 10 2.3333 nulbasis(A-ev(1)*eye(3)) = −0.6667 , so that v1 = 1.0000 3.0000 nulbasis(A-ev(2)*eye(3)) = −1.0000 1.0000 2.0000 nulbasis(A-ev(3)*eye(3)) = 0.0000 1.0000 4.0000 3.0000 2.0000 . 7 −2 . 3 , so that v = 2 3 −1 . 1 2 , so that v = 0 . 3 1 7 3 2 4t 3t Hence the general solution is x(t ) = c1 −2 e + c2 −1 e + c3 0 e 2t . The origin is a repellor, 3 1 1 because all eigenvalues are positive. All trajectories tend away from the origin. Copyright © 2016 Pearson Education, Ltd. 5.7 30 −11 17. [M] A = 6 64 −23 15 • Solutions 5-61 23 −9 . The eigenvalues of A are: 4 5.0000 + 2.0000i ev = eig(A)= 5.0000 − 2.0000i . 1.0000 23 − 34i 7.6667 - 11.3333i nulbasis(A-ev(1)*eye(3)) = -3.0000 + 4.6667i , so that v1 = −9 + 14i . 3 1.0000 23 + 34i 7.6667 + 11.3333i nulbasis (A-ev(2)*eye(3)) = -3.0000 − 4.6667i , so that v 2 = −9 − 14i . 3 1.0000 −3.0000 −3 nulbasis (A-ev(3)*eye(3)) = 1.0000 , so that v 3 = 1 . 1 1.0000 23 − 34i 23 + 34i −3 (5+ 2 i ) t (5− 2 i ) t Hence the general complex solution is x(t )= c1 −9 + 14i e + c2 −9 − 14i e + c3 1 et . 3 3 1 Rewriting the first eigenfunction yields 23 − 34i 23cos 2t + 34sin 2t 23sin 2t − 34cos 2t −9 + 14i e5t (cos 2t + i sin 2t ) = −9cos 2t − 14sin 2t e5t + i −9sin 2t + 14cos 2t e5t . 3 3cos 2t 3sin 2t Hence the general real solution is 23cos 2t + 34sin 2t 23sin 2t − 34cos 2t −3 5t 5t x(t ) = c1 −9cos 2t − 14sin 2t e + c2 −9sin 2t + 14cos 2t e + c3 1 et , where c1 , c2 , and c3 are 1 3cos 2t 3sin 2t real. The origin is a repellor, because the real parts of all eigenvalues are positive. All trajectories spiral away from the origin. 53 18. [M] A = 90 20 −30 −52 −10 −2 −3 . The eigenvalues of A are: 2 −7.0000 ev = eig(A)= 5.0000 + 1.0000i . 5.0000 − 1.0000i Copyright © 2016 Pearson Education, Ltd. 5-62 CHAPTER 5 • Eigenvalues and Eigenvectors 0.5000 1 nulbasis(A-ev(1)*eye(3)) = 1.0000 , so that v1 = 2 . 0 0.0000 0.6000 + 0.2000i nulbasis(A-ev(2)*eye(3)) = 0.9000 + 0.3000i , so that v= 2 1.0000 0.6000 - 0.20000 nulbasis(A-ev(3)*eye(3)) = 0.9000 − 0.3000i , 1.0000 6 + 2i 9 + 3i . 10 6 − 2i 9 − 3i . so that v= 3 10 1 6 + 2i 6 − 2i −7 t (5+ i ) t Hence the general complex solution is x= + c3 9 − 3i e(5−i )t . (t ) c1 2 e + c2 9 + 3i e 0 10 10 Rewriting the second eigenfunction yields 6 + 2i 6cos t − 2sin t 6sin t + 2cos t 9 + 3i e5t (cos t + i sin t ) = 9cos t − 3sin t e5t + i 9sin t + 3cos t e5t . 10 10cos t 10sin t 1 6cos t − 2sin t 6sin t + 2cos t −7 t 5t Hence the general real solution is x(t ) = c1 2 e + c2 9cos t − 3sin t e + c3 9sin t + 3cos t e5t , 0 10cos t 10sin t where c1 , c2 , and c3 are real. When c= 0 the trajectories tend toward the origin, and in other c= 2 3 cases the trajectories spiral away from the origin. 19. [M] Substitute R1 = 1/ 5, R2 = 1/ 3, C1 = 4, and C2 = 3 into the formula for A given in Example 1, and −2 use a matrix program to find the eigenvalues and eigenvectors: = A 1 1 −3 λ1 = −.5 : v1 = , λ 2 =−2.5 : v 2 = . The general solution is thus 2 2 3 / 4 , −1 1 −3 c1 4 4 1 −3 . By a = x(t ) c1 e −.5t + c2 e −2.5t. The condition x(0) = implies that = 2 c2 4 2 4 2 2 v (t ) 5 1 −.5t 1 −3 −2.5t matrix program, c1 = 5/ 2 and c2 =−1/ 2, so that 1 = . x= (t ) e − e 2 2 2 2 v2 (t ) 20. [M] Substitute R1 = 1 / 15, R2 = 1 / 3, C1 = 9 and C2 = 2 into the formula for A given in Example 1, −2 and use a matrix program to find the eigenvalues and eigenvectors: = A 3 / 2 1 −2 λ1 =−1 : v1 = , λ 2 =−2.5 : v 2 = . The general solution is thus 3 3 Copyright © 2016 Pearson Education, Ltd. 1 / 3 ,. −3 / 2 5.7 • Solutions 5-63 1 −2 c1 3 1 −2 3 . By a x(t ) c1 e −t + c2 e −2.5t. The condition x(0) = implies that = = 3 c2 3 3 3 3 3 v (t ) 5 1 −t 2 −2 −2.5t matrix program, c1 = 5/ 3 and c2 =−2 / 3, so that 1 = . x= e − e (t ) 3 3 3 3 v2 (t ) −1 21. [M] A = 5 −8 . Using a matrix program we find that an eigenvalue of A is −3 + 6i with −5 2 + 6i corresponding eigenvector v = . The conjugates of these form the second 5 eigenvalue-eigenvector pair. The general complex solution is 2 + 6i ( −3+ 6i )t 2 − 6i ( −3−6i )t , where c1 and c2 are arbitrary complex numbers. = x(t ) c1 + c2 e e 5 5 Rewriting the first eigenfunction and taking its real and imaginary parts, we have 2 + 6i −3t = ve( −3+ 6i )t e (cos 6t + i sin 6t ) 5 2cos 6t − 6sin 6t −3t 2sin 6t + 6cos 6t −3t = e + i e 5cos 6t 5sin 6t 2cos 6t − 6sin 6t −3t 2sin 6t + 6cos 6t −3t The general real solution has the form x(t ) c1 = e + c2 e , 5cos 6t 5sin 6t 0 where c1 and c2 now are real numbers. To satisfy the initial condition x(0) = , we solve 15 2 6 0 3, c2 = −1. We now have c1 + c2 = to get c1 = 5 0 15 −20sin 6t iL (t ) 2cos 6t − 6sin 6t −3t 2sin 6t + 6cos 6t −3t −3t 3 x(t ) = e − e = v (t ) = e . 5cos 6t 5sin 6t 15cos 6t − 5sin 6t C 0 22. [M] A = −.4 2 . Using a matrix program we find that an eigenvalue of A is −.4 + .8i with −.8 −1 − 2i corresponding eigenvector v = . The conjugates of these form the second eigenvalue 1 −1 − 2i ( −.4+.8i )t −1 + 2i ( −.4−.8i )t eigenvector pair. The general complex solution is x(t ) c1 , = + c2 e e 1 1 where c1 and c2 are arbitrary complex numbers. Rewriting the first eigenfunction and taking its real and imaginary parts, we have −1 − 2i −.4t = ve( −.4+.8i )t e (cos .8t + i sin .8t ) 1 − cos .8t + 2sin .8t −.4t − sin .8t − 2cos .8t −.4t = e + i e cos .8t sin .8t Copyright © 2016 Pearson Education, Ltd. 5-64 CHAPTER 5 • Eigenvalues and Eigenvectors − cos .8t + 2sin .8t −.4t − sin .8t − 2cos .8t −.4t The general real solution has the form x(t ) c1 = e + c2 e , cos .8t sin .8t 0 where c1 and c2 now are real numbers. To satisfy the initial condition x(0) = , we solve 12 −1 −2 0 12, c2 = −6. We now have c1 + c2 = to get c1 = 1 0 12 30sin .8t iL (t ) −.4t − cos .8t + 2sin .8t −.4t − sin .8t − 2cos .8t −.4t 12 e − 6 e = x(t ) = v (t ) = e cos .8t sin .8t 12cos .8t − 6sin .8t C 5.8 SOLUTIONS 1. The vectors in the given sequence approach an eigenvector v1. The last vector in the sequence, 1 x4 = , is probably the best estimate for v1. To compute an estimate for λ1 , examine .3326 4.9978 Ax 4 = . This vector is approximately λ1 v1. From the first entry in this vector, an estimate 1.6652 of λ1 is 4.9978. 2. The vectors in the given sequence approach an eigenvector v1. The last vector in the sequence, −.2520 x4 = , is probably the best estimate for v1. To compute an estimate for λ1 , examine 1 −1.2536 Ax 4 = . This vector is approximately λ1 v1. From the second entry in this vector, an 5.0064 estimate of λ1 is 5.0064. 3. The vectors in the given sequence approach an eigenvector v1. The last vector in the sequence, .5188 x4 = , is probably the best estimate for v1. To compute an estimate for λ1 , examine 1 .4594 Ax 4 = . This vector is approximately λ1 v1. From the second entry in this vector, an estimate .9075 of λ1 is .9075. 4. The vectors in the given sequence approach an eigenvector v1. The last vector in the sequence, 1 x4 = , is probably the best estimate for v1. To compute an estimate for λ1 , examine .7502 Copyright © 2016 Pearson Education, Ltd. 5.8 • Solutions 5-65 −.4012 Ax 4 = . This vector is approximately λ1 v1. From the first entry in this vector, an estimate −.3009 of λ1 is −.4012. 24991 5. Since A5 x = is an estimate for an eigenvector, the vector −31241 1 24991 −.7999 is a vector with a 1 in its second entry that is close to an v= − = 1 31241 −31241 4.0015 eigenvector of A. To estimate the dominant eigenvalue λ1 of A, compute Av = . From the −5.0020 second entry in this vector, an estimate of λ1 is −5.0020. −2045 1 −2045 −.4996 is an estimate for an eigenvector,= the vector v = is 6. Since A5 x = 1 4093 4093 4093 a vector with a 1 in its second entry that is close to an eigenvector of A. To estimate the dominant −2.0008 eigenvalue λ1 of A, compute Av = . From the second entry in this vector, an estimate of 4.0024 λ1 is 4.0024. 6 7 1 7. [M] = A , x= 0 0 . The data in the table below was calculated using Mathematica, which 8 5 carried more digits than shown here. k 0 1 2 3 4 5 xk 1 0 .75 1 1 .9565 .9932 1 1 .9990 .9998 1 Ax k 6 8 11.5 11.0 12.6957 12.7826 12.9592 12.9456 12.9927 12.9948 12.9990 12.9987 µk 8 11.5 12.7826 12.9592 12.9948 12.9990 The actual eigenvalue is 13. 2 1 1 8. [M] = A , x= 0 . The data in the table below was calculated using Mathematica, which 4 5 0 carried more digits than shown here. k 0 1 2 3 4 5 xk 1 0 .5 1 .2857 1 .2558 1 .2510 1 .2502 1 Copyright © 2016 Pearson Education, Ltd. 5-66 CHAPTER 5 • Eigenvalues and Eigenvectors Ax k 2 4 2 7 1.5714 6.1429 1.5116 6.0233 1.5019 6.0039 1.5003 6.0006 µk 4 7 6.1429 6.0233 6.0039 6.0006 The actual eigenvalue is 6. 0 12 8 1 9. [M] A = 1 −2 1 , x 0 = 0 . The data in the table below was calculated using Mathematica, 0 0 3 0 which carried more digits than shown here. k 0 1 2 3 4 5 6 xk 1 0 0 1 .125 0 1 .0938 .0469 1 .1004 .0328 1 .0991 .0359 1 .0994 .0353 1 .0993 .0354 Ax k 8 1 0 8 .75 .375 8.5625 .8594 .2812 8.3942 .8321 .3011 8.4304 .8376 .2974 8.4233 .8366 .2981 8.4246 .8368 .2979 µk 8 8 8.5625 8.3942 8.4304 8.4233 8.4246 Thus µ5 = 8.4233 and µ 6 = 8.4246. The actual eigenvalue is (7 + 97) / 2, or 8.42443 to five decimal places. 1 2 −2 1 10.= [M] A 1 1 = 9 , x0 0 . The data in the table below was calculated using Mathematica, 0 1 0 9 which carried more digits than shown here. k 0 1 2 3 4 5 6 xk 1 0 0 1 1 0 1 .6667 .3333 .3571 1 .7857 .0932 1 .9576 .0183 1 .9904 .0038 1 .9982 Ax k 1 1 0 3 2 1 1.6667 4.6667 3.6667 .7857 8.4286 8.0714 .1780 9.7119 9.6186 .0375 9.9319 9.9136 .0075 9.9872 9.9834 µk 1 3 4.6667 8.4286 9.7119 9.9319 9.9872 Thus µ5 = 9.9319 and µ 6 = 9.9872. The actual eigenvalue is 10. Copyright © 2016 Pearson Education, Ltd. 5.8 • Solutions 5-67 5 2 1 11. [M]= A ,= x0 . The data in the table below was calculated using Mathematica, which 2 2 0 carried more digits than shown here. k 0 1 2 3 4 xk 1 0 1 .4 1 .4828 1 .4971 1 .4995 Ax k 5 2 5.8 2.8 5.9655 2.9655 5.9942 2.9942 5.9990 2.9990 µk 5 5.8 5.9655 5.9942 5.9990 R(x k ) 5 5.9655 5.9990 5.99997 5.9999993 The actual eigenvalue is 6. The bottom two columns of the table show that R (x k ) estimates the eigenvalue more accurately than µ k . −3 2 1 12. [M] = A = , x 0 . The data in the table below was calculated using Mathematica, 2 0 0 which carried more digits than shown here. k 0 1 2 3 4 xk 1 0 −1 .6667 1 −.4615 −1 .5098 1 −.4976 Ax k −3 2 4.3333 −2.0000 −3.9231 2.0000 4.0196 −2.0000 −3.9951 2.0000 µk −3 −4.3333 −3.9231 −4.0196 −3.9951 R(x k ) −3 −3.9231 −3.9951 −3.9997 −3.99998 The actual eigenvalue is −4. The bottom two columns of the table show that R (x k ) estimates the eigenvalue more accurately than µ k . 13. If the eigenvalues close to 4 and −4 have different absolute values, then one of these is a strictly dominant eigenvalue, so the power method will work. But the power method depends on powers of the quotients λ 2 /λ1 and λ 3 /λ1 going to zero. If | λ 2 /λ1 | is close to 1, its powers will go to zero slowly, and the power method will converge slowly. 14. If the eigenvalues close to 4 and −4 have the same absolute value, then neither of these is a strictly dominant eigenvalue, so the power method will not work. However, the inverse power method may still be used. If the initial estimate is chosen near the eigenvalue close to 4, then the inverse power method should produce a sequence that estimates the eigenvalue close to 4. Copyright © 2016 Pearson Education, Ltd. 5-68 CHAPTER 5 • Eigenvalues and Eigenvectors 15. Suppose Ax = λx, with x ≠ 0. For any α , Ax − α Ix = (λ − α )x. If α is not an eigenvalue of A, then A − α I is invertible and λ − α is not 0; hence x= ( A − aaaa I ) −1 (λ − )x and (λ − ) −1 x= ( A − I ) −1 x This last equation shows that x is an eigenvector of ( A − αI ) −1 corresponding to the eigenvalue (λ − α ) −1. 16. Suppose that µ is an eigenvalue of ( A − αI ) −1 with corresponding eigenvector x. Since ( A − α I ) −1 x = µ x, x = ( A − α I )( µ x) = A( µ x) − (α I )( µ x) = µ ( Ax) − αµ x , solving this equation for 1 Ax, we find that Ax = (αµ x + x) = µ corresponding eigenvector x. 1 α + x . Thus λ= α + (1/µ ) is an eigenvalue of A with µ 10 17. [M] A = −8 −4 −8 13 5 −4 1 4 , x0 =0 ,α =3.3. The data in the table below was calculated using 4 0 Mathematica, which carried more digits than shown here. k 0 1 2 xk 1 0 0 1 .7873 .0908 1 .7870 .0957 yk 26.0552 20.5128 2.3669 47.1975 37.1436 4.5187 47.1233 37.0866 4.5083 µk 26.0552 47.1975 47.1233 νk 3.3384 3.32119 3.3212209 Thus an estimate for the eigenvalue to four decimal places is 3.3212. The actual eigenvalue is (25 − 337) / 2, or 3.3212201 to seven decimal places. 8 18. [M] A = 1 0 0 12 1 −2 1 , x0 =0 ,α =−1.4. The data in the table below was calculated using 0 3 0 Mathematica, which carried more digits than shown here. k 0 1 2 3 4 xk 1 0 0 1 .3646 −.7813 1 .3734 −.7854 1 .3729 −.7854 1 .3729 −.7854 Copyright © 2016 Pearson Education, Ltd. 5.8 • Solutions yk 40 14.5833 −31.25 −38.125 −14.2361 29.9479 −41.1134 −15.3300 32.2888 −40.9243 −15.2608 32.1407 −40.9358 −15.2650 32.1497 µk 40 −38.125 −41.1134 −40.9243 −40.9358 νk −1.375 −1.42623 −1.42432 −1.42444 −1.42443 Thus an estimate for the eigenvalue to four decimal places is −1.4244. The actual eigenvalue is (7 − 97) / 2, or −1.424429 to six decimal places. 10 7 19. [M] A = 8 7 7 8 7 5 6 5 = , x0 6 10 9 5 9 10 1 0 . 0 0 (a) The data in the table below was calculated using Mathematica (with α = 0 ), which carried more digits than shown here. k 0 1 2 3 xk 1 0 0 0 1 .7 .8 .7 .988679 .709434 1 .932075 .961467 .691491 1 .942201 Ax k 10 7 8 7 26.2 18.8 26.5 24.7 29.3774 21.1283 30.5547 28.7887 29.0505 20.8987 30.3205 28.6097 µk 10 26.5 30.5547 30.3205 k 4 5 6 7 xk .958115 .689261 1 .943578 .957691 .688978 1 .943755 .957637 .688942 1 .943778 .957630 .688938 1 .943781 Ax k 29.0110 20.8710 30.2927 28.5889 29.0060 20.8675 30.2892 28.5863 29.0054 20.8671 30.2887 28.5859 29.0053 20.8670 30.2887 28.5859 µk 30.2927 30.2892 30.2887 30.2887 Copyright © 2016 Pearson Education, Ltd. 5-69 5-70 CHAPTER 5 • Eigenvalues and Eigenvectors Thus an estimate for the eigenvalue to four decimal places is 30.2887. The actual eigenvalue is 30.2886853 to seven decimal places. An estimate for the corresponding eigenvector is .957630 .688938 . 1 .943781 (b) The data in the table below was calculated using Mathematica (with α = 0 ), which carried more digits than shown here. k 0 1 2 3 4 xk 1 0 0 0 −.609756 1 −.243902 .146341 −.604007 1 −.251051 .148899 −.603973 1 −.251134 .148953 −.603972 1 −.251135 .148953 yk 25 −41 10 −6 −59.5610 98.6098 −24.7561 14.6829 −59.5041 98.5211 −24.7420 14.6750 −59.5044 98.5217 −24.7423 14.6751 −59.5044 98.5217 −24.7423 14.6751 µk −41 98.6098 98.5211 98.5217 98.5217 νk −.0243902 .0101410 .0101501 .0101500 .0101500 Thus an estimate for the eigenvalue to five decimal places is .01015. The actual eigenvalue is .01015005 to eight decimal places. An estimate for the corresponding eigenvector is −.603972 1 . −.251135 .148953 3 2 1 2 2 12 13 11 , x 20. [M] A = = 0 −2 3 0 2 5 7 2 4 1 0 . 0 0 (a) The data in the table below was calculated using Mathematica, which carried more digits than shown here. k 0 1 2 3 4 xk 1 0 0 0 .25 .5 −.5 1 .159091 1 .272727 .181818 .187023 1 .170483 .442748 .184166 1 .180439 .402197 Copyright © 2016 Pearson Education, Ltd. 5.8 • Solutions Ax k 1 2 −2 4 1.75 11 3 2 3.34091 17.8636 3.04545 7.90909 3.58397 19.4606 3.51145 7.82697 3.52988 19.1382 3.43606 7.80413 µk 4 11 17.8636 19.4606 19.1382 k 5 6 7 8 9 xk .184441 1 .179539 .407778 .184414 1 .179622 .407021 .184417 1 .179615 .407121 .184416 1 .179615 .407108 .184416 1 .179615 .407110 Ax k 3.53861 19.1884 3.44667 7.81010 3.53732 19.1811 3.44521 7.80905 3.53750 19.1822 3.44541 7.80921 3.53748 19.1820 3.44538 7.80919 3.53748 19.1811 3.44539 7.80919 µk 19.1884 19.1811 19.1822 19.1820 19.1820 5-71 Thus an estimate for the eigenvalue to four decimal places is 19.1820. The actual eigenvalue is 19.1820368 to seven decimal places. An estimate for the corresponding eigenvector is .184416 1 . .179615 .407110 (b) The data in the table below was calculated using Mathematica, which carried more digits than shown here. k 0 1 2 xk 1 0 0 0 1 .226087 −.921739 .660870 1 .222577 −.917970 .660496 yk 115 26 −106 76 81.7304 18.1913 −75.0261 53.9826 81.9314 18.2387 −75.2125 54.1143 µk 115 81.7304 81.9314 νk .00869565 .0122353 .0122053 Copyright © 2016 Pearson Education, Ltd. 5-72 CHAPTER 5 • Eigenvalues and Eigenvectors Thus an estimate for the eigenvalue to four decimal places is .0122. The actual eigenvalue is .01220556 to eight decimal places. An estimate for the corresponding eigenvector is 1 .222577 . −.917970 .660496 .8 21. = a. A 0 0 .5 = , x . Here is the sequence Ak x for k = 1, …5 : .2 .5 .4 .32 .256 .2048 .16384 5 4 .1 , .02 , .004 , .0008 , .00016 . Notice that A x is approximately .8( A x). Conclusion: If the eigenvalues of A are all less than 1 in magnitude, and if x ≠ 0, then Ak x is approximately an eigenvector for large k. 1 0 .5 b. = A = , x . Here is the sequence Ak x for k = 1, …5 : 0 .8 .5 .5 .5 .5 .5 .5 .5 k .4 , .32 , .256 , .2048 , .16384 . Notice that A x seems to be converging to 0 . Conclusion: If the strictly dominant eigenvalue of A is 1, and if x has a component in the direction of the corresponding eigenvector, then { Ak x} will converge to a multiple of that eigenvector. 8 0 .5 c. = A = , x . Here is the sequence Ak x for k = 1,…5 : 0 2 .5 4 32 256 2048 16384 k 1 , 2 , 4 , , 16 . Notice that the distance of A x from either eigenvector of 8 A is increasing rapidly as k increases. Conclusion: If the eigenvalues of A are all greater than 1 in magnitude, and if x is not an eigenvector, then the distance from Ak x to the nearest eigenvector will increase as k → ∞. Chapter 5 SUPPLEMENTARY EXERCISES 1. a. True. If A is invertible and if Ax = 1 ⋅ x for some nonzero x, then left-multiply by A−1 to obtain x = A−1x, which may be rewritten as A−1x = 1 ⋅ x. Since x is nonzero, this shows 1 is an eigenvalue of A−1. b. False. If A is row equivalent to the identity matrix, then A is invertible. The matrix in Example 4 of Section 5.3 shows that an invertible matrix need not be diagonalizable. Also, see Exercise 31 in Section 5.3. c. True. If A contains a row or column of zeros, then A is not row equivalent to the identity matrix and thus is not invertible. By the Invertible Matrix Theorem (as stated in Section 5.2), 0 is an eigenvalue of A. Copyright © 2016 Pearson Education, Ltd. Chapter 5 • Supplementary Exercises 5-73 d. False. Consider a diagonal matrix D whose eigenvalues are 1 and 3, that is, its diagonal entries are 1 and 3. Then D 2 is a diagonal matrix whose eigenvalues (diagonal entries) are 1 and 9. In general, the eigenvalues of A2 are the squares of the eigenvalues of A. 2 e. True. Suppose a nonzero vector x satisfies Ax = λ x, then A= x A( A = x) A(λ = x) λ= Ax λ2 x This shows that x is also an eigenvector for A2 . f. True. Suppose a nonzero vector x satisfies Ax = λ x, then left-multiply by A−1 to obtain −1 A−1x, which = x A= (λ x) λ A−1x. Since A is invertible, the eigenvalue λ is not zero. So λ −1x = g. h. i. j. k. shows that x is also an eigenvector of A−1. False. Zero is an eigenvalue of each singular square matrix. True. By definition, an eigenvector must be nonzero. False.See Example 4 of Section 5.1. True. This follows from Theorem 4 in Section 5.2 False. Let A be the 3 × 3 matrix in Example 3 of Section 5.3. Then A is similar to a diagonal matrix D. The eigenvectors of D are the columns of I 3 , but the eigenvectors of A are entirely different. 2 0 1 0 l. False. Let A = . Then e1 = and e 2 = are eigenvectors of A, but e1 + e 2 is not. 0 0 3 1 (Actually, it can be shown that if two eigenvectors of A correspond to distinct eigenvalues, then their sum cannot be an eigenvector.) m. False. All the diagonal entries of an upper triangular matrix are the eigenvalues of the matrix (Theorem 1 in Section 5.1). A diagonal entry may be zero. n. True. Matrices A and AT have the same characteristic polynomial, because det( AT −= λI ) det( A − λ= I )T det( A − λI ), by the determinant transpose property. o. False. Counterexample: Let A be the 5 × 5 identity matrix. p. True. For example, let A be the matrix that rotates vectors through π/ 2 radians about the origin. Then Ax is not a multiple of x when x is nonzero. q. False. If A is a diagonal matrix with 0 on the diagonal, then the columns of A are not linearly independent. r. True. If Ax = λ1x and Ax = λ2 x, then λ1x = λ2 x and (λ1 − λ2 )x = 0. If x ≠ 0, then λ1 must equal λ2 . s. False. Let A be a singular matrix that is diagonalizable. (For instance, let A be a diagonal matrix with 0 on the diagonal.) Then, by Theorem 8 in Section 5.4, the transformation x Ax is represented by a diagonal matrix relative to a coordinate system determined by eigenvectors of A. t. True. By definition of matrix multiplication, = A AI = A[e1 e2 e= n ] [ Ae1 Ae2 Ae n ] If Ae j = d j e j for j =, 1 …, n, then A is a diagonal matrix with diagonal entries d1 , …, d n . u. True. If B = PDP −1 , where D is a diagonal matrix, and if A = QBQ −1 , then = A Q= ( PDP −1 )Q −1 (QP ) D( PQ) −1 , which shows that A is diagonalizable. Copyright © 2016 Pearson Education, Ltd. 5-74 CHAPTER 5 • Eigenvalues and Eigenvectors v. True. Since B is invertible, AB is similar to B ( AB ) B −1 , which equals BA. w. False. Having n linearly independent eigenvectors makes an n × n matrix diagonalizable (by the Diagonalization Theorem 5 in Section 5.3), but not necessarily invertible. One of the eigenvalues of the matrix could be zero. x. True. If A is diagonalizable, then by the Diagonalization Theorem, A has n linearly independent eigenvectors v1 , …, v n in R n . By the Basis Theorem, {v1 , …, v n } spans R n . This means that each vector in R n can be written as a linear combination of v1 , …, v n . 2. Suppose Bx ≠ 0 and ABx = λx for some λ . Then A( Bx) = λx. Left-multiply each side by B and obtain BA( Bx) = B (λx ) = λ ( Bx). This equation says that Bx is an eigenvector of BA, because Bx ≠ 0. 3. a. Suppose Ax = λx, with x ≠ 0. Then (5I − A)x= 5x − Ax= 5x − λx= (5 − λ )x. The eigenvalue is 5 − λ . b. (5 I − 3 A + A2 )x= 5x − 3 Ax + A( Ax= ) 5x − 3(λx) + λ 2 x= (5 − 3λ + λ 2 )x. The eigenvalue is 5 − 3λ + λ 2. 4. Assume that Ax = λx for some nonzero vector x. The desired statement is true for m = 1, by the assumption about λ . Suppose that for some k ≥ 1, the statement holds when m = k . That is, suppose that Ak x = λ k x. Then= Ak +1x A= ( Ak x) A(λk x) by the induction hypothesis. Continuing, k +1 k A= Ax λ k +1x, because x is an eigenvector of A corresponding to A. Since x is nonzero, this x λ= equation shows that λk +1 is an eigenvalue of Ak +1 , with corresponding eigenvector x. Thus the desired statement is true when m= k + 1. By the principle of induction, the statement is true for each positive integer m. 5. Suppose Ax = λx, with x ≠ 0. Then p ( A)x= (c0 I + c1 A + c2 A2 + …+ cn An )x = c0 x + c1 Ax + c2 A2 x + …+ cn An x = c0 x + c1λx + c2 λ 2 x + …+ cn λ n x = p (λ)x So p (λ) is an eigenvalue of p ( A). 6. a. If A = PDP −1 , then Ak = PD k P −1 , and B = 5 I − 3 A + A2 = 5 PIP −1 − 3PDP −1 + PD 2 P −1 = P (5 I −3D + D2 ) P−1 Since D is diagonal, so is 5 I − 3D + D 2 . Thus B is similar to a diagonal matrix. n −1 −1 2 −1 b. p ( A) = c0 I + c1 PDP + c2 PD P + + cn PD P = P (c0 I + c1D + c2 D2 + + cn Dn ) P−1 = Pp ( D) P−1 This shows that p ( A) is diagonalizable, because p ( D) is a linear combination of diagonal matrices and hence is diagonal. In fact, because D is diagonal, it is easy to see that Copyright © 2016 Pearson Education, Ltd. Chapter 5 p (2) p( D) = 0 • Supplementary Exercises 5-75 0 . p (7) 7. If A = PDP −1 , then p ( A) = Pp ( D) P −1 , as shown in Exercise 6. If the ( j, j ) entry in D is λ , then the ( j, j ) entry in D k is λ k , and so the ( j, j ) entry in p ( D) is p (λ ). If p is the characteristic polynomial of A, then p (λ) =0 for each diagonal entry of D, because these entries in D are the eigenvalues of A. Thus p ( D) is the zero matrix. Thus p ( A) = P ⋅ 0 ⋅ P −1 = 0. 8. a . The characteristic polynomial of A is 3−λ p(λ ) = det(A − λ I) = det 2 4 = (3 − λ )2 − 8 = λ 2 − 6λ + 1 . 3−λ Hence,
17 p(A) = A − 6A + I = 12 2 24 3 −6 17 2 4 1 0 0 0 + = . 3 0 1 0 0 The characteristic polynomial of B is 4−λ p(λ ) = det(B − λ I) = det 0 0 3 4−λ 0 Hence, 0 −3 −2 3 0 3 p(B) = (4I −B) = 0 0 −3 = 0 0 0 0 0 2 3 = (4 − λ )3 . 4−λ −3 −2 0 0 9 0 0 0 −3 0 0 0 = 0 0 0 0 0 0 0 0 0 0 0 . 0 b . Since A2 − 6A + I = 0, we have A2 = 6A − I A3 = AA2 = A(6A − I) = 6A2 − A = 6(6A − I) − A = 35A − 6I Morever, I = 6A − A2 = A(6I − A) implies that A−1 = 6I − A. 9. a . If λ is an eigenvalue of an n×n diagonalizable matrix A, then A = PDP−1 for an invertible matrix P and an n×n diagonal matrix D whose diagonal entries are the eigenvalues of A. If the multiplicity of λ is n, then λ must appear in every diagonal entry of D. That is, D = λ I . In this case, A = P(λ I)P−1 = λ PIP−1 = λ PP−1 = λ I . 2 b . Since the matrix A = 1 0 0 0 2 0 is triangular, its eigenvalues are on the diagonal. 1 2 Thus 2 is an eigenvalue with multiplicity 3. If the 3×3 matrix A were diagonalizable, then A would be 2I , by part (a). This is not the case, so A is not diagonalizable. Copyright © 2016 Pearson Education, Ltd. 5-76 CHAPTER 5 • Eigenvalues and Eigenvectors 10. To show that Ak tends to the zero matrix, it suffices to show that each column of Ak can be made as close to the zero vector as desired by taking k sufficiently large. The jth column of A is Ae j , where e j is the jth column of the identity matrix. Since A is diagonalizable, there is a basis for n consisting of eigenvectors v1 ,…, v n , corresponding to eigenvalues λ1 ,…,λ n . So there exist scalars c1 , …, cn , such that e j= c1 v1 + …+ cn v n (an eigenvector decomposition of e j ) . Then, for k = 1, 2,…, the vector Ak e j = c1 (λ1 ) k v1 + + cn (λ n ) k v n . If the eigenvalues are all less than 1 in absolute value, then their kth powers all tend to zero. So the equation shows that Ak e j tends to the zero vector, as desired. 11. a. Take x in H. Then x = cu for some scalar c. So Ax = A(cu) =c( Au) =c(λu) = (cλ)u, which shows that Ax is in H. b. Let x be a nonzero vector in K. Since K is one-dimensional, K must be the set of all scalar multiples of x. If K is invariant under A, then Ax is in K and hence Ax is a multiple of x. Thus x is an eigenvector of A. 12. Let U and V be echelon forms of A and B, obtained with r and s row interchanges, respectively, and no scaling. Then det A = (−1) r det U and det B = (−1) s det V Using first the row operations that reduce A to U, we can reduce G to a matrix of the form U Y G′ = . Then, using the row operations that reduce B to V, we can further reduce G′ to 0 B U G′′ = 0 Y . There will be r + s row interchanges, and so V A X U Y U Y Since = (−1) r + s det det G = det is upper triangular, its determinant 0 B 0 V 0 V equals the product of the diagonal entries, and since U and V are upper triangular, this product also equals (det U ) (det V ). Thus det G = (−1) r + s (det U )(det V ) = (det A)(det B ) . For any scalar λ , the matrix G − λI has the same partitioned form as G, with A − λI and B − λI as its diagonal blocks. (Here I represents various identity matrices of appropriate sizes.) Hence the result about det G shows that det(G −= λI ) det( A − λI ) ⋅ det( B − λI ) 13. By Exercise 12, the eigenvalues of A are the eigenvalues of the matrix [3] together with the 5 −2 5 eigenvalues of . The only eigenvalue of [3] is 3, while the eigenvalues of 3 −4 −4 1 and 7. Thus the eigenvalues of A are 1, 3, and 7.
