Students’ Solutions Manual for Applied Linear Algebra by Peter J.Olver and Chehrzad Shakiban Second Edition Undergraduate Texts in Mathematics Springer, New York, 2018. ISBN 978–3–319–91040–6 To the Student These solutions are a resource for students studying the second edition of our text Applied Linear Algebra, published by Springer in 2018. An expanded solutions manual is available for registered instructors of courses adopting it as the textbook. Using the Manual The material taught in this book requires an active engagement with the exercises, and we urge you not to read the solutions in advance. Rather, you should use the ones in this manual as a means of verifying that your solution is correct. (It is our hope that all solutions appearing here are correct; errors should be reported to the authors.) If you get stuck on an exercise, try skimming the solution to get a hint for how to proceed, but then work out the exercise yourself. The more you can do on your own, the more you will learn. Please note: for students taking a course based on Applied Linear Algebra, copying solutions from this Manual can place you in violation of academic honesty. In particular, many solutions here just provide the final answer, and for full credit one must also supply an explanation of how this is found. Acknowledgements We thank a number of people, who are named in the text, for corrections to the solutions manuals that accompanied the first edition. Of course, as authors, we take full respon- sibility for all errors that may yet appear. We encourage readers to inform us of any misprints, errors, and unclear explanations that they may find, and will accordingly up- date this manual on a timely basis. Corrections will be posted on the text’s dedicated web site: http://www.math.umn.edu/∼olver/ala2.html Peter J. Olver Cheri Shakiban University of Minnesota University of St. Thomas [email protected] [email protected] Minnesota, September 2018 Table of Contents Chapter1. LinearAlgebraicSystems. 1 Chapter2. VectorSpacesandBases . 11 Chapter3. InnerProductsandNorms . 19 Chapter4. Orthogonality. 28 Chapter5. MinimizationandLeastSquares. 36 Chapter6. Equilibrium . 44 Chapter7. Linearity. 49 Chapter8. EigenvaluesandSingularValues. 58 Chapter9. Iteration . 70 Chapter10. Dynamics . 81 Students’ Solutions Manual for Chapter 1: Linear Algebraic Systems 1.1.1. (b) Reduce the system to 6u + v = 5, 5 v = 5 ; then use Back Substitution to solve − 2 2 for u = 1, v = 1. − (d) Reduce the system to 2u v + 2w = 2, 3 v + 4w = 2, w = 0; then solve for − − 2 − u = 1 , v = 4 , w = 0. 3 − 3 (f ) Reduce the system to x + z 2w = 3, y + 3w = 1, 4z 16w = 4, 6w = 6; then − − − − − − solve for x = 2, y = 2, z = 3, w = 1. − 1.1.3. (a) With Forward Substitution, we just start with the top equation and work down. ♥ Thus 2x = 6 so x = 3. Plugging this into the second equation gives 12 + 3y = 3, and so − − y = 3. Plugging the values of x and y in the third equation yields 3 + 4( 3) z = 7, and − − − − so z = 22. − 0 1.2.1. (a) 3 4, (b) 7, (c) 6, (d) ( 2 0 1 2), (e) 2 . × − 6 − 1 2 3 1 2 3 4 1 1.2.2. Examples: (a) 4 5 6 , (c) 4 5 6 7 , (e) 2 . 7 8 9 7 8 9 3 3 6 1 u 5 1.2.4. (b) A = , x = , b = ; 3 2 ! v ! 5 ! − 2 1 2 u 2 − (d) A = 1 1 3 , x = v , b = 1 ; − − 3 0 2 w 1 − 1 0 1 2 x 3 − − 2 1 2 1 y 5 (f ) A = − − , x = , b = − . 0 6 4 2 z 2 − − 1 3 2 1 w 1 − 1.2.5. (b) u + w = 1, u + v = 1, v + w = 2. The solution is u = 2, v = 1, w = 1. − − − (c) 3x x = 1, 2x x = 0, x + x 3x = 1. 1 − 3 − 1 − 2 1 2 − 3 The solution is x = 1 , x = 2 , x = 2 . 