−2 are 3 3 4 14. By Exercise 12, the eigenvalues of A are the eigenvalues of the matrix together with the 4 3 1 2 3 4 . The eigenvalues of are −1 and 7, while the eigenvalues of eigenvalues of 4 3 4 3 1 2 are −1 and −5. Thus the eigenvalues of A are − 1, − 5, and 7, and the eigenvalue −1 has 4 3 multiplicity 2. Copyright © 2016 Pearson Education, Ltd. Chapter 5 • Supplementary Exercises 5-77 15. Replace b by b −λ in the determinant formula from Exercise 16 in Chapter 3 Supplementary Exercises. det(A −λ I) = (na + b −λ )(b −λ )n−1 This determinant is zero only if na +b −λ = 0 or b −λ = 0. Thus λ is an eigenvalue of A if and only if λ = na + b or λ = b. From the formula for det(A −λ I) above, the algebraic multiplicity is 1 for na + b and n − 1 for b. 16. Since the first matrix has a = 7, b = −9, and n = 4, it has eigenvalues (4)(7) − 9 = 19 and −9 with algebraic multiplicities 1 and 3 respectively. Since the second matrix has a = −2, b = 9, and n = 5, it has eigenvalues (5)(−2) + 9 = −1 and 9 with algebraic multiplicities 1 and 4 respectively. 17. The characteristic polynomial is det(A − λ I) = (a − λ )(d − λ ) − bc = λ 2 − (a + d)λ + (ad − bc) = λ 2 − (trA)λ + det A . Using the quadratic formula to solve the characteristic equation gives us: p tr A ± (tr A)2 − 4 det A λ= 2 Therefore, eigenvalues of A are both real if and only if the discriminant is nonnegative, that is, (trA)2 − 4 det A ≥ 0 or (trA)2 ≥ 4 det A. Moreover, if (trA)2 = 4 det A, A has only one eigenvalue. 18. The eigenvalues of A are 1 and .6. Use this to factor A and Ak . −1 A = 2 −3 1 2 0 0 1 2 ⋅ .6 4 −2 −1 Ak = 2 −3 1k 2 0 ⋅ k .6 = 1 −1 4 2 = 1 −2 + 6(.6) 4 4−4(.6)k 1 −2 → 4 4 0 1 2 4 −2 2 −3 2 −2 ⋅ (.6) k k 3 −1 3 −1 k −(.6) 3 −3+3(.6)k 6−2(.6)k −3 as k → ∞ 6 1 −λ ; det(C p − λ I) = det = λ 2 + a1 λ + a0 = p(λ ). −a0 −a1 − λ 0 1 3 −λ 1 0 1 − λ det(C p − λ I) = det 0 −λ 1 = (−λ ) det −2 3−λ 0 −2 3−λ 0 1 19. C p = −a0 −a1 1 0 20. C p = 0 0 0 −2
= (−λ )(λ 2 − 3λ + 2) = −2λ + 3λ 2 − λ 3 = −p(λ ) Copyright © 2016 Pearson Education, Ltd. 5-78 CHAPTER 5 • Eigenvalues and Eigenvectors Since p(t) = t(t 2 − 3t + 2) = t(t − 1)(t − 2), the eigenvalues of its companion matrix are 0, 1 and 2, all three with multiplicity 1. 21. If p is a polynomial of order 2, then Exercise 19 shows that the characteristic polynomial of C p is det(C p − λ I) = (−1)2 p(λ ), so the result is true for n = 2. Suppose the result is true for n = k for some k ≥ 2, and consider a polynomial p of degree k + 1. Then expanding det(C p − λ I) by cofactors down the first column, the determinant of C p − λ I equals −λ 1 0 ··· 0 0 0 −λ 1 ··· 0 0 . .. + (−1)k+1 a . (−λ ) det . . 0 0 0 0 ··· −λ 1 −a1 −a2 −a3 · · · −ak−1 −ak − λ The k×k matrix shown is Cq − λ I , where q(t) = a1 + a2t + · · · + ak t k−1 + t k . By the induction assumption, the determinant of Cq − λ I is (−1)k q(λ ). Thus det(C p − λ I) = (−1)k+1 a0 + (−λ )(−1)k q(λ ) = (−1)k+1 a0 + λ (a1 + · · · + ak λ k−1 + λ k ) = (−1)k+1 p(λ ) So the formula holds for n = k + 1 when it holds for n = k. By the principle of induction, the formula for det(C p − λ I) is true for all n ≥ 2. 22. a. C p = 0 0 −a 0 1 0 − a1 0 1 − a2 b. Since λ is a zero of p, a0 + a1λ + a2 λ 2 + λ 3 =0 and − a0 − a1λ − a2 λ 2 = λ 3 . Thus 1 Cp λ 2 λ = − a0 λ λ2 − a1λ − a2λ2 λ 2 = λ 3 λ . That is, C p (1,λ,λ 2= ) λ (1,λ,λ 2 ), which shows that (1,λ,λ 2 ) is an eigenvector of C p corresponding to the eigenvalue λ . 23. From Exercise 22, the columns of the Vandermonde matrix V are eigenvectors of C p , corresponding to the eigenvalues λ1 ,λ 2 ,λ 3 (the roots of the polynomial p). Since these eigenvalues are distinct, the eigenvectors from a linearly independent set, by Theorem 2 in Section 5.1. Thus V has linearly independent columns and hence is invertible, by the Invertible Matrix Theorem. Finally, since the columns of V are eigenvectors of C p , the Diagonalization Theorem (Theorem 5 in Section 5.3) shows that V −1C pV is diagonal. Copyright © 2016 Pearson Education, Ltd. Chapter 5 • Supplementary Exercises 5-79 24. [M] The MATLAB command roots (p) requires as input a row vector p whose entries are the coefficients of a polynomial, with the highest order coefficient listed first. MATLAB constructs a companion matrix C p whose characteristic polynomial is p, so the roots of p are the eigenvalues of C p . The numerical values of the eigenvalues (roots) are found by the same QR algorithm used by the command eig(A). 25. [M] The MATLAB command [P D]= eig(A) produces a matrix P, whose condition number is 1.6 ×10 , and a diagonal matrix D, whose entries are almost 2, 2, 1. However, the exact eigenvalues of A are 2, 2, 1, and A is not diagonalizable. 8 26. [M] This matrix may cause the same sort of trouble as the matrix in Exercise 25. A matrix program that computes eigenvalues by an interative process may indicate that A has four distinct eigenvalues, all close to zero. However, the only eigenvalue is 0, with multiplicity 4, because A4 = 0. Copyright © 2016 Pearson Education, Ltd. 6.1 SOLUTIONS Notes: The first half of this section is computational and is easily learned. The second half concerns the concepts of orthogonality and orthogonal complements, which are essential for later work. Theorem 3 is an important general fact, but is needed only for Supplementary Exercise 13 at the end of the chapter and in Section 7.4. The optional material on angles is not used later. Exercises 27–31 concern facts used later. −1 4 v ⋅u 8 2 2 = . 1. Since u = and v = , u ⋅ u =(−1) + 2 =5 , v ⋅ u = 4(–1) + 6(2) = 8, and u ⋅u 5 2 6 3 6 2 2 2 2. Since w = −1 and x = −2 , w ⋅ w = 3 + (−1) + (−5) = 35 , x ⋅ w = 6(3) + (–2)(–1) + 3(–5) = 5, 3 −5 x⋅w 5 1 and = = . w ⋅ w 35 7 3 1 2 2 2 w= 3. Since w = −1 , w ⋅ w = 3 + (−1) + (−5) = 35 , and w⋅w −5 3/ 35 −1/ 35 . −1/ 7 −1 −1/ 5 1 2 2 u= 4. Since u = , u ⋅ u =(−1) + 2 =5 and . u ⋅u 2 2 / 5 −1 4 5. Since u = and v = , u ⋅ v = (–1)(4) + 2(6) = 8, v ⋅ v = 42 + 62 = 52, and 2 6 2 4 8 /13 u⋅v . = v = 13 6 12 /13 v⋅v 6-1 Copyright © 2016 Pearson Education, Ltd. 6-2 CHAPTER 6 • Orthogonality and Least Squares 6 3 2 2 2 6. Since x = −2 and w = −1 , x ⋅ w = 6(3) + (–2)(–1) + 3(–5) = 5, x ⋅ x= 6 + (−2) + 3 = 49, and −5 3 6 30 / 49 5 x⋅w x = −2 = −10 / 49 . 49 x⋅x 3 15 / 49 3 7. Since w = −1 , || w=|| −5 6 8. Since x = −2 , || x =|| 3 w ⋅ w= x ⋅ x= 32 + (−1) 2 + (−5) 2= 62 + (−2) 2 + 32= 35. 49= 7. 9. A unit vector in the direction of the given vector is −30 1 −30 −3/ 5 = = 2 40 50 40 4 / 5 (−30) + 40 1 2 10. A unit vector in the direction of the given vector is −6 4 = 2 2 2 (−6) + 4 + (−3) −3 1 −6 1 4 = 61 −3 −6 / 61 4 / 61 −3 61 11. A unit vector in the direction of the given vector is 7 / 4 1/ 2 = 2 2 2 (7 / 4) + (1/ 2) + 1 1 1 7 / 4 1 1/ 2 = 69 /16 1 7 / 69 2 / 69 4 / 69 12. A unit vector in the direction of the given vector is 8 / 3 1 = (8 / 3) 2 + 22 2 8 / 3 4 / 5 1 = 100 / 9 2 3/ 5 10 −1 2 2 2 13. Since x = and y = , || x − y ||= [10 − (−1)] + [−3 − (−5)] = 125 and −3 −5 dist (= x, y ) = 125 5 5. Copyright © 2016 Pearson Education, Ltd. 6.1 0 14. Since u = −5 and z = 2 dist (u= , z) • Solutions 6-3 −4 −1 , || u − z ||2 = [0 − (−4)]2 + [−5 − (−1)]2 + [2 − 8]2 = 68 and 8 = 68 2 17. 15. Since a ⋅ b = 8(–2) + (–5)( –3) = –1 ≠ 0, a and b are not orthogonal. 16. Since u ⋅ v = 12(2) + (3)( –3) + (–5)(3) = 0, u and v are orthogonal. 17. Since u ⋅ v = 3(–4) + 2(1) + (–5)( –2) + 0(6) = 0, u and v are orthogonal. 18. Since y ⋅ z = (–3)(1) + 7(–8) + 4(15) + 0(–7) = 1 ≠ 0, y and z are not orthogonal. 19. a. True. See the definition of || v ||. b. True. See Theorem 1(c). c. True. See the discussion of Figure 5. 1 1 d. False. Counterexample: . 0 0 e. True. See the box following Example 6. 20. a. b. c. d. e. True. See Example 1 and Theorem 1(a). False. The absolute value sign is missing. See the box before Example 2. True. See the defintion of orthogonal complement. True. See the Pythagorean Theorem. True. See Theorem 3. (u + v) w =+ (u v )w = u w+v w= u ⋅ w + v ⋅ w . The second and 21. Theorem 1(b): (u + v ) ⋅ w = third equalities used Theorems 3(b) and 2(c), respectively, from Section 2.1. Theorem 1(c): T T T T T (cu) ⋅ v= (cu)T v= c(uT v)= c(u ⋅ v) . The second equality used Theorems 3(c) and 2(d), respectively, from Section 2.1. 22. Since u ⋅ u is the sum of the squares of the entries in u, u ⋅ u ≥ 0. The sum of squares of numbers is zero if and only if all the numbers are themselves zero. 2 2 2 2 23. One computes that u ⋅ v = 2(–7) + (–5)( –4) + (–1)6 = 0, || u || = u ⋅ u = 2 + (−5) + (−1) = 30, || v ||2 = v ⋅ v = (−7) 2 + (−4) 2 + 62 = 101, and || u + v ||2 = (u + v) ⋅ (u + v) = (2 + (−7)) 2 + (−5 + (−4)) 2 + (−1 + 6) 2 =131. 2 2 2 24. One computes that || u + v || = (u + v ) ⋅ (u + v ) = u ⋅ u + 2u ⋅ v + v ⋅ v = || u || +2u ⋅ v + || v || and || u − v ||2 = (u − v) ⋅ (u − v) = u ⋅ u − 2u ⋅ v + v ⋅ v = || u ||2 −2u ⋅ v + || v ||2 , so || u + v ||2 + || u − v= ||2 || u ||2 +2u ⋅ v + || v ||2 + || u ||2 −2u ⋅ v + || v= ||2 2 || u ||2 +2 || v ||2 . Copyright © 2016 Pearson Education, Ltd. 6-4 CHAPTER 6 • Orthogonality and Least Squares a x 25. When v = , the set H of all vectors that are orthogonal to v is the subspace of vectors whose b y entries satisfy ax + by = 0. If a ≠ 0, then x = – (b/a)y with y a free variable, and H is a line through −b the origin. A natural choice for a basis for H in this case is . If a = 0 and b ≠ 0, then by = 0. a Since b ≠ 0, y = 0 and x is a free variable. The subspace H is again a line through the origin. A 1 −b natural choice for a basis for H in this case is , but is still a basis for H since a = 0 0 a and b ≠ 0. If a = 0 and b = 0, then H = 2 since the equation 0x + 0y = 0 places no restrictions on x or y. 26. Theorem 2 in Chapter 4 may be used to show that W is a subspace of 3 , because W is the null space of the 1 × 3 matrix uT . Geometrically, W is a plane through the origin. 27. If y is orthogonal to u and v, then y ⋅ u = y ⋅ v = 0, and hence by a property of the inner product, y ⋅ (u + v) = y ⋅ u + y ⋅ v = 0 + 0 = 0. Thus y is orthogonal to u + v. 28. An arbitrary w in Span{u, v} has the form = w c1u + c2 v . If y is orthogonal to u and v, then u ⋅ y = v ⋅ y = 0. By Theorem 1(b) and 1(c), w ⋅ y = (c1u + c2 v ) ⋅ y = c1 (u ⋅ y ) + c2 ( v ⋅ y ) = 0 + 0 = 0 w c1 v1 + …+ c p v p . If x is orthogonal to each v j , then by 29. A typical vector in W has the form = Theorems 1(b) and 1(c), w= ⋅ x (c1 v1 + …+ c p v p = ⋅ x) 0 ) ⋅ x c1 ( v1 ⋅ x) + …+ c p ( v p = So x is orthogonal to each w in W. 30. a. If z is in W ⊥ , u is in W, and c is any scalar, then (cz) ⋅ u = c(z ⋅ u) = c 0 = 0. Since u is any element of W, c z is in W ⊥ . b. Let z1 and z 2 be in W ⊥ . Then for any u in W, (z1 + z 2 ) ⋅ u = z1 ⋅ u + z 2 ⋅ u = 0 + 0 = 0. Thus z1 + z 2 is in W ⊥ . c. Since 0 is orthogonal to every vector, 0 is in W ⊥ . Thus W ⊥ is a subspace. 31. Suppose that x is in W and W ⊥ . Since x is in W ⊥ , x is orthogonal to every vector in W, including x itself. So x ⋅ x = 0, which happens only when x = 0. 32. [M] 0 for i ≠ j. a. One computes that ||= a1 || || = a 2 || || = a3 || || = a 4 || 1 and that ai ⋅ a j = b. Answers will vary, but it should be that || Au || = || u || and || Av || = || v ||. c. Answers will again vary, but the cosines should be equal. d. A conjecture is that multiplying by A does not change the lengths of vectors or the angles between vectors. Copyright © 2016 Pearson Education, Ltd. 6.2 • Solutions 6-5 33. [M] Answers to the calculations will vary, but will demonstrate that the mapping x⋅v n x T ( x) = v (for v ≠ 0) is a linear transformation. To confirm this, let x and y be in , and v⋅v let c be any scalar. Then x⋅v y⋅v (x + y ) ⋅ v (x ⋅ v) + (y ⋅ v) = T (x + y ) = T ( x) + T ( y ) v v = v + v = v⋅v v⋅v v⋅v v⋅v and (cx) ⋅ v c(x ⋅ v) x⋅v = T (cx) = v = v c= v cT (x) v ⋅ v v ⋅ v v⋅v 1 −5 −1 4 34. [M] One finds that N = = 1 0 , R 0 −1 0 3 1 0 0 0 1 0 5 1 0 0 0 1 −1/ 3 −4 / 3 . 1/ 3 The row-column rule for computing RN produces the 3 × 2 zero matrix, which shows that the rows of R are orthogonal to the columns of N. This is expected by Theorem 3 since each row of R is in Row A and each column of N is in Nul A. 6.2 SOLUTIONS Notes: The nonsquare matrices in Theorems 6 and 7 are needed for the QR factorization in Section 6.4. It is important to emphasize that the term orthogonal matrix applies only to certain square matrices. The subsection on orthogonal projections not only sets the stage for the general case in Section 6.3, it also provides what is needed for the orthogonal diagonalization exercises in Section 7.1, because none of the eigenspaces there have dimension greater than 2. For this reason, the Gram-Schmidt process (Section 6.4) is not really needed in Chapter 7. Exercises 13 and 14 are good preparation for Section 6.3. −1 3 1. Since 4 ⋅ −4 = 2 ≠ 0, the set is not orthogonal. −3 −7 1 0 2. Since −2 ⋅ 1 = 1 2 1 −5 −2 ⋅ −2 = 1 1 0 −5 1 ⋅ −2 = 0, the set is orthogonal. 2 1 −6 3 −30 ≠ 0, the set is not orthogonal. 3. Since −3 ⋅ 1 = 9 −1 Copyright © 2016 Pearson Education, Ltd. 6-6 CHAPTER 6 • Orthogonality and Least Squares 2 0 4. Since −5 ⋅ 0 = −3 0 2 4 −5 ⋅ −2 = −3 6 0 4 0 ⋅ −2 = 0, the set is orthogonal. 0 6 3 −1 −2 3 5. Since ⋅ = 1 −3 3 4 3 3 −2 8 ⋅ = 1 7 3 0 −1 3 3 8 ⋅ = 0, the set is orthogonal. −3 7 4 0 −4 3 1 3 6. Since ⋅ = −32 ≠ 0, the set is not orthogonal. −3 5 8 −1 7. Since u1 ⋅ u 2 = 12 − 12 = 0, {u1 , u 2 } is an orthogonal set. Since the vectors are non-zero, u1 and u 2 are linearly independent by Theorem 4. Two such vectors in 2 automatically form a basis for 2 . So {u1 , u 2 } is an orthogonal basis for 2 . By Theorem 5, x⋅u x ⋅ u2 1 x =1 u1 + u2 = 3u1 + u 2 u1 ⋅ u1 u2 ⋅ u2 2 8. Since u1 ⋅ u 2 =−6 + 6 =0, {u1 , u 2 } is an orthogonal set. Since the vectors are non-zero, u1 and u 2 are linearly independent by Theorem 4. Two such vectors in 2 automatically form a basis for 2 . So {u1 , u 2 } is an orthogonal basis for 2 . By Theorem 5, x⋅u x ⋅ u2 3 3 − u1 + u 2 x = 1 u1 + u2 = 2 4 u1 ⋅ u1 u2 ⋅ u2 9. Since u1 ⋅ u 2 = u1 ⋅ u3 = u 2 ⋅ u3 = 0, {u1 , u 2 , u3 } is an orthogonal set. Since the vectors are non-zero, u1 , u 2 , and u3 are linearly independent by Theorem 4. Three such vectors in automatically 3 form a basis for 3 . So {u1 , u 2 , u3 } is an orthogonal basis for 3 . By Theorem 5, x ⋅ u3 x ⋅ u1 x ⋅ u2 5 3 x= u1 + u2 + u3 = u1 − u 2 + 2u3 2 2 u1 ⋅ u1 u2 ⋅ u2 u3 ⋅ u3 10. Since u1 ⋅ u 2 = u1 ⋅ u3 = u 2 ⋅ u3 = 0, {u1 , u 2 , u3 } is an orthogonal set. Since the vectors are non-zero, u1 , u 2 , and u3 are linearly independent by Theorem 4. Three such vectors in automatically 3 form a basis for 3 . So {u1 , u 2 , u3 } is an orthogonal basis for 3 . By Theorem 5, x= x ⋅ u3 x ⋅ u1 x ⋅ u2 4 1 1 u1 + u2 + u3 = u1 + u 2 + u3 u1 ⋅ u1 u2 ⋅ u2 u3 ⋅ u3 3 3 3 Copyright © 2016 Pearson Education, Ltd. 6.2 • Solutions 6-7 1 −4 11. Let y = and u = . The orthogonal projection of y onto the line through u and the origin is 7 2 −2 y ⋅u 1 yˆ = u = u . the orthogonal projection of y onto u, and this vector is= u ⋅u 2 1 −1 1 12. Let y = and u = . The orthogonal projection of y onto the line through u and the origin is 3 −1 2 / 5 2 y ⋅u − u= the orthogonal projection of y onto u, and this vector is yˆ = u = −6 / 5 . 5 u ⋅u −4 / 5 y ⋅u 13 − u= 13. The orthogonal projection of y onto u is yˆ = u = 7 / 5 . The component of y u ⋅u 65 14/5 −4/5 14/5 orthogonal to u is y − yˆ = 8/5 . Thus y = yˆ + ( y − yˆ )= 7/5 + 8/5 . 14 / 5 2 y ⋅u = u = u . The component of y orthogonal 5 u ⋅u 2 / 5 14/5 −4/5 −4/5 to u is y − yˆ = 28/5 . Thus y = yˆ + ( y − yˆ )= 2/5 + 28/5 . yˆ 14. The orthogonal projection of y onto u is= 15. The distance from y to the line through u and the origin is ||y – yˆ ||. One computes that y − yˆ = y − 3 3 8 3/ 5 y ⋅u u= − = , so || y − yˆ ||= 9/25 +16/25 = 1 is the desired distance. u ⋅u 1 10 6 −4 / 5 16. The distance from y to the line through u and the origin is ||y – yˆ ||. One computes that y − yˆ = y − −3 1 −6 y ⋅u u = − 3 = , so || y − yˆ ||= u ⋅u 9 2 3 36 + 9 = 3 5 is the desired distance. 1/ 3 −1/ 2 0 . Since u ⋅ v = 0, {u, v} is an orthogonal set. However, || u ||2 = u ⋅ u = 1/ 3 17. Let u = 1/ 3 , v = 1/ 3 1/ 2 and || v || = v ⋅ v = 1/ 2, so {u, v} is not an orthonormal set. The vectors u and v may be normalized to form the orthonormal set 2 u v , = || u || || v || 0 18. Let u = 1 , v = 0 3 / 3 − 2 / 2 0 3 / 3 , 3 / 3 2 / 2 0 −1 . Since u ⋅ v = –1 ≠ 0, {u, v} is not an orthogonal set. 0 Copyright © 2016 Pearson Education, Ltd. 6-8 CHAPTER 6 • Orthogonality and Least Squares −.6 .8 2 19. Let u = , v = . Since u ⋅ v = 0, {u, v} is an orthogonal set. Also, || u || = u ⋅ u = 1 and .8 .6 || v ||2 = v ⋅ v = 1, so {u, v} is an orthonormal set. 1/ 3 −2 / 3 2 20. Let u = 1/ 3 , v = 2 / 3 . Since u ⋅ v = 0, {u, v} is an orthogonal set. However, || u || = u ⋅ u = 1 2 / 3 0 2 and || v || = v ⋅ v = 5 / 9, so {u, v} is not an orthonormal set. The vectors u and v may be normalized u v , to form the orthonormal set = || u || || v || 1/ 10 21. Let u = 3/ 20 , v = 3/ 20 3/ 10 −1/ 20 , and w = −1/ 20 −2 / 3 1/ 5 1/ 3 , 2 / 5 . 2 / 3 0 0 −1/ 2 . Since u ⋅ v = u ⋅ w = v ⋅ w = 0, {u, v, w} is an 1/ 2 2 2 2 orthogonal set. Also, || u || = u ⋅ u = 1, || v || = v ⋅ v = 1, and || w || = w ⋅ w = 1, so {u, v, w} is an orthonormal set. 1/ 18 1/ 2 −2 / 3 0 , and w = 1/ 3 . Since u ⋅ v = u ⋅ w = v ⋅ w = 0, {u, v, w} is an 22. Let u = 4 / 18 , v = −1/ 2 −2 / 3 1/ 18 2 2 2 orthogonal set. Also, || u || = u ⋅ u = 1, || v || = v ⋅ v = 1, and || w || = w ⋅ w = 1, so {u, v, w} is an orthonormal set. 23. a. b. c. d. e. True. For example, the vectors u and y in Example 3 are linearly independent but not orthogonal. True. The formulas for the weights are given in Theorem 5. False. See the paragraph following Example 5. False. The matrix must also be square. See the paragraph before Example 7. False. See Example 4. The distance is ||y – yˆ ||. 24. a. True. But every orthogonal set of nonzero vectors is linearly independent. See Theorem 4. b. False. To be orthonormal, the vectors is S must be unit vectors as well as being orthogonal to each other. c. True. See Theorem 7(a). d. True. See the paragraph before Example 3. e. True. See the paragraph before Example 7. 25. To prove part (b), note that (Ux) ⋅ (U y ) = (Ux)T (U y ) = xT U T U y = xT y = x⋅y Copyright © 2016 Pearson Education, Ltd. 6.2 • Solutions 6-9 because U T U = I . If y = x in part (b), (Ux) ⋅ (Ux) = x ⋅ x, which implies part (a). Part (c) of the Theorem follows immediately fom part (b). 26. A set of n nonzero orthogonal vectors must be linearly independent by Theorem 4, so if such a set spans W it is a basis for W. Thus W is an n-dimensional subspace of n , and W = n . 27. If U has orthonormal columns, then U T U = I by Theorem 6. If U is also a square matrix, then the equation U T U = I implies that U is invertible by the Invertible Matrix Theorem. T −1 28. If U is an n × n orthogonal matrix, then = I UU = UU T . Since U is the transpose of U , Theorem 6 applied to U T says that U T has orthogonal columns. In particular, the columns of U T are linearly independent and hence form a basis for n by the Invertible Matrix Theorem. That is, the rows of U form a basis (an orthonormal basis) for n . 29. Since U and V are orthogonal, each is invertible. By Theorem 6 in Section 2.2, UV is invertible and 1 −1 T T (UV ) −1 V −= (UV )T , where the final equality holds by Theorem 3 in Section 2.1. Thus U V= U = UV is an orthogonal matrix. 30. If U is an orthogonal matrix, its columns are orthonormal. Interchanging the columns does not change their orthonormality, so the new matrix – say, V – still has orthonormal columns. By Theorem 6, V T V = I . Since V is square, V T = V −1 by the Invertible Matrix Theorem. 31. Suppose that yˆ = y ⋅u u . Replacing u by cu with c ≠ 0 gives u ⋅u c(y ⋅ u) c 2 (y ⋅ u) y ⋅ (cu) y ⋅u = (cu) = ( c ) = u u = u yˆ 2 2 (cu) ⋅ (cu) u ⋅u c (u ⋅ u) c (u ⋅ u) So yˆ does not depend on the choice of a nonzero u in the line L used in the formula. 32. If v1 ⋅ v 2 = 0 , then by Theorem 1(c) in Section 6.1, (c1 v1 ) ⋅ (c2 v 2 ) = c1[ v1 ⋅ (c2 v 2 )] = c1c2 ( v1 ⋅ v 2 ) = c1c2 0 = 0 x⋅u u . For any vectors x and y in n and any u ⋅u scalars c and d, the properties of the inner product (Theorem 1) show that 33. Let L = Span{u}, where u is nonzero, and let T (x) = (cx + dy ) ⋅ u T (cx + dy ) = u u ⋅u cx ⋅ u + dy ⋅ u = u u ⋅u cx ⋅ u dy ⋅ u = u+ u u ⋅u u ⋅u = cT (x) + dT (y ) Thus T is a linear transformation. Another approach is to view T as the composition of the following three linear mappings: x a = x ⋅ v, a b = a / v ⋅ v, and b bv. Copyright © 2016 Pearson Education, Ltd. 6-10 CHAPTER 6 • Orthogonality and Least Squares 34. Let L = Span{u}, where u is nonzero, and let = T ( x) refl = 2projL y − y . By Exercise 33, the Ly mapping y projL y is linear. Thus for any vectors y and z in n and any scalars c and d, T (c y += d z ) 2 projL (c y + d z ) − (c y + d z ) = 2(c projL y + d projL z ) − c y − d z = 2c projL y − c y + 2d projL z − d z = c(2 projL y − y ) + d (2 projL z − z ) = cT (y ) + dT (z ) Thus T is a linear transformation. T 35. [M] One can compute that A A = 100 I 4 . Since the off-diagonal entries in AT A are zero, the columns of A are orthogonal. 36. [M] T a. One computes that U U = I 4 , while 82 0 −20 1 8 UU T = 100 6 20 24 0 0 42 24 0 −20 6 20 −32 −20 24 58 20 0 32 0 6 8 0 20 82 24 −20 6 0 6 −20 0 24 18 0 −8 20 20 6 32 −20 0 58 0 24 24 20 0 6 −8 0 18 −20 0 −32 6 0 20 24 −20 42 The matrices U T U and UU T are of different sizes and look nothing like each other. b. Answers will vary. The vector p = UU y is in Col U because p = U (U y ) . Since the columns of U are simply scaled versions of the columns of A, Col U = Col A. Thus each p is in Col A. c. One computes that U T z = 0 . d. From (c), z is orthogonal to each column of A. By Exercise 29 in Section 6.1, z must be T ⊥ orthogonal to every vector in Col A; that is, z is in (Col A) . Copyright © 2016 Pearson Education, Ltd. T 6.3 • Solutions 6.3 6-11 SOLUTIONS Notes: Example 1 seems to help students understand Theorem 8. Theorem 8 is needed for the GramSchmidt process (but only for a subspace that itself has an orthogonal basis). Theorems 8 and 9 are needed for the discussions of least squares in Sections 6.5 and 6.6. Theorem 10 is used with the QR factorization to provide a good numerical method for solving least squares problems, in Section 6.5. Exercises 19 and 20 lead naturally into consideration of the Gram-Schmidt process. 10 −6 x ⋅ u4 72 1. The vector in Span{u 4 } is = = u4 u 4 2= u 4 . Since −2 36 u4 ⋅ u4 2 10 10 0 −8 −6 −2 x ⋅ u4 x ⋅ u4 u 4 = − = is in x = c1u1 + c2u 2 + c3u3 + u 4 , the vector x − 2 −2 4 u4 ⋅ u4 u4 ⋅ u4 0 2 −2 Span{u1 , u 2 , u3 }. 2 4 v ⋅ u1 14 v ⋅ u1 2. The vector in Span{u1} is = u1 = u1 2= u1 . Since= x u1 + c2u 2 + c3u3 + c4u 4 , 2 u1 ⋅ u1 7 u1 ⋅ u1 2 4 2 5 4 v ⋅ u1 the vector v − u1 = − = −3 2 u1 ⋅ u1 3 2 2 1 is in Span{u , u , u }. 2 3 4 −5 1 3. Since u1 ⋅ u 2 =−1 + 1 + 0 =0, {u1 , u 2 } is an orthogonal set. The orthogonal projection of y onto 1 −1 −1 y ⋅ u1 y ⋅ u2 3 5 3 5 u1 + u 2 = u1 + u 2 = 1 + 1 = 4 . Span{u1 , u 2 } is yˆ = 2 2 2 2 u1 ⋅ u1 u2 ⋅ u2 0 0 0 4. Since u1 ⋅ u 2 =−12 + 12 + 0 =0, {u1 , u 2 } is an orthogonal set. The orthogonal projection of y onto 3 −4 6 y ⋅ u1 y ⋅ u2 30 15 6 3 u1 + u 2 = u1 − u 2 = 4 − 3 = 3 . Span{u1 , u 2 } is yˆ = 25 25 5 5 u1 ⋅ u1 u2 ⋅ u2 0 0 0 Copyright © 2016 Pearson Education, Ltd. 6-12 CHAPTER 6 • Orthogonality and Least Squares 5. Since u1 ⋅ u 2 = 3 + 1 − 4 = 0, {u1 , u 2 } is an orthogonal set. The orthogonal projection of y onto 3 1 −1 y ⋅ u1 y ⋅ u2 7 15 1 5 u1 + u2 = u1 − u 2 = −1 − −1 = 2 . Span{u1 , u 2 } is yˆ = u1 ⋅ u1 u2 ⋅ u2 14 6 2 2 2 −2 6 6. Since u1 ⋅ u 2 = 0 − 1 + 1 = 0, {u1 , u 2 } is an orthogonal set. The orthogonal projection of y onto −4 0 6 y ⋅ u1 y ⋅ u2 27 5 3 5 u1 + u2 = − u1 + u 2 = − −1 + 1 = Span{u1 , u 2 } is yˆ = 4 . 18 2 2 2 u1 ⋅ u1 u2 ⋅ u2 1 1 1 7. Since u1 ⋅ u 2 = 5 + 3 − 8 = 0, {u1 , u 2 } is an orthogonal set. By the Orthogonal Decomposition 10 / 3 −7 / 3 y ⋅ u1 y ⋅ u2 2 u1 + u 2 = 0u1 + u 2 = 2 / 3 , z = y − yˆ = 7 / 3 and y = yˆ + z, where Theorem, yˆ = 3 u1 ⋅ u1 u2 ⋅ u2 8 / 3 7 / 3 yˆ is in W and z is in W ⊥ . 8. Since u1 ⋅ u 2 =−1 + 3 − 2 =0, {u1 , u 2 } is an orthogonal set. By the Orthogonal Decomposition 3/ 2 −5 / 2 y ⋅ u1 y ⋅ u2 1 u1 + u 2 = 2u1 + u 2 = 7 / 2 , z = y − yˆ = 1/ 2 and y = yˆ + z, where yˆ Theorem, yˆ = 2 u1 ⋅ u1 u2 ⋅ u2 1 2 is in W and z is in W ⊥ . 9. Since u1 ⋅ u 2 = u1 ⋅ u3 = u 2 ⋅ u3 = 0, {u1 , u 2 , u3 } is an orthogonal set. By the Orthogonal Decomposition Theorem, 2 2 4 −1 y ⋅ u3 y ⋅ u1 y ⋅ u2 2 2 ˆy = ˆ , z = y − y = and y = yˆ + z, where u1 + u2 + u3 = 2u1 + u 2 − u3 = 0 3 3 3 u1 ⋅ u1 u2 ⋅ u2 u3 ⋅ u3 0 −1 yˆ is in W and z is in W ⊥ . 10. Since u1 ⋅ u 2 = u1 ⋅ u3 = u 2 ⋅ u3 = 0, {u1 , u 2 , u3 } is an orthogonal set. By the Orthogonal Decomposition Theorem, 5 −2 2 2 y ⋅ u3 y ⋅ u1 y ⋅ u2 1 14 5 yˆ = u1 + u2 + u3 = u1 + u 2 − u3 = , z = y − yˆ = and y = yˆ + z, 3 2 3 3 3 u1 ⋅ u1 u2 ⋅ u2 u3 ⋅ u3 6 0 where yˆ is in W and z is in W ⊥ . Copyright © 2016 Pearson Education, Ltd. 6.3 • Solutions 6-13 11. Note that v1 and v 2 are orthogonal. The Best Approximation Theorem says that yˆ , which is the orthogonal projection of y onto W = Span{v1 , v 2 }, is the closest point to y in W. This vector is 3 −1 y ⋅ v1 y ⋅ v2 1 3 yˆ = v1 + v 2 = v1 + v 2 = . 1 2 2 v1 ⋅ v1 v2 ⋅ v2 −1 12. Note that v1 and v 2 are orthogonal. The Best Approximation Theorem says that yˆ , which is the orthogonal projection of y onto W = Span{v1 , v 2 }, is the closest point to y in W. This vector is −1 −5 y ⋅ v1 y ⋅ v2 yˆ = v1 + v 2 = 3v1 + 1v 2 = . −3 v1 ⋅ v1 v2 ⋅ v2 9 13. Note that v1 and v 2 are orthogonal. By the Best Approximation Theorem, the closest point in −1 −3 z ⋅ v1 z ⋅ v2 2 7 v1 + v 2 = v1 − v 2 = . Span{v1 , v 2 } to z is zˆ = −2 3 3 v1 ⋅ v1 v2 ⋅ v2 3 14. Note that v1 and v 2 are orthogonal. By the Best Approximation Theorem, the closest point in 1 0 z ⋅ v1 z ⋅ v2 1 . v1 + v 2 = v1 + 0 v 2 = Span{v1 , v 2 } to z is zˆ = −1/ 2 v1 ⋅ v1 v2 ⋅ v2 2 −3/ 2 15. The distance from the point y in to a subspace W is defined as the distance from y to the closest point in W. Since the closest point in W to y is yˆ = projW y , the desired distance is || y – yˆ ||. One 3 3 2 computes that yˆ = −9 , y − yˆ = 0 , and || y − yˆ || = −1 6 40 = 2 10. 16. The distance from the point y in to a subspace W is defined as the distance from y to the closest point in W. Since the closest point in W to y is yˆ = projW y , the desired distance is || y – yˆ ||. One 4 −1 −5 ˆ , y − y= ˆ computes that y= −3 9 4 4 , and || y – yˆ || = 8. 4 4 Copyright © 2016 Pearson Education, Ltd. 6-14 CHAPTER 6 • Orthogonality and Least Squares 1 17. a. U U = 0 T 8/9 0 T , UU = −2 / 9 1 2 / 9 −2 / 9 5/9 4/9 2 / 9 4 / 9 . 5 / 9 T b. Since U U = I 2 , the columns of U form an orthonormal basis for W, and by Theorem 10 8 / 9 −2 / 9 2 / 9 4 2 −2 / 9 4 . projW y = 5 / 9 4 / 9 8 = UU y = 2 / 9 4 / 9 5 / 9 1 5 T = 18. a. U T U [1=] 1 / 10 T 1, UU= −3 / 10 −3 / 10 9 / 10 b. Since U T U = 1, {u1} forms an orthonormal basis for W, and by Theorem 10 1/10 T y = = proj W y UU −3/10 −3/10 7 −2 = . 9 /10 9 6 19. By the Orthogonal Decomposition Theorem, u3 is the sum of a vector in W = Span{u1 , u 2 } and a vector v orthogonal to W. This exercise asks for the vector v: 0 0 0 1 1 v = u3 − projW u3 = u3 − − u1 + u 2 = 0 − −2 / 5 = 2 / 5 . Any multiple of the vector v will 15 3 1 4 / 5 1/ 5 also be in W ⊥ . 20. By the Orthogonal Decomposition Theorem, u 4 is the sum of a vector in W = Span{u1 , u 2 } and a vector v orthogonal to W. This exercise asks for the vector v: 0 0 0 1 1 4 / 5 . Any multiple of the vector v will v= u 4 − projW u 4 = u 4 − u1 − u 2 = 1 − 1/ 5 = 6 30 0 −2 / 5 2 / 5 also be in W ⊥ . 21. a. True. See the calculations for z 2 in Example 1 or the box after Example 6 in Section 6.1. b. True. See the Orthogonal Decomposition Theorem. c. False. See the last paragraph in the proof of Theorem 8, or see the second paragraph after the statement of Theorem 9. d. True. See the box before the Best Approximation Theorem. e. True. Theorem 10 applies to the column space W of U because the columns of U are linearly independent and hence form a basis for W. 22. a. b. c. d. True. See the proof of the Orthogonal Decomposition Theorem. True. See the subsection “A Geometric Interpretation of the Orthogonal Projection.” True. The orthgonal decomposition in Theorem 8 is unique. False. The Best Approximation Theorem says that the best approximation to y is projW y. Copyright © 2016 Pearson Education, Ltd. 6.3 • Solutions 6-15 e. False. This statement is only true if x is in the column space of U. If n > p, then the column space of U will not be all of n , so the statement cannot be true for all x in n . 23. By the Orthogonal Decomposition Theorem, each x in n can be written uniquely as x = p + u, with ⊥ ⊥ p in Row A and u in (Row A) . By Theorem 3 in Section 6.1, (Row A) = Nul A, so u is in Nul A. Next, suppose Ax = b is consistent. Let x be a solution and write x = p + u as above. Then Ap = A(x – u) = Ax – Au = b – 0 = b, so the equation Ax = b has at least one solution p in Row A. Finally, suppose that p and p1 are both in Row A and both satisfy Ax = b. Then p − p1 is in Nul A = (Row A) ⊥ , since A(p − p1 ) = Ap − Ap1 = b − b = 0 . The equations p = p1 + (p − p1 ) and ⊥ p = p + 0 both then decompose p as the sum of a vector in Row A and a vector in (Row A) . By the uniqueness of the orthogonal decomposition (Theorem 8), p = p1 , and p is unique. 24. a. By hypothesis, the vectors w1 , …, w p are pairwise orthogonal, and the vectors v1 , …, v q are 0 for any i pairwise orthogonal. Since w i is in W for any i and v j is in W ⊥ for any j, w i ⋅ v j = and j. Thus {w1 ,…, w p , v1 ,…, v q } forms an orthogonal set. b. For any y in n , write y = yˆ + z as in the Orthogonal Decomposition Theorem, with yˆ in W and z in W ⊥ . Then there exist scalars c1 ,…, c p and d1 ,…, d q such that n y = yˆ + z = c1w1 + …+ c p w p + d1 v1 + …+ d q v q . Thus the set {w1 ,…, w p , v1 ,…, v q } spans . c. The set {w1 ,…, w p , v1 ,…, v q } is linearly independent by (a) and spans n by (b), and is thus a ⊥ basis for n . Hence dimW + dimW = p + q = dim n . T 25. [M] Since U U = I 4 , U has orthonormal columns by Theorem 6 in Section 6.2. The closest point to 1.2 .4 1.2 1.2 Τ y in Col U is the orthogonal projection yˆ of y onto Col U. From Theorem 10, = yˆ UU = y .4 1.2 .4 .4 Copyright © 2016 Pearson Education, Ltd. 6-16 CHAPTER 6 • Orthogonality and Least Squares 26.[M] The distance from b to Col U is || b – bˆ ||, where bˆ = UU Τb. One computes that .2 .8 .92 .08 .44 .56 1 0 112 Τ . which is 2.1166 to four decimal places. = = ,= , || b= − bˆ || bˆ UU b b − bˆ −.2 −.8 5 −.44 −.56 .6 −1.6 −.92 −.08 6.4 SOLUTIONS Notes: The QR factorization encapsulates the essential outcome of the Gram-Schmidt process, just as the LU factorization describes the result of a row reduction process. For practical use of linear algebra, the factorizations are more important than the algorithms that produce them. In fact, the Gram-Schmidt process is not the appropriate way to compute the QR factorization. For that reason, one should consider deemphasizing the hand calculation of the Gram-Schmidt process, even though it provides easy exam questions. The Gram-Schmidt process is used in Sections 6.7 and 6.8, in connection with various sets of orthogonal polynomials. The process is mentioned in Sections 7.1 and 7.4, but the one-dimensional projection constructed in Section 6.2 will suffice. The QR factorization is used in an optional subsection of Section 6.5, and it is needed in Supplementary Exercise 7 of Chapter 7 to produce the Cholesky factorization of a positive definite matrix. −1 x 2 ⋅ v1 v1 =x 2 − 3v1 = 5 . Thus an orthogonal basis for W 1. Set v1 = x1 and compute that v 2 =x 2 − v1 ⋅ v1 −3 3 −1 is 0 , 5 . −1 −3 Copyright © 2016 Pearson Education, Ltd. 6.4 • Solutions 6-17 5 x 2 ⋅ v1 1 v1 =x 2 − v1 = 4 . Thus an orthogonal basis for W 2. Set v1 = x1 and compute that v 2 =x 2 − v1 ⋅ v1 2 −8 0 5 is 4 , 4 . 2 −8 3 x 2 ⋅ v1 1 v1 =x 2 − v1 =3/ 2 . Thus an orthogonal basis for 3. Set v1 = x1 and compute that v 2 =x 2 − v1 ⋅ v1 2 3/ 2 W is 2 3 −5 , 3/ 2 1 3/ 2 . 3 x 2 ⋅ v1 v1= x 2 − (−2) v1= 6 . Thus an orthogonal basis for 4. Set v1 = x1 and compute that v 2 = x 2 − v1 ⋅ v1 3 3 3 W is −4 , 6 . 5 3 5 1 x ⋅v 5. Set v1 = x1 and compute that v 2 =x 2 − 2 1 v1 =x 2 − 2 v1 = . Thus an orthogonal basis for W −4 v1 ⋅ v1 −1 is 1 5 −4 1 , 0 −4 1 −1 . 4 6 x ⋅v 6. Set v1 = x1 and compute that v 2= x 2 − 2 1 v1= x 2 − (−3) v1= . Thus an orthogonal basis for −3 v1 ⋅ v1 0 3 4 6 −1 W is , . 2 −3 −1 0 Copyright © 2016 Pearson Education, Ltd. 6-18 CHAPTER 6 • Orthogonality and Least Squares || v 2 || 7. Since || v1 ||= 30 and= v1 v , 2 = || v1 || || v 2 || 2 / 30 2 / 6 −5 / 30 , 1/ 6 . 1/ 30 1/ 6 v 2 || 8. Since || v1 ||= 50 and || = v1 v , 2 = || v1 || || v 2 || = 27 / 2 3 6 / 2, an orthonormal basis for W is = 54 3 6, an orthonormal basis for W is 3/ 50 1/ 6 −4 / 50 , 2 / 6 . 5 / 50 1/ 6 9. Call the columns of the matrix x1 , x 2 , and x3 and perform the Gram-Schmidt process on these vectors: v1 = x1 1 3 x ⋅v v 2= x 2 − 2 1 v1= x 2 − (−2) v1= 3 v1 ⋅ v1 −1 −3 1 x ⋅v x ⋅v 3 1 v 3 = x3 − 3 1 v1 − 3 2 v 2 = x3 − v1 − − v 2 = 1 2 v1 ⋅ v1 v2 ⋅ v2 2 3 Thus an orthogonal basis for W is 3 1 −3 1 3 1 , , −1 3 1 3 −1 3 . 10. Call the columns of the matrix x1 , x 2 , and x3 and perform the Gram-Schmidt process on these vectors: v1 = x1 3 1 x ⋅v v 2= x 2 − 2 1 v1= x 2 − (−3) v1= 1 v1 ⋅ v1 −1 −1 −1 x ⋅v x ⋅v 1 5 v 3 =x3 − 3 1 v1 − 3 2 v 2 =x3 − v1 − v 2 = 3 v1 ⋅ v1 v2 ⋅ v2 2 2 −1 Copyright © 2016 Pearson Education, Ltd. 6.4 • Solutions Thus an orthogonal basis for W is −1 3 −1 3 1 −1 , , 1 1 3 1 −1 −1 . 11. Call the columns of the matrix x1 , x 2 , and x3 and perform the Gram-Schmidt process on these vectors: v1 = x1 3 0 x ⋅v v 2= x 2 − 2 1 v1= x 2 − (−1) v1= 3 v1 ⋅ v1 −3 3 2 0 x ⋅v x ⋅v 1 v 3 = x3 − 3 1 v1 − 3 2 v 2 = x3 − 4 v1 − − v 2 = 2 v1 ⋅ v1 v2 ⋅ v2 3 2 −2 Thus an orthogonal basis for W is 1 3 2 −1 0 0 −1 , 3 , 2 1 −3 2 1 3 −2 . 12. Call the columns of the matrix x1 , x 2 , and x3 and perform the Gram-Schmidt process on these vectors: v1 = x1 −1 1 x 2 ⋅ v1 v 2 =x 2 − v1 =x 2 − 4 v1 = 2 v1 ⋅ v1 1 1 3 3 x ⋅v x ⋅v 7 3 v 3 =x3 − 3 1 v1 − 3 2 v 2 =x3 − v1 − v 2 = 0 v1 ⋅ v1 v2 ⋅ v2 2 2 −3 3 Copyright © 2016 Pearson Education, Ltd. 6-19 6-20 CHAPTER 6 • Orthogonality and Least Squares Thus an orthogonal basis for W is 1 −1 3 −1 1 3 0 , 2 , 0 1 1 −3 1 1 3 5/ 6 T 13. Since A and Q are given,= R Q= A −1/ 6 −2 / 7 T 14. Since A and Q are given, R Q= A = 5/ 7 1/ 6 5/ 6 5/ 7 2/7 . −3/ 6 1/ 6 5 1/ 6 1 3/ 6 −3 1 9 7 6 = −5 0 5 12 . 6 2/7 −4 / 7 −2 4 / 7 5 2 / 7 2 4 3 7 7 = −2 0 6 7 . 7 15. The columns of Q will be normalized versions of the vectors v1 , v 2 , and v 3 found in Exercise 11. 1/ −1/ Q −1/ Thus= 1/ 1/ 5 1/ 2 5 0 5 1/ 2 5 −1/ 2 5 1/ 2 1/ 2 0 T , R Q= A 1/ 2 = 1/ 2 −1/ 2 5 0 0 4 5 −2 . 4 − 5 6 0 16. The columns of Q will be normalized versions of the vectors v1 , v 2 , and v 3 found in Exercise 12. 1/ 2 −1 / 2 Q Thus= 0 1/ 2 1/ 2 −1 / (2 2) 1 / (2 2) 1/ 2 1 / (2 2) 1 / (2 2) 1 / 2 1 / 2 T , R Q= A 0 = −1 / 2 1 / 2 2 0 0 8 2 2 0 7 3 2 . 6 17. a. False. Scaling was used in Example 2, but the scale factor was nonzero. b. True. See (1) in the statement of Theorem 11. c. True. See the solution of Example 4. 18. a. False. The three orthogonal vectors must be nonzero to be a basis for a three-dimensional subspace. (This was the case in Step 3 of the solution of Example 2.) b. True. If x is not in a subspace W, then x cannot equal projW x , because projW x is in W. This idea was used for v k +1 in the proof of Theorem 11. c. True. See Theorem 12. Copyright © 2016 Pearson Education, Ltd. 6.4 • Solutions 6-21 19. Suppose that x satisfies Rx = 0; then QRx = Q0 = 0, and Ax = 0. Since the columns of A are linearly independent, x must be 0. This fact, in turn, shows that the columns of R are linearly indepedent. Since R is square, it is invertible by the Invertible Matrix Theorem. 20. If y is in Col A, then y = Ax for some x. Then y = QRx = Q(Rx), which shows that y is a linear combination of the columns of Q using the entries in Rx as weights. Conversly, suppose that y = Qx −1 for some x. Since R is invertible, the equation A = QR implies that Q = AR . So −1 = y AR = x A( R −1x), which shows that y is in Col A. 21. Denote the columns of Q by {q1 ,…, q n } . Note that n ≤ m, because A is m × n and has linearly independent columns. The columns of Q can be extended to an orthonormal basis for m as follows. Let f1 be the first vector in the standard basis for m that is not= in Wn Span{q1 ,…, q n }, let u1= f1 − projWn f1 , and let q n+1 = u1 / || u1 || . Then {q1 ,…, q n , q n+1} is an orthonormal basis for = Wn +1 Span{q1 ,…, q n , q n +1}. Next let f 2 be the first vector in the standard basis for that is not m in Wn +1 , let u 2= f 2 − projWn+1 f 2 , and let q n+ 2 = u 2 / || u 2 || . Then {q1 ,…, q n , q n +1 , q n + 2 } is an orthogonal basis for Wn + 2 Span{q1 ,…, q n , q n +1 , q n + 2 }. This process will continue until m – n vectors = have been added to the original n vectors, and {q1 ,…, q n , q n +1 ,…, q m } is an orthonormal basis for = Q0 [q n +1 n . Let … q m ] and Q1 = [Q Q0 ] . Then, using partitioned matrix multiplication, R Q1 = = A. QR O 22. We may assume that {u1 ,…, u p } is an orthonormal basis for W, by normalizing the vectors in the original basis given for W, if necessary. Let U be the matrix whose columns are u1 ,…, u p . Then, by n T (x) proj = Theorem 10 in Section 6.3,= W x (UU ) x for x in . Thus T is a matrix transformation and hence is a linear transformation, as was shown in Section 1.8. T 23. Given A = QR, partition A = [ A1 A2 ] , where A1 has p columns. Partition Q as Q = [Q1 R11 where Q1 has p columns, and partition R as R = O Q2 ] R12 , where R11 is a p × p matrix. Then R22 R12 R Q2 ] 11 = [Q1R11 Q1R12 + Q2 R22 ] O R22 Thus A1 = Q1R11. The matrix Q1 has orthonormal columns because its columns come from Q. The matrix R11 is square and upper triangular due to its position within the upper triangular matrix R. The diagonal entries of R11 are positive because they are diagonal entries of R. Thus Q1 R11 is a QR factorization of A1 . 24. [M] Call the columns of the matrix x1 , x 2 , x3 , and x 4 and perform the Gram-Schmidt process on these vectors: = A [ A1 A= = 2 ] QR [Q1 v1 = x1 Copyright © 2016 Pearson Education, Ltd. 6-22 CHAPTER 6 • Orthogonality and Least Squares 3 3 x ⋅v v 2= x 2 − 2 1 v1= x 2 − (−1) v1= −3 v1 ⋅ v1 0 3 6 0 x ⋅v x ⋅v 1 4 v 3 = x3 − 3 1 v1 − 3 2 v 2 = x3 − − v1 − − v 2 = 6 v1 ⋅ v1 v2 ⋅ v2 2 3 6 0 0 5 x ⋅v x ⋅v x ⋅v 1 1 v 4 = x 4 − 4 1 v1 − 4 2 v 2 − 4 3 v 3= x 4 − v1 − (−1) v 2 − − v 3 = 0 2 v1 ⋅ v1 v2 ⋅ v2 v3 ⋅ v3 2 0 −5 Thus an orthogonal basis for W is −10 3 6 0 2 3 0 5 −6 , −3 , 6 , 0 16 0 6 0 2 3 0 −5 . 25. [M] The columns of Q will be normalized versions of the vectors v1 , v 2 , and v 3 found in Exercise 24. Thus −1/ 2 1/10 Q −3/10 = 4/5 1/10 1/ 2 1/ 3 1/ 2 0 −1/ 2 1/ 3 0 1/ 3 1/ 2 0 0 1/ 2 T , R Q= A 0 = 0 −1/ 2 20 0 0 0 −20 6 −10 −8 0 6 3 0 0 26. [M] In MATLAB, when A has n columns, suitable commands are Q = A(:,1)/norm(A(:,1)) % The first column of Q for j=2: n v=A(:,j) – Q*(Q’*A(:,j)) Q(:,j)=v/norm(v) % Add a new column to Q end Copyright © 2016 Pearson Education, Ltd. 10 −6 −3 3 5 2 6.5 6.5 • Solutions 6-23 SOLUTIONS Notes: This is a core section – the basic geometric principles in this section provide the foundation for all the applications in Sections 6.6–6.8. Yet this section need not take a full day. Each example provides a stopping place. Theorem 13 and Example 1 are all that is needed for Section 6.6. Theorem 15, however, gives an illustration of why the QR factorization is important. Example 4 is related to Exercise 17 in Section 6.6. 1. To find the normal equations and to find xˆ , compute −1 = A A 2 T 2 −1 2 −1 = 2 −3 −3 3 −1 3 4 2 −1 −4 6 −11 −1 T = 1 ; A b = . −11 22 3 11 2 −3 2 −11 x1 −4 = . 22 x2 11 6 T T a. The normal equations are ( A A)x = A b : −11 6 b. = Compute xˆ (= A A) A b −11 −1 T T −1 −11 −4 1 22 11 −4 1 33 3 = = . = 22 11 11 11 6 11 11 22 2 2. To find the normal equations and to find xˆ , compute 2 = A A 1 T 2 1 −5 2 8 12 2 −2 2 −24 T −2= 0 = 8 . 8 10 ; A b = 3 1 0 3 −2 1 2 3 −2 0 12 T T a. The normal equations are ( A A)x = A b : 8 12 b. = Compute xˆ (= A A) A b 8 −1 T T 8 x1 −24 = . 10 x2 −2 −1 8 −24 1 10 = 10 −2 56 −8 −8 −24 1 −224 −4 = = . 12 −2 56 168 3 3. To find the normal equations and to find xˆ , compute 1 −2 3 2 6 6 1 −1 0 2 −1 1 −1 0 2 1 6 T T ; A b = . = A A = = 2 3 5 0 3 6 42 2 3 5 −4 −6 −2 −2 5 2 2 6 T T a. The normal equations are ( A A)x = A b : 6 6 b. Compute xˆ = (Α Α) Α b = 6 T −1 = T −1 6 x1 6 = . 42 x2 −6 6 6 1 42 = 42 −6 216 −6 −6 6 . 6 −6 1 288 4 / 3 = 216 −72 −1/ 3 Copyright © 2016 Pearson Education, Ltd. 6-24 CHAPTER 6 • Orthogonality and Least Squares 4. To find the normal equations and to find xˆ , compute 3 1 5 1 1 6 1 1 1 3 3 1 T T 1 = A A = . 1 −1 = 3 11 ; A b = 3 −1 1 14 3 −1 1 1 0 1 3 3 x1 6 T T a. The normal equations are ( A A)x = A b : = . 3 11 x2 14 3 b. Compute xˆ = (ΑT Α)−1 ΑT b = 3 −1 3 6 1 11 = 11 14 24 −3 −3 6 1 24 1 = = . 3 14 24 24 1 5. To find the least squares solutions to Ax = b, compute and row reduce the augmented matrix for the 1 5 4 2 2 14 1 0 T T T T 4 ∼ 0 1 −1 −3 , so all vectors of the A A b 2 2 0 system A Ax = A b : A= 2 0 2 10 0 0 0 0 5 −1 form xˆ = −3 + x3 1 are the least-squares solutions of Ax = b. 0 1 6. To find the least squares solutions to Ax = b, compute and row reduce the augmented matrix for the 1 5 6 3 3 27 1 0 T T T T A A b 3 3 0 12 ∼ 0 1 −1 −1 , so all vectors of the system A Ax = A b : A= 3 0 3 15 0 0 0 0 5 −1 form xˆ = −1 + x3 1 are the least-squares solutions of Ax = b. 0 1 1 −1 7. From Exercise 3, A = 0 2 1 −1 Axˆ − b = 0 2 || Axˆ − b ||= −2 3 1 2 4/3 , b = , and xˆ = . Since −4 3 −1/3 5 2 −2 3 2 3 −1 2 4/3 1 −2 1 −3 , the least squares error is − = − = 3 −1/3 −4 −1 −4 3 5 2 1 2 −1 20 = 2 5. 1 A 1 8. From Exercise 4,= 1 3 5 1 −1 , b = 1 , and xˆ = . Since 1 0 1 Copyright © 2016 Pearson Education, Ltd. 6.5 3 1 5 4 5 −1 1 ˆ Axˆ − b = 1 −1 1 − 1 = 0 − 1 = −1 , the least squares error is || Ax − b ||= 1 0 2 0 2 1 • Solutions 6-25 6. 9. a. Because the columns a1 and a 2 of A are orthogonal, the method of Example 4 may be used to find bˆ , the orthogonal projection of b onto Col A: 1 5 1 ˆb = b ⋅ a1 a + b ⋅ a 2 a = 2 a + 1 a = 2 3 + 1 1 = 1 . 1 2 1 2 a1 ⋅ a1 a2 ⋅ a2 7 7 7 7 −2 4 0 b. The vector xˆ contains the weights which must be placed on a1 and a 2 to produce bˆ . These 2/7 weights are easily read from the above equation, so xˆ = . 1/7 10. a. Because the columns a1 and a 2 of A are orthogonal, the method of Example 4 may be used to find bˆ , the orthogonal projection of b onto Col A: 1 2 4 ⋅ ⋅ b a b a 1 1 1 2 bˆ = a1 + a 2 =3a1 + a 2 =3 −1 + 4 = −1 . a1 ⋅ a1 a2 ⋅ a2 2 2 1 2 4 b. The vector xˆ contains the weights which must be placed on a1 and a 2 to produce bˆ . These 3 weights are easily read from the above equation, so xˆ = . 1/ 2 11. a. Because the columns a1 , a 2 and a3 of A are orthogonal, the method of Example 4 may be used to find bˆ , the orthogonal projection of b onto Col A: b ⋅ a3 b ⋅ a1 b ⋅ a2 2 1 bˆ = a1 + a2 + a3 = a1 + 0a 2 + a3 3 3 a1 ⋅ a1 a2 ⋅ a2 a3 ⋅ a3 4 0 1 3 1 −5 2 1 1 1 . = +0 + = 1 3 0 4 3 6 1 −1 −5 −1 b. The vector xˆ contains the weights which must be placed on a1 , a 2 , and a3 to produce bˆ . These 2 / 3 xˆ 0 . weights are easily read from the above equation, so= 1/ 3 12. a. Because the columns a1 , a 2 and a3 of A are orthogonal, the method of Example 4 may be used to find bˆ , the orthogonal projection of b onto Col A: Copyright © 2016 Pearson Education, Ltd. 6-26 CHAPTER 6 • Orthogonality and Least Squares ˆ b ⋅ a1 a + b ⋅ a 2 a + b ⋅ a3 a= 1 a + 14 a + − 5 a b= 1 2 3 1 2 3 3 3 a1 ⋅ a1 a2 ⋅ a2 a3 ⋅ a3 3 1 1 0 5 1 0 2 1 14 5 −1 = + − = 3 0 3 1 3 1 3 −1 1 −1 6 b. The vector xˆ contains the weights which must be placed on a1 , a 2 , and a3 to produce bˆ . These 1/3 weights are easily read from the above equation, so xˆ = 14/3 . −5/3 11 0 2 , || b − Au ||=40 ; 13. One computes that Au = −11 , b − Au = 11 −6 7 4 3 , || b − Av ||=29 . Av = −12 , b − Av = 7 −2 Since Av is closer to b than Au is, Au is not the closest point in Col A to b. Thus u cannot be a leastsquares solution of Ax = b. 3 2 −4 , || b − Au ||=24 ; 14. One computes that Au = 8 , b − Au = 2 2 7 Av= 2 , b − Av= 8 −2 2 , || b − Av=|| −4 24 . Since Au and Av are equally close to b, and the orthogonal projection is the unique closest point in Col A to b, neither Au nor Av can be the closest point in Col A to b. Thus neither u nor v can be a least-squares solution of Ax = b. 3 T 15. The least squares solution satisfies Rxˆ = Q b. Since R = 0 5 7 and QT b = , the augmented 1 −1 3 R QT b matrix for the system may be row reduced to find = 0 4 xˆ = is the least squares solution of Ax = b. −1 Copyright © 2016 Pearson Education, Ltd. 5 1 7 1 ∼ −1 0 0 1 4 and so −1 6.5 2 T 16. The least squares solution satisfies Rxˆ = Q b. Since R = 0 6-27 3 17 / 2 and QT b = , the augmented 5 9 / 2 2 R QT b = matrix for the system may be row reduced to find 0 2.9 so xˆ = is the least squares solution of Ax = b. .9 17. a. b. c. d. e. • Solutions 3 17 / 2 1 ∼ 5 9 / 2 0 0 1 2.9 , and .9 True. See the beginning of the section. The distance from Ax to b is || Ax – b ||. True. See the comments about equation (1). False. The inequality points in the wrong direction. See the definition of a least-squares solution. True. See Theorem 13. True. See Theorem 14. 18. a. True. See the paragraph following the definition of a least-squares solution. b. False. If xˆ is the least-squares solution, then A xˆ is the point in the column space of A closest to b. See Figure 1 and the paragraph preceding it. c. True. See the discussion following equation (1). d. False. The formula applies only when the columns of A are linearly independent. See Theorem 14. e. False. See the comments after Example 4. f. False. See the Numerical Note. T T 19. a. If Ax = 0, then AT= Ax A= 0 0. This shows that Nul A is contained in Nul A A. T 2 T b. If AT Ax = 0, then xT AT = Ax x= 0 0. So ( Ax) ( Ax) = 0, which means that || Ax || = 0, and T hence Ax = 0. This shows that Nul A A is contained in Nul A. T 20. Suppose that Ax = 0. Then AT= Ax A= 0 0. Since AT A is invertible, x must be 0. Hence the columns of A are linearly independent. 21. a. If A has linearly independent columns, then the equation Ax = 0 has only the trivial solution. By Exercise 19, the equation AT Ax = 0 also has only the trivial solution. Since AT A is a square matrix, it must be invertible by the Invertible Matrix Theorem. b. Since the n linearly independent columns of A belong to m , m could not be less than n. c. The n linearly independent columns of A form a basis for Col A, so the rank of A is n. 22. Note that AT A has n columns because A does. Then by the Rank Theorem and Exercise 19, rank AT A = n − dim Nul AT A = n − dim Nul A = rank A Axˆ =( A AT A)−1 AT b. The matrix A( AT A) −1 AT is sometimes called the hat23. By Theorem 14, bˆ = matrix in statistics. 24. Since in this case AT A = I , the normal equations give xˆ = AT b. Copyright © 2016 Pearson Education, Ltd. 6-28 CHAPTER 6 • Orthogonality and Least Squares 2 2 x 6 25. The normal equations are = , whose solution is the set of all (x, y) such that x + y = 3. 2 2 y 6 The solutions correspond to the points on the line midway between the lines x + y = 2 and x + y = 4. 2 / 2, a= a2 ≈ .353535 and a1 = .5. Using .707 as an 0 2 / 2 , a= a2 ≈ .35355339 , a1 = .5. 0 26. [M] Using .7 as an approximation for approximation for 6.6 SOLUTIONS Notes: This section is a valuable reference for any person who works with data that requires statistical analysis. Many graduate fields require such work. Science students in particular will benefit from Example 1. The general linear model and the subsequent examples are aimed at students who may take a multivariate statistics course. That may include more students than one might expect. 1 0 1 1 , y 1. The design matrix X and the observation vector= y are X = 1 2 1 3 1 1 , and one can compute 2 2 6 T 4 6 ˆ .9 −1 T = = XT X X y . The least-squares line = , X y = , β ( X T X )= y β 0 + β1 x is thus 6 14 11 .4 y = .9 + .4x. 1 1 1 2 , y 2. The design matrix X and the observation vector= y are X = 1 4 1 5 0 1 , and one can compute 2 3 4 12 T 6 ˆ −.6 −1 T XT X X y . The least-squares line = = = , X y = , β ( X T X )= y β 0 + β1 x is 12 46 25 .7 thus y = –.6 + .7x. 1 −1 1 0 3. The design matrix X and the observation vector y are = X = ,y 1 1 2 1 0 1 , 2 4 4 2 T 7 ˆ 1.1 −1 T = XT X X y . The least-squares , X y = , β ( X T X )= and one can compute= 2 6 10 1.3 line = y β 0 + β1 x is thus y = 1.1 + 1.3x. Copyright © 2016 Pearson Education, Ltd. 6.6 • Solutions 6-29 1 2 3 1 3 , y 2 , and one can compute 4. The design matrix X and the observation vector= y are X = 1 5 1 1 6 0 4 16 T 6 ˆ 4.3 −1 T = XT X = , X y = , β ( X T X )= X y . The least-squares line = y β 0 + β1 x is 16 74 17 −.7 thus y = 4.3 – .7x. 5. If two data points have different x-coordinates, then the two columns of the design matrix X cannot be multiples of each other and hence are linearly independent. By Theorem 14 in Section 6.5, the normal equations have a unique solution. 6. If the columns of X were linearly dependent, then the same dependence relation would hold for the vectors in 3 formed from the top three entries in each column. That is, the columns of the matrix 1 x1 x12 2 1 x2 x2 would also be linearly dependent, and so this matrix (called a Vandermonde matrix) 2 1 x3 x3 would be noninvertible. Note that the determinant of this matrix is ( x2 − x1 )( x3 − x1 )( x3 − x2 ) ≠ 0 since x1 , x2 , and x3 are distinct. Thus this matrix is invertible, which means that the columns of X are in fact linearly independent. By Theorem 14 in Section 6.5, the normal equations have a unique solution. 7. a. The model that produces the correct least-squares fit is y = Xβ + where 1 1 1 1.8 2 4 2 2.7 β1 = X 3 = , and 3 . 9 , y = 3.4 , β = β 2 4 4 16 3.8 5 5 25 3.9 1.76 b. [M] One computes that (to two decimal places) βˆ = , so the desired least-squares equation −.20 2 = is y 1.76 x − .20 x . 8. a. The model that produces the correct least-squares fit is y = Xβ + ϵ where x1 x12 x13 y1 β1 1 = , y = , β β 2 = = X , and 2 3 yn β3 n xn xn xn Copyright © 2016 Pearson Education, Ltd. 6-30 CHAPTER 6 • Orthogonality and Least Squares 16 64 4 6 36 216 8 64 512 10 100 1000 b. [M] For the given data, X = and y = 12 144 1728 14 196 2744 16 256 4096 18 324 5832 1.58 2.08 2.5 2.8 , so 3.1 3.4 3.8 4.32 .5132 βˆ = ( X X ) X y = −.03348 , and the least-squares curve is .001016 T −1 T y= .5132 x − .03348 x 2 + .001016 x3 . 9. The model that produces the correct least-squares fit is y = Xβ + ϵ where 1 cos 1 sin 1 7.9 A = X cos 2 sin = 2 , y = 5.4 , β = , and 2 B 3 cos 3 sin 3 −.9 The model that produces the correct least-squares fit is y = Xβ + ϵ where e −.02(10) e −.07(10) 1 21.34 −.02(11) −.07(11) e e 2 20.68 M A −.07(12) 3 , = , β = X e −.02(12) e= , , and = y 20.05 M B e −.02(14) e −.07(14) 4 18.87 5 18.30 e −.02(15) e −.07(15) 10. a. 19.94 b. [M] One computes that (to two decimal places) βˆ = , so the desired least-squares 10.10 −.02t = + 10.10e −.07t . equation is y 19.94e 11. [M] The model that produces the correct least-squares fit is y = Xβ + ϵ where 3 cos .88 1 1 3 1 2.3 cos 1.1 2 2.3 β X 1 1.65 cos= , , , and β = y = = 3 . One computes that (to two decimal 1.42 1.65 e 4 1 1.25 cos 1.77 1.25 5 1 1.01 cos 2.14 1.01 1.45 places) βˆ = . Since e = .811 < 1 the orbit is an ellipse. The equation r = β / (1 – e cos ϑ) .811 produces r = 1.33 when ϑ = 4.6. Copyright © 2016 Pearson Education, Ltd. 6.6 • Solutions 6-31 12. [M] The model that produces the correct least-squares fit is y = Xβ + , where 1 1 3.78 91 1 4.11 98 2 β 0 3 . One computes that (to two decimal places) , y = X 1 4.39 , and = = 103 , β = β1 4 1 4.73 110 5 1 4.88 112 18.56 βˆ = , so the desired least-squares equation is p = 18.56 + 19.24 ln w. When w = 100, p ≈ 107 19.24 millimeters of mercury. 13. [M] a. The model that produces the correct least-squares fit is y = Xβ + ϵ where 0 0 1 0 1 0 0 1 1 1 8.8 1 2 1 22 23 2 29.9 1 3 32 33 3 62.0 42 43 1 4 4 104.7 2 3 β0 1 5 5 5 5 159.1 β 1 1 6 62 63 , y = X = = , and 6 . One computes that (to four 222.0 , β = β 2 1 7 7 294.5 72 73 β3 82 83 1 8 8 380.4 9 471.1 92 93 1 9 1 10 102 103 10 571.7 11 686.8 1 11 112 113 809.2 12 2 3 1 12 12 12 −.8558 4.7025 , so the desired least-squares polynomial is decimal places) βˆ = 5.5554 −.0274 y (t ) = −.8558 + 4.7025t + 5.5554t 2 − .0274t 3 . b. The velocity v(t) is the derivative of the position function y(t), so v(t ) =4.7025 + 11.1108t − .0822t 2 , and v(4.5) = 53.0 ft/sec. 14. Write the design matrix as [1 x ]. Since the residual vector ϵ = y – X βˆ is orthogonal to Col X, 0 = 1 ⋅ = 1 ⋅ (y − X βˆ ) = 1T y − (1T X ) βˆ =( y1 + …+ yn ) − n βˆ ∑ x βˆ0 =∑ y − nβˆ0 − βˆ1 ∑ x =ny − nβˆ0 − nβˆ1 x 1 y βˆ0 + βˆ1 x . This equation may be solved for y to find = Copyright © 2016 Pearson Education, Ltd. 6-32 CHAPTER 6 • Orthogonality and Least Squares 1 x1 1 … 1 = 15. Notice X X = x1 … xn 1 x n T y1 n x ∑ 1 … 1 ∑y T = . The = 2 ; X y ∑ x ∑ x x1 … xn y ∑ xy n T T equations (7) in the text follow immediately from the normal equations X X β = X y. 16. The determinant of the coefficient matrix of the equations in (7) is n∑ x 2 − (∑ x) 2 . Using the 2 × 2 βˆ0 1 formula for the inverse of the coefficient matrix, = 2 2 ˆ β1 n∑ x − (∑ x) Hence βˆ0 ∑ x2 −∑ x −∑ x ∑ y . n ∑ xy (∑ x 2 )(∑ y ) − (∑ x)(∑ xy ) ˆ n∑ xy − (∑ x)(∑ y ) , β1 = . n ∑ x 2 − (∑ x) 2 n ∑ x 2 − (∑ x) 2 Note: A simple algebraic calculation shows that formula for βˆ once βˆ is known 0 n βˆ0 , which provides a simple ∑ y − (∑ x) βˆ1 = 1 17. a. The mean of the data in Example 1 is x = 5.5, so the data in mean-deviation form are 1 1 (–3.5, 1), (–.5, 2), (1.5, 3), (2.5, 3), and the associated design matrix is X = 1 1 of X are orthogonal because the entries in the second column sum to 0. −3.5 −.5 . The columns 1.5 2.5 0 β 0 9 4 T T b. The normal equations are X X β = X y , or = . One computes that 0 21 β1 7.5 9 / 4 (9 / 4) + (5/14) x* = (9 / 4) + (5/14)( x − 5.5). βˆ = , so the desired least-squares line is y = 5 /14 1 x1 1 … 1 = 18. Since X X = x1 … xn 1 x n T n ∑ x ∑ x , 2 ∑ x X T X is a diagonal matrix when ∑ x = 0. 19. The residual vector ϵ = y – X βˆ is orthogonal to Col X, while yˆ =X βˆ is in Col X. Since ϵ and yˆ are thus orthogonal, apply the Pythagorean Theorem to these vectors to obtain SS(T) = || y ||2 = || yˆ + ||2 = || yˆ ||2 + || ||2 = || X βˆ ||2 + || y − X βˆ ||2 = SS(R) + SS(E) . T T 20. Since βˆ satisfies the normal equations, X X βˆ = X y , and T T ˆ )T ( X βˆ ) βˆ= || X βˆ ||2 ( X β= = X X βˆ βˆ T X T y . Since || X βˆ ||2 = SS(R) and = yT y ||= y ||2 SS(T) , SS(T) − SS(R) = yT y − βˆ T X T y . Exercise 19 shows that SS(E) = Copyright © 2016 Pearson Education, Ltd. 6.7 6.7 • Solutions 6-33 SOLUTIONS Notes: The three types of inner products described here (in Examples 1, 2, and 7) are matched by examples in Section 6.8. It is possible to spend just one day on selected portions of both sections. Example 1 matches the weighted least squares in Section 6.8. Examples 2–6 are applied to trend analysis in Seciton 6.8. This material is aimed at students who have not had much calculus or who intend to take more than one course in statistics. For students who have seen some calculus, Example 7 is needed to develop the Fourier series in Section 6.8. Example 8 is used to motivate the inner product on C[a, b]. The Cauchy-Schwarz and triangle inequalities are not used here, but they should be part of the training of every mathematics student. 1. The inner product is 〈 x= , y〉 4 x1 y1 + 5 x2 y2 . Let x = (1, 1), y = (5, –1). 2 2 a. Since || x || = 〈 x, x〉 = 9, || x || = 3. Since || y || = 〈 y, y〉 = 105, || y ||= 105. Finally, | 〈 x, y〉 |2= 152= 225. b. A vector z is orthogonal to y if and only if 〈x, y〉 = 0, that is, 20 z1 − 5 z2 = 0, or 4 z1 = z2 . Thus 1 all multiples of are orthogonal to y. 4 2. The inner product is 〈 x= , y〉 4 x1 y1 + 5 x2 y2 . Let x = (3, –2), y = (–2, 1). Compute that 1156 . || x ||2 = 〈 x, x〉 = 56, || y ||2 = 〈 y, y〉 = 21, || x ||2 || y ||2 = 56 ⋅ 21 = 1176 , 〈x, y〉 = –34, and | 〈 x, y〉 |2 = 2 2 2 Thus | 〈 x, y〉 | ≤ || x || || y || , as the Cauchy-Schwarz inequality predicts. 3. The inner product is 〈 p, q〉 = p(–1)q(–1) + p(0)q(0) + p(1)q(1), so 2 〈 4 + t ,5 − 4t = 〉 3(1) + 4(5) + 5(1) = 28 . 2 2 4. The inner product is 〈 p, q〉 = p(–1)q(–1) + p(0)q(0) + p(1)q(1), so 〈3t − t , 3 + 2t 〉 = (−4)(5) + 0(3) + 2(5) = −10. 5. The inner product is 〈 p, q〉 = p(–1)q(–1) + p(0)q(0) + p(1)q(1), so 〈 p, p〉 = 〈 4 + t , 4 + t 〉 = 32 + 42 + 52 = 50 and || p ||= 〈 q, q〉 = 〈5 − 4t 2 ,5 − 4t 2 〉 = 12 + 52 + 12 = 27 and || q ||= 〈 p, p〉 = 50 = 5 2 . Likewise 〈 q, q〉 = 27 = 3 3 . 2 2 6. The inner product is 〈 p, q〉 = p(–1)q(–1) + p(0)q(0) + p(1)q(1), so 〈 p, p〉 = 〈3t − t ,3t − t 〉 = (−4) 2 + 02 + 22 = 20 and || p ||= 52 + 32 + 52 = 59 and || q ||= 〈 p, p〉 = 〈 q, q〉 = 20 = 2 5. Likewise 〈 q, q〉 = 〈3 + 2t 2 ,3 + 2t 2 〉 = 59. 7. The orthogonal projection qˆ of q onto the subspace spanned by p is 〈 q, p〉 28 56 14 qˆ = p= (4 + t ) = + t. 〈 p, p〉 50 25 25 Copyright © 2016 Pearson Education, Ltd. 6-34 CHAPTER 6 • Orthogonality and Least Squares 8. The orthogonal projection qˆ of q onto the subspace spanned by p is 〈 q, p〉 10 3 1 qˆ = p= − (3t − t 2 ) = − t + t2 . 〈 p, p〉 20 2 2 9. The inner product is 〈p, q〉 = p(–3)q(–3) + p(–1)q(–1) + p(1)q(1) + p(3)q(3). a. The orthogonal projection pˆ 2 of p2 onto the subspace spanned by p0 and p1 is 〈 p2 , p0 〉 〈p , p 〉 20 0 pˆ 2 = p0 + 2 1 p1 = (1) + t = 5 . 〈 p0 , p0 〉 〈 p1 , p1 〉 4 20 2 b. The vector q= p2 − pˆ 2= t − 5 will be orthogonal to both p0 and p1 and { p0 , p1 , q} will be an orthogonal basis for Span{ p0 , p1 , p2 }. The vector of values for q at (–3, –1, 1, 3) is (4, –4, –4, 4), = q (1/ 4)(t 2 − 5). so scaling by 1/4 yields the new vector 3 10. The best approximation to p = t by vectors in W = Span{ p0 , p1 , q} will be 〈 p, p0 〉 〈 p, p1 〉 0 164 0 t 2 − 5 41 〈 p, q〉 projW p = (t ) + pˆ = p0 + p1 + q = (1) + = t. 4 20 4 4 5 〈 p0 , p0 〉 〈 p1 , p1 〉 〈 q, q〉 11. The orthogonal projection of p = t onto W = Span{ p0 , p1 , p2 } will be 〈 p, p0 〉 〈 p, p1 〉 〈 p, p2 〉 0 34 0 17 (1) + (t ) + (t 2 − 2)= pˆ= projW p= p0 + p1 + p2= t. 5 10 14 5 〈 p0 , p0 〉 〈 p1 , p1 〉 〈 p2 , p2 〉 3 p − projW p = t − (17 / 5)t will make { p0 , p1 , p2 , p3 } an 12. Let W = Span{ p0 , p1 , p2 }. The vector p3 = orthogonal basis for the subspace 3 of 4. The vector of values for p3 at (–2, –1, 0, 1, 2) is 3 3 (–6/5, 12/5, 0, –12/5, 6/5), so scaling by 5/6 yields the new vector p3 = (5 / 6)(t − (17 / 5)t ) = (5/ 6)t 3 − (17 / 6)t. 13. Suppose that A is invertible and that 〈u, v〉 = (Au) ⋅ (Av) for u and v in n . Check each axiom in the definition of an inner product space, using the properties of the dot product. i. 〈u, v〉 = (Au) ⋅ (Av) = (Av) ⋅ (Au) = 〈v, u〉 ii. 〈u + v, w〉 = (A(u + v)) ⋅ (Aw) = (Au + Av) ⋅ (Aw) = (Au) ⋅ (Aw) + (Av) ⋅ (Aw) = 〈u, w〉 + 〈v, w〉 iii. 〈c u, v〉 = (A( cu)) ⋅ (Av) = (c(Au)) ⋅ (Av) = c((Au) ⋅ (Av)) = c〈u, v〉 , u〉 ( Au) ⋅ ( A= u) || Au ||2 ≥ 0, and this quantity is zero if and only if the vector Au is 0. But iv. 〈u= Au = 0 if and only u = 0 because A is invertible. 14. Suppose that T is a one-to-one linear transformation from a vector space V into n and that 〈u, v〉 = T(u) ⋅ T(v) for u and v in n . Check each axiom in the definition of an inner product space, using the properties of the dot product and T. The linearity of T is used often in the following. i. 〈u, v〉 = T(u) ⋅ T(v) = T(v) ⋅ T(u) = 〈v, u〉 ii. 〈u + v, w〉 = T(u + v) ⋅ T(w) = (T(u) + T(v)) ⋅ T(w) = T(u) ⋅ T(w) + T(v) ⋅ T(w) = 〈u, w〉 + 〈v, w〉 iii. 〈cu, v〉 = T(cu) ⋅ T(v) = (cT(u)) ⋅ T(v) = c(T(u) ⋅ T(v)) = c〈u, v〉 u〉 T (u) ⋅ T (= u) || T (u) ||2 ≥ 0, and this quantity is zero if and only if u = 0 since T is a oneiv. 〈u,= to-one transformation. Copyright © 2016 Pearson Education, Ltd. 6.7 • Solutions 6-35 15. Using Axioms 1 and 3, 〈u, c v〉 = 〈c v, u〉 = c〈v, u〉 = c〈u, v〉. 16. Using Axioms 1, 2 and 3, || u − v ||2 =〈u − v, u − v〉 =〈u, u − v〉 − 〈 v, u − v〉 = 〈u, u〉 − 〈u, v〉 − 〈 v, u〉 + 〈 v, v〉 = 〈u, u〉 − 2〈u, v〉 + 〈 v, v〉 = || u ||2 −2〈u, v〉 + || v ||2 u ||2 ||= v ||2 1 and 〈u, v〉 = 0. So || u − v ||2 = 2. Since {u, v} is orthonormal, ||= 17. Following the method in Exercise 16, || u + v ||2 =〈u + v, u + v〉 =〈u, u + v〉 + 〈 v, u + v〉 = 〈u, u〉 + 〈u, v〉 + 〈 v, u〉 + 〈 v, v〉 = 〈u, u〉 + 2〈u, v〉 + 〈 v, v〉 = || u ||2 + 2〈u, v〉 + || v ||2 2 2 4 u, v〉 , and dividing by 4 gives the Subtracting these results, one finds that || u + v || − || u − v || =〈 desired identity. v ||2 || u ||2 −2〈u, v〉 + || v ||2 and || u + v ||2 = 18. In Exercises 16 and 17, it has been shown that || u − = ||2 2 || u ||2 + 2 || v ||2 . || u ||2 + 2〈u, v〉 + || v ||2 . Adding these two results gives || u + v ||2 + || u − v= b a 2 2 19. let u = and v = . Then || u || = a + b, || v || = a + b, and 〈u, v〉 =2 ab . Since a and b are a b =|| nonnegative, || u a + b , || v=|| a + b . Plugging these values into the Cauchy-Schwarz inequality gives 2 ab =| 〈u, v〉 | ≤ || u || || v || = a + b a + b =a + b . Dividing both sides of this equation by 2 gives the desired inequality. 20. The Cauchy-Schwarz inequality may be altered by dividing both sides of the inequality by 2 and then 2 a || u ||2 || v ||2 〈 u, v 〉 ≤ squaring both sides of the inequality. The result is . Now let u = and 4 2 b 1 2 a 2 + b 2 , || v ||2 = 2 , and 〈u, v〉 = a + b. Plugging these values into the v = . Then || u ||= 1 inequality above yields the desired inequality. 2 3 21. The inner product is 〈 f , g 〉 =∫ f (t ) g (t )dt. Let f (t ) = 1 − 3t , g (t ) = t − t . Then 1 0 〈 f , g= 〉 1 ∫0 (1 − 3t 2 )(t − t 3 ) = dt 1 ∫0 3t 5 − 4t 3 + t = dt 0 . ) t 3 − t 2 . Then 22. The inner product is 〈 f , g 〉 =∫ f (t ) g (t ) dt. Let f (t) = 5t – 3, g (t= 1 0 〈 f= , g〉 1 ∫0 (5t − 3)(t − t= )dt 3 2 1 ∫0 5t 4 − 8t 3 + 3= t 2 dt 0 . Copyright © 2016 Pearson Education, Ltd. 6-36 CHAPTER 6 • Orthogonality and Least Squares 1 23. The inner product is 〈 f , g 〉 =∫ f (t ) g (t ) dt , so 〈 f , = f〉 0 || f ||= ) = dt 2 2 1 4 1 6 − 2t 5 + t 4= dt 1/105, and ∫0 9t − 6t 2 + 1 = dt 4 / 5, and 〈 f , f 〉 = 2 / 5. 1 24. The inner product is 〈 f , g 〉 =∫ f (t ) g (t ) dt , so 〈 g , = g〉 0 || g ||= 1 ∫0 (1 − 3t 1 ∫0 (t 3 − t 2 ) 2= dt ∫0 t 〈 g , g 〉 = 1/ 105. 1 25. The inner product is 〈 f , g 〉 =∫ f (t ) g (t )dt. Then 1 and t are orthogonal because = 〈1, t 〉 −1 1 t dt ∫−= 1 0. 2 So 1 and t can be in an orthogonal basis for Span{1, t , t }. By the Gram-Schmidt process, the third basis element in the orthogonal basis can be t 2 − = 〈1,1〉 1 1 dt ∫−= 1 t 2 , t〉 2, and 〈= 1 t dt ∫−= 1 3 〈t 2 ,1〉 〈t 2 , t 〉 1− t . Since 〈= t 2 ,1〉 〈1,1〉 〈t , t 〉 1 t dt ∫−= 1 2 2 / 3, 0, the third basis element can be written as t − (1/ 3). This 2 2 element can be scaled by 3, which gives the orthogonal basis as {1, t , 3t − 1}. 2 26. The inner product is 〈 f , g 〉 =∫ f (t ) g (t )dt. Then 1 and t are orthogonal because = 〈1, t 〉 −2 2 t dt ∫−= 2 0. 2 So 1 and t can be in an orthogonal basis for Span{1, t , t }. By the Gram-Schmidt process, the third basis element in the orthogonal basis can be t 2 − = 〈1, 1〉 2 1 dt ∫−= 2 t 2 , t〉 4, and 〈= 2 t dt ∫−= 2 3 〈t 2 ,1〉 〈t 2 , t 〉 1− t . Since 〈= t 2 , 1〉 〈1,1〉 〈t , t 〉 2 t dt ∫−= 2 2 16 / 3, 0, the third basis element can be written as t − (4 / 3). This 2 2 element can be scaled by 3, which gives the orthogonal basis as {1, t , 3t − 4}. 27. [M] The new orthogonal polynomials are multiples of −17t + 5t 3 and 72 − 155t 2 + 35t 4 . These polynomials may be scaled so that their values at –2, –1, 0, 1, and 2 are small integers. 2 28. [M] The orthogonal basis is f 0 (t ) = 1, f1 (t ) = cos t , f 2 (t ) = cos t − (1/ 2) = (1/ 2)cos 2 t , and f3 (t ) = cos3t − (3/ 4)cos t = (1/ 4)cos 3t. 6.8 SOLUTIONS Notes: The connections between this section and Section 6.7 are described in the notes for that section. For my junior-senior class, I spend three days on the following topics: Theorems 13 and 15 in Section 6.5, plus Examples 1, 3, and 5; Example 1 in Section 6.6; Examples 2 and 3 in Section 6.7, with the motivation for the definite integral; and Fourier series in Section 6.8. 1. The weighting matrix W, design matrix X, parameter vector β, and observation vector y are: 1 0 0 0 0 1 −2 0 0 2 0 0 0 1 −1 0 β 0 = W 0 0 2 0 = , y 2 . 0 , X 1 = 0 , β = β1 1 0 0 0 2 0 1 4 0 0 0 0 1 1 4 2 Copyright © 2016 Pearson Education, Ltd. 6.8 • Solutions 6-37 The design matrix X and the observation vector y are scaled by W: 1 −2 2 −2 WX = = 2 0 , Wy 2 2 1 2 0 0 4 . 