1 5 2 − 5 3 − 5 10000 00000 01000 00000 1.2.6. (a) I= 00100 , O= 00000 . 00010 00000 00001 00000 (b) I+O=I , IO=OI =O. No, it does not. 1 c 2018 Peter J. Olver and Chehrzad Shakiban 2 Students’ Solutions Manual: Chapter 1 1 11 9 3 6 0 1.2.7. (b) undefined, (c) , (f ) 3 12 12 , 1 4 2 ! − − − 7 8 8 1.2.9. 1, 6, 11, 16. 1 0 0 1.2.10. (a) 0 0 0 . 0 0 1 − 1.2.11. (a) True. x y 1.2.14. B = where x,y are arbitrary. 0 x ! 1.2.17. A On p =Om p, Ol m A =Ol n. × × × × 1 0 0 0 0 0 1.2.19. False: for example, = . 0 0 ! 1 0 ! 0 0 ! 1.2.22. (a) A must be a square matrix. (b) By associativity, AA2 = AAA = A2 A = A3. 1 1 1.2.25. (a) X = − . 3 2 ! − 1.2.29. (a) The ith entry of A z is 1 a + 1 a + + 1 a = a + + a , which is ♦ i1 i2 · · · in i1 · · · in 1 the ith row sum. (b) Each row of W has n 1 entries equal to and one entry equal − n 1 n 1 1 n to − and so its row sums are (n 1) + − = 0. Therefore, by part (a), W z = 0. n − n n Consequently, the row sums of B = AW are the entries of B z = AW z = A0 = 0, and the result follows. 1.2.34. (a) This follows by direct computation. (b) (i) ♥ 2 1 1 2 2 1 2 4 1 0 1 4 − − = − ( 1 2)+ (1 0)= − + = − . 3 2 ! 1 0 ! 3 ! − 2 ! 3 6 ! 2 0 ! 5 6 ! − − 1.3.1. (b) u = 1, v = 1; (d) x = 11 , x = 10 , x = 2 ; − 1 3 2 − 3 3 − 3 (f ) a = 1 , b = 0, c = 4 , d = 2 . 3 3 − 3 1 7 4 2R1+R2 1 7 4 1.3.2. (a) . Back Substitution yields x2 = 2, x1 = 10. 2 9 2 ! 0 5 10 ! −→ − − − 1 2 1 0 1 2 1 0 3 1 2 1 0 − 4R1+R3 − 2 R2+R3 − c ( ) 0 2 8 8 0 2 8 8 0 2 8 8 . − −→ − −→ − 4 5 9 9 0 3 13 9 0 0 1 3 − − − − Back Substitution yields z = 3, y = 16, x = 29. c 2018 Peter J. Olver and Chehrzad Shakiban Students’ Solutions Manual: Chapter 1 3 1 0 2 0 1 1 0 2 0 1 − − − − 0 1 0 1 2 0 1 0 1 2 (e) reduces to . − − 0 3 2 0 0 0 0 2 3 6 − − 4 0 0 7 5 0 0 0 5 15 − − − Back Substitution yields x = 3, x = 3 , x = 1, x = 4. 4 − 3 − 2 2 − 1 − 1.3.3. (a) 3x + 2y = 2, 4x 3y = 1; solution: x = 4, y = 5. − − − − 2 1 2 1 1.3.4. (a) Regular: 7 . 1 4 ! −→ 0 2 ! 1 2 3 1 2 3 − − (d) Not regular: 2 4 1 0 0 5 . − − −→ 3 1 2 0 5 7 − − i 1+i 1 i 1+i 1 1.3.5. (a) − − − − ; 1 i 1 3 i ! −→ 0 1 2 i 1 2 i ! − − − − use Back Substitution to obtain the solution y = 1, x = 1 2 i . − aij, i>j, aij, i<j, aij, i = j, 1.3.11. (a) Set lij = , uij = dij = ♦ 0, i j, 0, i j, 0, i = j. ≤ ≥ 6 0 0 0 3 0 0 0 1 1 − (b) L = 1 0 0 , D = 0 4 0 , U = 0 0 2 . − 2 0 0 0 0 5 0 0 0 − 1.3.15. (a) Add 2 times the second row to the first row of a 2 n matrix. − × (c) Add 5 times the third row to the second row of a 3 n matrix. − × 1 0 0 0 1 0 0 3 0 1 0 0 0 1 0 0 1.3.16. (a) , (c) . 0 0 1 0 0 0 1 0 0 0 1 1 0 0 0 1 1.3.20. (a) Upper triangular; (b) both upper and lower unitriangular; (d) lower unitriangular. 1 0 0 1 1 1 1 0 1 3 − − 1.3.21. (a) L = , U = , (c) L = 1 1 0 , U = 0 2 0 , 1 1 ! 0 3 ! − − 1 0 1 0 0 3 1 0 0 1 0 0 − (e) L = 2 1 0 , U = 0 3 0 . − − 1 1 1 0 0 2 − − 1.3.25. (a) a = 0 for all i = j; (c) a = 0 for all i>j and a = 1 for all i. ij 6 ij ii 1 1 1.3.27. False. For instance is regular. Only if the zero appear in the (1, 1) position 1 0 ! does it automatically preclude regularity of the matrix. c 2018 Peter J. Olver and Chehrzad Shakiban 4 Students’ Solutions Manual: Chapter 1 0 1 1 − 1.3.32. (a) x = −2 , (c) x = 1 , (e) x = 1 . ! −5 3 0 2 5 9 1 0 1 3 11 1 11 1.3.33. (a) L = , U = − ; x1 = − , x2 = , x3 = . 3 1 ! 0 11 ! 2 1 ! 3 − 11 11 1 0 0 9 2 1 − − 1 2 − (c) L = 2 , U = 1 1 ; x = 2 , x = 9 . 3 1 0 0 3 3 1 2 − − − 2 5 1 3 1 1 0 0 − 9 3 − 3 5 1 1 0 00 1 0 1 0 4 14 − 1 5 0 1 00 0 2 3 1 4 14 (e) L = 3 , U = 7 −7 ; x1 = − , x2 = − .