8 4 0 14 28 = , (WX )T Wy and find that Further compute . (WX )T WX = 0 16 24 0 28 2 1/14 = βˆ ((WX )T= WX ) −1 (WX )T Wy = . Thus the weighted least-squares line is 0 1/16 24 3/ 2 y = 2 + (3/2)x. 2. Let X be the original design matrix, and let y be the original observation vector. Let W be the weighting matrix for the first method. Then 2W is the weighting matrix for the second method. The weighted least-squares by the first method is equivalent to the ordinary least-squares for an equation T T whose normal equation is (WX ) WX βˆ = (WX ) Wy , while the second method is equivalent to the T T ordinary least-squares for an equation whose normal equation is (2WX ) (2W ) X βˆ = (2WX ) (2W ) y. T T Since the second equation can be written as 4(WX ) WX βˆ = 4(WX ) Wy , it has the same solutions as the first equation). ) t 2 − 2, 3. From Example 2 and the statement of the problem, p0 (t ) = 1, p1 (t ) = t , p2 (t= = p3 (t ) (5 / 6)t 3 − (17 / 6)t , and g = (3, 5, 5, 4, 3). The cubic trend function for g is the orthogonal projection pˆ of g onto the subspace spanned by p0 , p1 , p2 , and p3 : pˆ = 〈 g , p0 〉 〈 g , p3 〉 〈 g , p1 〉 〈 g , p2 〉 p0 + p1 + p2 + p3 〈 p0 , p0 〉 〈 p1 , p1 〉 〈 p2 , p2 〉 〈 p3 , p3 〉 = 20 2 5 17 −1 −7 2 (1) + t + t − 2 + t3 − t 5 10 14 10 6 6 =4 − ( ) 1 1 1 5 17 2 1 1 t − t 2 − 2 + t 3 − t =5 − t − t 2 + t 3 10 2 5 6 6 3 2 6 ( ) This polynomial happens to fit the data exactly. 4. The inner product is 〈 p, q〉 = p(–5)q(–5) + p(–3)q(–3) + p(–1)q(–1) + p(1)q(1) + p(3)q(3) + p(5)q(5). 2 a. Begin with the basis {1, t , t } for 2. Since 1 and t are orthogonal, let p0 (t ) = 1 and p1 (t ) = t. 〈t 2 ,1〉 〈t 2 , t 〉 70 2 35 =t − . The t =t 2 − 1− 〈1,1〉 〈t , t 〉 6 3 vector of values for p2 is (40/3, –8/3, –32/3, –32/3, –8/3, 40/3), so scaling by 3/8 yields the new Then the Gram-Schmidt process gives p2 (t ) =t 2 − 2 2 function p2 = (3/ 8)(t − (35 / 3)) = (3/ 8)t − (35 / 8). b. The data vector is g = (1, 1, 4, 4, 6, 8). The quadratic trend function for g is the orthogonal projection pˆ of g onto the subspace spanned by p0 , p1 and p2 : Copyright © 2016 Pearson Education, Ltd. 6-38 CHAPTER 6 pˆ = • Orthogonality and Least Squares 〈 g , p0 〉 〈 g , p1 〉 〈 g , p2 〉 24 50 6 3 35 (1) + t + t 2 − p0 + p1 + p2 = 6 70 84 8 8 〈 p0 , p0 〉 〈 p1 , p1 〉 〈 p2 , p2 〉 5 1 3 35 59 5 3 2 =4 + t + t 2 − = + t + t 7 14 8 8 16 7 112 5. The inner product is 〈 f , g 〉 =∫ 2π 0 〈sin= mt , sin nt 〉 2π ∫0 f (t ) g (t )dt. Let m ≠ n. Then = sin mt sin nt dt 1 2π cos((m − n)t ) − cos((= m + n)t )dt 0 . Thus sin mt and 2 ∫0 sin nt are orthogonal. 6. The inner product is 〈 f , g 〉 =∫ 2π 0 〈sin = mt ,cos nt 〉 2π ∫0 f (t ) g (t )dt. Let m and n be positive integers. Then sin mt= cos nt dt 1 2π sin((m + n)t ) + sin((= m − n)t )dt 0 . Thus sin mt and 2 ∫0 cos nt are orthogonal. 7. The inner product is 〈 f , g 〉 =∫ 2π 0 f (t ) g (t )dt. Let k be a positive integer. Then 2π 1 2π 1 + cos 2kt dt = π and 0 2 ∫0 2π 1 2π || sin kt ||2 = 〈sin kt ,sin kt 〉 = ∫ sin 2 kt dt = ∫ 1 − cos 2kt dt = π . 0 2 0 || cos kt ||2 = 〈 cos kt ,cos kt 〉 = ∫ cos 2 kt dt = 8. Let f(t) = t – 1. The Fourier coefficients for f are: a0 1 1 2π 1 = f (t ) dt = ∫ 0 2 2π 2π 2π ∫0 t − 1 dt =−1 + π 1 2π 1 2π and for k > 0, ak = ∫ f (t )cos kt dt = ∫ (t − 1)cos kt dt = 0 , and π π 0 0 1 2π 1 2π 2 bk = f (t )sin kt dt = − . The third-order Fourier approximation to f is (t − 1)sin kt dt = ∫ ∫ 0 0 k π π a0 2 + b1sin t + b2sin 2t + b3sin 3t =−1 + π − 2 sin t − sin 2t − sin 3t . thus 2 3 a0 1 1 2π 1 f (= t ) dt 9. Let f(t) = 2π – t. The Fourier coefficients for f are: = ∫ 0 2 2π 2π 1 2π 1 2π for k > 0, ak = ∫ f (t ) cos kt dt = ∫ (2π − t ) cos kt dt =0 and π 1 2π π 0 2π −= t dt π and 0 2π 2 (2π − t ) sin kt dt = .The third-order Fourier approximation to f is k a0 2 + b1sin t + b2sin 2t + b3sin 3t = π + 2 sin t + sin 2t + sin 3t . thus 2 3 π ∫0 bk = 1 2π ∫0 π ∫0 f (t ) sin kt dt = 1 for 0 ≤ t < π . The Fourier coefficients for f are: 10. Let f (t ) = −1 for π ≤ t < 2π a0 1 1 2π 1 π 1 2π = f (t ) dt = ∫ dt − dt =0 , and for k > 0, ∫ 0 0 2 2π 2π 2π ∫π Copyright © 2016 Pearson Education, Ltd. 6.8 • Solutions 6-39 1 2π 1 π 1 2π ak = ∫ f (t ) cos kt dt = ∫ cos kt dt − ∫ cos kt dt =0 and π 0 π 0 π π 4 /(kπ ) 1 2π 1 π 1 2π bk = ∫ f (t ) sin kt dt = ∫ sin kt dt − ∫ sin kt dt = 0 0 π 0 π π π for k odd . for k even 4 4 sin 3t . The third-order Fourier approximation to f is thus b1sin t + b3sin 3t = sin t + π 3π 1 1 − cos 2t . The expression on 2 2 the right is in the subspace spanned by the trigonometric polynomials of order 3 or less, so this expression is the third-order Fourier approximation to sin 2 t . 2 11. The trigonometric identity cos 2t = 1 − 2 sin t shows that . sin 2t= 3 1 cos t + cos 3t . The 4 4 expression on the right is in the subspace spanned by the trigonometric polynomials of order 3 or less, so this expression is the third-order Fourier approximation to cos3t. cos 3t 4 cos3t − 3 cos t shows that = cos3t 12. The trigonometric identity = 13. Let f and g be in C [0, 2π] and let m be a nonnegative integer. Then the linearity of the inner product shows that 〈( f + g), cos mt〉 = 〈 f, cos mt〉 + 〈g, cos mt〉 and 〈( f + g), sin mt〉 = 〈 f, sin mt〉 + 〈 g, sin mt〉. Dividing these identities respectively by 〈cos mt, cos mt〉 and 〈sin mt, sin mt〉 shows that the Fourier coefficients am and bm for f + g are the sums of the corresponding Fourier coefficients of f and of g. 14. Note that g and h are both in the subspace H spanned by the trigonometric polynomials of order 2 or less. Since h is the second-order Fourier approximation to f, it is closer to f than any other function in the subspace H. 15. [M] The weighting matrix W is the 13 × 13 diagonal matrix with diagonal entries 1, 1, 1, .9, .9, .8, .7, .6, .5, .4, .3, .2, .1. The design matrix X, parameter vector β, and observation vector y are: 0 0 1 0 1 0.0 1 1 1 8.8 2 3 1 2 2 2 29.9 1 3 32 33 62.0 42 43 1 4 104.7 β 2 3 0 5 5 1 5 159.1 β 1 62 63 , β = , y 222.0 . X 1 6 = = β 2 1 7 294.5 72 73 β 3 380.4 2 3 1 8 8 8 471.1 2 3 9 9 1 9 1 10 102 103 571.7 686.8 1 11 112 113 809.2 2 3 1 12 12 12 Copyright © 2016 Pearson Education, Ltd. 6-40 CHAPTER 6 • Orthogonality and Least Squares The design matrix X and the observation vector y are scaled by W: 1.0 1.0 1.0 .9 .9 .8 WX = .7 .6 .5 .4 .3 .2 .1 0.0 1.0 2.0 2.7 3.6 4.0 4.2 4.2 4.0 3.6 3.0 2.2 1.2 0.0 1.0 4.0 8.1 14.4 20.0 25.2 29.4 32.0 32.4 30.0 24.2 14.4 0.0 0.00 8.80 1.0 29.90 8.0 24.3 55.80 94.23 57.6 100.0 127.28 151.2 , Wy = 155.40 205.8 176.70 190.20 256.0 188.44 291.6 300.0 171.51 137.36 266.2 172.8 80.92 Further compute 22.23 120.77 797.19 6.66 22.23 120.77 797.19 5956.13 (WX )T WX = , (WX )T Wy 120.77 797.19 5956.13 48490.23 797.19 5956.13 48490.23 420477.17 747.844 4815.438 35420.468 285262.440 −0.2685 3.6095 T T −1 ˆ . and find that β ((= WX ) WX ) (WX ) Wy = 5.8576 −0.0477 g (t ) −.2685 + 3.6095t + 5.8576t − .0477t . The Thus the weighted least-squares cubic is y == velocity at t = 4.5 seconds is g'(4.5) = 53.4 ft./sec. This is about 0.7% faster than the estimate obtained in Exercise 13 of Section 6.6. 2 1 16. [M] Let f (t ) = −1 3 for 0 ≤ t < π . The Fourier coefficients for f have already been found to be for π ≤ t < 2π 4 /(kπ ) 0 ak = 0 for all k ≥ 0 and bk = for k odd . Thus for k even 4 4 4 4 4 f 4 (t ) = sin t + sin 3t and f5 (t ) = sin t + sin 3t + sin 5t . A graph of f 4 over the interval π π 3π 3π 5π [0, 2π] is Copyright © 2016 Pearson Education, Ltd. 6.8 A graph of f5 over the interval [0, 2π] is A graph of f5 over the interval [–2π, 2π] is Copyright © 2016 Pearson Education, Ltd. • Solutions 6-41 6-42 CHAPTER 6 Chapter 6 1. • Orthogonality and Least Squares SUPPLEMENTARY EXERCISES a. False. The length of the zero vector is zero. b. True. By the displayed equation before Example 2 in Section 6.1, with c = –1, || –x || = || (–1)x || =| –1 ||| x || = || x ||. c. True. This is the definition of distance. d. False. This equation would be true if r|| v || were replaced by | r ||| v ||. e. False. Orthogonal nonzero vectors are linearly independent. f. True. If x ⋅ u = 0 and x ⋅ v = 0, then x ⋅ (u – v) = x ⋅ u – x ⋅ v = 0. g. True. This is the “only if” part of the Pythagorean Theorem in Section 6.1. h. True. This is the “only if” part of the Pythagorean Theorem in Section 6.1 where v is replaced 2 2 by –v, because || −v || is the same as || v || . i. False. The orthogonal projection of y onto u is a scalar multiple of u, not y (except when y itself is already a multiple of u). j. True. The orthogonal projection of any vector y onto W is always a vector in W. k. True. This is a special case of the statement in the box following Example 6 in Section 6.1 (and proved in Exercise 30 of Section 6.1). l. False. The zero vector is in both W and W ⊥ . 0, then (ci v i ) ⋅ (c j v j ) = ci c j ( v i ⋅ v j ) = ci c j 0= 0. m. True. See Exercise 32 in Section 6.2. If v i ⋅ v j = n. False. This statement is true only for a square matrix. See Theorem 10 in Section 6.3. o. False. An orthogonal matrix is square and has orthonormal columns. p. True. See Exercises 27 and 28 in Section 6.2. If U has orthonormal columns, then U T U = I . If U is also square, then the Invertible Matrix Theorem shows that U is invertible and U −1 = U T . In this case, UU T = I , which shows that the columns of U T are orthonormal; that is, the rows of U are orthonormal. q. True. By the Orthogonal Decomposition Theorem, the vectors projW v and v − projW v are orthogonal, so the stated equality follows from the Pythagorean Theorem. r. False. A least-squares solution is a vector xˆ (not A xˆ ) such that A xˆ is the closest point to b in Col A. ( AΤ A)−1 AΤb describes the solution of the normal equations, not the s. False. The equation xˆ = matrix form of the normal equations. Furthermore, this equation makes sense only when AT A is invertible. 2. If {v1 , v 2 } is an orthonormal set and= x c1 v1 + c2 v 2 , then the vectors c1 v1 and c2 v 2 are orthogonal (Exercise 32 in Section 6.2). By the Pythagorean Theorem and properties of the norm || x ||2 = || c1 v1 + c2 v 2 ||2 = || c1 v1 ||2 + || c2 v 2 ||2 = (c1 || v1 ||)2 + (c2 || v 2 ||)2 = | c1 |2 + | c2 |2 So the stated equality holds for p = 2. Now suppose the equality holds for p = k, with k ≥ 2. Let {v1 ,…, v k +1} be an orthonormal set, and consider = x c1 v1 + … + ck v k + ck +1 v k += u k + ck +1 v k +1 , 1 Copyright © 2016 Pearson Education, Ltd. Chapter 6 • Supplementary Exercises 6-43 0 for j where = u k c1 v1 + … + ck v k . Observe that u k and ck +1 v k +1 are orthogonal because v j ⋅ v k +1 = = 1,…,k. By the Pythagorean Theorem and the assumption that the stated equality holds for k, and 2 2 = || ck +1 v k +1 ||2 | c= | ck +1 |2 , because k +1 | || v k +1 || || x= ||2 || u k + ck +1 v k +1= ||2 || u k ||2 + || ck +1 v k +1= ||2 | c1 |2 +…+ | ck +1 |2 Thus the truth of the equality for p = k implies its truth for p = k + 1. By the principle of induction, the equality is true for all integers p ≥ 2. 3. Given x and an orthonormal set {v1 ,…, v p } in n , let xˆ be the orthogonal projection of x onto the subspace spanned by v1 ,…, v p . By Theorem 10 in Section 6.3, xˆ= (x ⋅ v1 ) v1 + …+ (x ⋅ v p ) v p . By Exercise 2, || xˆ ||2= | x ⋅ v1 |2 +…+ | x ⋅ v p |2 . Bessel’s inequality follows from the fact that || xˆ ||2 ≤ || x ||2 , which is noted before the proof of the Cauchy-Schwarz inequality in Section 6.7. 4. By parts (a) and (c) of Theorem 7 in Section 6.2, {Uv1 ,…,Uv k } is an orthonormal set in n . Since there are n vectors in this linearly independent set, the set is a basis for n . 5. Solution for Exercise 8 in ISM. 6. If Ux = λx for some x ≠ 0, then by Theorem 7(a) in Section 6.2 and by a property of the norm, || x || = || Ux || = || λx || = | λ ||| x ||, which shows that | λ | = 1, because x ≠ 0. 7. a . Let u be a unit vector, and let Q = I − 2uuT . Since (uuT )T = (uT )T uT = uuT , QT = (I − 2uuT )T = I − 2(uuT )T = I − 2uuT = Q Then QT Q = Q2 = (I − 2uuT )2 = I − 2uuT − 2uuT + 4(uuT )(uuT ) Since u is a unit vector, uT u = u · u = 1, so (uuT )(uuT ) = u(uT u)uT = uuT , and QT Q = I − 4uuT + 4uuT = I Thus Q is an orthogonal matrix. b . Let v = cu in Span{u}. We have Qv = Q(cu) = cQu = c(I − 2uuT )u = cu − 2c(uuT )u = cu − 2cu(uT u) = cu − 2cu = −cu = −v Let v in (Span{u})⊥ . We have uT v = u · v = 0 and Qv = (I − 2uuT )v = v − 2(uuT )v = v − 2u(uT v) = v 8. We need to find a unit vector u in Rn such that Q = I −2uuT satisfies Qv = e1 and Qe1 = v. Since Q2 = I (see Exercise 13 in the Supplementary Exercises for Chapter 2), we need to consider only Qe1 = v. Since Qe1 = e1 −2uuT e1 = e1 −2u1 u, the unit vector u is a multiple of v − e1 and we can choose u = (v − e1 )/kv − e1 k. Copyright © 2016 Pearson Education, Ltd. 6-44 CHAPTER 6 • Orthogonality and Least Squares 9. a . The best approximation to z in Rn by vectors in W = Span{v1, v2, . . ., vp} is ˆz = projW z. If we set A = [v1 v2 · · · v p], then zˆ is in Col A. Thus there is a vector, say, xˆ in R p , with Aˆx = zˆ . So, xˆ is a least-squares solution of Ax = z. The normal equations may be solved to find xˆ , and then zˆ may be found by computing Aˆx . 2 4 b . Let A = −5 −1 . To find the normal equations and to find xˆ , compute 1 2 2 4 2 −5 1 30 15 T A A= −5 −1 = 4 −1 2 15 21 1 2 6 2 −5 1 −15 T A z= 7 = 4 −1 2 33 8 The solution of the normal equations (ATA)x = AT z is −1 1 30 15 −15 21 −15 −15 −2 T −1 T xˆ = (A A) A z = = = 15 21 33 30 33 3 405 −15 8 Hence, the best approximation to z by vectors in Span{v1 , v2 } is zˆ = Aˆx = 7 . 4 Alternately, we can first find an orthogonal basis {u1 , u2 } for W , for example the one 2 3 found in Exercise 3 in Section 6.4, namely u1 = −5 and u2 = 3/2 and compute 1 3/3 8 z · u2 z · u1 −15 81 1 u = u + u = − u1 + 3 u2 = 7 . u + zˆ = projW z = 30 1 27 2 2 u1 · u1 1 u2 · u2 2 4 10. By Theorem 14 in Section 6.5, the equation Ax = b has a unique least-squares solution for each b in Rm . This solution is given by xˆ = (ATA)−1 AT b. If we take b = c1 b1 + c2 b2 , linearity of matrix multiplication gives us (ATA)−1 AT (c1 b1 + c2 b2 ) = c1 (ATA)−1 AT b1 + c2 (ATA)−1 AT b2 = c1 xˆ 1 + c2 xˆ 2 vT 1 −2 5 T v= 1 −2 5 . Then the given set of equations is T 1 −2 5 v Ax = b, and the set of all least-squares solutions coincides with the set of solutions of the normal equations AT Ax = AT b . The column-row expansions of AT A and AT b give x a 11. Let x = y , b = b , v = z c 1 −2 , and A = 5 AT A = vvT + vvT + vvT = 3vvT , AT b = av + bv + cv = (a + b + c) v Copyright © 2016 Pearson Education, Ltd. Chapter 6 • Supplementary Exercises 6-45 T T AT Ax 3(vv = )x 3v(v = x) 3( vT x)v since vT x is a scalar, and the normal equations have Thus= become 3(v x) v = (a + b + c)v, so 3( v x) = a + b + c, or v x = (a + b + c) / 3. Computing vT x gives the equation x – 2y + 5z = (a + b + c)/3 which must be satisfied by all least-squares solutions to Ax = b. T T T 12. The equation (1) in the exercise has been written as Vλ = b, where V is a single nonzero column vector v, and b = Av. The least-squares solution λˆ of Vλ = b is the exact solution of the normal equations V T V λ =V T b. In the original notation, this equation is vT vλ =vT Av. Since vT v is T T nonzero, the least squares solution λˆ is v Av /( v v). This expression is the Rayleigh quotient discussed in the Exercises for Section 5.8. 13. a. The row-column calculation of Au shows that each row of A is orthogonal to every u in Nul A. So ⊥ ⊥ each row of A is in (Nul A) . Since (Nul A) is a subspace, it must contain all linear ⊥ combinations of the rows of A; hence (Nul A) contains Row A. b. If rank A = r, then dim Nul A = n – r by the Rank Theorem. By Exercsie 24(c) in Section 6.3, dimNul A + dim(Nul A) ⊥ = n, so dim(Nul A) ⊥ must be r. But Row A is an r-dimensional ⊥ ⊥ subspace of (Nul A) by the Rank Theorem and part (a). Therefore, Row A = (Nul A) . T T ⊥ T c. Replace A by AT in part (b) and conclude that Row A = (Nul A ) . Since Row A = Col A, Col A = (Nul AT ) ⊥ . 14. a . Since P is invertible, its columns are linearly independent by the Invertible Matrix Theorem. Therefore, by Theorem 12 in Section 6.4, P can be factored as P = QS where Q is an n×n orthogonal matrix and S is an n×n upper triangular invertible matrix with positive entries on its diagonal. Hence A = (QS)D(QS)−1 = QSDS−1 Q−1 = Q(SDS−1)QT The inverse of an upper triangular matrix being upper triangular (see Example 5 in Section 2.4) and the product of upper triangular matrices being also upper triangular, we conclude that SDS−1 is upper triangular. Thus setting U = Q and R = SDS−1 we obtain T a (real) Schur factorization A = U RU . 3 4 2 0 −1 b . We have A = PDP where P = and D = (see Exercise 4 in Sec1 1 0 1 3 4 tion 5.3). Since the columns x1 = and x2 = of P are not orthogonal we 1 1 3 1 3 1 1 1 can consider v1 = and v2 = and define Q = √10 v1 v2 = √10 . 1 −3 1 −3 3 1 3 4 10 13 1 1 T √ √ Hence, S = Q P = 10 = 10 and 1 −3 1 1 0 1 10 13 2 0 1 1 −13 20 −130 2 −13 −1 1 1 √ SDS = √10 = 10 = 0 1 0 1 10 0 10 0 10 0 1 Thus, we obtain a (real) Schur factorization A = U RU T where 1 3 1 2 −13 −1 U =Q= √ and R = SDS = 3 0 1 10 1 Copyright © 2016 Pearson Education, Ltd. 6-46 CHAPTER 6 • Orthogonality and Least Squares 1 1 2 5 0 −1 c . Here have A = PDP where P = 1 0 −1 and D = 0 1 1 −1 0 0 0 1 1 0 of P cise 5 in Section 5.3). The columns x1 = 1 and x2 = 1 −1 and 2 1 1 1 x ·x x ·x 1 2 2 x3 − 3 1 x1 − 3 2 x2 = −1 − 1 − 0 = −2 x1 · x1 x2 · x2 3 2 3 0 1 −1 1 is orthogonal to Span{x1 , x2 }. Hence, we can write P = QS where √ √ √ √ 1/ 2 1/ 6 3 0 1/ 3 √ √ 1/√3 0 −2/ 6 and S = QT P = 0 2 Q= √ √ √ 1/ 3 −1/ 2 1/ 6 0 0 and obtain a (real) Schur factorization A√= U RU T where √ √ 1/ 2 1/√6 5 1/√3 −1 0√ −2/√6 and R = SDS = 0 U = Q = 1 / √3 1/ 3 −1/ 2 1/ 6 0 0 0 (see Exer1 are orthogonal √ 1/ 3 √ 2 √ 4/ 6 √ 0 − 2 1 0 0 1 15. Let A = U RU T . By hypothesis U is orthogonal, hence U T = U −1 and A − λ I = U RU T − λ I = U RU T − λ U IU T = U (R − λ I)U T Therefore det(A − λ I) = det U (R − λ I)U T = det(U ) det(R − λ I) det(U T ) = det(R − λ I) and A and R have the same characteristic polynomial. By hypothesis R is upper triangular, hence its eigenvalues are its n real diagonal entries. Given that A has the same eigenvalues as R, A has n real eigenvalues, counting multiplicities. 16.= a. If U [u1 u2 … u n ] , then AU = [λ1u1 Au 2 … Au n ]. Since u1 is a unit vector and λ1U u1 = λ1e1. u 2 ,…, u n are orthogonal to u1 , the first column of U T AU is U (λ1u1 ) = T T b. From (a), λ1 0 U T AU = 0 * * A1 * * View U T AU as a 2 × 2 block upper triangular matrix, with A1 as the (2, 2)-block. Then from Supplementary Exercise 12 in Chapter 5, det(U T AU − λ I n ) = det(λ1 − λI1 ) ⋅ det( A1 − λ I n −1 ) = (λ1 − λ ) ⋅ det( A1 − λ I n −1 ) This shows that the eigenvalues of U T AU , namely, λ1 ,…, λ n , consist of λ1 and the eigenvalues of A1 . So the eigenvalues of A1 are λ 2 ,…, λ n . Copyright © 2016 Pearson Education, Ltd. Chapter 6 • Supplementary Exercises 6-47 −4 17. [M] Compute that || ∆x ||/|| x || = .4618 and cond(A) × (|| ∆b || / || b ||) =3363 × (1.548 × 10 ) =.5206 . In this case, || ∆x ||/|| x || is almost the same as cond(A) × || ∆b ||/|| b ||. 18. [M] Compute that || ∆x ||/|| x || = .00212 and cond(A) × (|| ∆b ||/|| b ||) = 3363 × (.00212) ≈ 7.130. In this case, || ∆x ||/|| x || is almost the same as || ∆b ||/|| b ||, even though the large condition number suggests that || ∆x ||/|| x || could be much larger. −4 −8 || x || 7.178×10 and cond(A) × (|| ∆b || / || b ||) = 23683 × (2.832 × 10 ) = 19. [M] Compute that || ∆x || /= 6.707. Observe that the relative change in x is much smaller than the relative change in b. In fact the theoretical bound on the relative change in x is 6.707 (to four significant figures). This exercise shows that even when a condition number is large, the relative error in the solution need not be as large as you suspect. −5 20. [M] Compute that || ∆x ||/|| x || = .2597 and cond(A) × (|| ∆b || / || b ||) = 23683 × (1.097 × 10 ) = .2598 . This calculation shows that the relative change in x, for this particular b and ∆b, should not exceed .2598. In this case, the theoretical maximum change is almost acheived. Copyright © 2016 Pearson Education, Ltd. 7.1 SOLUTIONS Notes: Students can profit by reviewing Section 5.3 (focusing on the Diagonalization Theorem) before working on this section. Theorems 1 and 2 and the calculations in Examples 2 and 3 are important for the sections that follow. Note that symmetric matrix means real symmetric matrix, because all matrices in the text have real entries, as mentioned at the beginning of this chapter. The exercises in this section have been constructed so that mastery of the Gram-Schmidt process is not needed. Theorem 2 is easily proved for the 2 × 2 case: ( ) a b 1 , then λ= a + d ± ( a − d ) 2 + 4b 2 . If A = c d 2 If b = 0 there is nothing to prove. Otherwise, there are two distinct eigenvalues, so A must be d − λ diagonalizable. In each case, an eigenvector for λ is . −b 5 3 T = A = 1. Since A , the matrix is symmetric. − 5 7 3 −5 AT , the matrix is symmetric. 2.= Since A = −5 −3 2 = A 3. Since 4 3 ≠ AT , the matrix is not symmetric. 4 0 4. Since = A 8 3 8 0 2 −6 5. Since A = 2 0 3 −4 ≠ AT , the matrix is not symmetric. 0 2 −6 2 0 2 =AT , the matrix is symmetric. −6 7-1 Copyright © 2016 Pearson Education, Ltd. 7-2 CHAPTER 7 • Symmetric Matrices and Quadratic Forms 6. Since A is not a square matrix A ≠ AT and the matrix is not symmetric. .6 7. Let P = .8 .8 .8 .6 .8 1 0 .6 PT P = , and compute that= = I 2 . Since P is a square −.6 .8 −.6 .8 −.6 0 1 .8 .6 −1 T P= matrix, P is orthogonal and P= .8 −.6 . 1 1 1 P , and compute that PT= 8. Let P = 1 1 −1 orthogonal. −4 / 5 9. Let P = 3/5 1 1 −1 1 1 2 0 = 2 I 2 ≠ I 2 . Thus P is not −1 0 2 3 / 5 −4 / 5 PT P , and compute that= 4 / 5 3/5 3 / 5 −4 / 5 3 / 5 1 0 = = I 2 . Since 4 / 5 3 / 5 4 / 5 0 1 −4 / 5 −1 T P is a square matrix, P is orthogonal and P= P= 3/5 1/ 3 10. Let P 2 / 3 = 2 / 3 1/ 3 P P 2 / 3 = 2 / 3 T 3 / 5 . 4 / 5 2 / 3 −2 / 3 , and compute that 1 / 3 2/3 1/ 3 −2 / 3 2 / 3 1 / 3 −2 / 3 2 / 3 1 / 3 2 / 3 2/3 1/ 3 −2 / 3 1/ 3 P is orthogonal and P= P= 2 / 3 2 / 3 −1 T 2/3 1/ 3 −2 / 3 2/3 1/ 3 −2 / 3 2 / 3 / 3 −2= 1 / 3 1 0 0 0 0 1= 0 I 3 . Since P is a square matrix, 0 1 2 / 3 −2 / 3 . 1 / 3 2/3 1 / 3 2 / 3 = 1 / 3 −2 / 3 , and compute that 11. Let P 0 5 / 3 −4 / 3 −2 / 3 0 5 / 3 2 / 3 2/3 1 / 3 29 / 9 2 / 3 T −16 / 9 2/3 1 / 3 −4 / 3 0 1 / 3 −2 / 3 = P P= 1 / 3 −2 / 3 −2 / 3 5 / 3 −4 / 3 −2 / 3 −8 / 9 −16 / 9 21 / 9 8/9 orthogonal. Copyright © 2016 Pearson Education, Ltd. −8 / 9 8 / 9 ≠ I 3 . Thus P is not 1 7.1 .5 .5 12. Let P = .5 .5 .5 .5 PT P = −.5 −.5 .5 .5 −.5 −.5 .5 .5 .5 .5 −.5 .5 −.5 .5 .5 −.5 −.5 .5 7-3 −.5 .5 , and compute that .5 −.5 .5 .5 −.5 .5 .5 .5 −.5 .5 .5 −.5 −.5 .5 .5 .5 = .5 −.5 −.5 .5 −.5 −.5 .5 .5 T −1 matrix, P is orthogonal and P= P= −.5 −.5 3 13. Let A = 1 • Solutions .5 .5 .5 .5 1 0 0 0 0 1 0 0 = I 4 . Since P is a square 0 0 1 0 0 0 0 1 .5 −.5 −.5 .5 .5 −.5 . .5 −.5 1 . Then the characteristic polynomial of A is 3 (3 − λ )2 − 1 = λ 2 − 6λ + 8 = (λ − 4)(λ − 2), so the eigenvalues of A are 4 and 2. For λ = 4 , one 1/ 2 1 computes that a basis for the eigenspace is , which can be normalized to get u1 = . For 1 1/ 2 −1 λ = 2 one computes that a basis for the eigenspace is , which can be normalized to get 1 −1/ 2 . Let P u2 = = 1/ 2 u1 u 2 ] [= 1 / 2 1 / 2 4 −1 / 2 and D = 1 / 2 0 0 . Then P orthogonally 2 diagonalizes A, and A = PDP −1 . 1 −5 2 2 14. Let A = . Then the characteristic polynomial of A is (1 − λ ) − 25 = λ − 2λ − 24 5 1 − = (λ − 6)(λ + 4), so the eigenvalues of A are 6 and –4. For λ = 6 , one computes that a basis for the −1 / 2 −1 eigenspace is , which can be normalized to get u1 = . For λ = –4 , one computes that 1 1 / 2 1 / 2 1 a basis for the eigenspace is , which can be normalized to get u 2 = . Let 1 1 / 2 −1 / 2 1 / 2 0 6 = P [= u1 u 2 ] = and D . Then P orthogonally diagonalizes A, and 0 −4 1 / 2 1 / 2 A = PDP −1. 3 4 2 15. Let A = . Then the characteristic polynomial of A is (3 − λ )(9 − λ ) − 16 = λ − 12λ + 11 4 9 = (λ − 11)(λ − 1) , so the eigenvalues of A are 11 and 1. For λ = 11 , one computes that a basis for Copyright © 2016 Pearson Education, Ltd. 7-4 CHAPTER 7 • Symmetric Matrices and Quadratic Forms 1 / 5 1 the eigenspace is , which can be normalized to get u1 = . For λ = 2 2 / 5 −2 / −2 a basis for the eigenspace is , which can be normalized to get u 2 = 1 1 / = P u1 u 2 ] [= 1/ 5 2 / 5 −2 / 5 11 = and D 1 / 5 0 1 , one computes that 5 . Let 5 0 . Then P orthogonally diagonalizes A, and 1 A = PDP −1. 6 −2 . Then the characteristic polynomial of A is (6 − λ )(9 − λ ) − 4 = λ 2 − 15λ + 50 16. Let A = 9 −2 = (λ − 5)(λ − 10) , so the eigenvalues of A are 5 and 10. For λ = 5 , one computes that a basis for the 2 / 5 2 eigenspace is , which can be normalized to get u1 = . For λ = 10 , one computes that a 1 1 / 5 −1 / 5 −1 basis for the eigenspace is , which can be normalized to get u 2 = . Let 2 2 / 5 2 / 5 −1 / 5 5 0 = P [= u1 u 2 ] and D = . Then P orthogonally diagonalizes A, and 2 / 5 0 10 1 / 5 A = PDP −1 . 1 17. Let A = 1 5 1 5 1 5 1 . The eigenvalues of A are -4, 4, and 7. For λ = −4 , one computes that a basis 1 −1 / 2 −1 for the eigenspace is 0 , which can be normalized to get u1 = 0 . For λ = 4 , one 1/ 2 1 1/ 6 1 computes that a basis for the eigenspace is −2 , which can be normalized to get u 2 = −2 / 6 . 1 1/ 6 1 For λ = 7 , one computes that a basis for the eigenspace is 1 , which can be normalized to get 1 1 / 3 −1 / 2 u3 = 1 / 3= 0 u 2 u3 ] . Let P [ u1 = 1 / 3 1 / 2 orthogonally diagonalizes A, and A = PDP −1 . 1/ 6 −2 / 6 1/ 6 1 / 3 −4 1 / 3 and D = 0 0 1 / 3 Copyright © 2016 Pearson Education, Ltd. 0 4 0 0 0 . Then P 7 7.1 1 −6 −6 18. Let A = 2 4 −2 • Solutions 7-5 4 −2 . The eigenvalues of A are −3 , −6 and 9. For λ = −3 , one computes that a −3 1 1 / 3 basis for the eigenspace is 2 , which can be normalized to get u1 = 2 / 3 . For λ = −6 , one 2 2 / 3 −2 / 3 −2 computes that a basis for the eigenspace is −1 , which can be normalized to get u 2 = −1 / 3 . For 2 2 / 3 2 λ = 9 , one computes that a basis for the eigenspace is −2 , which can be normalized to get 1 0 0 −3 2 / 3 2 / 3 1 / 3 −2 / 3 D 0 −6 0 . Then P = u3 = −2 / 3 . Let P= [ u1 u 2 u3 ]= 2 / 3 −1 / 3 −2 / 3 and 0 9 0 1 / 3 2/3 1 / 3 2 / 3 orthogonally diagonalizes A, and A = PDP −1 . 3 19. Let A = −2 4 −2 6 2 the eigenspace is 4 2 . The eigenvalues of A are 7 and –2. For λ = 7 , one computes that a basis for 3 . This basis may be converted via orthogonal projection to an −1 4 orthogonal basis for the eigenspace: 2 , 2 . These vectors can be normalized to get 0 5 −1 1 2 , 0 0 1 −1/ 5 4 / 45 −2 u1 = 2 / 5 , u 2 = 2 / 45 . For λ = –2 , one computes that a basis for the eigenspace is −1 , 0 5 / 45 2 −1 / 5 4 / 45 −2 / 3 −2 / 3 Let P [= which can be normalized to get u3 = = u1 u 2 u3 ] 2 / 5 2 / 45 −1 / 3 1/ 3 . − 0 5 / 45 2 / 3 2 / 3 0 7 0 0 . Then P orthogonally diagonalizes A, and A = PDP −1 . and D = 0 7 0 0 −2 Copyright © 2016 Pearson Education, Ltd. 7-6 CHAPTER 7 • Symmetric Matrices and Quadratic Forms 5 20. Let A 8 = − 4 8 5 −4 −4 −4 . The eigenvalues of A are −3 and 15. For λ = −3 , one computes that a −1 2 −1 basis for the eigenspace is −1 , 2 which is orthogonal and can be normalized to get 2 2 2 / 3 −1 / 3 2 {u1 , u2=} −1 / 3 , 2 / 3 . For λ = 15, one computes that a basis for the eigenspace is 2 , 2 / 3 2 / 3 − 1 2 / 3 2 / 3 2 / 3 −1 / 3 which can be normalized to get u3 = 2 / 3 . Let P = [ u1 u 2 u3 ] = −1 / 3 2/3 2 / 3 and − 1 / 3 2 / 3 2 / 3 1 / 3 − −3 D 0 = 0 4 3 21. Let A = 1 1 0 0 −3 0 . Then P orthogonally diagonalizes A, and A = PDP −1 . 0 15 3 4 1 1 1 1 4 3 1 1 . The eigenvalues of A are 1, 5, and 9. For λ = 1 , one computes that a basis 3 4 −1 0 0 1 for the eigenspace is , , which is an orthogonal set and can be normalized to get 0 −1 0 1 −1 / 2 0 1 / 2 0 , . {u1 , u2 } = For λ = 5 , one computes that a basis for the eigenspace is 0 −1 / 2 0 1 / 2 −1 / 2 −1 / 2 . For λ = 9 , one computes that a basis for the which can be normalized to get u3 = 1 / 2 1 / 2 eigenspace is 1 1 1 1 1 / 2 1 / 2 . Let . This vector can be normalized to get u 4 = 1 / 2 1 / 2 Copyright © 2016 Pearson Education, Ltd. −1 −1 , 1 1 7.1 −1 / 2 1/ 2 = P [= u1 u 2 u3 u 4 ] 0 0 0 0 −1 / 2 1/ 2 1 / 2 −1 / 2 1 / 2 = and D 1 / 2 1 / 2 1 / 2 1 / 2 −1 / 2 1 0 0 0 0 1 0 0 0 0 5 0 • Solutions 7-7 0 0 . Then P 0 9 orthogonally diagonalizes A, and A = PDP −1 . 4 0 22. Let A = 1 0 0 4 0 1 1 0 4 0 the eigenspace is 0 1 . The eigenvalues of A are 3 and 5. For λ = 3 , one computes that a basis for 0 4 −1 0 0 −1 , 1 0 0 1 . This basis is an orthogonal basis for the eigenspace, and these −1 / 2 0 0 −1 / 2 , . vectors can be normalized to get {u1 , u 2 } = For λ = 5 , one computes that a 1 / 2 0 0 1 / 2 1 0 0 1 basis for the eigenspace is , , which is orthogonal and can be normalized to get 1 0 0 1 1 / 2 0 0 1 / 2 , . Let P = {u3 , u4 } = 0 1 / 2 0 1 / 2 3 0 and D = 0 0 0 3 0 0 4 −1 −1 23. Let A = 4 − 1 −1 0 0 5 0 −1 / 2 0 u1 u 2 u3 u 4 ] [= 1/ 2 0 0 1/ 2 −1 / 2 0 0 1/ 2 1/ 2 0 0 1/ 2 0 1 / 2 0 0 . Then P orthogonally diagonalizes A, and A = PDP −1 . 0 5 −1 1 4 −1 −1 . Since each row of A sums to 2, A 1 = 4 1 − 1 −1 4 −1 −1 1 2 1 2 = 2 1 and −1 1 = 4 1 2 1 1 1 is an eigenvector of A with corresponding eigenvalue λ = 2 . The eigenvector may be 1 Copyright © 2016 Pearson Education, Ltd. 7-8 CHAPTER 7 • Symmetric Matrices and Quadratic Forms 1/ 3 normalized to get u1 = 1/ 3 . For λ = 5 , one computes that a basis for the eigenspace is 1/ 3 −1 −1 1 , 0 , so λ = 5 is an eigenvalue of A. This basis may be converted via orthogonal 0 1 −1 −1 projection to an orthogonal basis 1 , −1 for the eigenspace, and these vectors can be 0 2 −1/ 2 −1 / 6 normalized to get u 2 = 1/ 2 and u3 = −1 / 6 . Let 0 2 / 6 1 / 3 −1 / 2 −1 / 6 2 0 0 = P [= u1 u 2 u3 ] 1 / 3 1 / 2 −1 / 6 and D = 0 5 0 . Then P orthogonally 0 0 5 0 2 / 6 1 / 3 diagonalizes A, and A = PDP −1 . 2 −1 −1 24. Let A = 2 1 −1 1 −1 −1 −1 −1 . One may compute that A 0 = 0 , so v1 = 0 is an eigenvector of A 2 1 1 1 with associated eigenvalue λ1 = 1 . For λ1 = 1 , one computes that a basis for the eigenspace is −1 1 0 , 1 1 0 . This basis may be converted via orthogonal projection to an orthogonal basis for the −1 1 1 4 1 eigenspace: {v1 , v 3 } = 0 , 2 . Likewise one may compute that A −1 = −4 =4 −1 , so 1 1 1 4 1 1 v 2 = −1 is an eigenvector of A with associated eigenvalue 4. The eigenvectors v1 , v 2 , and v 3 1 1 / 3 1/ 6 −1 / 2 may be normalized to get the vectors u1 = 0 , u 2 = −1 / 3 , and u3 = 2 / 6 . Let 1/ 2 1 / 3 1 / 6 Copyright © 2016 Pearson Education, Ltd. 7.1 −1 / 2 1 / 3 = P [ u1 = u 2 u3 ] 0 −1 / 3 1/ 3 1 / 2 diagonalizes A, and A = PDP −1 . 25. a. b. c. d. 1/ 6 1 2 / 6 and D = 0 1 / 6 0 0 4 0 • Solutions 7-9 0 0 . Then P orthogonally 1 True. See Theorem 2 and the paragraph preceding the theorem. True. This is a particular case of the statement in Theorem 1, where u and v are nonzero. False. There are n real eigenvalues (Theorem 3), but they need not be distinct (Example 3). False. See the paragraph following formula (2), in which each u is a unit vector. 26. a. False. See Theorem 2. b. True. See the displayed equation in the paragraph before Theorem 2. c. False. An orthogonal matrix can be symmetric (and hence orthogonally diagonalizable), but not every orthogonal matrix is symmetric. See the matrix P in Example 2. d. False. See Theorem 3(b). T ) ⋅ y ( Ax= )T y xT= AT y x= Ay x ⋅ ( Ay ) , since 27. Let A be an n × n symmetric matrix. Then ( Ax= AT = A . T T TT BT AB)T B= A B BT AB , and BT AB is symmetric. Applying this result 28. Since A is symmetric, (= T T TT T = B BBT , so BBT is symmetric. ) B= with A = I gives BT B is symmetric. Finally, ( BB 29. Since A is orthogonally diagonalizable, A = PDP −1 , where P is orthogonal and D is diagonal. Since A = A−1 (= PDP −1 ) −1 PD −1 P −1 . Notice that D −1 is a diagonal matrix, so A−1 is is invertible, orthogonally diagonalizable. 30. If A and B are orthogonally diagonalizable, then A and B are symmetric by Theorem 2. If AB = BA, T T = )T ( BA = )T A= B AB . So AB is symmetric and hence is orthogonally diagonalizable by then ( AB Theorem 2. 31. The Diagonalization Theorem of Section 5.3 says that the columns of P are linearly independent eigenvectors corresponding to the eigenvalues of A listed on the diagonal of D. So P has exactly k columns of eigenvectors corresponding to λ. These k columns form a basis for the eigenspace. −1 32. If A = PRP −1 , then P AP = R . Since P is orthogonal, R = PT AP . Hence T T TT = R T (= PT AP)T P= A P PT AP = R, which shows that R is symmetric. Since R is also upper triangular, its entries above the diagonal must be zeros to match the zeros below the diagonal. Thus R is a diagonal matrix. 33. It has previously been found that A is orthogonally diagonalized by P, where −1 / 2 −1 / 6 1 / 3 8 0 0 = P [ u1 = u 2 u3 ] 1 / 2 −1 / 6 1 / 3 and D = 0 6 0 . Thus the spectral 0 0 3 0 2 / 6 1 / 3 Copyright © 2016 Pearson Education, Ltd. 7-10 CHAPTER 7 • Symmetric Matrices and Quadratic Forms decomposition of A is A =λ1u1u1T + λ2 u 2 u 2T + λ3u3u3T =8u1u1T + 6u 2 u 2T + 3u3u3T −1/ 2 1/ 2 0 1/ 2 8 −1/ 2 = 0 0 1/ 6 0 + 6 1/ 6 −2 / 6 0 1/ 6 1/ 6 −2 / 6 −2 / 6 1/ 3 −2 / 6 + 3 1/ 3 1/ 3 4 / 6 1/ 3 1/ 3 . 1/ 3 1/ 3 1/ 3 1/ 3 34. It has previously been found that A is orthogonally diagonalized by P, where 1 / 2 −1 / 18 −2 / 3 0 7 0 0 . = P [= u1 u 2 u3 ] 0 4 / 18 −1 / 3 and D = 0 7 0 0 − 2 1 / 18 2 / 3 1 / 2 Thus the spectral decomposition of A is A = λ1u1u1T + λ2 u 2 u 2T + λ3u3u3T = 7u1u1T + 7u 2 u 2T − 2u3u3T 1/ 2 = 7 0 1/ 2 0 0 0 1/ 2 1/18 0 + 7 −4 /18 1/ 2 −1/18 −4 /18 16 /18 4 /18 −1/18 4/9 4 /18 − 2 2 / 9 1/18 −4 / 9 2/9 1/ 9 −2 / 9 −4 / 9 −2 / 9 4 / 9 n = )x u= (u x) (u x)u, because uT x is a scalar. So Bx = (x ⋅ u)u. Since 35. a. Given x in = , bx (uu u is a unit vector, Bx is the orthogonal projection of x onto u. T T T T BT (uuT= )T uTT= uT uu = B, B is a symmetric matrix. Also, b. Since = T T = B 2 (uuT )(uu= ) u(uT u)= uT uu = B because uT u = 1. Bu (uuT= )u u(uT= u) u= (1) u , so u is an eigenvector of B with corresponding c. Since uT u = 1 , = eigenvalue 1. 36. Given any y in , let yˆ = By and z = y – yˆ . Suppose that BT = B and B 2 = B . Then n BT= B BB = B. T T T T T a. Since z ⋅ yˆ = (y − yˆ ) ⋅ ( By ) = y ⋅ ( By ) − yˆ ⋅ ( By ) = y By − ( By ) By = y By − y B By = 0 , z is orthogonal to yˆ. b. Any vector in W = Col B has the form Bu for some u. Noting that B is symmetric, Exercise 28 gives ( y – yˆ ) ⋅ (Bu) = [B(y – yˆ )] ⋅ u = [By – BBy] ⋅ u = 0, since B 2 = B. So y – yˆ is in W ⊥ , and the decomposition y = yˆ + (y – yˆ ) expresses y as the sum of a vector in W and a vector in W ⊥ . By the Orthogonal Decomposition Theorem in Section 6.3, this decomposition is unique, and so yˆ must be projW y. Copyright © 2016 Pearson Education, Ltd. 7.1 6 2 37. [M] Let A = 9 −6 2 6 −6 9 9 −6 6 2 • Solutions 7-11 −6 9 . The eigenvalues of A are 19, 11, 5, and –11. For λ = 19 , one 2 6 −1 −1/ 2 1 1/ 2 . For computes that a basis for the eigenspace is , which can be normalized to get u1 = −1 −1/ 2 1 1/ 2 1 1 λ = 11 , one computes that a basis for the eigenspace is , which can be normalized to get 1 1 1/ 2 1 1/ 2 1 u 2 = . For λ = 5, one computes that a basis for the eigenspace is , which can be 1/ 2 −1 −1 1/ 2 1 1/ 2 1/ 2 . For λ = –11 , one computes that a basis for the eigenspace is −1 , normalized to get u3 = −1 −1/ 2 1 −1/ 2 1/ 2 −1/ 2 . Let which can be normalized to get u 4 = −1/ 2 1/ 2 0 1/ 2 1/ 2 −1/ 2 1/ 2 19 0 0 1/ 2 1/ 2 0 11 0 0 1/ 2 −1/ 2 . Then P and D P [u = = = u u u 1 2 3 4] −1/ 2 1/ 2 −1/ 2 −1/ 2 0 0 5 0 1/ 2 1/ 2 1/ 2 −1/ 2 0 0 0 −11 orthogonally diagonalizes A, and A = PDP −1 . .63 −.18 38. [M] Let A = −.06 −.04 −.18 .84 −.04 .12 −.06 −.04 .72 −.12 −.04 .12 . The eigenvalues of A are .5, .55, .8, and 1. For λ = .5 , −.12 .66 4 .8 2 .4 one computes that a basis for the eigenspace is , which can be normalized to get u1 = . For 2 .4 1 .2 Copyright © 2016 Pearson Education, Ltd. 7-12 CHAPTER 7 • Symmetric Matrices and Quadratic Forms −1 −2 λ = .55 , one computes that a basis for the eigenspace is , which can be normalized to get 2 4 2 −.2 −1 −.4 u 2 = . For λ = .8 , one computes that a basis for the eigenspace is , which can be −4 .4 2 .8 .4 −2 −.2 4 normalized to get u3 = . For λ = 1 , one computes that a basis for the eigenspace is , −.8 −1 2 .4 .4 −.4 .8 −.2 −.4 .4 −.4 −.2 .8 .8 which can be normalized to get = = u u u u 4 = . Let P [u 1 2 3 4] .4 −.2 .4 −.8 −.2 .8 .4 .4 .2 .4 .5 0 and D = 0 0 0 .55 0 0 .31 .58 39. [M] Let A = .08 .44 0 0 .8 0 .58 −.56 .44 −.58 0 0 . Then P orthogonally diagonalizes A, and A = PDP −1 . 0 1 .08 .44 .19 −.08 .44 −.58 . The eigenvalues of A are .75, 0, and –1.25. For λ = .75 , −.08 .31 1 3 0 2 one computes that a basis for the eigenspace is , . This basis may be converted via 0 2 1 0 1 3 4 0 orthogonal projection to the orthogonal basis , . These vectors can be normalized to get 0 4 1 −3 1/ u1 = 1/ 3/ 2 0 4/ , u2 = 0 4/ 2 −3/ 50 50 . For λ = 0 , one computes that a basis for the eigenspace is 50 50 Copyright © 2016 Pearson Education, Ltd. −2 −1 , 4 2 7.1 • Solutions 7-13 −2 / 5 −1 / 5 . For λ = –1.25 , one computes that a basis for the which can be normalized to get u3 = 4 / 5 2 / 5 −2 −2 / 5 4 4 / 5 . Let eigenspace is , which can be normalized to get u 4 = −1 −1 / 5 2 2 / 5 1 / 2 3 / 50 −2 / 5 −2 / 5 0 4 / 50 −1 / 5 4 / 5 P= [ u1 u 2 u3 u 4 = ] P= and 0 4 / 50 4 / 5 1 / 5 − 1 / 2 −3 / 50 2 / 5 2 / 5 .75 0 D= 0 0 0 .75 0 0 8 2 40. [M] Let A = 2 −6 9 0 0 0 0 0 0 . Then P orthogonally diagonalizes A, and A = PDP −1 . 0 −1.25 2 8 2 −6 9 2 2 8 −6 9 −6 −6 −6 24 9 9 9 9 . The eigenvalues of A are 6, 30, –30, and 15. For λ = 6, 9 −21 1 −1 −1 0 one computes that a basis for the eigenspace is 0 , 1 . This basis may be converted via 0 0 0 0 1 1 −1 1 orthogonal projection to the orthogonal basis 0 , −2 . These vectors can be normalized to get 0 0 0 0 1/ −1/ u1 = 1/ 2 1/ 2 = u , 0 2 −2 / 0 0 6 6 6 . For λ = 30 , one computes that a basis for the eigenspace is 0 0 Copyright © 2016 Pearson Education, Ltd. 1 1 1 , −3 0 7-14 CHAPTER 7 • Symmetric Matrices and Quadratic Forms 1/ 1/ which can be normalized to get u3 = 1/ −3/ 12 12 12 . For λ = –30 , one computes that a basis for the 12 0 1/ 20 1 1 1/ 20 eigenspace is 1 , which can be normalized to get u 4 = 1/ 20 . For λ = 15 , one computes 1 1/ 20 −4 −4 / 20 1/ 5 1 1 1/ 5 that a basis for the eigenspace is 1 , which can be normalized to get u5 = 1/ 5 . Let 1 1/ 5 1 1/ 5 1/ 2 1/ 6 1/ 12 1/ 20 1/ 5 1/ 6 1/ 12 1/ 20 1/ 5 −1/ 2 P [u = u 2 u3 u 4 u5 ] 0 −2 / 6 1/ 12 1/ 20 1/ 5 and 1 0 0 −3/ 12 1/ 20 1/ 5 0 0 0 −4 / 20 1/ 5 0 0 0 6 0 0 6 0 0 0 D = 0 0 30 0 0 . Then P orthogonally diagonalizes A, and A = PDP −1 . 0 −30 0 0 0 0 0 0 0 15 7.2 SOLUTIONS Notes: This section can provide a good conclusion to the course, because the mathematics here is widely used in applications. For instance, Exercises 23 and 24 can be used to develop the second derivative test for functions of two variables. However, if time permits, some interesting applications still lie ahead. Theorem 4 is used to prove Theorem 6 in Section 7.3, which in turn is used to develop the singular value decomposition. Copyright © 2016 Pearson Education, Ltd. 7.2 1. a. xT Ax = [ x1 • Solutions 7-15 5 1 / 3 x1 2 x2 ] =5 x12 + x1 x2 + x22 1 x2 3 1 / 3 6 T 2 2 b. When x = , x Ax = 5(6) + (2 / 3)(6)(1) + (1) = 185. 1 1 T 2 2 c. When x = , x Ax = 5(1) + (2 / 3)(1)(3) + (3) = 16. 3 2. a. x Ax = [ x1 T x2 3 x3 ] 2 0 2 2 1 0 x1 1 x2 = 3x12 + 2 x22 + 4 x1 x2 + 2 x2 x3 0 x3 −2 T 2 2 b. When x = −1 , x Ax = 3( −2) + 2( −1) + 4( −2)( −1) + 2( −1)(5) = 12. 5 1 / 2 T 2 2 c. When x = 1 / 2 , x Ax = 3(1 / 2) + 2(1 / 2) + 4(1 / 2)(1 / 2) + 2(1 / 2)(1 / 2) = 11 / 2. 1 / 2 3 3. a. The matrix of the quadratic form is −2 3 b. The matrix of the quadratic form is 1 5 4. a. The matrix of the quadratic form is 8 0 b. The matrix of the quadratic form is 1 3 5. a. The matrix of the quadratic form is −3 4 0 b. The matrix of the quadratic form is 3 2 3 6. a. The matrix of the quadratic form is 2 − 3 −2 . 5 1 . 0 8 . −5 1 . 0 −3 2 −2 3 0 −5 2 −2 0 4 −2 . −5 2 −5 . 0 −3 0 . 5 Copyright © 2016 Pearson Education, Ltd. 7-16 CHAPTER 7 • Symmetric Matrices and Quadratic Forms 0 b. The matrix of the quadratic form is −1 0 −1 0 2 0 2 . 4 1 5 7. The matrix of the quadratic form is A = . The eigenvalues of A are 6 and –4. An eigenvector 5 1 1/ 2 1 −1 for λ = 6 is , which may be normalized to u1 = . An eigenvector for λ = –4 is , 1 1 1/ 2 −1/ 2 −1 which may be normalized to u 2 = . Then A = PDP , where 1/ 2 1/ 2 −1/ 2 0 6 = P [u = u2 ] and D = . The desired change of variable is x = Py, and 1 0 −4 1/ 2 1/ 2 T Ax ( Py )T A( P= y ) yT PT AP = y yT D = y 6 y12 − 4 y22 . the new quadratic form is x = 9 −4 4 8. The matrix of the quadratic form is A = −4 7 0 . The eigenvalues of A are 3, 9, and 15. An 4 0 11 −2 −2 / 3 eigenvector for λ = 3 is −2 , which may be normalized to u1 = −2 / 3 . An eigenvector for λ = 9 1 1/ 3 2 −1 −1/ 3 is 2 , which may be normalized to u 2 = 2 / 3 . An eigenvector for λ = 15 is −1 , which may 2 2 2 / 3 2 / 3 2 / 3 −2 / 3 −1/ 3 −1 be normalized to u3 = −1/ 3 . Then A = PDP , where P = 2 / 3 −1/ 3 [u1 u 2 u3 ] = −2 / 3 2 / 3 1/ 3 2/3 2 / 3 0 3 0 and D = 0 9 0 . The desired change of variable is x = Py, and the new quadratic form is 0 0 15 xT Ax = ( Py )T A( Py ) = yT PT APy = yT Dy = 3 y12 + 9 y22 + 15 y32 . 4 9. The matrix of the quadratic form is A = −2 −2 . The eigenvalues of A are 6 and 2, so the 4 −1 quadratic form is positive definite. An eigenvector for λ = 6 is , which may be normalized to 1 1 / 2 −1 / 2 1 u1 = . Then . An eigenvector for λ = 2 is , which may be normalized to u 2 = 1 1 / 2 1 / 2 Copyright © 2016 Pearson Education, Ltd. 7.2 −1 / 2 1 / 2 6 and D = 1 / 2 0 1 / 2 variable is x = Py, and the new quadratic form is A = PDP −1 , where = P u1 u 2 ] [= • Solutions 7-17 0 . The desired change of 2 xT = Ax ( Py )T A( P= y ) y T PT AP = y yT D = y 6 y12 + 2 y22 . 2 10. The matrix of the quadratic form is A = 3 3 . The eigenvalues of A are −7 and 3 , so the −6 −1 quadratic form is indefinite. An eigenvector for λ = −7 is , which may be normalized to 3 −1 / 10 3 / 10 3 u1 = . An eigenvector for λ = 3 is , which may be normalized to u 2 = . 1 3 / 10 1 / 10 −1 / 10 3 / 10 −7 and D = 0 3 / 10 1 / 10 change of variable is x = Py, and the new quadratic form is Then A = PDP −1 , where = P u1 u 2 ] [= 0 . The desired 3 xT Ax = ( Py )T A( Py ) = y T PT APy = y T Dy = −7 y12 + 3 y22 2 11. The matrix of the quadratic form is A = −2 −2 . The eigenvalues of A are −2 and 3, so the −1 1 quadratic form is indefinite. An eigenvector for λ = −2 is , which may be normalized to 2 1 / 5 −2 / 5 −2 u1 = . An eigenvector for λ = 3 is , which may be normalized to u 2 = . 1 2 / 5 1 / 5 1 / 5 −2 / 5 −2 and D = 1 / 5 0 2 / 5 of variable is x = Py, and the new quadratic form is Then A = PDP −1 , where = P u1 u 2 ] [= 0 . The desired change 3 xT Ax = ( Py )T A( Py ) = y T PT APy = y T Dy = −2 y12 + 3 y22 . −1 12. The matrix of the quadratic form is A = −1 −1 . The eigenvalues of A are –2 and 0, so the −1 1 quadratic form is negative semidefinite. An eigenvector for λ = −2 is , which may be 1 1 / 2 −1 normalized to u1 = . An eigenvector for λ = 0 is , which may be normalized to 1 1 / 2 1 / 2 −1 / 2 −1 / 2 −2 −1 = P [= u1 u 2 ] u2 = and D = . Then A = PDP , where 1 / 2 0 1 / 2 1 / 2 desired change of variable is x = Py, and the new quadratic form is xT Ax = ( Py )T A( Py ) = y T PT APy = y T Dy = −2 y12 . Copyright © 2016 Pearson Education, Ltd. 0 . The 0 7-18 CHAPTER 7 • Symmetric Matrices and Quadratic Forms 1 13. The matrix of the quadratic form is A = −3 −3 . The eigenvalues of A are 10 and 0, so the 9 1 quadratic form is positive semidefinite. An eigenvector for λ = 10 is , which may be −3 1/ 10 3 normalized to u1 = . An eigenvector for λ = 0 is , which may be normalized to 1 −3/ 10 3/ 10 1/ 10 3/ 10 10 −1 = P [u = u2 = u2 ] . Then A = PDP , where and D = 1 0 1/ 10 −3/ 10 1/ 10 desired change of variable is x = Py, and the new quadratic form is 0 . The 0 T T x= Ax ( Py )T A= ( Py ) yT PT= APy y= Dy 10 y12 3 14. The matrix of the quadratic form is A = 2 2 . The eigenvalues of A are −1 and 4, so the 0 −1 quadratic form is indefinite. An eigenvector for λ = −1 is , which may be normalized to 2 −1 / 5 2 / 5 2 u1 = . An eigenvector for λ = 4 is , which may be normalized to u 2 = . Then 1 2 / 5 1 / 5 −1 / 5 2 / 5 −1 and D = 0 2 / 5 1 / 5 variable is x = Py, and the new quadratic form is A = PDP −1 , where = P u1 u 2 ] [= 0 . The desired change of 4 ( Py )T A( Py ) = xT Ax = y T PT APy = y T Dy = − y12 + 4 y22 −3 2 15. [M] The matrix of the quadratic form is A = 2 2 2 −7 0 0 2 0 −10 3 2 0 . The eigenvalues of A are 3 −10 −13, − 9, − 7 and −1 so the quadratic form is negative definite. The corresponding eigenvectors 3 0 −1 0 1 −2 1 0 may be computed: λ = −1: , λ = −7 : , λ = −9 : , λ = −13 : . These 1 1 1 −1 1 1 1 1 eigenvectors may be normalized to form the columns of P, and A = PDP −1 , where 3 / 12 0 −1 / 2 0 0 0 0 −1 1/ 2 0 0 −7 0 0 1 / 12 −2 / 6 . The desired change of P = and D 0 0 9 0 − 1/ 6 1 / 2 −1 / 2 1 / 12 0 0 −13 1 / 12 0 1/ 6 1/ 2 1 / 2 variable is x = Py, and the new quadratic form is xT Ax = ( Py )T A( Py ) = y T PT APy = y T Dy = − y12 − 7 y22 − 9 y32 − 13 y42 Copyright © 2016 Pearson Education, Ltd. 7.2 • Solutions 7-19 4 4 0 −3 4 4 3 0 . The eigenvalues of A are −1 and 9, 16. [M] The matrix of the quadratic form is A = 0 3 4 4 4 −3 0 4 so the quadratic form is indefinite. The corresponding eigenvectors may be computed: 4 5 4 −5 −5 −4 5 −4 λ= , 9: , −1: , λ = . These bases may be converted via orthogonal 3 0 3 0 0 3 0 3 projections and scalings to orthogonal bases for the respective eigenspaces: 4 3 4 −3 0 0 −5 5 λ= 9 : , . Normalize these vectors to form the columns of −1: , , λ = 3 −4 3 4 0 5 0 5 4 / 50 3 / 50 4 / 50 −3 / 50 0 5 / 50 0 −5 / 50 −1 P, and A = PDP , where P = , and 3 / 50 − 4 / 50 3 / 50 4 / 50 0 5 / 50 0 5 / 50 0 0 0 −1 0 −1 0 0 . The desired change of variable is x = Py, and the new quadratic form is D= 0 0 9 0 0 0 9 0 ( Py )T A( Py ) = xT Ax = y T PT APy = y T Dy = −1 y12 − 1 y22 + 9 y32 + 9 y42 . 11 8 0 −6 8 11 6 0 . The eigenvalues of A are 21 and 17. [M] The matrix of the quadratic form is A = 0 6 11 8 8 11 −6 0 1, so the quadratic form is positive definite. The corresponding eigenvectors may be computed: 4 5 4 −5 −5 −4 5 −4 λ 1:= , = , λ 21: , . These bases may be converted via orthogonal 3 0 3 0 0 3 0 3 projections and scalings to orthogonal bases for the respective eigenspaces: 4 3 4 −3 0 −5 0 5 λ 1:= , = , λ 21: , . Normalize the vectors to form the columns of P, 3 −4 3 4 0 5 0 5 Copyright © 2016 Pearson Education, Ltd. 7-20 CHAPTER 7 • Symmetric Matrices and Quadratic Forms 4 / 50 3 / 50 4 / 50 −3 / 50 1 0 0 5 / 50 0 −5 / 50 −1 and A = PDP , where P = and D = 0 3 / 50 − 4 / 50 3 / 50 4 / 50 0 0 5 / 50 0 5 / 50 The desired change of variable is x = Py, and the new quadratic form is xT Ax = ( Py )T A( Py ) = y T PT APy = y T Dy = y12 + y22 + 21 y32 + 21 y42 . 0 1 0 0 0 0 21 0 0 0 . 0 21 2 −3 −3 −3 −3 2 −3 −3 . The eigenvalues of A are −7, 1, 18. [M] The matrix of the quadratic form is A = −3 −3 0 −1 0 −3 −3 −1 and 5, so the quadratic form is indefinite. The corresponding eigenvectors may be computed: −1 −1 1 0 1 0 −1 1 λ= 1: , λ = 5 : , . These eigenvectors may be normalized to −7 : , λ = 1 −1 0 1 0 1 1 1 1 / 2 0 −1 / 2 −1 / 2 0 1 / 2 −1 / 2 1 / 2 −1 form the columns of P, and A = PDP , where P = and 1 / 2 − 1 / 2 0 1 / 2 1 / 2 1/ 2 0 1 / 2 −7 0 D= 0 0 0 1 0 0 0 0 5 0 0 0 . The desired change of variable is x = Py, and the new quadratic form is 0 5 xT Ax = ( Py )T A( Py ) = y T PT APy = y T Dy = −7 y12 + y22 + 5 y32 + 5 y42 . 1 , the largest 19. Since 8 is larger than 5, the x2 term should be as large as possible. Since x1 + x2 = value that x2 can take is 1, and x1 = 0 when x2 = 1 . Thus the largest value the quadratic form can 2 2 2 take when xT x = 1 is 5(0) + 8(1) = 8. 2 20. Since 5 is larger in absolute value than –3, the x1 term should be as large as possible. Since x1 + x2 = 1 , the largest value that x1 can take is 1, and x2 = 0 when x1 = 1 . Thus the largest value 2 2 the quadratic form can take when xT x = 1 is 5(1) – 3(0) = 5. 21. a. True. See the definition before Example 1, even though a nonsymmetric matrix could be used to compute values of a quadratic form. b. True. See the paragraph following Example 3. c. True. The columns of P in Theorem 4 are eigenvectors of A. See the Diagonalization Theorem in Section 5.3. Copyright © 2016 Pearson Education, Ltd. 7.2 • Solutions 7-21 d. False. Q(x) = 0 when x = 0. e. True. See Theorem 5(a). f. True. See the Numerical Note after Example 6. 22. a. False. See the paragraph before Example 1. b. False. The matrix P must be orthogonal and make PT AP diagonal. See the paragraph before Example 4. c. False. There are also “degenerate” cases: a single point, two intersecting lines, or no points at all. See the subsection “A Geometric View of Principal Axes.” d. True. See the definition before Theorem 5. e. False. See Theorem 5(b). If xT Ax has only negative values for x ≠ 0, then xT Ax is negative definite. 23. The characteristic polynomial of A may be written in two ways: b a − λ det( A − λI ) = det = λ 2 − ( a + d )λ + ad − b2 and d − λ b (λ − λ1 )(λ − λ2 ) = λ 2 − (λ1 + λ2 )λ + λ1λ2 . The coefficients in these polynomials may be equated to 2 obtain λ1 + λ2 =a + d and λ1λ2 = ad − b = det A . 24. If det A > 0, then by Exercise 23, λ1λ2 > 0 , so that λ1 and λ2 have the same sign; also, ad= det A + b > 0 . 2 a. If det A > 0 and a > 0, then d > 0 also, since ad > 0. By Exercise 23, λ1 + λ2 = a + d > 0 . Since λ1 and λ2 have the same sign, they are both positive. So Q is positive definite by Theorem 5. b. If det A > 0 and a < 0, then d < 0 also, since ad > 0. By Exercise 23, λ1 + λ2 = a + d < 0 . Since λ1 and λ2 have the same sign, they are both negative. So Q is negative definite by Theorem 5. c. If det A < 0, then by Exercise 23, λ1λ2 < 0 . Thus indefinite by Theorem 5. λ1 and λ2 have opposite signs. So Q is T T T Bx ( Bx)= Bx || Bx || ≥ 0 , so 25. Exercise 28 in Section 7.1 showed that BT B is symmetric. Also x B= the quadratic form is positive semidefinite, and the matrix BT B is positive semidefinite. Suppose that B is square and invertible. Then if xT BT Bx = 0, || Bx || = 0 and Bx = 0. Since B is invertible, x = 0. Thus if x ≠ 0, xT BT Bx > 0 and BT B is positive definite. 26. Let A = PDPT , where PT = P −1. The eigenvalues of A are all positive: denote them λ1 ,…, λ n . Let C 2 be the diagonal matrix with λ1 ,…, λ n on its diagonal. Then = D C= C T C . If B = PCPT , then B is positive definite because its eigenvalues are the positive numbers on the diagonal of C. Also T T T = BT B ( PCPT )T ( PCP = ) ( PTT C T PT )( PCP = ) PC T= CPT PDP = A since PT P = I . 27. Since the eigenvalues of A and B are all positive, the quadratic forms xT Ax and xT Bx are positive T T T definite by Theorem 5. Let x ≠ 0. Then xT Ax > 0 and xT Bx > 0 , so x ( A + B )x = x Ax + x Bx > 0 , and the quadratic form x ( A + B )x is positive definite. Note that A + B is also a symmetric matrix. Thus by Theorem 5 all the eigenvalues of A + B must be positive. T Copyright © 2016 Pearson Education, Ltd. 7-22 CHAPTER 7 • Symmetric Matrices and Quadratic Forms 28. The eigenvalues of A are all positive by Theorem 5. Since the eigenvalues of A−1 are the reciprocals of the eigenvalues of A (see Exercise 25 in Section 5.1), the eigenvalues of A−1 are all positive. Note that A−1 is also a symmetric matrix. By Theorem 5, the quadratic form xT A−1x is positive definite. 7.3 SOLUTIONS Notes: Theorem 6 is the main result needed in the next two sections. Theorem 7 is mentioned in Example 2 of Section 7.4. Theorem 8 is needed at the very end of Section 7.5. The economic principles in Example 6 may be familiar to students who have had a course in macroeconomics. 2 0 5 1. The matrix of the quadratic form on the= left is A 2 6 −2 . The equality of the quadratic 0 −2 7 forms implies that the eigenvalues of A are 9, 6, and 3. An eigenvector may be calculated for each 1 / 3 2 / 3 −2 / 3 eigenvalue and normalized: 2 / 3 , λ 6= : 1 / 3 , λ 3 : 2 / 3 . A desired change λ 9 := = 2 / 3 2 / 3 1 / 3 − 1/ 3 2 / 3 − 2 / 3 . of variable is x = Py, where P = 2 / 3 1/ 3 2 / 3 −2 / 3 2 / 3 1/ 3 3 3 1 2. The matrix of the quadratic form on the left is A = 3 3 1 . The equality of the quadratic forms 1 1 5 implies that the eigenvalues of A are 7, 4, and 0. An eigenvector may be calculated for each 1 / 3 −1 / 6 −1 / 2 eigenvalue and normalized: λ = 7 : 1 / 3 , λ = 4 : −1 / 6 , λ = 0 : 1 / 2 . A desired 0 1 / 3 2 / 6 1 / 3 change of variable is x = Py, where = P 1 / 3 1 / 3 −1 / 6 −1 / 6 2/ 6 −1 / 2 1 / 2 . 0 Copyright © 2016 Pearson Education, Ltd. 7.3 • Solutions 7-23 3. a. By Theorem 6, the maximum value of xT Ax subject to the constraint xT x = 1 is the greatest eigenvalue λ1 of A. By Exercise 1, λ1 = 9. b. By Theorem 6, the maximum value of xT Ax subject to the constraint xT x = 1 occurs at a unit −1 / 3 eigenvector u corresponding to the greatest eigenvalue λ1 of A. By Exercise 1, u =± −2 / 3 . 2 / 3 c. By Theorem 7, the maximum value of xT Ax subject to the constraints xT x = 1 and xT u = 0 is the second greatest eigenvalue λ2 of A. By Exercise 1, λ2 = 6. 4. a. By Theorem 6, the maximum value of xT Ax subject to the constraint xT x = 1 is the greatest eigenvalue λ1 of A. By Exercise 2, λ1 = 7. b. By Theorem 6, the maximum value of xT Ax subject to the constraint xT x = 1 occurs at a unit 1/ 3 eigenvector u corresponding to the greatest eigenvalue λ1 of A. By Exercise 2, u = ± 1/ 3 . 1/ 3 c. By Theorem 7, the maximum value of xT Ax subject to the constraints xT x = 1 and xT u = 0 is the second greatest eigenvalue λ2 of A. By Exercise 2, λ2 = 4. 1 5. The matrix of the quadratic form is A = −5 −5 . The eigenvalues of A are λ1 = 6 and λ2 = −4. 1 a. By Theorem 6, the maximum value of xT Ax subject to the constraint xT x = 1 is the greatest eigenvalue λ1 of A, which is 6. b. By Theorem 6, the maximum value of xT Ax subject to the constraint xT x = 1 occurs at a unit −1 eigenvector u corresponding to the greatest eigenvalue λ1 of A. One may compute that is 1 −1/ 2 an eigenvector corresponding to λ1 = 6, so u = ± . 1/ 2 c. By Theorem 7, the maximum value of xT Ax subject to the constraints xT x = 1 and xT u = 0 is the second greatest eigenvalue λ2 of A, which is −4 . 3 6. The matrix of the quadratic form is A = 4 4 . The eigenvalues of A are λ1 = 11 and λ2 = 1. 9 a. By Theorem 6, the maximum value of xT Ax subject to the constraint xT x = 1 is the greatest eigenvalue λ1 of A, which is 11. Copyright © 2016 Pearson Education, Ltd. 7-24 CHAPTER 7 • Symmetric Matrices and Quadratic Forms b. By Theorem 6, the maximum value of xT Ax subject to the constraint xT x = 1 occurs at a unit 1 eigenvector u corresponding to the greatest eigenvalue λ1 of A. One may compute that is 2 1 / 5 an eigenvector corresponding to λ1 = 11, so u = ± . 2 / 5 c. By Theorem 7, the maximum value of xT Ax subject to the constraints xT x = 1 and xT u = 0 is the second greatest eigenvalue λ2 of A, which is 1. 7. The eigenvalues of the matrix of the quadratic form are λ1 = 2, λ2 = −1, and λ3 = −4. By Theorem 6, the maximum value of xT Ax subject to the constraint xT x = 1 occurs at a unit eigenvector u 1/ 2 corresponding to the greatest eigenvalue λ1 of A. One may compute that 1 is an eigenvector 1 1/ 3 corresponding to λ1 = 2, so u = ± 2 / 3 . 2 / 3 8. The eigenvalues of the matrix of the quadratic form are λ1 = 9, and λ2 = −3. By Theorem 6, the maximum value of xT Ax subject to the constraint xT x = 1 occurs at a unit eigenvector u −1 −2 corresponding to the greatest eigenvalue λ1 of A. One may compute that 0 and 1 are linearly 1 0 independent eigenvectors corresponding to λ1 = 9, so u can be any unit vector that is a linear −2 −1 combination of 0 and 1 . Alternatively, u can be any unit vector which is orthogonal to the 0 1 1 eigenspace corresponding to the eigenvalue λ2 = −3. Since multiples of 2 are eigenvectors 1 1 corresponding to λ2 = −3, u can be any unit vector orthogonal to 2 . 1 9. This is equivalent to finding the maximum value of xT Ax subject to the constraint xT x = 1. By Theorem 6, this value is the greatest eigenvalue λ1 of the matrix of the quadratic form. The matrix of 7 the quadratic form is A = −1 −1 , and the eigenvalues of A are λ1= 5 + 5, λ2= 5 − 5. Thus 3 the desired constrained maximum value is λ1= 5 + 5. Copyright © 2016 Pearson Education, Ltd. 7.3 • Solutions 7-25 10. This is equivalent to finding the maximum value of xT Ax subject to the constraint xT x = 1 . By Theorem 6, this value is the greatest eigenvalue λ1 of the matrix of the quadratic form. The matrix of −3 the quadratic form is A = −1 −1 , and the eigenvalues of A are λ1 = 1 + 17, λ2 = 1 − 17. Thus 5 the desired constrained maximum value is λ1 = 1 + 17. T xT Ax x= (3x) 11. Since x is an eigenvector of A corresponding to the eigenvalue 3, Ax = 3x, and = 3(= || x ||2 3 since x is a unit vector. xT x) 3= T x T Ax x= (λ x ) λ= ( x T x ) λ since xT x = 1 . 12. Let x be a unit eigenvector for the eigenvalue λ. Then= So λ must satisfy m ≤ λ ≤ M. T 13. If m = M, then let t = (1 – 0)m + 0M = m and x = u n . Theorem 6 shows that u n Au n = m. Now suppose that m < M, and let t be between m and M. Then 0 ≤ t – m ≤ M – m and 0 ≤ (t – m)/(M – m) ≤ 1. Let α = (t – m)/(M – m), and let x =1 − α u n + α u1. The vectors 1 − α u n and α u1 are orthogonal because they are eigenvectors for different eigenvalues (or one of them is 0). By the Pythagorean Theorem xT x =|| x ||2 = || 1 − α u n ||2 + || α u1 ||2 = |1 − α ||| u n ||2 + | α ||| u1 ||2 = (1 − α ) + α = 1 , since u n and u1 are unit vectors and 0 ≤ α ≤ 1. Also, since u n and u1 are orthogonal, xT Ax =( 1 − α u n + α u1 )T A( 1 − α u n + α u1 ) =( 1 − α u n + α u1 )T (m 1 − α u n + M α u1 ) |1 α | muTn u n + | α | M u1T u1 =− (1 α )m + α M = =− t Thus the quadratic form xT Ax assumes every value between m and M for a suitable unit vector x. 0 3/ 2 3 / 2 0 14. [M] The matrix of the quadratic form is A = 5 / 2 7 / 2 7 / 2 5 / 2 λ1 = 15 / 2, λ2 = −1 / 2, λ3 = −5 / 2, and λ4 = −9 / 2. 5/ 2 7/2 0 3/ 2 7 / 2 5 / 2 . The eigenvalues of A are 3 / 2 0 a. By Theorem 6, the maximum value of xT Ax subject to the constraint xT x = 1 is the greatest eigenvalue λ1 of A, which is 15/2. Copyright © 2016 Pearson Education, Ltd. 7-26 CHAPTER 7 • Symmetric Matrices and Quadratic Forms b. By Theorem 6, the maximum value of xT Ax subject to the constraint xT x = 1 occurs at a unit 1 1 eigenvector u corresponding to the greatest eigenvalue λ1 of A. One may compute that is an 1 1 1/ 2 1/ 2 eigenvector corresponding to λ1 = 15 / 2, so u = ± . 1/ 2 1/ 2 c. By Theorem 7, the maximum value of xT Ax subject to the constraints xT x = 1 and xT u = 0 is the second greatest eigenvalue λ2 of A, which is −1 / 2 . 4 −3 15. [M] The matrix of the quadratic form is A = −5 −5 λ2 = 3, λ3 = 1, and λ4 = −9. −3 0 −3 −3 −5 −3 0 −1 −5 −3 . The eigenvalues of A are λ1 = 9, −1 0 a. By Theorem 6, the maximum value of xT Ax subject to the constraint xT x = 1 is the greatest eigenvalue λ1 of A, which is 9. b. By Theorem 6, the maximum value of xT Ax subject to the constraint xT x = 1 occurs at a unit −2 0 eigenvector u corresponding to the greatest eigenvalue λ1 of A. One may compute that is 1 1 −2 / 6 0 an eigenvector corresponding to λ1 = 9, so u = ± . 1/ 6 1/ 6 c. By Theorem 7, the maximum value of xT Ax subject to the constraints xT x = 1 and xT u = 0 is the second greatest eigenvalue λ2 of A, which is 3. −6 −2 16. [M] The matrix of the quadratic form is A = −2 −2 λ1 = −4, λ2 = −10, λ3 = −12, and λ4 = −16. −2 −10 0 0 −2 0 −13 3 −2 0 . The eigenvalues of A are 3 −13 a. By Theorem 6, the maximum value of xT Ax subject to the constraint xT x = 1 is the greatest eigenvalue λ1 of A, which is –4. Copyright © 2016 Pearson Education, Ltd. 7.4 • Solutions 7-27 b. By Theorem 6, the maximum value of xT Ax subject to the constraint xT x = 1 occurs at a unit −3 1 eigenvector u corresponding to the greatest eigenvalue λ1 of A. One may compute that is 1 1 −3/ 12 1/ 12 an eigenvector corresponding to λ1 = −4, so u = ± . 1/ 12 1/ 12 c. By Theorem 7, the maximum value of xT Ax subject to the constraints xT x = 1 and xT u = 0 is the second greatest eigenvalue λ2 of A, which is –10. 0 1 / 2 17. [M] The matrix of the quadratic form is A = 3 / 2 15 λ1 = 17, λ2 = 13, λ3 = −14, and λ4 = −16. 1/ 2 0 15 3/ 2 3/ 2 15 0 1/ 2 15 3 / 2 . The eigenvalues of A are 1 / 2 0 a. By Theorem 6, the maximum value of xT Ax subject to the constraint xT x = 1 is the greatest eigenvalue λ1 of A, which is 17. b. By Theorem 6, the maximum value of xT Ax subject to the constraint xT x = 1 occurs at a unit 1 1 eigenvector u corresponding to the greatest eigenvalue λ1 of A. One may compute that is an 1 1 1/ 2 1/ 2 eigenvector corresponding to λ1 = 17, so u = ± . 1/ 2 1/ 2 c. By Theorem 7, the maximum value of xT Ax subject to the constraints xT x = 1 and xT u = 0 is the second greatest eigenvalue λ2 of A, which is 13. 7.4 SOLUTIONS Notes: This section presents a modern topic of great importance in applications, particularly in computer calculations. An understanding of the singular value decomposition is essential for advanced work in science and engineering that requires matrix computations. Moreover, the singular value decomposition explains much about the structure of matrix transformations. The SVD does for an arbitrary matrix almost what an orthogonal decomposition does for a symmetric matrix. Copyright © 2016 Pearson Education, Ltd. 7-28 CHAPTER 7 1 1. Let A = 0 • Symmetric Matrices and Quadratic Forms 0 1 . Then AT A = −3 0 0 , and the eigenvalues of AT A are seen to be (in decreasing 9 order) λ1 = 9 and λ2 = 1. Thus the singular values of A are σ= 1 −3 2. Let A = 0 0 9 . Then AT A = 0 0 3 4 . Then AT A = 2 6 0 73 . Then AT A = 3 24 = 9 3 and σ= 2 = 16 4 and σ= 2 0 4 . Then AT A = 0 0 = 1 1. 24 , and the eigenvalues of AT A are seen to be (in decreasing 9 σ1 order) λ1 = 81 and λ2 = 1. Thus the singular values of A are= −2 5. Let A = 0 = 0 0. 6 , and the eigenvalues of AT A are (in decreasing order) 13 σ1 λ1 = 16 and λ2 = 1. Thus the singular values of A are= 3 4. Let A = 8 = 1 1. 0 , and the eigenvalues of AT A are (in decreasing order) 0 = λ1 = 9 and λ2 = 0. Thus the singular values of A are σ 1 2 3. Let A = 0 = 9 3 and σ= 2 = 81 9 and σ= 2 = 1 1. 0 , and the eigenvalues of AT A are seen to be (in decreasing 0 1 order) λ1 = 4 and λ2 = 0. Associated unit eigenvectors may be computed: λ1 = 4 : , 0 1 0 0 = = 4 2 and λ2 = 0 : . Thus one choice for V is V = 1 . The singular values of A are σ 0 1 1 2 0 1 −1 σ= = 0 0. Thus the matrix Σ is Σ = = u1 = Av1 . Because 2 . Next compute σ1 0 0 0 Av2 = 0, the only column found for U so far is u1. The other column of U is found by extending {u1} 0 −1 0 to an orthonormal basis for 2 . An easy choice is u2 = . Let U = . Thus 0 1 1 −1 A =U Σ V T = 0 −3 6. Let A = 0 0 2 1 0 0 1 0 0 0 9 . Then AT A = −2 0 0 . 1 0 , and the eigenvalues of AT A are seen to be (in decreasing 4 1 order) λ1 = 9 and λ2 = 4. Associated unit eigenvectors may be computed: λ1 = 9 : , 0 1 0 0 = = 9 3 and λ2 = 4 : . Thus one choice for V is V = 1 . The singular values of A are σ 0 1 1 3 0 1 −1 σ= = 4 2. Thus the matrix Σ is Σ = = u1 = Av , 2 . Next compute σ 1 1 0 0 2 Copyright © 2016 Pearson Education, Ltd. 7.4 −1 0 1 = Av 2 . Since {u1 , u 2 } is a basis for 2 , let U = σ2 0 −1 0 3 0 1 0 −1 A =U Σ V T = . 0 −1 0 2 0 1 = u2 2 7. Let A = 2 −1 8 . Then AT A = 2 2 • Solutions 7-29 0 . Thus −1 2 , and the characteristic polynomial of AT A is 5 λ 2 − 13λ + 36 = (λ − 9)(λ − 4), and the eigenvalues of AT A are (in decreasing order) λ1 = 9 and 2 / 5 −1 / 5 computed: λ1 9= λ2 = 4. Associated unit eigenvectors may be = : , λ2 4 : . Thus 2 / 5 1 / 5 2 / 5 −1/ 5 9 3 and = one choice for V is V = . The singular values of A are σ= 1 2 / 5 1/ 5 1 / 5 3 0 1 σ= = 4 2. Thus the matrix Σ is Σ = . Next compute = u1 = Av1 , 2 σ1 0 2 2 / 5 = u2 −2 / 5 1/ 5 1 2 = Av 2 . Since {u1 , u 2 } is a basis for , let U = σ2 1 / 5 2 / 5 1/ 5 A =U Σ V T = 2 / 5 4 8. Let A = 0 −2 / 5 3 1/ 5 0 6 16 . Then AT A = 4 24 0 2 / 5 2 −1/ 5 −2 / 5 . Thus 1/ 5 1/ 5 . 2 / 5 24 , and the characteristic polynomial of AT A is 52 λ 2 − 68λ + 256 =(λ − 64)(λ − 4), and the eigenvalues of AT A are (in decreasing order) λ1 = 64 and 1 / 5 −2 / 5 computed: λ1 64 λ2 = 4. Associated unit eigenvectors may be = = : , λ2 4 : . 2 / 5 1 / 5 1/ 5 −2 / 5 σ1 = 64 8 and Thus one choice for V is V = . The singular values of A are= 1/ 5 2 / 5 2 / 5 8 0 1 σ= = 4 2. Thus the matrix Σ is Σ = . Next compute = u1 = Av1 , 2 σ1 0 2 1 / 5 = u2 −1 / 5 2 / 5 1 2 = Av 2 . Since {u1 , u 2 } is a basis for , let U = σ2 1/ 5 2 / 5 2 / 5 A =U Σ V T = 1 / 5 −1 / 5 8 2 / 5 0 0 1 / 5 2 −2 / 5 2 / 5 . 1 / 5 Copyright © 2016 Pearson Education, Ltd. −1/ 5 . Thus 2 / 5 7-30 CHAPTER 7 3 9. Let A = 0 1 • Symmetric Matrices and Quadratic Forms −3 10 T 0 . Then A A = −8 1 −8 , and the characteristic polynomial of AT A is 10 λ 2 − 20λ + 36 = (λ − 18)(λ − 2), and the eigenvalues of AT A are (in decreasing order) λ1 = 18 and −1 / 2 1 / 2 be computed: λ1 18 λ2 = 2. Associated unit eigenvectors may= = : , λ2 2 : . 1 / 2 1 / 2 −1 / 2 1 / 2 σ1 = 18 3 2 and Thus one choice for V is V = . The singular values of A are= 1 / 2 1 / 2 3 2 0 1 σ 2 = 2. Thus the matrix Σ is Σ = 0 Av = u1 = 2 . Next compute σ1 1 0 0 0 1 Av 2 0 . Since {u1 , u 2 } is not a basis for 3 , we need a unit vector = u2 = σ2 1 to both u1 and u 2 . The vector u3 must satisfy the set of equations u1 x = 0 0 − x1 + 0 x2 + 0 x3 = 0 equivalent to the linear equations = , so x = 1 , and u3 0 x1 + 0 x2 + x3 = 0 0 T −1 U = 0 0 7 10. Let A = 5 0 0 0 1 0 −1 T 1 . Thus A =U Σ V = 0 0 0 1 72 T 5 . Then A A = 32 0 0 0 1 0 3 2 1 0 0 0 0 −1 / 2 2 1/ 2 0 −1 0 , 0 u3 that is orthogonal and u 2 x = 0. These are 0 1 . Therefore let 0 T 1/ 2 . 1 / 2 32 , and the characteristic polynomial of AT A is 26 λ 2 − 100λ + 900 =(λ − 90)(λ − 10) , and the eigenvalues of AT A are (in decreasing order) λ1 = 90 and λ2 = 10. Associated unit eigenvectors may be computed: 2 / 5 = = λ 90 : , λ 10 : 1 / 5 −1 / 5 2 / 5 . Thus one choice for V is V = 2 / 5 1 / 5 −1 / 5 . The 2 / 5 3 10 0 90 3 10 and σ 2 = 10. Thus the matrix Σ is Σ = σ1 = singular values of A are= 0 10 . 0 0 1 / 2 −1 / 2 1 1 Next compute = u1 = Av 2 1 / 2 . Since {u1 , u 2 } is not a basis for Av1 1 / 2= , u2 = σ1 σ2 0 0 Copyright © 2016 Pearson Education, Ltd. 7.4 • Solutions 7-31 3 , we need a unit vector u3 that is orthogonal to both u1 and u 2 . The vector u3 must satisfy the T T set of equations u1 x = 0 and u 2 x = 0. These are equivalent to the linear equations − 1 x1 + 2 1 x1 + 2 1 x2 + 0 x3 = 0 2 = , so x 1 x2 + 0 x3 = 0 2 1 / 2 A =U Σ V =1 / 2 0 T −3 11. Let A 6 = 6 −1 / 2 1/ 2 0 0 = 0 , and u3 1 0 3 10 0 0 1 0 1 81 T −2 . Then A A = −27 −2 1 / 2 0 0 Therefore let U = 1 / 2 0 1 0 2/ 5 10 −1 / 5 0 −1 / 2 1/ 2 0 0 0 . Thus 1 1 / 5 . 2 / 5 −27 , and the characteristic polynomial of AT A is 9 λ 2 − 90λ = λ (λ − 90), and the eigenvalues of AT A are (in decreasing order) λ1 = 90 and λ2 = 0. 3 / 10 1 / 10 Associated unit eigenvectors may= be computed: λ1 90 = : , λ2 0 : . Thus one −1 / 10 3 / 10 3/ 10 1/ 10 σ1 = 90 3 10 and choice for V is V = . The singular values of A are= −1/ 10 3/ 10 3 10 0 −1/ 3 1 0 0 . Next compute σ= = 0 0. Thus the matrix Σ is Σ = Av1 2 / 3 . Because = u1 = 2 σ 1 2 / 3 0 0 Av2 = 0, the only column found for U so far is u1. The other columns of U can be found by extending {u1} to an orthonormal basis for 3 . In this case, we need two orthogonal unit vectors u2 and u3 that are orthogonal to u1. Each vector must satisfy the equation u1 x = 0, which is equivalent to the equation − x1 + 2 x2 + 2 x3 = 0. An orthonormal basis for the solution set of this equation is 2/3 2 / 3 2 / 3 2 / 3 −1/ 3 Therefore, let U 2 / 3 . Thus u2 = = −1/ 3 , u3 = 2 / 3 . 2 / 3 −1/ 3 2 / 3 −1/ 3 2 / 3 −1/ 3 2 / 3 T −1/ 3 A =U Σ V = 2 / 3 2 / 3 T 2/3 −1/ 3 2/3 2 / 3 3 10 2 / 3 0 0 −1/ 3 1 1 2 12. Let A = 0 1 . Then AT A = 0 −1 1 0 3/ 10 0 1/ 10 0 −1/ 10 . 3/ 10 0 , and the eigenvalues of AT A are seen to be (in decreasing 3 0 order) λ1 = 3 and λ2 = 2. Associated unit eigenvectors may be computed: λ1 = 3 : , 1 Copyright © 2016 Pearson Education, Ltd. 7-32 CHAPTER 7 • Symmetric Matrices and Quadratic Forms 0 1 1 λ2 = 2 : . Thus one choice for V is V = . The singular values of A are σ 1 = 3 and 1 0 0 3 σ 2 = 2. Thus the matrix Σ is Σ = 0 0 0 1 = u1 = Av1 2 . Next compute σ 1 0 1 / 3 1 / 3 , 1 / 3 1/ 2 1 Av 2 0 . Since {u1 , u 2 } is not a basis for 3 , we need a unit vector u3 that is = u2 = σ2 −1 / 2 T T orthogonal to both u1 and u 2 . The vector u3 must satisfy the set of equations u1 x = 0 and u 2 x = 0. 1/ 6 1 x1 + x2 + x3 = 0 −2 , and u = These are equivalent to the linear equations , so x = −2 / 6 . 3 x1 + 0 x2 − x3 = 0 1 1 / 6 1/ 3 Therefore let U 1/ 3 = 1/ 3 1/ 3 A =U Σ V T =1/ 3 1/ 3 1/ 2 0 −1/ 2 1/ 2 0 −1/ 2 1/ 6 −2 / 6 . Thus 1/ 6 1/ 6 3 −2 / 6 0 1/ 6 0 0 0 2 1 0 1 . 0 2 3 2 8 17 T , ATT= T AT AA = . Then A = 2 , and the eigenvalues of 3 −2 8 17 2 −2 ATT AT are seen to be (in decreasing order) λ1 = 25 and λ 2 = 9. Associated unit eigenvectors may 3 13. Let A = 2 2 3 1 / 2 −1 / 2 be computed: λ 25 = = : , λ 9 : . Thus one choice for V is 1 / 2 1 / 2 1/ 2 −1/ 2 T σ1 = 25 5 and σ= = 9 3. Thus the V = . The singular values of A are= 2 1/ 2 1/ 2 1/ 2 −1/ 18 1 T = A v 2 1/ 18 . Since 2 , u2 1/ = σ2 0 − 4 / 18 3 {u1 , u 2 } is not a basis for , we need a unit vector u3 that is orthogonal to both u1 and u 2 . The 5 matrix Σ is Σ =0 0 0 1 T u1 = = A v1 3 . Next compute σ1 0 T T vector u3 must satisfy the set of equations u1 x = 0 and u 2 x = 0. These are equivalent to the linear Copyright © 2016 Pearson Education, Ltd. 7.4 −2 x1 + x2 + 0 x3 = 0 equations = , so x = 2 , and u3 − x1 + x2 − 4 x3 = 0 1 1/ 2 U = 1/ 2 0 −1/ 18 1/ 18 −4 / 18 1/ 2 AT =U Σ V T =1/ 2 0 • Solutions 7-33 −2 / 3 2 / 3 . Therefore let 1 / 3 −2 / 3 2 / 3 . Thus 1/ 3 −1/ 18 1/ 18 −4 / 18 1/ 2 taking transposes: = A 1/ 2 −2 / 3 5 2 / 3 0 1/ 3 0 −1/ 2 5 1/ 2 0 0 1/ 2 3 −1/ 2 0 0 3 2 / 5 T 14. From Exercise 7, A= U Σ V with V = 1/ 5 1/ 2 . An SVD for A is computed by 1/ 2 1/ 2 1/ 2 0 −1/ 18 1/ 18 0 2/3 −2 / 3 0 −4 / 18 . 1/ 3 −1/ 5 . Since the first column of V is a unit 2 / 5 eigenvector associated with the greatest eigenvalue λ1 of AT A, so the first column of V is a unit vector at which || Ax || is maximized. 15. a. Since A has 2 nonzero singular values, rank A = 2. b. By Example 6,= {u1 , u 2 } for Nul A. .40 −.78 .37 , −.33 −.84 −.52 } is a basis for Col A and {v 3= .58 −.58 .58 is a basis 16. a. Since A has 2 nonzero singular values, rank A = 2. −.86 −.11 b. By Example 6, {u1 , u 2 } = .31 , .68 is a basis for Col A and .41 −.73 .65 −.34 .08 .42 {v 3 , v 4 } = , is a basis for Nul A. −.16 −.84 −.73 −.08 T = 1 det = I det U= U 17. First note that the determinant of an orthogonal matrix is ±1, because (det U T )(det U ) = (det U ) 2 . Suppose that A is square and A= U Σ V T . Then Σ is square, and det A = (det U )(det Σ)(det V T ) = ±det Σ = ±σ 1 …σ n . Copyright © 2016 Pearson Education, Ltd. 7-34 CHAPTER 7 • Symmetric Matrices and Quadratic Forms T −1 18. Let A =U Σ V =U Σ V . Since A is square and invertible, rank A = n, and all of the entries on the −1 (U Σ V −1 ) −1 = V Σ −1 U −1 = V Σ −1 U T . diagonal of Σ must be nonzero. So A = 19. Since U and V are orthogonal matrices, AT A= (U ΣV T )T U ΣV T = V ΣT U T U ΣV T = V (ΣT Σ)V T = V (ΣT Σ)V −1 If σ 1,…,σ r are the diagonal entries in Σ, then ΣT Σ is a diagonal matrix with diagonal entries σ 1 ,…,σ r and possibly some zeros. Thus V diagonalizes AT A and the columns of V are 2 2 eigenvectors of AT A by the Diagonalization Theorem in Section 5.3. Likewise AAT= U Σ V T (U Σ V T )T= U Σ V T V ΣT U T= U (Σ ΣT )U T= U (Σ ΣT )U −1 so U diagonalizes AAT and the columns of U must be eigenvectors of AAT . Moreover, the Diagonalization Theorem states that σ 1 ,…,σ r are the nonzero eigenvalues of AT A . Hence σ 1,…,σ r are the nonzero singular values of A. 2 2 20. Let A= U ΣV T . The matrix PU is orthogonal, because P and U are both orthogonal. (See Exercise 29 PA ( PU ) Σ V has the form required for a singular value in Section 6.2). So the equation= decomposition. By Exercise 19, the diagonal entries in Σ are the singular values of PA. T 21. The right singular vector v1 is an eigenvector for the largest eigenvector λ1 of AT A. By Theorem 7 T T in Section 7.3, the second largest eigenvalue λ2 is the maximum of x ( A A)x over all unit vectors orthogonal to v1 . Since x ( A A)x = || Ax || , the square root of λ2 , which is the second largest singular value of A, is the maximum of || Ax || over all unit vectors orthogonal to v1. T T 2 22. If A is positive definite, then A = PDPT , where P is an orthogonal matrix and D is a diagonal matrix. The diagonal entries of D are positive because they are the eigenvalues of a positive definite matrix. Since P is an orthogonal matrix, PPT = I and the square matrix PT is invertible. Moreover, ( PT ) −1= ( P −1 ) −1= P= ( PT )T , so PT is an orthogonal matrix. Thus the factorization A = PDPT has the properties that make it a singular value decomposition. = UΣ 23. From the proof of Theorem 10, T of the product (U Σ)V [σ 1u1 … σ r ur 0 … 0]. The column-row expansion v1T T T shows that A= (U Σ)V T= (U Σ) = σ 1u1 v1 + …+ σ r u r v r , where r is T v n the rank of A. 0 = AT σ 1 v1u1T + …+ σ r v r uTr . Then since uTi u j = 24. From Exercise 23, 1 for i ≠ j , for i = j T T T = AT u j (σ 1 v1u1T + …+ σ = (σ= σ= r v r u r )u j j v j u j )u j j v j (u j u j ) σ j v j Copyright © 2016 Pearson Education, Ltd. 7.4 • Solutions 7-35 25. Consider the SVD for the standard matrix A of T, say A= U ΣV T . Let = B {v1 ,…, v n } and = C {u1 ,…, u m } be bases for and n m constructed respectively from the columns of V and U. Since the columns of V are orthogonal, V v j = e j , where e j is the jth column of the n × n identity T matrix. To find the matrix of T relative to B and C, compute T ( v j ) = Av j =U ΣV T v j =U Σe j =U σ j e j =σ jUe j =σ j u j , so [T ( v j )]C = σ j e j . Formula (4) in the discussion at the beginning of Section 5.4 shows that the “diagonal” matrix Σ is the matrix of T relative to B and C. −18 2 26. [M] Let A = −14 −2 13 19 11 21 −4 −4 −12 4 4 528 −392 12 T . Then A A = 224 8 8 −176 −392 1092 −176 536 224 −176 192 −128 −176 536 , and the −128 288 eigenvalues of AT A are found to be (in decreasing order) λ1 = 1600, λ2 = 400, λ3 = 100, and −.4 .8 .4 .8 .4 −.2 λ4 = 0. Associated unit eigenvectors may be computed: λ1 : , λ2 : , λ3 : , −.2 .4 −.8 .4 .2 .4 .4 −.2 −.4 .8 −.2 .8 .4 −.2 −.4 −.4 . The singular values of A are λ4 : .Thus one choice for V is V = −.2 .4 −.8 .4 .4 .4 .8 .4 .2 .8 40 0 σ 1 = 40, σ 1 = 20, σ 3 = 10, and σ 4 = 0. Thus the matrix Σ is Σ = 0 0 0 20 0 0 0 0 10 0 0 0 . Next compute 0 0 .5 −.5 −.5 .5 .5 .5 1 1 , u , u 2 1= Av1 = Av 2 = Av 3 . Because Av4 = 0, only three columns = u1 = = 3 .5 −.5 .5 σ1 σ2 σ3 .5 .5 −.5 of U have been found so far. The last column of U can be found by extending {u1, u2, u3} to an T T orthonormal basis for 4 . The vector u4 must satisfy the set of equations u1 x = 0, u 2 x = 0, and −1 x1 + x2 + x3 + x4 = 0 −1 uT3 x = 0. These are equivalent to the linear equations − x1 + x2 − x3 + x4 = 0, so x = , 1 − x1 + x2 + x3 − x4 = 0 1 Copyright © 2016 Pearson Education, Ltd. 7-36 CHAPTER 7 • Symmetric Matrices and Quadratic Forms .5 −.5 .5 −.5 and u 4 = . Therefore, let U = .5 .5 .5 .5 .5 .5 A =U ΣV T = .5 .5 −.5 .5 −.5 .5 −.5 .5 .5 −.5 −.5 40 −.5 0 .5 0 .5 0 −.5 .5 −.5 .5 −.5 .5 .5 −.5 0 20 0 0 0 0 10 0 −.5 −.5 . Thus .5 .5 0 −.4 0 .8 0 .4 0 −.2 .8 .4 −.2 −.4 −.2 .4 −.8 .4 .4 .2 . .4 .8 14 −38 −8 41 −32 −4 −32 118 74 −3 −92 4 . Then AT A = −38 10 −52 , and the −3 121 2 14 −92 10 81 −72 −8 −8 74 −52 −72 100 eigenvalues of AT A are found to be (in decreasing order) λ1 = 270.87, λ2 = 147.85, λ3 = 23.73, λ4 = 18.55, and λ5 = 0. Associated unit eigenvectors may be computed: 6 2 27. [M] Let A = 0 −1 −8 7 −1 −2 −4 −5 −8 4 5 −6 2 4 −.10 .61 λ1 : −.21 , λ2 : −.52 .55 −.39 .29 .84 , λ3 : −.14 −.19 −.10 .61 V = −.21 −.52 .55 −.74 −.27 −.07 .38 .49 −.39 .29 .84 −.14 −.19 −.74 −.27 −.07 , λ4 : .38 .49 .41 −.50 .45 , λ5 : −.23 .58 −.36 −.48 −.19 .Thus one choice for V is −.72 −.29 −.36 −.48 −.19 . The nonzero singular values of A are σ 1 = 16.46, −.72 −.29 0 0 0 16.46 0 12.16 0 0 σ 1 = 12.16, σ 3 = 4.87, and σ 4 = 4.31. Thus the matrix Σ is Σ = 0 0 4.87 0 0 0 0 4.31 .41 −.50 .45 −.23 .58 −.57 −.65 −.42 .63 −.24 −.68 1 1 1 , Next compute Av Av1 ,u Av 2 , u3 = = u1 = = = = .07 2 σ 2 −.63 σ 3 3 .53 σ1 −.51 .34 −.29 −.57 −.65 .27 .63 −.24 −.29 1 . Since {u1 , u 2 , u3 , u 4 } is a basis for 4, let U = Av 4 u4 = = .07 −.63 −.56 σ4 .34 −.51 −.73 Thus A= U ΣV T Copyright © 2016 Pearson Education, Ltd. −.42 −.68 .53 −.29 0 0 . 0 0 .27 −.29 . −.56 −.73 7.5 • Solutions 7-37 .61 −.21 −.52 .55 −.10 0 −.39 .29 .84 −.14 −.19 0 −.74 −.27 −.07 .38 .49 . 0 .45 −.23 .58 .41 −.50 0 −.36 −.48 −.19 −.72 −.29 0 −7 −7 52 4 102 −43 −27 −6 1 11 9 30 −33 −88 −43 . Then AT A = , and the eigenvalues of 28. [M] Let A = 7 −5 10 19 −27 −33 279 335 2 3 −1 −1 52 −16 335 492 AT A are found to be (in decreasing order) λ1 = 749.979, λ2 = 146.201, λ3 = 6.82061, and λ4 = .000001 The singular values of A are thus σ 1 = 27.3857, σ 2 = 12.0914, σ 3 = 2.61163, and σ 4 = .001156 The condition number σ 1 / σ 4 ≈ 23683 −.57 .63 = .07 −.51 −.65 −.24 −.63 .34 −.42 −.68 .53 −.29 5 6 29. [M] Let A = 7 9 8 .27 16.46 −.29 0 −.56 0 −.73 0 0 12.16 0 0 0 0 4.87 0 0 0 0 4.31 9 255 168 90 160 168 111 60 104 −8 9 . Then AT A = 90 60 34 39 −5 160 104 39 415 47 4 30 8 178 eigenvalues of AT A are found to be (in decreasing order) λ1 = 672.589, λ2 3 4 5 6 5 1 2 3 4 2 7 8 10 −9 11 47 30 8 , and the 178 267 = 280.745, −7 λ5 1.428 × 10 . The singular values of A are thus σ 1 = 25.9343, λ3 = 127.503, λ4 = 1.163, and= σ 2 = 16.7554, σ 3 = 11.2917, σ 4 = 1.07853, and σ 5 = .000377928. The condition number σ 1 / σ 5 = 68,622. 7.5 SOLUTIONS Notes: The application presented here has turned out to be of interest to a wide variety of students, including engineers. I cover this in Course Syllabus 3 described in the front matter of the text, but I only have time to mention the idea briefly to my other classes. 19 1. The matrix of observations is X = 12 22 6 6 9 3 2 15 13 20 and the sample mean is 5 1 72 12 = . The mean-deviation form B is obtained by subtracting M from each column of X, 6 60 10 8 7 10 −6 −9 −10 . The sample covariance matrix is so B = 5 3 −5 2 −4 −1 = M = S 1 1 430 −135 86 = BBT = 80 −27 6 −1 5 −135 −27 . 16 Copyright © 2016 Pearson Education, Ltd. 7-38 CHAPTER 7 • Symmetric Matrices and Quadratic Forms 1 5 2 6 7 3 2. The matrix of observations is X = and the sample mean is 3 11 6 8 15 11 1 24 4 = M = . The mean-deviation form B is obtained by subtracting M from each column of X, 6 54 9 2 3 −1 −3 1 −2 . The sample covariance matrix is so B = 2 −6 2 −3 −1 6 1 1 28 40 5.6 8 = S = BBT = . 6 −1 5 40 90 8 18 3. The principal components of the data are the unit eigenvectors of the sample covariance matrix S. 86 −27 are λ1 = 95.2041 and One computes that (in descending order) the eigenvalues of S = 16 −27 −2.93348 λ2 = 6.79593. One further computes that corresponding eigenvectors are v1 = and 1 .340892 v2 = . These vectors may be normalized to find the principal components, which are 1 .946515 .322659 u1 = for λ1 = 95.2041 and u 2 = for λ2 = 6.79593. −.322659 .946515 4. The principal components of the data are the unit eigenvectors of the sample covariance matrix S. 5.6 8 One computes that (in descending order) the eigenvalues of S = are λ1 = 21.9213 and 8 18 .490158 λ2 = 1.67874. One further computes that corresponding eigenvectors are v1 = and 1 −2.04016 v2 = . These vectors may be normalized to find the principal components, which are 1 .44013 −.897934 u1 = for λ1 = 21.9213 and u 2 = for λ2 = 1.67874. .897934 .44013 164.12 5. [M] The largest eigenvalue of S = 32.73 81.04 32.73 539.44 249.13 81.04 249.13 is λ1 = 677.497, and the first 189.11 .129554 principal component of the data is the unit eigenvector corresponding to λ1 , which is u1 = .874423 .467547 . The fraction of the total variance that is contained in this component is λ= 677.497 / (164.12 + 539.44 += 189.11) .758956 so 75.8956% of the variance of the data 1 / tr( S ) is contained in the first principal component. Copyright © 2016 Pearson Education, Ltd. 7.5 29.64 6. [M] The largest eigenvalue of S = 18.38 5.00 18.38 20.82 14.06 • Solutions 7-39 5.00 14.06 is λ1 = 51.6957, and the first principal 29.21 .615525 component of the data is the unit eigenvector corresponding to λ1 , which is u1 = .599424 . Thus .511683 one choice for the new variable is y1 = .615525 x1 + .599424 x2 + .511683 x3 . The fraction of the total variance that is contained in this component is = + 29.21) .648872, so 64.8872% of the variance of the data is λ 51.6957 / (29.64 + 20.82= 1 / tr( S ) explained by y1. .946515 7. Since the unit eigenvector corresponding to λ1 = 95.2041 is u1 = , one choice for the −.322659 new variable is y1 .946515 x1 − .322659 x2 . The fraction of the total variance that is contained in this = component is λ= 95.2041 / (86 = + 16) .933374, so 93.3374% of the variance of the data is 1 / tr( S ) explained by y1. .44013 8. Since the unit eigenvector corresponding to λ1 = 21.9213 is u1 = , one choice for the new .897934 variable is y1 .44013 x1 + .897934 x2 . The fraction of the total variance that is contained in this = component is λ= 21.9213 / (5.6 = + 18) .928869, so 92.8869% of the variance of the data is 1 / tr( S ) explained by y1. 5 9. The largest eigenvalue of S = 2 0 2 6 2 0 2 is λ1 = 9, and the first principal component of the data is 7 1/ 3 the unit eigenvector corresponding to λ1 , which is u1 = 2 / 3 . Thus one choice for y is 2 / 3 y =(1/ 3) x1 + (2 / 3) x2 + (2 / 3) x3 , and the variance of y is λ1 = 9. 5 10. [M] The largest eigenvalue of S = 4 2 4 11 4 2 4 is λ1 = 15, and the first principal component of the 5 1/ 6 data is the unit eigenvector corresponding to λ1 , which is u1 = 2 / 6 . Thus one choice for y is 1/ 6 y =(1/ 6) x1 + (2 / 6) x2 + (1/ 6) x3 , and the variance of y is λ1 = 15. Copyright © 2016 Pearson Education, Ltd. 7-40 CHAPTER 7 • Symmetric Matrices and Quadratic Forms 11. a. If w is the vector in N with a 1 in each position, then [ X1 the X k are in mean-deviation form. Then [ Y1 … PT X1 YN ] w = … PT X N w = PT [ X1 … … XN ]w = X1 + …+ X = 0 since N XN ]w = PT 0 = 0 Thus Y1 + … + YN = 0, and the Yk are in mean-deviation form. b. By part a., the covariance matrix S Y of Y1 ,…, YN is 1 [ Y1 … YN ][ Y1 … YN ]T N −1 1 = PT [ X1 … X N ] ( PT [ X1 … N −1 SY = 1 = PT [ X1 … X N ][ X1 … N −1 since the X k are in mean-deviation form. X N ] )T T T XN ] = P P SP 12. By Exercise 11, the change of variables X = PY changes the covariance matrix S of X into the T covariance matrix PT SP of Y. The total variance of the data as described by Y is tr( P SP ). However, since PT SP is similar to S, they have the same trace (by Exercise 25 in Section 5.4). Thus the total variance of the data is unchanged by the change of variables X = PY. ˆ= X − M. Let ˆ ˆ be the = B X … X 13. Let M be the sample mean for the data, and let X k k 1 N matrix of observations in mean-deviation form. By the row-column expansion of BBT , the sample covariance matrix is S= 1 BBT N −1 1 ˆ X = 1 N −1 = Chapter 7 … ˆT X 1 ˆ X N X ˆT N 1 N ˆ ˆT 1 N X X = ( X k − M )( X k − M )T ∑ ∑ k k N − 1 k =1 N −1 k =1 SUPPLEMENTARY EXERCISES 1. a. True. This is just part of Theorem 2 in Section 7.1. The proof appears just before the statement of the theorem. 0 −1 . b. False. A counterexample is A = 0 1 c. True. This is proved in the first part of the proof of Theorem 6 in Section 7.3. It is also a consequence of Theorem 7 in Section 6.2. Copyright © 2016 Pearson Education, Ltd. Chapter 7 • Supplementary Exercises 7-41 d. False. The principal axes of xT Ax are the columns of any orthogonal matrix P that diagonalizes A. Note: When A has an eigenvalue whose eigenspace has dimension greater than 1, the principal axes are not uniquely determined. 1 −1 . The columns here are orthogonal but not e. False. A counterexample is P = 1 1 orthonormal. f. False. See Example 6 in Section 7.2. 0 2 1 and x = . Then xT Ax= 2 > 0 , but xT Ax is an g. False. A counterexample is A = 0 −3 0 indefinite quadratic form. h. True. This is basically the Principal Axes Theorem from Section 7.2. Any quadratic form can be written as xT Ax for some symmetric matrix A. i. False. See Example 3 in Section 7.3. j. False. The maximum value must be computed over the set of unit vectors. Without a restriction on the norm of x, the values of xT Ax can be made as large as desired. k. False. Any orthogonal change of variable x = Py changes a positive definite quadratic form into another positive definite quadratic form. Proof: By Theorem 5 of Section 7.2., the classification of a quadratic form is determined by the eigenvalues of the matrix of the form. Given a form xT Ax, the matrix of the new quadratic form is P −1 AP, which is similar to A and thus has the same eigenvalues as A. l. False. The term “definite eigenvalue” is undefined and therefore meaningless. T T )T A( Pu) u= P APu uT P −1 APu . = xT Ax ( Pu= m. True. If x = Pu, then 1 −1 n. False. A counterexample is U = . The columns of U must be orthonormal to make 1 −1 UU T x the orthogonal projection of x onto Col U. o. True. This follows from the discussion in Example 2 of Section 7.4., which refers to a proof given in Example 1. T p. True. Theorem 10 in Section 7.4 writes the decomposition in the form U Σ V , where U and V are orthogonal matrices. In this case, V T is also an orthogonal matrix. Proof: Since V is T −1 −1 T ) (V= ) (V T )T , and since V is square = orthogonal, V is invertible and V −1 = V T . Then (V and invertible, V T is an orthogonal matrix. 2 q. False. A counterexample is A = 0 values of AT A are 4 and 1. 0 . The singular values of A are 2 and 1, but the singular 1 2. a. Each term in the expansion of A is symmetric by Exercise 35 in Section 7.1. The fact that ( B + C )T =BT + C T implies that any sum of symmetric matrices is symmetric, so A is symmetric. T b. Since u1 u1 = 1 and uTj u1 = 0 for j ≠ 1, Copyright © 2016 Pearson Education, Ltd. 7-42 CHAPTER 7 • Symmetric Matrices and Quadratic Forms T T = Au1 (λ1u1u1T )u1 + …+ (λ= u n (uTn u1 ) λ1u1 n u n u n )u1 λ1u1 (u1 u1 ) + …+ λ n= Since u1 ≠ 0 , λ1 is an eigenvalue of A. A similar argument shows that λ j is an eigenvalue of A for j = 2, …, n. 3. If rank A = r, then dim Nul A = n – r by the Rank Theorem. So 0 is an eigenvalue of A with multiplicity n – r, and of the n terms in the spectral decomposition of A exactly n – r are zero. The remaining r terms (which correspond to nonzero eigenvalues) are all rank 1 matrices, as mentioned in the discussion of the spectral decomposition. = A) ⊥ Nul = AT Nul A since AT = A. 4. a. By Theorem 3 in Section 6.1, (Col b. Let y be in n . By the Orthogonal Decomposition Theorem in Section 6.3, y = yˆ + z, where yˆ is ⊥ in Col A and z is in (Col A) . By part a., z is in Nul A. −1 = v λ= Av A(λ −1 v), which shows that v is a linear 5. If Av = λv for some nonzero λ, then combination of the columns of A. 6. We need to f nd an 3×3 upper triangular matrix R with positive diagonal entries such thatA = RT R. 9 27 27 82 18 51 Since 18 r11 0 0 r11 r12 r13 51 = r12 r22 0 0 r22 r23 49 r13 r23 r33 0 0 r33 2 r11 r11 r12 r11 r13 = r11 r12 r122 + r222 r12 r13 + r22 r23 r11 r13 r12 r13 + r22 r23 r123 + r223 + r323 we obtain the following equations r121 = 9 r11 r12 = 27 r11 r13 = 18 r122 + r222 = 82 r12 r13 + r22 r23 = 51 r123 + r223 + r323 = 49 Solving these equations under the requirement that the diagonal entries are positive gives us A = RT R where 3 9 6 R = 0 1 −3 0 0 2 7. a . If A admits a Cholesky factorization, A = RT R, where R is upper triangular with positive diagonal entries, then det R = r11 r22 · · · rnn > 0 and R is invertible. Hence A is positive def nite by Exercise 25 in Section 7.2. b. Suppose that A is positive definite Then by Exercise 26 in Section 7.2, A = BT B for some positive definit matrix B. Since the eigenvalues of B are positive, 0 is not an eigenvalue of B and B is invertible. Thus the columns of B are linearly independent. By Theorem 12 in Section 6.4, B = QR for some n×n matrix Q with orthonormal columns and some upper triangular matrix R with positive entries on its diagonal. Since Q is a Copyright © 2016 Pearson Education, Ltd. Chapter 7 • Supplementary Exercises 7-43 square matrix, QT Q = I , and A = BT B = (QR)T (QR) = RT QT QR = RT R and R has the required properties. 3 9 6 1 −3 . Hence A is positive def nite. c . By exercise 6, A = RT R, where R = 0 0 0 2 8. a . Suppose that A is positive definite, and consider a Cholesky factorization of A = RT R with R upper triangular and having positive entries on its diagonal. Let D be the diagonal matrix whose diagonal entries are the entries on the diagonal of R. Since right-multiplication by a diagonal matrix scales the columns of the matrix on its left, the matrix L = RT D−1 is lower triangular with 1’s on its diagonal. If U = DR, then A = RT D−1DR = LU . b . Suppose that A has an LU factorization, A = LU , where U has positive pivots on its √ √ diagonal. Let D be the diagonal matrix with u11 , . . . , unn on its diagonal. Since left-multiplication by a diagonal matrix scales the rows of the matrix on its right, the matrix V = (D2 )−1U is upper triangular with 1’s on its diagonal and we have A = LD2V . Since A is symmetric, AT = (LD2V )T = V T D2 LT = A and L = V T . If R = DV = D−1U , then A = V T DDV = (DV )T (DV ) = RT R. c . We have 9 27 18 9 27 18 9 27 18 A = 27 82 51 ∼ 0 1 −3 ∼ 0 1 −3 = U 18 51 49 0 −3 13 0 0 4 and 1 0 0 L = 3 1 0 2 −3 1 3 0 0 3 9 6 If we set D = 0 1 0 , then R = D−1U = 0 1 −3 as in Exercise 6. 0 0 2 0 0 2 9. a. If A is an m×n matrix and x is in Rn , then xT AT Ax = (Ax)T (Ax) = kAxk2 ≥ 0. Thus G = AT A is positive semidefinite. By Exercise 22 in Section 6.5, rankAT A = rankA. b. If the columns of A are linearly independent then rankA = n. Hence rankG = n and G is invertible. Let b in Rm . We have projCol A b = Aˆx where xˆ is a least-squares solution of Ax = b. Since the normal equations AT Ax = AT b can be written as Gx = c, where c = AT b in Rn and G is invertible, we obtain xˆ = G−1 c. Thus projCol A b = AG−1 AT b is the orthogonal projection of b onto Col A. Copyright © 2016 Pearson Education, Ltd. 7-44 CHAPTER 7 • Symmetric Matrices and Quadratic Forms 10. If rank G = r, then dim Nul G = n – r by the Rank Theorem. Hence 0 is an eigenvalue of G with multiplicity n – r, and the spectral decomposition of G is = G λ1u1u1T + …+ λ r u r uTr Also λ1 ,…, λ r are positive because G is positive semidefinite. Thus = G ( λ1 u1 )( λ1 u1 T ) + …+ ( λr ur )( λr ur T ) By the column-row expansion of a matrix product, G = BBT where B is the n × r matrix = B λ1 u1 … λ r u r . Finally, G = AT A for A = BT . 11. a . LetA =UΣVT be a singular value decomposition of A. Since U is orthogonal, UTU = I and A = UΣUTUVT = PQ whereP = UΣUT = UΣU−1 and Q = UVT . Since Σ is symmetric, P is symmetric, and P has nonnegative eigenvalues because it is similar to Σ, which is diagonal with nonnegative diagonal entries. Thus P is positive semidefinite. The matrix Q is orthogonal since it is the product of orthogonal matrices. b . The SVD of A found in Exercise 7 in Section 7.4 is A = UΣVT where √ √ 1/√5 −2/√5 U= , 2/ 5 1/ 5
3 Σ= 0 From part (a) we obtain A = PQ where 0 2
and 11/5 2/5 P = U ΣU = 2/5 14/5 4/5 −3/5 T Q = UV = 3/5 4/5 T V=
√ 2/√5 1/ 5 √ −1/√5 2/ 5
12. a. Because the columns of Vr are orthonormal, −1 T T = AA+ y (U r DVrT= )(Vr D −1U rT )y (U = r DD U r ) y U rU r y T Since U rU r y is the orthogonal projection of y onto Col U r by Theorem 10 in Section 6.3, and + since Col U r = Col A by (5) in Example 6 of Section 7.4, AA y is the orthogonal projection of y onto Col A. b. Because the columns of U r are orthonormal, T −1 T T = A+ Ax (Vr D −1U r= )(U r DVrT )x (V= r D DVr ) x VrVr x T Since VrVr x is the orthogonal projection of x onto Col Vr by Theorem 10 in Section 6.3, and since Col Vr = Row A by (8) in Example 6 of Section 7.4, A+ Ax is the orthogonal projection of x onto Row A. c. Using the reduced singular value decomposition, the definition of A+ , and the associativity of matrix multiplication gives: T −1 T T −1 T T = AA+ A (U = r DVr )(Vr D U r )(U r DVr ) (U r DD U r )(U r DVr ) −1 T = U r DD= DVrT U= A r DVr T T −1 T −1 T −1 −1 T = A+ AA+ (V= r D U r )(U r DVr )(Vr D U r ) (Vr D DVr )(Vr D U r ) −1 T −1 T = Vr D −1 DD = U r V= A+ r D Ur Copyright © 2016 Pearson Education, Ltd. Chapter 7 • Supplementary Exercises 7-45 + 13. a. If b = Ax, then = x + A= b A+ Ax. By Exercise 12(a), x + is the orthogonal projection of x onto Row A. + + x) ( AA+ A)= x A= x b. b. From part (a) and Exercise 12(c), Ax= A( A A= c. Let Au = b. Since x + is the orthogonal projection of x onto Row A, the Pythagorean Theorem ||2 || x + ||2 + || u − x + ||2 ≥ || x + ||2 , with equality only if u = x + . shows that || u = 14. The least-squares solutions of Ax = b are precisely the solutions of Ax = bˆ , where bˆ is the orthogonal projection of b onto Col A. From Exercise 13, the minimum length solution of Ax = bˆ is A+ bˆ , so A+ bˆ is the minimum length least-squares solution of Ax = b. However, bˆ = AA+ b by Exercise 12(a) and hence A+ bˆ = A+ AA+ b = A + b by Exercise 12(c). Thus A+ b is the minimum length least-squares solution of Ax = b. T 15. [M] The reduced SVD of A is A = U r DVr , where Ur .253758 −.034804 .966641 .185205 −.786338 −.589382 = , D .125107 −.398296 .570709 .570709 .125107 −.398296 −.313388 −.313388 and Vr = −.633380 .633380 .035148 .009549 .009549 .023005 −.023005 .999379 9.84443 0 0 0 2.62466 0 0 0 , 1.09467 .633795 .633795 −.313529 .313529 .002322 + −1 T So the pseudoinverse A = Vr D U r may be calculated, as well as the solution xˆ = A+ b for the system Ax = b: A+ .325 .325 −.05 −.35 −.05 −.35 .325 .325 = .15 −.175 −.175 , xˆ −.05 .175 .175 .05 −.15 .10 −.30 −.150 −.150 .7 .7 −.8 .8 .6 Row reducing the augmented matrix for the system AT z = xˆ shows that this system has a solution, so 0 −1 0 1 T xˆ is in Col A = Row A. A basis for Nul A is {a1 , a 2 } = 1 , 0 , and an arbitrary element of 1 0 0 0 (131/50) + 2c 2 + 2d 2 . Nul A is = u ca1 + da 2 . One computes that || xˆ ||= 131/50 , while || xˆ + u ||= Thus if u ≠ 0, || xˆ || < || xˆ + u ||, which confirms that xˆ is the minimum length solution to Ax = b. Copyright © 2016 Pearson Education, Ltd. 7-46 CHAPTER 7 • Symmetric Matrices and Quadratic Forms T 16. [M] The reduced SVD of A is A = U r DVr , where Ur .936307 .095396 −.337977 .591763 .290230 −.752053 = ,D −.231428 −.062526 −.206232 −.694283 −.187578 −.618696 −.690099 0 and Vr = .341800 .637916 0 .721920 0 .387156 .573534 0 12.9536 0 0 0 1.44553 0 0 0 , .337763 .050939 0 −.856320 .513928 0 + −1 T So the pseudoinverse A = Vr D U r may be calculated, as well as the solution xˆ = A+ b for the system Ax = b: 0 −.05 −.15 .5 0 0 0 0 A+ = = 0 2 .5 1.5 , xˆ .5 −1 −.35 −1.05 0 0 0 0 2.3 0 5.0 −.9 0 Row reducing the augmented matrix for the system AT z = xˆ shows that this system has a solution, so 0 0 1 0 T xˆ is in Col A = Row A . A basis for Nul A is {a1 , a 2 } = 0 , 0 , and an arbitrary element of 0 0 0 1 311/10 , while || xˆ + u ||= (311/10) + c 2 + d 2 . Nul A is = u ca1 + da 2 . One computes that || xˆ ||= Thus if u ≠ 0, || xˆ || < || xˆ + u ||, which confirms that xˆ is the minimum length solution to Ax = b. Copyright © 2016 Pearson Education, Ltd. The Geometry of Vector Spaces 8.1 SOLUTIONS Notes. This section introduces a special kinds of linear combination used to describe the sets created when a subspace is shifted away from the origin. An affine combination is a linear combination in which the coefficients sum to one. Theorems 1, 3, and 4 connect affine combinations directly to linear combinations. There are several approaches to solving many of the exercises in this section, and some of the alternatives are presented here. 1 −2 0 3 5 1.= v1 = , v2 = , v3 = 4, v 4 = ,y 2 2 4 7 3 −3 −1 2 4 − v1 , v3 = − v1 , v 4 = − v1 , y = − v1 v2 = 0 2 5 1 Solve c 2 (v 2 − v 1 ) + c 3 (v 3 − v 1 ) + c 4 (v 4 − v 1 ) = y − v 1 by row reducing the augmented matrix. 4 −3 −1 2 4 −3 −1 2 0 2 5 1 � 0 1 2.5 .5 � −3 0 4.5 4.5 1 0 −1.5 −1.5 0 1 2.5 .5 � 0 1 2.5 .5 The general solution is c 2 = 1.5c 4 −1.5, c 3 = −2.5c 4 + .5, with c 4 free. When c 4 = 0, y − v 1 = −1.5(v 2 − v 1 ) + .5(v 3 − v 1 ) and y = 2v 1 − 1.5v 2 + .5v 3 If c 4 = 1, then c 2 = 0 and y − v 1 = −2(v 3 − v 1 ) + 1(v 4 − v 1 ) and y = 2v 1 − 2v 3 + v 4 If c 4 = 3, then y − v 1 = 3(v 2 − v 1 ) − 7(v 3 − v 1 ) + 3(v 4 − v 1 ) and y = 2v 1 + 3v 2 − 7v 3 + 3v 4 Of course, many other answers are possible. Note that in all cases, the weights in the linear combination sum to one. 1 −1 3 5 −2 2 4 2. = v1 = , v2 = , v3 = , y , so v 2 = − v1 , v3= − v1 , and y= − v1 1 2 2 7 1 1 6 Solve c 2 (v 2 – v 1 ) + c 3 (v 3 – v 1 ) = y – v 1 by row reducing the augmented matrix: Copyright © 2016 Pearson Education, Ltd. 8-1 8-2 CHAPTER 8 −2 1 2 1 • The Geometry of Vector Spaces 4 1 −1 −2 1 0 2 ~ ~ 6 8 0 2 0 1 4 The general solution is c 2 = 2 and c 3 = 4, so y – v 1 = 2(v 2 – v 1 ) + 4(v 3 – v 1 ) and y = –5v 1 + 2v 2 + 4v 3 . The weights sum to one, so this is an affine sum. 3. 4. 5. Row reduce the augmented matrix [v 2 – v 1 v 3 – v 1 y – v 1 ] to find a solution for writing y – v 1 in terms of v 2 – v 1 and v 3 – v 1 . Then solve for y to get y = −3v 1 + 2v 2 + 2v 3 . The weights sum to one, so this is an affine sum. Row reduce the augmented matrix [v 2 – v 1 v 3 – v 1 y – v 1 ] to find a solution for writing y – v 1 in terms of v 2 – v 1 and v 3 – v 1 . Then solve for y to get y = 2.6v 1 − .4v 2 − 1.2v 3 . The weights sum to one, so this is an affine sum. Since {b 1 , b 2 , b 3 } is an orthogonal basis, use Theorem 5 from Section 6.2 to write pj= p j � b1 b1 � b1 b1 + p j � b2 b2 � b2 b2 + p j � b3 b3 � b3 b3 a. p 1 = 3b 1 – b 2 – b 3 ∈ aff S since the coefficients sum to one. b. p 2 = 2b 1 + 0b 2 + b 3 ∉ aff S since the coefficients do not sum to one. c. p 3 = – b 1 + 2b 2 + 0b 3 ∈ aff S since the coefficients sum to one. 6. Since {b 1 , b 2 , b 3 } is an orthogonal basis, use Theorem 5 from Section 6.2 to write pj= p j � b1 b1 � b1 b1 + p j � b2 b2 � b2 b2 + p j � b3 b3 � b3 b3 a. p 1 = – 4b 1 +2b 2 +3b 3 ∈ aff S since the coefficients sum to one. b. p 2 = .2b 1 + .5b 2 + .3b 3 ∈ aff S since the coefficients sum to one. c. p 3 = b 1 + b 2 + b 3 ∉ aff S since the coefficients do not sum to one. 1 0 7. The matrix [v 1 v 2 v 3 p 1 p 2 p 3 ] row reduces to 0 0 0 1 0 0 0 2 2 2 0 1 −4 2 . 1 −1 3 2 0 0 0 −5 Parts a., b., and c. use columns 4, 5, and 6, respectively, as the “augmented” column. a. p 1 = 2v 1 + v 2 − v 3 , so p 1 is in Span S. The weights do not sum to one, so p 1 ∉ aff S. b. p 2 = 2v 1 − 4v 2 + 3v 3 , so p 2 is in Span S. The weights sum to one, so p 2 ∈ aff S. c. p 3 ∉ Span S because 0 ¹ −5, so p 3 cannot possibly be in aff S. Copyright © 2016 Pearson Education, Ltd. 8.1 1 0 8. The matrix [v 1 v 2 v 3 p 1 p 2 p 3 ] row reduces to 0 0 0 1 0 0 0 3 0 −1 1 1 0 0 • Solutions 8-3 0 −2 0 6 . 0 −3 1 0 Parts a., b., and c. use columns 4, 5, and 6, respectively, as the “augmented” column. a. p 1 = 3v 1 − v 2 + v 3 , so p 1 is in Span S. The weights do not sum to one, so p 1 ∉ aff S. b. p 2 ∉ Span S because 0 ¹ 1 (column 5 is the augmented column), so p 2 cannot possibly be in aff S. c. p 3 = −2v 1 + 6v 2 − 3v 3 , so p 3 is in Span S. The weights sum to one, so p 3 ∈ aff S. 9. Choose v 1 and v 2 to be any two points on the line x = x 3 u + p. For example, take x 3 = 0 and x 3 = 1 −3 1 to get v1 = and v 2 = respectively. Other answers are possible. 0 −2 10. Choose v 1 and v 2 to be any two points on the line x = x 3 u + p. For example, take x 3 = 0 and x 3 = 1 1 6 v1 = −3 and v 2 = −2 respectively. Other answers are possible. to get 4 2 11. a. b. c. d. True. See the definition at the beginning of this section. False. The weights in the linear combination must sum to one. See the definition. True. See equation (1). False. A flat is a translate of a subspace. See the definition prior to Theorem 3. e. True. A hyperplane in 3 has dimension 2, so it is a plane. See the definition prior to Theorem 3. 12. a. False. If S = {x}, then aff S = {x}. See the definition at the beginning of this section. b. True. Theorem 2. c. True. See the definition prior to Theorem 3. d. False. A flat of dimension 2 is called a hyperplane only if the flat is considered a subset of 3. In general, a hyperplane is a flat of dimension n − 1. See the definition prior to Theorem 3. e. True. A flat through the origin is a subspace translated by the 0 vector. 13. Span {v 2 − v 1 , v 3 − v 1 } is a plane if and only if {v 2 − v 1 , v 3 − v 1 } is linearly independent. Suppose c 2 and c 3 satisfy c 2 (v 2 − v 1 ) + c 3 (v 3 − v 1 ) = 0. Then c 2 v 2 + c 3 v 3 − (c 2 + c 3 )v 1 = 0. Then c 2 = c 3 = 0, because {v 1 , v 2 , v 3 } is a linearly independent set. This shows that {v 2 − v 1 , v 3 − v 1 } is a linearly independent set. Thus, Span {v 2 − v 1 , v 3 − v 1 } is a plane in 3. 14. Since {v 1 , v 2 , v 3 } is a basis for 3, the set W = Span {v 2 − v 1 , v 3 − v 1 } is a plane in 3, by Exercise 13. Thus, W + v 1 is a plane parallel to W that contains v 1 . Since v 2 = (v 2 − v 1 ) + v 1 , W + v 1 contains v 2 . Similarly, W + v 1 contains v 3 . Finally, Theorem 1 shows that aff {v 1 , v 2 , v 3 } is the plane W + v 1 that contains v 1 , v 2 , and v 3 . 15. Let S = {x : Ax = b}. To show that S is affine, it suffices to show that S is a flat, by Theorem 3. Let W = {x : Ax = 0}. Then W is a subspace of n, by Theorem 2 in Section 4.2 (or Theorem 12 Copyright © 2016 Pearson Education, Ltd. 8-4 CHAPTER 8 • The Geometry of Vector Spaces in Section 2.8). Since S = W + p, where p satisfies Ap = b, by Theorem 6 in Section 1.5, S is a translate of W, and hence S is a flat. 16. Suppose p, q ∈ S and t ∈ . Then, by properties of the dot product (Theorem 1 in Section 6.1), [(1 − t)p + t q] ⋅ v = (1 − t)(p ⋅ v) + t (q ⋅ v) = (1 − t)k + t k = k Thus, [(1 − t)p + t q] ∈ S, by definition of S. This shows that S is an affine set. 17. A suitable set consists of any three vectors that are not collinear and have 5 as their third entry. If 5 is their third entry, they lie in the plane x 3 = 5. If the vectors are not collinear, their affine hull 1 0 1 cannot be a line, so it must be the plane. For example use S = 0 , 1 , 1 . 5 5 5 18. A suitable set consists of any four vectors that lie in the plane 2x 1 + x 2 − 3x 3 = 12 and are not collinear. If the vectors are not collinear, their affine hull cannot be a line, so it must be the plane. 6 0 0 3 For example use S = 0 , 12 , 0 , 3 . 0 0 −4 −1 19. If p, q ∈ f (S), then there exist r, s ∈ S such that f (r) = p and f (s) = q. Given any t ∈ , we must show that z = (1 − t)p + t q is in f (S). Since f is linear, z = (1 − t)p + t q = (1 − t) f (r) + t f (s) = f ((1 − t)r + t s) Since S is affine, (1 − t)r + t s ∈ S. Thus, z is in S and f (S) is affine. 20. Given an affine set T, let S = {x ∈ n : f (x) ∈ T}. Consider x, y ∈ S and t ∈ . Then f ((1 − t)x + t y) = (1 − t) f (x) + t f (y) But f (x) ∈ T and f (y) ∈ T, so (1 − t) f (x) + t f (y) ∈ T because T is an affine set. It follows that [(1 − t)x + t y] ∈ S. This is true for all x, y ∈ S and t ∈ , so S is an affine set. 21. Since B is affine, Theorem 1 implies that B contains all affine combinations of points of B. Hence B contains all affine combinations of points of A. That is, aff A Ì B. 22. Since B ⊂ aff B, we have A ⊂ B ⊂ aff B. But aff B is an affine set, so Exercise 21 implies aff A ⊂ aff B. 23. Since A Ì (A È B), it follows from Exercise 22 that aff A Ì aff (A È B). Similarly, aff B Ì aff (A È B), so [aff A È aff B] Ì aff (A È B). 24. One possibility is to let A ={(0, 1), (0, 2)} and B = {(1, 0), (2, 0)}. Then (aff A) È (aff B) consists of the two coordinate axes, but aff (A È B) = 2. 25. Since (A Ç B) ⊆ A, it follows from Exercise 22 that aff (A Ç B) Ì aff A. Similarly, aff (A Ç B) Ì aff B, so aff (A Ç B) Ì (aff A Ç aff B). 26. One possibility is to let A = {(0, 0), (0, 1)} and B = {(0, 2), (0, 3)}. Then both aff A and aff B are equal to the x-axis. But A Ç B = ∅, so aff (A Ç B) = ∅. Copyright © 2016 Pearson Education, Ltd. 8.2 8.2 • Solutions 8-5 SOLUTIONS Notes: Affine dependence and independence are developed in this section. Theorem 5 links affine independence to linear independence. This material has important applications to computer graphics. 3 0 2 −3 −1 = v1 = , v 2 = , v3 . Then v 2 = − v1 , v3 = − v1 . Since v 3 − v 1 is a multiple 1. Let −3 6 0 9 3 of v 2 − v 1 , these two points are linearly dependent. By Theorem 5, {v 1 , v 2 , v 3 } is affinely dependent. Note that (v 2 − v 1 ) − 3(v 3 − v 1 ) = 0. A rearrangement produces the affine dependence relation 2v 1 + v 2 − 3v 3 = 0. (Note that the weights sum to zero.) Geometrically, v 1 , v 2 , and v 3 are not collinear. 2 5 −3 3 −5 v1 , = v 2 ,= v3 . v 2 −= v1 , v3 −= v1 . Since v 3 − v 1 and v 2 − v 1 are not 2. = 1 4 −2 3 −3 multiples, they are linearly independent. By Theorem 5, {v 1 , v 2 , v 3 } is affinely independent. 3. The set is affinely independent. If the points are called v 1 , v 2 , v 3 , and v 4 , then row reduction of [v 1 v 2 v 3 v 4 ] shows that {v 1 , v 2 , v 3 } is a basis for 3 and v 4 = 16v 1 + 5v 2 − 3v 3 . Since there is unique way to write v 4 in terms of the basis vectors, and the weights in the linear combination do not sum to one, v 4 is not an affine combination of the first three vectors. 2 3 0 4. Name the points v 1 , v 2 , v 3 , and v 4 . Then v 2 − v1 =− 8 , v3 − v1 =− 7 , v 4 − v1 = 2 . To study 4 −9 −6 the linear independence of these points, row reduce the augmented matrix for Ax = 0: 3 0 0 2 3 0 0 2 3 0 0 1 0 −.6 0 2 −8 −7 2 0 ~ 0 5 2 0 ~ 0 5 2 0 ~ 0 1 .4 0 . The first three columns 4 −9 −6 0 0 −15 −6 0 0 0 0 0 0 0 0 0 are linearly dependent, so {v 1 , v 2 , v 3 , v 4 } is affinely dependent, by Theorem 5. To find the affine dependence relation, write the general solution of this system: x 1 = .6x 3 , x 2 = −.4x 3 , with x 3 free. Set x 3 = 5, for instance. Then x 1 = 3, x 2 = −2, and x 3 = 5. Thus, 3(v 2 − v 1 ) − 2(v 3 − v 1 ) + 5(v 4 − v 1 ) = 0. Rearrange to obtain −6v 1 + 3v 2 − 2v 3 + 5v 4 = 0. Alternative solution: Name the points v 1 , v 2 , v 3 , and v 4 . Use Theorem 5(d) and study the homogeneous forms of the points. The first step is to move the bottom row of ones (in the augmented matrix) to the top to simplify the arithmetic: [ v 1 v 2 v 3 1 1 1 1 1 −2 0 1 −2 0 ~ v 4 ] ~ 5 −3 −2 7 0 3 7 −6 −3 0 0 1 0 0 0 1.2 0 −.6 1 .4 0 0 Thus, x 1 + 1.2x 4 = 0, x 2 − .6x 4 = 0, and x 3 + .4x 4 = 0, with x 4 free. Take x 4 = 5, for example, and get x 1 = −6, x 2 = 3, and x 3 = −2 . An affine dependence relation is −6v 1 + 3v 2 − 2v 3 + 5v 4 = 0. 5. − 4v 1 + 5v 2 – 4v 3 + 3v 4 = 0 is an affine dependence relation. It can be found by row reducing the matrix [ v 1 v 2 v 3 v 4 ] , and proceeding as in the solution to Exercise 4. Copyright © 2016 Pearson Education, Ltd. 8-6 CHAPTER 8 • The Geometry of Vector Spaces 6. The set is affinely independent, as the following calculation with homogeneous forms shows: [ v 1 v 2 v 3 1 1 1 0 v 4 ] ~ 3 −1 1 −2 1 2 5 2 1 1 3 0 ~ 5 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 Alternative solution: Row reduction of [v 1 v 2 v 3 v 4 ] shows that {v 1 , v 2 , v 3 } is a basis for 3 and v 4 = −2v 1 + 1.5v 2 + 2.5v 3 , but the weights in the linear combination do not sum to one, so this v 4 is not an affine combination of the basis vectors and hence the set is affinely independent. Note: A potential exam question might be to change the last entry of v 4 from 0 to 1 and again ask if the set is affinely independent. Notice that row reduction of this new set of vectors [v 1 v 2 v 3 v 4 ] shows that {v 1 , v 2 , v 3 } is a basis for 3 and v 4 = −3v 1 + v 2 + 3v 3 is an affine combination of the basis. 7. Denote the given points as v 1 , v 2 , v 3 , and p. Row reduce the augmented matrix for the equation x 1 ṽ 1 + x 2 ṽ 2 + x 3 ṽ 3 = p.̃ Remember to move the bottom row of ones to the top as the first step to simplify the arithmetic by hand. [ v 1 v 2 v 3 1 1 p ] ~ −1 2 1 1 1 1 2 1 5 1 2 4 ~ 0 −2 −2 1 0 2 1 0 0 0 0 0 1 0 0 0 0 −2 0 4 1 −1 0 0 0 0 Thus, x 1 = −2, x 2 = 4, x 3 = −1, and p̃ = −2ṽ 1 + 4ṽ 2 − ṽ 3 , so p = −2v 1 + 4v 2 − v 3 , and the barycentric coordinates are (−2, 4, −1). Alternative solution: Another way that this problem can be solved is by “translating” it to the origin. That is, compute v 2 – v 1 , v 3 – v 1 , and p – v 1 , find weights c 2 and c 3 such that c 2 (v 2 – v 1 ) + c 3 (v 3 – v 1 ) = p – v 1 and then write p = (1 – c 2 – c 3 )v 1 + c 2 v 2 + c 3 v 3 . Here are the calculations for Exercise 7: 2 1 1 −1 v 2 − v1 = − = 0 2 1 1 [ v 2 − v1 1 2 , v −v = 3 1 −2 0 v3 − v1 p − v1 ] 1 1 2 −1 − = −2 2 0 1 0 3 , p− v = 1 − 4 −1 0 4 1 1 2 0 3 5 ~ ~ −2 − 4 − 4 0 1 0 −1 0 Thus p – v 1 = 4(v 2 – v 1 ) – 1(v 3 – v 1 ), and p = –2 v 1 + 4v 2 – v 3 . Copyright © 2016 Pearson Education, Ltd. 0 4 1 −1 0 0 0 0 5 1 4 −1 − = −2 2 2 1 4 5 − 4 1 8.2 • Solutions 8-7 8. Denote the given points as v 1 , v 2 , v 3 , and p. Row reduce the augmented matrix for the equation x 1 ṽ 1 + x 2 ṽ 2 + x 3 ṽ 3 = p̃. 1 1 1 1 1 0 0 2 0 1 1 −1 0 1 0 −1 [ v 1 v 2 v 3 p ] ~ 1 1 4 1 ~ 0 0 1 0 −2 0 −6 −4 0 0 0 0 1 2 5 0 0 0 0 0 Thus. p̃ = 2ṽ 1 − ṽ 2 + 0ṽ 3 , so p = 2v 1 − v 2 . The barycentric coordinates are (2 , −1, 0). Notice v 3 = 3v 1 + v 2 . 9. a. True. Theorem 5 uses the point v 1 for the translation, but the paragraph after the theorem points out that any one of the points in the set can be used for the translation. b. False, by (d) of Theorem 5. c. False. The weights in the linear combination must sum to zero, not one. See the definition at the beginning of this section. d. False. The only points that have barycentric coordinates determined by S belong to aff S. See the definition after Theorem 6. e. True. The barycentric coordinates have some zeros on the edges of the triangle and are only positive for interior points. See Example 6. 10. a. False. By Theorem 5, the set of homogeneous forms must be linearly dependent, too. b. True. If one statement in Theorem 5 is false, the other statements are false, too. c. False. Theorem 6 applies only when S is affinely independent. d. False. The color interpolation applies only to points whose barycentric coordinates are nonnegative, since the colors are formed by nonnegative combinations of red, green, and blue. See Example 5. e. True. See the discussion of Fig. 5. 11. When a set of five points is translated by subtracting, say, the first point, the new set of four points must be linearly dependent, by Theorem 8 in Section 1.7, because the four points are in 3. By Theorem 5, the original set of five points is affinely dependent. 12. Suppose v 1 , …, v p are in n and p ≥ n + 2. Since p − 1 ≥ n + 1, the points v 2 − v 1 , v 3 − v 1 , … , v p − v 1 are linearly dependent, by Theorem 8 in Section 1.7. By Theorem 5, {v 1 , v 2 , …, v p } is affinely dependent. 13. If {v 1 , v 2 } is affinely dependent, then there exist c 1 and c 2 , not both zero, such that c 1 + c 2 = 0, and c 1 v 1 + c 2 v 2 = 0. Then c 1 = – c 2 ≠ 0 and c 1 v 1 = – c 2 v 2 = c 1 v 2 , which implies that v 1 = v 2 . Conversely, if v 1 = v 2 , let c 1 = 1 and c 2 = −1. Then c 1 v 1 + c 2 v 2 = v 1 + (−1)v 1 = 0 and c 1 + c 2 = 0, which shows that {v 1 , v 2 } is affinely dependent. 14. Let S 1 consist of three (distinct) points on a line through the origin. The set is affinely dependent because the third point is on the line determined by the first two points. Let S 2 consist of two (distinct) points on a line through the origin. By Exercise 13, the set is affinely independent Copyright © 2016 Pearson Education, Ltd. 8-8 CHAPTER 8 • The Geometry of Vector Spaces because the two points are distinct. (A correct solution should include a justification for the sets presented.) 1 3 15. a. The vectors v 2 − v 1 = and v 3 − v 1 = are not multiples and hence are linearly 2 −2 independent. By Theorem 5, S is affinely independent. ( ) ( ) ( ) ( ) , − 85 , − 18 , p 4 ↔ 86 , − 85 , 87 , p5 ↔ b. p1 ↔ − 86 , 89 , 85 , p 2 ↔ 0, 12 , 12 , p3 ↔ 14 8 c. p 6 is (−, −, +), p 7 is (0, +, −), and p 8 is (+, +, −). ( 14 , 81 , 85 ) 1 4 16. a. The vectors v 2 − v 1 = and v 3 − v 1 = are not multiples and hence are linearly 4 2 independent. By Theorem 5, S is affinely independent. v2 p1 2 , 2 , 3) b. p1 ↔ (− 72 , 75 , 74 ), p 2 ↔ ( 72 , − 75 , 10 ), p ↔ ( 3 7 7 7 7 p5 p7 c. p 4 ↔ (+, −, −), p5 ↔ (+, +, −), p3 v3 p 6 ↔ (+, +, + ), p 7 ↔ (−,0, + ). p6 See the figure to the right. Actually, 19 5 12 , − 142 , − 143 ), p5 ↔ ( 14 , 14 , − 143 ), p 4 ↔ ( 14 9 2 3 , 14 , 14 ), p 7 ↔ (− 12 , 0, 32 ). p 6 ↔ ( 14 p2 v1 p4 17. Suppose S = {b 1 , …, b k } is an affinely independent set. Then (7) has a solution, because p is in aff S. Hence (8) has a solution. By Theorem 5, the homogeneous forms of the points in S are linearly independent. Thus (8) has a unique solution. Then (7) also has a unique solution, because (8) encodes both equations that appear in (7). The following argument mimics the proof of Theorem 7 in Section 4.4. If S = {b 1 , …, b k } is an affinely independent set, then scalars c 1 , …, c k exist that satisfy (7), by definition of aff S. Suppose x also has the representation x = d 1 b 1 + … + d k b k and d 1 + … + d k = 1 (7a) for scalars d 1 , …, d k . Then subtraction produces the equation 0 = x − x = (c 1 − d 1 )b 1 + … + (c k − d k )b k (7b) The weights in (7b) sum to zero because the c’s and the d’s separately sum to one. This is impossible, unless each weight in (8) is zero, because S is an affinely independent set. This proves that c i = d i for i = 1, …, k. x x a 0 0 0 x y z x y z 18. Let p = y . Then y= a 0 + b b + c 0 + 1 − a − b − c 0 So the barycentric coordi 0 0 c 0 z z nates are x/a, y/b, z/c, and 1 − x/a − y/b − z/c. This holds for any nonzero choices of a, b, and c. 19. If {p 1 , p 2 , p 3 } is an affinely dependent set, then there exist scalars c 1, c 2 , and c 3 , not all zero, such that c 1 p 1 + c 2 p 2 + c 3 p 3 = 0 and c 1 + c 2 + c 3 = 0. But then, applying the transformation f, Copyright © 2016 Pearson Education, Ltd. 8.2 • Solutions 8-9 c1 f (p1 ) + c2 f (p 2 ) + c3 f (p3 ) = f (c1p1 + c2p 2 + c3p3 ) = f (0) = 0 , since f is linear. This shows that { f (p1 ), f (p 2 ), f (p3 )} is also affinely dependent. 20. If the translated set {p 1 + q, p 2 + q, p 3 + q} were affinely dependent, then there would exist real numbers c 1, c 2 , and c 3 , not all zero and with c 1 + c 2 + c 3 = 0, such that c 1 (p 1 + q) + c 2 (p 2 + q) + c 3 (p 3 + q) = 0. But then, c 1 p 1 + c 2 p 2 + c 3 p 3 + (c 1 + c 2 + c 3 )q = 0. Since c 1 + c 2 + c 3 = 0, this implies c 1 p 1 + c 2 p 2 + c 3 p 3 = 0, which would make {p 1 , p 2 , p 3 } affinely dependent. But {p 1 , p 2 , p 3 } is affinely independent, so the translated set must in fact be affinely independent, too. a1 b1 c1 a1 a2 1 a1 b1 c1 , b ,= and c . Then det [ã b̃ c̃] = det a2 b2 c2 = det b1 b2 1 , 21. = Let a = a2 b2 c2 1 1 1 c1 c2 1 by using the transpose property of the determinant (Theorem 5 in Section 3.2). By Exercise 30 in Section 3.3, this determinant equals 2 times the area of the triangle with vertices at a, b, and c. 22. If p is on the line through a and b, then p is an affine combination of a and b, so p̃ is a linear combination of ã and b̃. Thus the columns of [ã b̃ p̃] are linearly dependent. So the determinant of this matrix is zero. r 23. If [ã b̃ c̃] s = p̃, then Cramer’s rule gives r = det [p̃ b̃ c̃] / det [ã b̃ c̃]. By Exercise 21, the t numerator of this quotient is twice the area of Δpbc, and the denominator is twice the area of Δabc. This proves the formula for r. The other formulas are proved using Cramer’s rule for s and t. 24. Let p = (1 − x)q + x a, where q is on the line segment from b to c. Then, because the determinant is a linear function of the first column when the other columns are fixed (Section 3.2), det [p̃ b̃ c̃] = det [(1 − x)q̃ + x ã b̃ c̃] = (1 − x) · det [q̃ b̃ c̃] + x · det [ã b̃ c̃]. Now, [q̃ b̃ c̃] is a singular matrix because q̃ is a linear combination of b̃ and c̃. So det [q̃ b̃ c̃] = 0 and det [p̃ b̃ c̃] = x · det [ã b̃ c̃] . 6 2 −1 3 . Solve v − v 25. v 2 − v1 =0 , v3 − v1 = 6 , and a − v1 =− [ 2 1 1 4 15 v3 − v1 4 3.1 15 4 3.1 15 6 2 −1.4 −1 1 1 0 6 −1.5 −3 ~ 0 6 −1.5 −3 ~ 0 2 − 0.5 −1 3.1 15 1 4 0 −22 −20 −91 0 −22 −20 −91 Copyright © 2016 Pearson Education, Ltd. c2 −b ] c3 =a − v1 . t 8-10 CHAPTER 8 • The Geometry of Vector Spaces 4.1 17 1 0 1 0 0 0.6 ~ 0 2 − 0.5 −1 ~ 0 2 0 1 0 0 −25.5 −102 4 0 0 1 Thus, c 2 = 0.6, c 3 = 0.5, and t = 4. 0 1.4 5.6 The intersection point is x(4) =a + 4b =0 + 4 1.5 = 6.0 and 9 −3.1 −3.4 1 7 3 5.6 6.0 . (1 − 0.6 − 0.5) v1 + 0.6 v 2 + 0.5 v3 = − 0.1 3 + 0.6 3 + 0.5 9 = x(4) = −6 −5 −2 −3.4 The first barycentric coordinate is negative, so the intersection point is not inside the triangle. 7 2 −1 2 . Solve 26. v 2 − v1 = 0 , v 3 − v1 = 8 , and a − v1 =− − 2 12 1 [ v 2 − v1 v3 − v1 c2 −b ] c3 =a − v1 . t 7 2 − 0.9 −1 1 −2 −3.7 −12 1 −2 −3.7 −12 0 8 −2 −2 ~ 0 −2 −2 ~ 0 4 −1 −1 8 3.7 12 25 83 0 29 87 −1 2 0 16 0 1 −2 −3.7 −12 1 −2 0 − 0.9 1 0 0 0.1 −1 −1 ~ 0 4 0 ~ 0 4 2 ~ 0 1 0 0.5 0 0 0 1 0 1 3 0 0 1 3 3 Thus, c 2 = 0.1, c 3 = 0.5, and t = 3. 0 0.9 2.7 The intersection point is x(3) =a + 3b =0 + 3 2.0 = 6.0 and 8 −3.7 −3.1 1 8 3 2.7 x(3) = (1 − 0.1 − 0.5) v1 + 0.1v 2 + 0.5 v3 = 0.4 2 + 0.1 2 + 0.5 10 = 6.0 . −4 −5 −2 −3.1 The barycentric coordinates are all positive, so the intersection point is inside the triangle. Copyright © 2016 Pearson Education, Ltd. 8.3 8.3 • Solutions 8-11 SOLUTIONS Notes: The notion of convexity is introduced in this section and has important applications in computer graphics. Bézier curves are introduced in Exercises 21-24 and explored in greater detail in Section 8.6. 0 V : 0 ≤ y < 1 is the vertical line segment from (0,0) to = 1. The set y (0,1) that includes (0,0) but not (0,1). The convex hull of S includes each line segment from a point in V to the point (2,0), as shown in the figure. The dashed line segment along the top of the shaded region indicates that this segment is not in conv S, because (0,1) is not in S. 1 2 2. a. Conv S includes all points p of the form 1/ 2 x 1/ 2 + t ( x − 1/ 2) p= (1 − t ) +t = , where 2 1/ x 2 − t (2 − 1/ x) 2 x ≥ 1/ 2 and 0 ≤ t ≤ 1. Notice that if t = a/x, then 2 4 1/ 2 + a − a / (2 x) 1/ 2 + a p( x) = and lim p( x) = , establishing that there are points 2 x →∞ 2 2 − 2a / x − a / x arbitrarily close to the line y = 2 in conv S. Since the curve y = 1/x is in S, the line segments between y = 2 and y = 1/x are also included in conv S, whenever x ≥ 1/ 2. b. Recall that for any integer n, sin( x + 2nπ ) = sin ( x ) . Then x x + 2nπ x + 2nπ t π= (1 − t ) +t = ∈ conv S. sin(x) sin( x + 2nπ ) sin( x) Notice that sin(x) is always a number between -1 and 1. For a fixed x and any real number r, an integer n and a number t (with 0 ≤ t ≤ 1 ) can be chosen so that r= x + 2nπ t. c. Conv S includes all points p of the form 2 0 x x (1 − t ) + t = p= t , where x ≥ 0 and 0 ≤ t ≤ 1. 0 x x 2 4 a p = lim Letting t = a/x, x →∞ 0 establishing that there are points arbitrarily close to y = 0 in the set. 3. From Exercise 5, Section 8.1, a. p 1 = 3b 1 – b 2 – b 3 ∉ conv S since some of the coefficients are negative. b. p 2 = 2b 1 + 0b 2 + b 3 ∉ conv S since the coefficients do not sum to one. c. p 3 = – b 1 + 2b 2 + 0b 3 ∉ conv S since some of the coefficients are negative. 4. From Exercise 5, Section 8.1, a. p 1 = – 4b 1 +2b 2 +3b 3 ∉ conv S since some of the coefficients are negative. b. p 2 = .2b 1 + .5b 2 + .3b 3 ∈ conv S since the coefficients are nonnegative and sum to one. Copyright © 2016 Pearson Education, Ltd. 8-12 CHAPTER 8 • The Geometry of Vector Spaces c. p 3 = b 1 + b 2 + b 3 ∉ conv S since the coefficients do not sum to one. 5. Row reduce the matrix [ v 1 v 2 v 3 v 4 p 1 p 2 ] to obtain the barycentric coordinates p 1 = − 16 v1 + 13 v 2 + 32 v 3 + 16 v 4 , so p 1 ∉ conv S, and p 2 = 1 3 v1 + 13 v 2 + 16 v 3 + 16 v 4 , so p 2 ∈ conv S. 6. Let W be the subspace spanned by the orthogonal set S = {v 1 , v 2 , v 3 }. As in Example 1, the barycentric coordinates of the points p 1 , …, p 4 with respect to S are easy to compute, and they determine whether or not a point is in Span S, aff S, or conv S. projW p1 = a. = p ⋅v p1 ⋅ v1 p ⋅v v + 1 2 v + 1 3v v1 ⋅ v1 1 v 2 ⋅ v 2 2 v3 ⋅ v3 3 −1 2 − 3 = p1 2 5 2 2 0 −2 0 −2 1 1 1 = − + 0 2 −1 2 2 2 1 2 This shows that p 1 is in W = Span S. Also, since the coefficients sum to 1, p 1 is an aff S. However, p 1 is not in conv S, because the coefficients are not all nonnegative. 9 4 9 4 9 2 1 1 1 v1 + v 2 + v3 = p 2 . This shows that p 2 9 9 9 4 4 2 lies in Span S. Also, since the coefficients sum to 1, p 2 is in aff S. In fact, p 2 is in conv S, because the coefficients are also nonnegative. b. Similarly, projW p 2 = c. v1 + v2 + v3 = 9 9 18 v1 + v 2 − v3 = v1 + v 2 − 2 v3 = p3 . Thus p 3 is in Span S. However, since 9 9 9 the coefficients do not sum to one, p 3 is not in aff S and certainly not in conv S. projW p3 = 6 8 8 v1 + v 2 + v3 ≠ p 4 . Since projW p 4 is the closest point in Span S to p 4 , the 9 9 9 point p 4 is not in Span S. In particular, p 4 cannot be in aff S or conv S. d. projW p 4 = −1 2 4 2 3 2 0 = , v 2 = , v3 = , p1 = , p 2 = , p3 = , p 4 , T = {v 1 , v 2 , v 3 } 7. v1 = 0 3 1 1 2 0 2 a. Use an augmented matrix (with four augmented columns) to write the homogeneous forms of p 1 , …, p 4 in terms of the homogeneous forms of v 1 , v 2 , and v 3 , with the first step interchanging rows 1 and 3: v 1 v 2 1 1 1 1 1 1 1 1 0 0 v 3 p 1 p 2 p 3 p 4 ~ −1 2 4 2 3 2 0 ~ 0 1 0 0 3 1 1 2 0 2 0 0 1 The first four columns reveal that 1 3 v 1 + 16 v 2 + 1 2 v 3 = p 1 and 1 3 1 3 1 6 1 2 v1 + 16 v 2 + 0 1 2 1 2 1 2 1 2 1 −4 3 4 1 2 3 4 − 14 v3 = p1 . Thus column 4 contains the barycentric coordinates of p 1 relative to the triangle determined by T. Similarly, column 5 (as an augmented column) contains the barycentric coordinates of p 2 , column 6 Copyright © 2016 Pearson Education, Ltd. 8.3 • Solutions 8-13 contains the barycentric coordinates of p 3 , and column 7 contains the barycentric coordinates of p4. b. p 3 and p 4 are outside conv T, because in each case at least one of the barycentric coordinates is negative. p 1 is inside conv T, because all of its barycentric coordinates are positive. p 2 is on the edge v1v 2 of conv T, because its barycentric coordinates are nonnegative and its first coordinate is 0. 8. a. The barycentric coordinates of p 1 , p 2 , p 3 , and p 4 are, respectively, ( 23 , 0, 13 ) , and ( 139 , − 131 , 135 ). ( 1213 , 133 , − 132 ) , ( 138 , 132 , 133 ) , b. The point p 1 and p 4 are outside conv T since they each have a negative coordinate. The point p 2 is inside conv T since the coordinates are positive, and p 3 is on the edge v1v 3 of conv T. 9. The points p 1 and p 3 are outside the tetrahedron conv S since their barycentric coordinates contain negative numbers. The point p 2 is on the face containing the vertices v 2 , v 3 , and v 4 since its first barycentric coordinate is zero and the rest are positive. The point p 4 is inside conv S since all its barycentric coordinates are positive. The point p 5 is on the edge between v 1 and v 3 since the first and third barycentric coordinates are positive and the rest are zero. 10. The point q 1 is inside conv S because the barycentric coordinates are all positive. The point q 2 is outside conv S because it has one negative barycentric coordinate. The point q 4 is outside conv S for the same reason. The point q 3 is on the edge between v 2 and v 3 because ( 0, 34 , 14 , 0 ) shows that q 3 is a convex combination of v 2 and v 3 . The point q 5 is on the face containing the vertices v 1 , v 2 , and v 3 because ( 13 , 13 , 13 , 0 ) shows that q 5 is a convex combination of those vertices. 11. a. False. In order for y to be a convex combination, the c’s must also all be nonnegative. See the definition at the beginning of this section. b. False. If S is convex, then conv S is equal to S. See Theorem 7. c. False. For example, the union of two distinct points is not convex, but the individual points are. 12. a. True. See the definition prior to Theorem 7. b. True. Theorem 9. c. False. The points do not have to be distinct. For example, S might consist of two points in 5. A point in conv S would be a convex combination of these two points. Caratheodory’s Theorem requires n + 1 or fewer points. 13. If p, q ∈ f (S), then there exist r, s ∈ S such that f (r) = p and f (s) = q. The goal is to show that the line segment y = (1 − t)p + t q, for 0 ≤ t ≤ 1, is in f (S). Since f is linear, y = (1 − t)p + t q = (1 − t) f (r) + t f (s) = f ((1 − t)r + t s) Since S is convex, (1 − t)r + t s ∈ S for 0 ≤ t ≤ 1. Thus y ∈ f (S) and f (S) is convex. 14. Suppose r, s ∈ S and 0 ≤ t ≤ 1. Then, since f is a linear transformation, Copyright © 2016 Pearson Education, Ltd. 8-14 CHAPTER 8 • The Geometry of Vector Spaces f [(1 − t)r + t s ] = (1 − t) f (r) + t f (s) But f (r) ∈ T and f (s) ∈ T, so (1 − t) f (r) + t f (s) ∈ T since T is a convex set. It follows that (1 − t)r + t s ∈ S, because S consists of all points that f maps into T. This shows that S is convex. 15. It is straightforward to confirm the equations in the problem: (1) 1v 3 1 + 13 v 2 + 16 v 3 + 16 v 4 = p and (2) v 1 – v 2 + v 3 – v 4 = 0. Notice that the coefficients of v 1 and v 3 in equation (2) are positive. With the notation of the proof of Caratheodory’s Theorem, d 1 = 1 and d 3 = 1. The corresponding coefficients in equation (1) are c1 = 13 and c3 = 16 . The ratios of these coefficients are c1 / d1 = 13 and c3 / d3 = 1 6 . Use the smaller ratio to eliminate v 3 from equation (1). That is, add − 16 times equation (2) to equation (1): p = ( 13 − 16 ) v1 + ( 13 + 16 ) v 2 + ( 16 − 16 ) v 3 + ( 16 + 16 ) v 4 = 1v 6 1 + 12 v 2 + 13 v 4 To obtain the second combination, multiply equation (2) by –1 to reverse the signs so that d 2 and d 4 become positive. Repeating the analysis with these terms eliminates the v 4 term resulting in p = 12 v1 + 16 v 2 + 13 v 3 . −1 0 3 1 1 = , v 2 = , v3 = , v4 = , p It is straightforward to confirm the equa16. v1 = 0 3 1 −1 2 1 v + 72 v + 37 v + 1 v = tions in the problem: (1) 121 p and (2) 10v 1 – 6v 2 + 7v 3 – 11v 4 = 0. 1 121 2 121 3 11 4 Notice that the coefficients of v 1 and v 3 in equation (2) are positive. With the notation of the proof of Caratheodory’s Theorem, d 1 = 10 and d 3 = 7. The corresponding coefficients in equation (1) are 1 and c = 37 . The ratios of these coefficients are c / d = 1 ÷ 10 = 1 and c1 = 121 1 1 3 121 1210 121 c3 / d3 = 37 ÷ 7 121 = 37 847 1 . Use the smaller ratio to eliminate v 1 from equation (1). That is, add − 1210 times equation (2) to equation (1): 1 − 10 ) v + ( 72 + 6 ) v + ( 37 − 7 ) v + ( 1 + 11 ) v = p = (121 1210 1 121 1210 2 121 1210 3 11 1210 4 3v 5 2 3 v + 1 v + 10 3 10 4 To obtain the second combination, multiply equation (2) by –1 to reverse the signs so that d 2 and d 4 become positive. Repeating the analysis with these terms eliminates the v 4 term resulting in 1 + 10 ) v + ( 72 − 6 ) v + ( 37 + 7 ) v + ( 1 − 11 ) v = p = (121 121 1 121 121 2 121 121 3 11 121 4 1 v 11 1 6 v + 4 v + 11 2 11 3 17. Suppose A Ì B, where B is convex. Then, since B is convex, Theorem 7 implies that B contains all convex combinations of points of B. Hence B contains all convex combinations of points of A. That is, conv A Ì B. 18. Suppose A Ì B. Then A Ì B Ì conv B. Since conv B is convex, Exercise 17 shows that conv A Ì conv B. 19 a. Since A Ì (A È B), Exercise 18 shows that conv A Ì conv (A È B). Similarly, conv B Ì conv (A È B). Thus, [(conv A) È (conv B)] Ì conv (A È B). Copyright © 2016 Pearson Education, Ltd. 8.3 • Solutions 8-15 b. One possibility is to let A be two adjacent corners of a square and B be the other two corners. Then (conv A) È (conv B) consists of two opposite sides of the square, but conv (A È B) is the whole square. 20. a. Since (A Ç B) Ì A, Exercise 18 shows that conv (A Ç B) Ì conv A. Similarly, conv (A Ç B) Ì conv B. Thus, conv (A Ç B) Ì [(conv A) Ç (conv B)]. b. One possibility is to let A be a pair of opposite vertices of a square and let B be the other pair of opposite vertices. Then conv A and conv B are intersecting diagonals of the square. A Ç B is the empty set, so conv (A Ç B) must be empty, too. But conv A Ç conv B contains the single point where the diagonals intersect. So conv (A Ç B) is a proper subset of conv A Ç conv B. 21. p1 p0 p2 22. p1 p2 p0 23. g(t) = (1 − t)f 0 (t) + t f 1 (t) = (1 − t)[(1 − t)p 0 + t p 1 ] + t[(1 − t)p 1 + t p 2 ] = (1 − t)2p 0 + 2t(1 − t)p 1 + t 2p 2 . The sum of the weights in the linear combination for g is (1 − t)2 + 2t(1 − t) + t 2, which equals (1 − 2t + t2) + (2t − 2t2) + t 2 = 1. The weights are each between 0 and 1 when 0 ≤ t ≤ 1, so g(t) is in conv{p 0 , p 1 , p 2 }. 24. h(t) = (1 − t)g 1 (t) + t g 2 (t). Use the representation for g 1 (t) from Exercise 23, and the analogous representation for g 2 (t), based on the control points p 1 , p 2 , and p 3 , and obtain h(t) = (1 − t)[(1 − t)2p 0 + 2t(1 − t)p 1 + t 2p 2 ] + t [(1 − t)2p 1 + 2t(1 − t)p 2 + t 2p 3 ] = (1 − t)3p 0 + 2t(1 − 2t + t 2)p 1 + (t 2 − t 3)p 2 + t (1 − 2t + t 2) p 1 + 2t 2(1 − t)p 2 + t 3p 3 = (1 − 3t + 3t 2 − t 3)p 0 + (2t − 4t 2 + 2t 3)p 1 + (t 2 − t 3)p 2 + (t − 2t 2 + t 3)p 1 + (2t 2 − 2t 3)p 2 + t 3p 3 = (1 − 3t + 3t 2 − t 3)p 0 + (3t − 6t 2 + 3t 3)p 1 + (3t 2 − 3t 3)p 2 + t 3p 3 By inspection, the sum of the weights in this linear combination is 1, for all t. To show that the weights are nonnegative for 0 ≤ t ≤ 1, factor the coefficients and write h(t) = (1 − t)3p 0 + 3t(1 − t)2p 1 + 3t2(1 − t)p 2 + t 3p 3 for 0 ≤ t ≤ 1 Thus, h(t) is in the convex hull of the control points p 0 , p 1 , p 2 , and p 3 . Copyright © 2016 Pearson Education, Ltd. 8-16 CHAPTER 8 8.4 • The Geometry of Vector Spaces SOLUTIONS Notes: In this section lines and planes are generalized to higher dimensions using the notion of hyperplanes. Important topological ideas such as open, closed, and compact sets are introduced. −1 3 3 −1 4 = and v 2 . Then v 2 − v1 = − = . Choose n to be a vector orthogonal 1. Let v1 = 4 1 1 4 −3 3 to v 2 − v1 , for example let n = . Then f (x 1 , x 2 ) = 3x 1 + 4x 2 and d = f (v 1 ) = 3(–1) + 4(4) = 13. 4 This is easy to check by verifying that f (v 2 ) is also 13. 1 −2 −2 1 −3 = and v 2 . Then v 2 − v1= − = . Choose n to be a vector orthogonal 2. Let v1 = 4 −1 −1 4 −5 5 to v 2 − v1 , for example let n = . Then f (x 1 , x 2 ) = 5x 1 − 3x 2 and d = f (v 1 ) = 5(1) − 3(4) = −7 . −3 3. a. The set is open since it does not contain any of its boundary points. b. The set is closed since it contains all of its boundary points. c. The set is neither open nor closed since it contains some, but not all, of its boundary points. d. The set is closed since it contains all of its boundary points. e. The set is closed since it contains all of its boundary points. 4. a. The set is closed since it contains all of its boundary points. b. The set is open since it does not contain any of its boundary points. c. The set is neither open nor closed since it contains some, but not all, of its boundary points. d. The set is closed since it contains all of its boundary points. e. The set is open since it does not contain any of its boundary points. 5. a. The set is not compact since it is not closed, however it is convex. b. The set is compact since it is closed and bounded. It is also convex. c. The set is not compact since it is not closed, however it is convex. d. The set is not compact since it is not bounded. It is not convex. e. The set is not compact since it is not bounded, however it is convex. 6. a. The set is compact since it is closed and bounded. It is not convex. b. The set is not compact since it is not closed. It is not convex. c. The set is not compact since it is not closed, however it is convex. d. The set is not compact since it is not bounded. It is convex. e. The set is not compact since it is not closed. It is not convex. Copyright © 2016 Pearson Education, Ltd. 8.4 • Solutions 8-17 1 2 −1 a b and compute the translated points 7. a. Let v1 = 1 , v 2 = 4 , v3 = −2 , n = 3 1 5 c 1 −2 v 2 − v1 = 3 , v3 − v1 = −3 . −2 2 To solve the system of equations (v 2 – v 1 ) · n = 0 and (v 3 – v 1 ) · n = 0, reduce the augmented matrix for a system of two equations with three variables. a a [1 3 − 2] b = 0, [−2 − 3 2] b = 0. c c 1 3 −2 Row operations show that −2 −3 2 0 is n = 2 . 3 b. 0 1 ~ 0 0 0 0 3 −2 0 . A suitable normal vector 0 2 x2 + 3 x3 , so d = f (1, 1, 3) = 2 + 9 = 11. As a check, The linear functional is f ( x1 , x2 , x= 3) evaluate f at the other two points on the hyperplane: f (2, 4, 1) = 8 + 3 = 11 and f (–1 , –2 , 5) = – 4 + 15 = 11. 4 8. a. Find a vector in the null space of the transpose of [v 2 – v 1 v 3 – v 1 ]. For example, take n = 3 . −6 b. f (x) = 4x 1 + 3x 2 − 6x 3 , d = f (v 1 ) = −8 9. a. Find a vector in the null space of the transpose of [v 2 – v 1 v 3 – v 1 v 4 – v 1 ]. For example, take 3 −1 n = . 2 1 b. f (x) = 3x 1 − x 2 + 2x 3 + x 4 , d = f (v 1 ) = 5 10. a. Find a vector in the null space of the transpose of [v 2 – v 1 v 3 – v 1 v 4 – v 1 ]. For example, take −2 3 n = . −5 1 Copyright © 2016 Pearson Education, Ltd. 8-18 CHAPTER 8 • The Geometry of Vector Spaces b. f (x) = −2x 1 + 3x 2 − 5x 3 + x 4 , d = f (v 1 ) = 4 11. n ⋅ p = 2; n ⋅ 0 = 0 < 2; n ⋅ v 1 = 5 > 2; n ⋅ v 2 = –2 < 2; n ⋅ v 3 = 2. Hence v 2 is on the same side of H as 0, v 1 is on the other side, and v 3 is in H. 12. Let H = [ f : d ], where f (x 1 , x 2 , x 3 ) = 3x 1 + x 2 − 2x 3 . f (a 1 ) = − 5, f (a 2 ) = 4. f (a 3 ) =3, f (b 1 ) =7, f (b 2 ) = 4, and f (b 3 ) = 6. Choose d = 4 so that all the points in A are in or on one side of H and all the points in B are in or on the other side of H. There is no hyperplane parallel to H that strictly separates A and B because both sets have a point at which f takes on the value of 4. There may be (and in fact is) a hyperplane that is not parallel to H that strictly separates A and B. 13. H 1 = {x : n 1 ⋅ x = d 1 } and H 2 = {x : n 2 ⋅ x = d 2 }. Since p 1 ∈ H 1 , d 1 = n 1 ⋅ p 1 = 4. Similarly, d 2 = n 2 ⋅ p 2 = 22. Solve the simultaneous system [1 2 4 2]x = 4 and [2 3 1 5]x = 22: 1 2 2 3 4 1 2 4 ~ 5 22 1 0 0 −10 1 7 4 32 −1 −14 The general solution provides one set of vectors, p, v 1 , and v 2 . Other choices are possible. 32 10 − 4 −14 −7 + x + x 1 = x= p + x3 v1 + x4 v 2 , where = p 0 3 1 4 0 0 0 1 32 14 −= , v 0 1 0 10 −7 = , v 1 2 0 − 4 1 0 1 Then H 1 ∩ H 2 = {x : x = p + x 3 v 1 + x 4 v 2 }. 14. Since each of F 1 and F 2 can be described as the solution sets of A 1 x = b 1 and A 2 x = b 2 respectively, A b where A 1 and A 2 have rank 2, their intersection is described as the solution set to 1 x = 1 . A2 b 2 A1 ≤ 4 , the solution set will have dimensions 6 − 2 = 4, 6 − 3 = 3, or 6 − 4 = 2. A 2 Since 2 ≤ rank 15. f (x 1 , x 2 , x 3 , x 4 ) = Ax = x 1 – 3x 2 + 4x 3 – 2x 4 and d = b = 5 16. f (x 1 , x 2 , x 3 , x 4 , x 5 ) = Ax = 2x 1 + 5x 2 – 3x 3 + 6x 5 and d = b= 0 17. Since by Theorem 3 in Section 6.1, Row B = (Nul B) ⊥ , choose a nonzero vector n ∈ Nul B . For 1 example take n = −2 . Then f (x 1 , x 2 , x 3 ) = x 1 – 2x 2 + x 3 and d = 0 1 18. Since by Theorem 3, Section 6.1, Row B = (Nul B) ⊥ , choose a nonzero vector n ∈ Nul B . For −11 example take n = 4 . Then f (x 1 , x 2 , x 3 ) = –11x 1 + 4x 2 + x 3 and d = 0 1 19. Theorem 3 in Section 6.1 says that (Col B)⊥ = Nul BT. Since the two columns of B are clearly linear independent, the rank of B is 2, as is the rank of BT. So dim Nul BT = 1, by the Rank Theorem, since Copyright © 2016 Pearson Education, Ltd. 8.4 • Solutions 8-19 there are three columns in BT. This means that Nul BT is one-dimensional and any nonzero vector n in Nul BT will be orthogonal to H and can be used as its normal vector. Solve the linear system BTx = 0 by row reduction to find a basis for Nul BT: 1 0 4 −7 2 −6 0 1 ~ 0 0 0 5 1 −3 −5 0 ⇒ n = 3 0 1 Now, let f (x 1 , x 2 , x 3 ) = –5x 1 + 3x 2 + x 3 . Since the hyperplane H is a subspace, it goes through the origin and d must be 0. The solution is easy to check by evaluating f at each of the columns of B. 20. Since by Theorem 3, Section 6.1, Col B = (Nul BT) ⊥ , choose a nonzero vector n in Nul BT . For −6 example take n = 2 . Then f (x 1 , x 2 , x 3 ) = – 6x 1 + 2x 2 + x 3 and d = 0 1 21. a. False. A linear functional goes from n to . See the definition at the beginning of this section. b. False. See the discussion of (1) and (4). There is a 1×n matrix A such that f (x) = Ax for all x in n. Equivalently, there is a point n in n such that f (x) = n ⋅ x for all x in n. c. True. See the comments after the definition of strictly separate. d. False. See the sets in Figure 4. 22. a. True. See the statement after (3). b. False. The vector n must be nonzero. If n = 0, then the given set is empty if d ≠ 0 and the set is all of n if d = 0. c. False. Theorem 12 requires that the sets A and B be convex. For example, A could be the boundary of a circle and B could be the center of the circle. d. False. Some other hyperplane might strictly separate them. See the caution at the end of Example 8. 3 . −2 23. Notice that the side of the triangle closest to p is v 2 v 3 . A vector orthogonal to v 2 v 3 is n = Take f (x 1 , x 2 ) = 3x 1 – 2x 2 . Then f (v 2 ) = f (v 3 ) = 9 and f (p) = 10 so any d satisfying 9 < d < 10 will work. There are other possible answers. −2 . 3 24. Notice that the side of the triangle closest to p is v1v 3 A vector orthogonal to v1v 3 is n = Take f (x 1 , x 2 ) = − 2x 1 +3x 2 . Then f (v 1 ) = f (v 3 ) = 4 and f (p) = 5 so any d satisfying 4 < d < 5 will work. There are other possible answers. 25. Let L be the line segment from the center of B(0, 3) to the center of B(p, 1). This is on the line through the origin in the direction of p. The length of L is (42 + 12)1/2 ≈ 4.1231. This exceeds the sum of the radii of the two disks, so the disks do not touch. If the disks did touch, there would be no hyperplane (line) strictly separating them, but the line orthogonal to L through the point of tangency would (weakly) separate them. Since the disks are separated slightly, the hyperplane need not be Copyright © 2016 Pearson Education, Ltd. 8-20 CHAPTER 8 • The Geometry of Vector Spaces exactly perpendicular to L, but the easiest one to find is a hyperplane H whose normal vector is p. So define f by f (x) = p ⋅ x. To find d, evaluate f at any point on L that is between the two disks. If the disks were tangent, that point would be three-fourths of the distance between their centers, since the radii are 3 and 1. Since the disks are slightly separated, the distance is about 4.1231. Three-fourths of this distance is greater than 3, and one-fourth of this distance is greater than 1. A suitable value of d is f (q), where q = (.25)0 + (.75)p = (3, .75). So d = p ⋅ q = 4(3) + 1(.75) = 12.75. 26. The normal to the separating hyperplane has the direction of the line segment between p and q. So, 4 let n = p − q = . The distance between p and q is 20 , which is more than the sum of the radii −2 of the two balls. The large ball has center q. A point three-fourths of the distance from q to p will be greater than 3 units from q and greater than 1 unit from p. This point is 6 2 5.0 x = .75p + .25q = .75 + .25 = 1 3 1.5 x 17 . Compute n ⋅ x = 17. The desired hyperplane is : 4 x − 2 y = y 27. Exercise 2(a) in Section 8.3 gives one possibility. Or let S = {(x, y) : x2y 2 = 1 and y > 0}. Then conv S is the upper (open) half-plane. 28. One possibility is B = {(x, y) : x2y 2 = 1 and y > 0} and A = {(x, y) : | x | ≤ 1 and y = 0}. 29. Let x, y ∈ B( p, δ ) and suppose z = (1 − t) x + t y, where 0 ≤ t ≤ 1. Then ||z − p|| = || [(1 − t) x + t y] − p|| = ||(1 − t)(x − p) + t (y − p)|| ≤ (1 − t) ||x − p|| + t ||y − p|| < (1 − t) δ + tδ = δ where the first inequality comes from the Triangle Inequality (Theorem 17 in Section 6.7) and the second inequality follows from x, y ∈ B( p, δ ). It follows that z ∈ B( p, δ ) and B( p, δ ) is convex. 30. Let S be a bounded set. Then there exists a δ > 0 such that S Ì B(0, δ ). But B(0, δ ) is convex by Exercise 29, so Theorem 9 in Section 8.3 (or Exercise 17 in Section 8.3) implies that conv S Ì B( p, δ ) and conv S is bounded. 8.5 SOLUTIONS Notes: A polytope is the convex hull of a finite number of points. Polytopes and simplices are important in linear programming, which has numerous applications in engineering design and business management. The behavior of functions on polytopes is studied in this section. 1. Evaluate each linear functional at each of the three extreme points of S. Then select the extreme point(s) that give the maximum value of the functional. a. f (p 1 ) = 1, f (p 2 ) = –1, and f (p 3 ) = –3, so m = 1 at p 1 . Copyright © 2016 Pearson Education, Ltd. 8.5 • Solutions 8-21 b. f (p 1 ) = 1, f (p 2 ) = 5, and f (p 3 ) = 1, so m = 5 at p 2 . c. f (p 1 ) = –3, f (p 2 ) = –3, and f (p 3 ) = 5, so m = 5 at p 3 . 2. Evaluate each linear functional at each of the three extreme points of S. Then select the point(s) that give the maximum value of the functional. a. f (p 1 ) = − 1, f (p 2 ) = 3, and f (p 3 ) = 3, so m = 3 on the set conv {p 2 , p 3 }. b. f (p 1 ) = 1, f (p 2 ) = 1, and f (p 3 ) = − 1, so m = 1 on the set conv {p 1 , p 2 }. c. f (p 1 ) = –1, f (p 2 ) = –3, and f (p 3 ) = 0, so m = 0 at p 3 . 3. Evaluate each linear functional at each of the three extreme points of S. Then select the point(s) that give the minimum value of the functional. a. f (p 1 ) = 1, f (p 2 ) = –1, and f (p 3 ) = –3, so m = –3 at the point p 3 b. f (p 1 ) = 1, f (p 2 ) = 5, and f (p 3 ) = 1, so m = 1 on the set conv {p 1 , p 3 }. c. f (p 1 ) = –3, f (p 2 ) = –3, and f (p 3 ) = 5, so m = –3 on the set conv {p 1 , p 2 }. 4. Evaluate each linear functional at each of the three extreme points of S. Then select the point(s) that give the minimum value of the functional. a. f (p 1 ) = − 1, f (p 2 ) = 3, and f (p 3 ) = 3, so m = –1 at the point p 1 . b. f (p 1 ) = 1, f (p 2 ) = 1, and f (p 3 ) = − 1, so m = –1 at the point p 3 . c. f (p 1 ) = –1, f (p 2 ) = –3, and f (p 3 ) = 0, so m = –3 at the point p 2 . 5. The two inequalities are (a) x 1 + 2x 2 ≤ 10 and (b) 3x 1 + x 2 ≤ 15. Line (a) goes from (0,5) to (10,0). Line (b) goes from (0,15) to (5,0). One vertex is (0,0). The x 1 -intercepts (when x 2 = 0) are 10 and 5, so (5,0) is a vertex. The x 2 -intercepts (when x 1 = 0) are 5 and 15, so (0,5) is a vertex. The two lines 0 5 4 0 intersect at (4,3) so (4,3) is a vertex. The minimal representation is , , , 0 0 3 5 6. The two inequalities are (a) 2x 1 + 3x 2 ≤ 18 and (b) 4x 1 + x 2 ≤ 16. Line (a) goes from (0,6) to (9,0). Line (b) goes from (0,16) to (4,0). One vertex is (0,0). The x 1 -intercepts (when x 2 = 0) are 9 and 4, so (4,0) is a vertex. The x 2 -intercepts (when x 1 = 0) are 6 and 16, so (0,6) is a vertex. The two lines 0 4 3 0 intersect at (3,4) so (3,4) is a vertex. The minimal representation is , , , 0 0 4 6 7. The three inequalities are (a) x 1 + 3x 2 ≤ 18, (b) x 1 + x 2 ≤ 10, and (c) 4x 1 + x 2 ≤ 28. Line (a) goes from (0,6) to (18,0). Line (b) goes from (0,10) to (10,0). And line (c) goes from (0,28) to (7,0). One vertex is (0,0). The x 1 -intercepts (when x 2 = 0) are 18, 10, and 7, so (7,0) is a vertex. The x 2 - Copyright © 2016 Pearson Education, Ltd. 8-22 CHAPTER 8 • The Geometry of Vector Spaces intercepts (when x 1 = 0) are 6, 10, and 28, so (0,6) is a vertex. All three lines go through (6,4), so 0 7 6 0 (6,4) is a vertex. The minimal representation is , , , . 0 0 4 6 8. The three inequalities are (a) 2x 1 + x 2 ≤ 8, (b) x 1 + x 2 ≤ 6, and (c) x 1 + 2x 2 ≤ 7. Line (a) goes from (0,8) to (4,0). Line (b) goes from (0,6) to (6,0). And line (c) goes from (0,3.5) to (7,0). One vertex is (0,0). The x 1 -intercepts (when x 2 = 0) are 4, 6, and 7, so (4,0) is a vertex. The x 2 -intercepts (when x 1 = 0) are 8, 6, and 3.5, so (0,3.5) is a vertex. All three lines go through (3,2), so (3,2) is a vertex. The 0 4 3 0 minimal representation is , , , 0 0 2 3.5 9. The origin is an extreme point, but it is not a vertex. It is an extreme point since it is not in the interior of any line segment that lies in S. It is not a vertex since the only supporting hyperplane (line) containing the origin also contains the line segment from (0,0) to (3,0). 10. One possibility is a ray. It has an extreme point at one end. 11. One possibility is to let S be a square that includes part of the boundary but not all of it. For example, include just two adjacent edges. The convex hull of the profile P is a triangular region. S conv P = 12. a. f 0 (S 5) = 6, f 1 (S 5) = 15, f 2 (S 5) = 20, f 3 (S 5) = 15, f 4 (S 5) = 6, and 6 − 15 + 20 − 15 + 6 = 2. b. f0 f1 f2 f3 S 1 2 S 2 3 3 S 3 4 6 4 S 4 5 10 10 5 S 5 6 15 20 15 f4 6 n + 1 a a! fk (S n ) = is the binomial coefficient. , where = k + 1 b b!(a − b)! 13. a. To determine the number of k-faces of the 5-dimensional hypercube C 5, look at the pattern that is followed in building C 4 from C 3. For example, the 2-faces in C 4 include the 2-faces of C 3 and the 2-faces in the translated image of C 3. In addition, there are the 1-faces of C 3 that are “stretched” into 2-faces. In general, the number of k-faces in C n equals twice the number of kfaces in C n – 1 plus the number of (k – 1)-faces in C n – 1. Here is the pattern: f k (C n) = 2 f k (C n – 1 ) + f k – 1 (C n – 1). For k = 0, 1, …, 4, and n = 5, this gives f 0 (C 5) = 32, f 1 (C 5) = 80, f 2 (C 5) = 80, Copyright © 2016 Pearson Education, Ltd. 8.5 • Solutions 8-23 f 3 (C 5) = 40, and f 4 (C 5) = 10. These numbers satisfy Euler’s formula since, 32 − 80 + 80 − 40 + 10 = 2. a a! n n−k n = is the binomial coefficient. b. The general formula is f k (C ) 2= , where k b b !(a − b)! 14. a. X 1 is a line segment 0 v1 X 2 is a parallelogram v1 v2 b. f 0 (X 3) = 6, f 1 (X 3) = 12, f 2 (X 3) = 8. X 3 is an octahedron. c. f 0 (X 4 ) = 8, f 1 (X 4 ) = 24, f 2 (X 4 ) = 32, f 3 (X 4 ) = 16, 8 − 24 + 32 − 16 = 0 d. n f k ( X n ) = 2k + 1 , 0 ≤ k ≤ n −1, where k + 1 a a! is the binomial coefficient. = b b !(a − b)! 15. a. f 0 (P n ) = f 0 (Q) + 1 b. f k (P n ) = f k (Q) + f k − 1 (Q) c. f n − 1 (P n ) = f n − 2 (Q) + 1 16. a. b. c. d. True. See the definition at the beginning of this section. True. See the definition after Example 1. False. S must be compact. See Theorem 15. True. See the comment after Fig. 7. 17. a. False. It has six facets (faces). b. True. See Theorem 14. c. False. The maximum is always attained at some extreme point, but there may be other points that are not extreme points at which the maximum is attained. See Theorem 16. d. True. Follows from Euler’s formula with n = 2. 18. Let v be an extreme point of the convex set S and let T = {y ∈ S : y ≠ v}. If y and z are in T, then yz ⊆ S since S is convex. But since v is an extreme point of S, v ∉ yz , so yz Ì T . Thus T is convex. Conversely, suppose v ∈ S, but v is not an extreme point of S. Then there exist y and z in S such that v ∈ yz , with v ≠ y and v ≠ z. It follows that y and z are in T, but yz Ë T . Hence T is not convex. 19. Let S be convex and let x ∈ cS + dS, where c > 0 and d > 0. Then there exist s 1 and s 2 in S such that x = cs 1 + ds 2 . But then x =cs1 + ds 2 =(c + d ) c s1 + d s 2 . Since S is convex, c+d c+d c s + d s ∈ S and x ∈ (c + d)S. For the converse, pick a typical point in (c + d)S and show it c+d 1 c+d 2 is in cS + dS. Copyright © 2016 Pearson Education, Ltd. 8-24 CHAPTER 8 • The Geometry of Vector Spaces 20. For example, let S = {1, 2} in 1 . Then 2S = {2, 4}, 3S = {3, 6} and (2 + 3)S = {5, 10}. However, 2S + 3S = {2, 4} + {3, 6} = {2 + 3, 4 + 3, 2 + 6, 4 + 6} = {5, 7, 8, 10} ≠ (2 + 3)S. 21. Suppose A and B are convex. Let x, y ∈ A + B. Then there exist a, c ∈ A and b, d ∈ B such that x = a + b and y = c + d. For any t such that 0 ≤ t ≤ 1, we have w = (1 − t )x + ty = (1 − t )(a + b) + t (c + d) = [ (1 − t )a + tc ] + [ (1 − t )b + td ] But (1 − t)a + tc ∈ A since A is convex, and (1 − t)b + td ∈ B since B is convex. Thus w is in A + B, which shows that A + B is convex. 22. a. Since each edge belongs to two facets, kr is twice the number of edges: k r = 2e. Since each edge has two vertices, s v = 2e. b. v − e + r = 2, so 2e − e + 2e = 2 ⇒ s k 1+1 = s k 1+1 2 e c. A polygon must have at least three sides, so k ≥ 3. At least three edges meet at each vertex, so s ≥ 3. But both k and s cannot both be greater than 3, for then the left side of the equation in (b) could not exceed 1/2. 1 , so s = 3, 4, or 5. For these values, we get e = 6, 12, or 30, When k = 3, we get 1s − 16 = e corresponding to the tetrahedron, the octahedron, and the icosahedron, respectively. When s = 3, we get 1−1 = 1, e k 6 so k = 3, 4, or 5 and e = 6, 12, or 30, respectively. These values correspond to the tetrahedron, the cube, and the dodecahedron. 8.6 SOLUTIONS Notes: This section moves beyond lines and planes to the study of some of the curves that are used to model surfaces in engineering and computer aided design. Notice that these curves have a matrix representation. 1. The original curve is x(t) = (1 – t)3p 0 + 3t(1 – t)2p 1 + 3t 2(1 – t)p 2 + t 3p 3 (0 < t < 1). Since the curve is determined by its control points, it seems reasonable that to translate the curve, one should translate the control points. In this case, the new Bézier curve y(t) would have the equation y(t) = (1 – t)3(p 0 + b) + 3t(1 – t)2(p 1 + b) + 3t 2(1 – t)(p 2 + b) + t 3(p 3 + b) = (1 – t)3p 0 + 3t(1 – t)2p 1 + 3t 2(1 – t)p 2 + t 3p 3 + (1 – t)3b + 3t(1 – t)2b + 3t 2(1 – t)b + t 3b A routine algebraic calculation verifies that (1 – t)3 + 3t(1 – t)2 + 3t 2(1 – t) + t 3 = 1 for all t. Thus y(t) = x(t) + b for all t, and translation by b maps a Bézier curve into a Bézier curve. 2. a. Equation (15) reveals that each polynomial weight is nonnegative for 0 < t < 1, since 4 − 3t > 0. For the sum of the coefficients, use (15) with the first term expanded: 1 – 3t + 6t 2 − t 3. The 1 here plus the 4 and 1 in the coefficients of p 1 and p 2 , respectively, sum to 6, while the other terms sum to 0. This explains the 1/6 in the formula for x(t), which makes the coefficients sum to 1. Thus, x(t) is a convex combination of the control points for 0 < t < 1. b. Since the coefficients inside the brackets in equation (14) sum to 6, it follows that Copyright © 2016 Pearson Education, Ltd. 8.6 [6b=] • Solutions 8-25 (1 − t )3 b + (3t 3 − 6t 2 + 4)b + (−3t 3 + 3t 2 + 3t + 1)b + t 3b and hence x(t) + may be written in a similar form, with p i replaced by p i + b for each i. This shows that x(t) + b is a cubic B-spline with control points p i + b for i = 0, …, 3. = b 1 6 1 6 3. a. x' (t) = (–3 + 6t – 3t 2)p 0 + (3 –12t + 9t 2)p 1 + (6t – 9t 2)p 2 + 3t 2p 3 , so x' (0) = –3p 0 + 3p 1 =3(p 1 – p 0 ), and x' (1) = –3p 2 + 3p 3 = 3(p 3 – p 2 ). This shows that the tangent vector x' (0) points in the direction from p 0 to p 1 and is three times the length of p 1 – p 0 . Likewise, x' (1) points in the direction from p 2 to p 3 and is three times the length of p 3 – p 2 . In particular, x' (1) = 0 if and only if p 3 = p 2 . b. x'' (t) = (6 – 6t)p 0 + (–12 + 18t)p 1 + (6 – 18t)p 2 + 6tp 3 , so that x'' (0) = 6p 0 – 12p 1 + 6p 2 = 6(p 0 – p 1 ) + 6(p 2 – p 1 ) and x'' (1) = 6p 1 – 12p 2 + 6p 3 = 6(p 1 – p 2 ) + 6(p 3 – p 2 ) For a picture of x'' (0), construct a coordinate system with the origin at p 1 , temporarily, label p 0 as p 0 − p 1 , and label p 2 as p 2 − p 1 . Finally, construct a line from this new origin through the sum of p 0 − p 1 and p 2 − p 1 , extended out a bit. That line points in the direction of x'' (0). 0 = p1 p2 – p1 p0 – p1 w ( ) ( ) ( ) 4. a. x' (t) = 16 −3t 2 + 6t − 3 p 0 + 9t 2 − 12t p1 + −9t 2 + 6t + 3 p 2 + 3t 2p3 1 1 x' (0) = 2 ( p 2 − p 0 ) and x' (1) = 2 ( p3 − p1 ) (Verify that, in the first part of Fig. 10, a line drawn through p 0 and p 2 is parallel to the tangent line at the beginning of the B-spline.) When x' (0) and x' (1) are both zero, the figure collapses and the convex hull of the set of control points is the line segment between p 0 and p 3 , in which case x(t) is a straight line. Where does x(t) start? In this case, x(t) = 16 (−4t 3 + 6t 2 + 2)p 0 + (4t 3 − 6t 2 + 4)p3 1 2 2 1 x(0) = 3 p 0 + 3 p3 and x(1) = 3 p 0 + 3 p3 The curve begins closer to p 3 and finishes closer to p 0 . Could it turn around during its travel? Since x' (t) = 2t(1 − t)(p 0 − p 3 ), the curve travels in the direction p 0 − p 3 , so when x' (0) = x' (1) = 0, the curve always moves away from p 3 toward p 0 for 0 < t < 1. b. x'' (t) = (1 – t)p 0 + (–2 + 3t)p 1 + (1 – 3t)p 2 + tp 3 x'' (0) = p 0 – 2p 1 + p 2 = (p 0 − p 1 ) + (p 2 − p 1 ) and x'' (1) = p 1 – 2p 2 + p 3 = (p 1 − p 2 ) + (p 3 − p 2 ) For a picture of x'' (0), construct a coordinate system with the origin at p 1 , temporarily, label p 0 as p 0 − p 1 , and label p 2 as p 2 − p 1 . Finally, construct a line from this new origin to the sum of p 0 − p 1 and p 2 − p 1 . That segment represents x'' (0). Copyright © 2016 Pearson Education, Ltd. 8-26 CHAPTER 8 • The Geometry of Vector Spaces For a picture of x'' (1), construct a coordinate system with the origin at p 2 , temporarily, label p 1 as p 1 − p 2 , and label p 3 as p 3 − p 2 . Finally, construct a line from this new origin to the sum of p 1 − p 2 and p 3 − p 2 . That segment represents x'' (1). p1 – p2 w p2 = 0 p3 – p2 5. a. From Exercise 3(a) or equation (9) in the text, x' (1) = 3(p 3 – p 2 ) Use the formula for x'(0), with the control points from y(t), and obtain y' (0) = –3p 3 + 3p 4 = 3(p 4 – p 3 ) For C1 continuity, 3(p 3 – p 2 ) = 3(p 4 – p 3 ), so p 3 = (p 4 + p 2 )/2, and p 3 is the midpoint of the line segment from p 2 to p 4 . b. If x' (1) = y' (0) = 0, then p 2 = p 3 and p 3 = p 4 . Thus, the “line segment” from p 2 to p 4 is just the point p 3 . [Note: In this case, the combined curve is still C1 continuous, by definition. However, some choices of the other control points, p 0 , p 1 , p 5 , and p 6 can produce a curve with a visible “corner” at p 3 , in which case the curve is not G1 continuous at p 3 .] 6. a. With x(t) as in Exercise 2, x(0) = (p 0 + 4p 1 + p 2 )/6 and x(1) = (p 1 + 4p 2 + p 3 )/6 Use the formula for x(0), but with the shifted control points for y(t), and obtain y(0) = (p 1 + 4p 2 + p 3 )/6 This equals x(1), so the B-spline is G0 continuous at the join point. b. From Exercise 4(a), x' (1) = (p 3 – p 1 )/2 and x' (0) = (p 2 – p 0 )/2 Use the formula for x' (0) with the control points for y(t), and obtain y' (0) = (p 3 – p 1 )/2 = x' (1) Thus the B-spline is C1 continuous at the join point. 7. From Exercise 3(b), x'' (0) = 6(p 0 – p 1 ) + 6(p 2 – p 1 ) and x'' (1) = 6(p 1 – p 2 ) + 6(p 3 – p 2 ) Use x'' (0) with the control points for y(t), to get y'' (0) = 6(p 3 – p 4 ) + 6(p 5 – p 4 ) Set x'' (1) = y'' (0) and divide by 6, to get (p 1 – p 2 ) + (p 3 – p 2 ) = (p 3 – p 4 ) + (p 5 – p 4 ) (*) Since the curve is C1 continuous at p 3 , the point p 3 is the midpoint of the segment from p 2 to p 4 , by 1 (p + p ) , which leads to p – p = p – p . Substituting into (*) gives Exercise 5(a). Thus= p3 4 3 3 2 2 4 2 (p 1 – p 2 ) + (p 3 – p 2 ) = –(p 3 – p 2 ) + p 5 – p 4 (p 1 – p 2 ) + 2(p 3 – p 2 ) + p 4 = p 5 Finally, again from C1 continuity, p 4 = p 3 + p 3 – p 2 . Thus, p 5 = p 3 + (p 1 – p 2 ) + 3(p 3 – p 2 ) So p 4 and p 5 are uniquely determined by p 1 , p 2 , and p 3 . Only p 6 can be chosen arbitrarily. Copyright © 2016 Pearson Education, Ltd. 8.6 • Solutions 8-27 8. From Exercise 4(b), x'' (0) = p 0 – 2p 1 + p 2 and x'' (1) = p 1 – 2p 2 + p 3 . Use the formula for x'' (0), with the shifted control points for y(t), to get y'' (0) = p 1 – 2p 2 +2p 3 = x'' (1) 2 Thus the curve has C continuity at x(1). 9. Write a vector of the polynomial weights for x(t), expand the polynomial weights and factor the vector as M B u(t): 1 − 4t + 6t 2 − 4t 3 + t 4 6 −4 1 6 −4 1 1 1 −4 1 −4 t 2 3 4 4t − 12t + 12t − 4t 0 4 −12 12 −4 0 4 −12 12 −4 2 2 3 4 6 −12 6 6 −12 6 t , M B = 0 0 6t − 12t + 6t = 0 0 0 0 0 4 −4 0 4 −4 t 3 0 0 4t 3 − 4t 4 0 0 0 0 1 t 4 0 0 1 0 0 t4 10. Write a vector of the polynomial weights for x(t), expand the polynomial weights, taking care to write the terms in ascending powers of t, and factor the vector as M S u(t): 1 − 3t + 3t 2 − t 3 1 −3 3 −1 1 1 −3 3 −1 t 2 3 1 4 − 6t + 3t 1 4 0 −6 3 1 4 0 −6 3 = = M u(t), M = S S 6 1 + 3t + 3t 2 − 3t 3 6 1 3 3 −3 t 2 6 1 3 3 −3 3 0 0 0 1 t 0 0 0 1 3 t 11. a. True. See equation (2). b. False. Example 1 shows that the tangent vector x′(t) at p 0 is two times the directed line segment from p 0 to p 1 . c. True. See Example 2. 12. a. False. The essential properties are preserved under translations as well as linear transformations. See the comment after Figure 1. b. True. This is the definition of G0 continuity at a point. c. False. The Bézier basis matrix is a matrix of the polynomial coefficients of the control points. See the definition before equation (4). 13. a. From (12), q1 − q 0 = 12 (p1 − p 0 ) = 12 p1 − 12 p 0 . Since= q 0 p 0 ,= q1 12 (p1 + p 0 ) . b. From (13), 8(q 3 – q 2 ) = –p 0 – p 1 + p 2 + p 3 . So 8q 3 + p 0 + p 1 – p 2 – p 3 = 8q 2 . c. Use (8) to substitute for 8q 3 , and obtain 8q 2 = (p 0 + 3p 1 + 3p 2 + p 3 ) + p 0 + p 1 – p 2 – p 3 = 2p 0 + 4p 1 + 2p 2 Then divide by 8, regroup the terms, and use part (a) to obtain q 2 =14 p 0 + 12 p1 + 14 p 2 =( 14 p 0 + 14 p1 ) + ( 14 p1 + 14 p 2 ) =12 q1 + 14 (p1 + p 2 ) = 12 (q1 + 12 (p1 + p 2 )) 14. a. 3(r 3 − r 2 ) = z'(1), by (9) with z'(1) and r j in place of x'(1) and p j . Copyright © 2016 Pearson Education, Ltd. 8-28 CHAPTER 8 • The Geometry of Vector Spaces z'(1) = .5x'(1), by (11) with t = 1. .5x'(1) = (.5)3(p 3 − p 2 ), by (9). b. From part (a), 6(r 3 − r 2 ) = 3(p 3 − p 2 ), r3 − r2 = Since r 3 = p 3 , this equation becomes= r2 1 (p 3 2 1p 2 3 − 12 p 2 , and r3 − 12 p3 + 12 p 2 = r2 . + p2 ) . c. 3(r 1 − r 0 ) = z' (0), by (9) with z'(0) and r j in place of x'(0) and p j . z' (0) = .5x'(.5), by (11) with t = 0. d. Part (c) and (10) show that 3(r 1 − r 0 ) = 3 8 (−p 0 − p 1 + p 2 + p 3 ). Multiply by 8 3 and rearrange to obtain 8r 1 = −p 0 − p 1 + p 2 + p 3 + 8r 0 . e. From (8), 8r 0 = p 0 + 3p 1 + 3p 2 + p 3 . Substitute into the equation from part (d), and obtain 8r 1 = 2p 1 + 4p 2 + 2p 3 . Divide by 8 and use part (b) to obtain r1 = 1 4 p1 + 12 p 2 + 14 p3 = ( 14 p1 + 14 p 2 ) + 14 ( p 2 + p3 ) = Interchange the terms on the right, and obtain r 1 = 1 �1 ( p + p ) + 1 r 1 2 2 2 2 2 1 [r + 1 (p + p )] . 1 2 2 2 2 15. a. From (11), y'(1) = .5x'(.5) = z'(0). b. Observe that y'(1) = 3(q 3 – q 2 ). This follows from (9), with y(t) and its control points in place of x(t) and its control points. Similarly, for z(t) and its control points, z'(0) = 3(r 1 – r 0 ). By part (a) 3(q 3 – q 2 ) = 3(r 1 – r 0 ). Replace r 0 by q 3 , and obtain q 3 – q 2 = r 1 – q 3 , and hence q 3 = (q 2 + r 1 )/2. c. Set q 0 = p 0 and r 3 = p 3 . Compute q 1 = (p 0 + p 1 )/2 and r 2 = (p 2 + p 3 )/2. Compute m = (p 1 + p 2 )/2. Compute q 2 = (q 1 + m)/2 and r 1 = (m + r 2 )/2. Compute q 3 = (q 2 + r 1 )/2 and set r 0 = q 3 . 16. A Bézier curve is completely determined by its four control points. Two are given directly: p 0 = x(0) and p 3 = x(1). From equation (9), x' (0) = 3(p 1 – p 0 ) and x' (1) = 3(p 3 – p 2 ). Solving gives p1 = p0 + 1 3 x' (0) and p 2 = p 3 − 1 3 x' (1). 17. a. The quadratic curve is w(t) = (1 – t) 2 p 0 + 2t(1 − t)p 1 + t 2 p 2 . From Example 1, the tangent vectors at the endpoints are w' (0) = 2p 1 − 2p 0 and w' (1) = 2p 2 − 2p 1 . Denote the control points of x(t) by r 0 , r 1 , r 2 , and r 3 . Then r 0 = x(0) = w(0) = p 0 and r 3 = x(1) = w(1) = p 2 From equation (9) or Exercise 3(a) (using r i in place of p i ) and Example 1, –3r 0 + 3r 1 = x' (0) = w' (0) = 2p 1 − 2p 0 so −p 0 + r 1 = 2p1 − 2p 0 3 and p 0 + 2p1 (i) 3 Similarly, using the tangent data at t = 1, along with equation (9) and Example 1, yields –3r 2 + 3r 3 = x' (1) = w' (1) = 2p 2 − 2p 1 , r1 = −r 2 + p 2 = 2p 2 − 2p1 2p − 2p1 , r2 = p2 − 2 , and 3 3 Copyright © 2016 Pearson Education, Ltd. 8.6 • Solutions 2p1 + p 2 (ii) 3 b. Write the standard formula (7) in this section, with r i in place of p i for i = 1, …, 4, and then replace r 0 and r 3 by p 0 and p 2 , respectively: (iii) x(t) = (1 – 3t + 3t2 – t 3)p 0 + (3t – 6t2 + 3t 3)r 1 + (3t2 – 3t 3)r 2 + t 3p 2 Use the formulas (i) and (ii) for r 1 and r 2 to examine the second and third terms in (iii): (3t − 6t 2 + 3t 3 )r1 = 13 (3t − 6t 2 + 3t 3 )p 0 + 32 (3t − 6t 2 + 3t 3 )p1 r2 = =(t − 2t 2 + t 3 )p 0 + (2t − 4t 2 + 2t 3 )p1 (3t 2 − 3t 3 )r2 = 23 (3t 2 − 3t 3 )p1 + 13 (3t 2 − 3t 3 )p 2 = (2t 2 − 2t 3 )p1 + (t 2 − t 3 )p 2 When these two results are substituted in (iii), the coefficient of p 0 is (1 − 3t + 3t 2 − t 3) + (t − 2t 2 + t 3) = 1 − 2t + t 2 = (1 − t)2 The coefficient of p 1 is (2t − 4t 2 + 2t 3) + (2t 2 − 2t 3) = 2t − 2t 2 = 2t(1 − t) The coefficient of p 2 is (t 2 − t 3) + t 3 = t 2. So x(t) = (1 − t)2p 0 + 2t(1 − t)p 1 + t 2p 2 , which shows that x(t) is the quadratic Bézier curve w(t). p0 −3p 0 + 3p1 18. 3p 0 − 6p1 + 3p 2 −p 0 + 3p1 − 3p 2 + p3 Copyright © 2016 Pearson Education, Ltd. 8-29 |
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