Answer key for the book

Preface This Instructor?s Solutions Manual contains solutions for essentially all of the exercises in the text that are intended to be done by hand. Solutions to Matlab exercises are not included. The Student?s Solutions Manual that accompanies this text contains solutions for only selected odd-numbered exercises, including those exercises whose answers appear in the answer key. The solutions that appear in the students? manual are identical to those provided in this manual, and generally provide a more detailed solution than is available in the answer key. Although no pattern is strictly adhered to throughout the student manual, the solutions provided there are primarily to the computational exercises, whereas solutions that involve proof are generally not included. None of the solutions to the supplementary end-of-chapter exercises are included in the student manual. Contents Preface iii 1 Matrices and Systems of Equations 1 1.1 Introduction to Matrices and Systems of Linear Equations . . . . . . . . . . . . . . 1 1.2 Echelon Form and Gauss-Jordan Elimination . . . . . . . . . . . . . . . . . . . . . 6 1.3 Consistent Systems of Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . 11 1.4 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 1.5 Matrix Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 1.6 Algebraic Properties of Matrix Operations . . . . . . . . . . . . . . . . . . . . . . . 21 1.7 Linear Independence and Nonsing. Matrices . . . . . . . . . . . . . . . . . . . . . . 26 1.8 Data fltting, Numerical Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 1.9 Matrix Inverses and their Properties . . . . . . . . . . . . . . . . . . . . . . . . . . 32 1.10 Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 1.11 Conceptual Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 2 Vectors in 2-Space and 3-Space 43 2.1 Vectors in the Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 2.2 Vectors in Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 2.3 The Dot Product and the Cross Product . . . . . . . . . . . . . . . . . . . . . . . . 48 2.4 Lines and Planes in Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 2.5 Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 2.6 Conceptual Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 3 The Vector Space Rn 59 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 3.2 Vector Space Properties of Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 3.3 Examples of Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 3.4 Bases for Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 3.5 Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 3.6 Orthogonal Bases for Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 3.7 Linear Transformations from Rn to Rm . . . . . . . . . . . . . . . . . . . . . . . . 83 v vi CONTENTS 3.8 Least-Squares Solutions to Inconsistent Systems . . . . . . . . . . . . . . . . . . . . 89 3.9 Fitting Data and Least Squares Solutions . . . . . . . . . . . . . . . . . . . . . . . 92 3.10 Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 3.11 Conceptual Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 4 The Eigenvalue Problems 99 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 4.2 Determinants and the Eigenvalue Problem . . . . . . . . . . . . . . . . . . . . . . . 101 4.3 Elementary Operations and Determinants . . . . . . . . . . . . . . . . . . . . . . . 104 4.4 Eigenvalues and the Characteristic Polynomial . . . . . . . . . . . . . . . . . . . . 108 4.5 Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 4.6 Complex Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . . . . . . . . . . . 117 4.7 Similarity Transformations & Diagonalization . . . . . . . . . . . . . . . . . . . . . 121 4.8 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 4.9 Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 4.10 Conceptual Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 5 Vector Spaces and Linear Transformations 135 5.1 Introduction (No exercises) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 5.2 Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135 5.3 Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 5.4 Linear Independence, Bases, and Coordinates . . . . . . . . . . . . . . . . . . . . . 144 5.5 Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 5.6 Inner-products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150 5.7 Linear Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154 5.8 Operations with Linear Transformations . . . . . . . . . . . . . . . . . . . . . . . . 158 5.9 Matrix Representations for Linear Transformations . . . . . . . . . . . . . . . . . . 161 5.10 Change of Basis and Diagonalization . . . . . . . . . . . . . . . . . . . . . . . . . . 166 5.11 Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 5.12 Conceptual Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 6 Determinants 175 6.1 Introduction (No exercises) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175 6.2 Cofactor Expansion of Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . 175 6.3 Elementary Operations and Determinants . . . . . . . . . . . . . . . . . . . . . . . 178 6.4 Cramer?s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 6.5 Applications of Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186 6.6 Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 6.7 Conceptual Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 7 Eigenvalues and Applications 193 7.1 Quadratic Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 7.2 Systems of Difierential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 7.3 Transformation to Hessenberg Form . . . . . . . . . . . . . . . . . . . . . . . . . . 199 7.4 Eigenvalues of Hessenberg Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . 202 7.5 Householder Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206 7.6 QR Factorization & Least-Squares . . . . . . . . . . . . . . . . . . . . . . . . . . 208 7.7 Matrix Polynomials & The Cayley-Hamilton Theorem . . . . . . . . . . . . . . . . 211 7.8 Generalized Eigenvectors & Difi. Eqns. . . . . . . . . . . . . . . . . . . . . . . . . . 212 7.9 Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216 7.10 Conceptual Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216 Chapter 1 Matrices and Systems of Equations 1.1 Introduction to Matrices and Systems of Linear Equations 1. Linear. 2. Nonlinear. 3. Linear. 4. Nonlinear. 5. Nonlinear. 6. Linear. 7. x1 +3x2 = 7 4x1 ¡x2 = 2 1+3¢2 = 7 4¢1¡2 = 2 8. 6x1 ¡x2 + x3 = 14 x1 +2x2 +4x3 = 4 6¢2¡(¡1)+1 = 14 2+2¢(¡1)+4¢1 = 4 9. x1 + x2 = 0 3x1 +4x2 = ¡1 ¡x1 +2x2 = ¡3 1+(¡1) = 0 3¢1+4¢(¡1) = ¡1 ¡1+2¢(¡1) = ¡3 10. 3x2 = 9; 4x1 = 8; 3¢3 = 9 4¢2 = 8 11. Unique solution. 12. No Solution 13. Inflnitely many solutions. 2 CHAPTER 1. MATRICES AND SYSTEMS OF EQUATIONS 14. No solution. 15. (a) The planes do not intersect; that is, the planes are parallel. (b) The planes intersect in a line or the planes are coincident. 16. The planes intersect in the line x = (1¡t)=2; y = 2; z = t: 17. The planes intersect in the line x = 4¡3t; y = 2t¡1; z = t: 18. Coincident planes. 19. A = ? 2 1 6 4 3 8 ? : 20. C = ? 1 2 7 1 2 2 4 3 ? : 21. Q = 2 4 1 4 ¡3 2 1 1 3 2 1 3 5: 22. x1 +2x2 +7x3 = 12x 1 +2x2 +4x3 = 3 23. 2x1 + x2 = 6 4x1 + 3x2 = 8 ; x1 + 4x2 = ¡3 2x1 + x2 = 1 3x1 + 2x2 = 1 24. A = ? 1 ¡1 1 1 ? ; B = ? 1 ¡1 ¡1 1 1 3 ? : 25. A = ? 1 1 ¡1 2 0 ¡1 ? ; B = ? 1 1 ¡1 2 2 0 ¡1 1 ? : 26. A = 2 4 1 3 ¡1 2 5 1 1 1 1 3 5; B = 2 4 1 3 ¡1 1 2 5 1 5 1 1 1 3 3 5: 27. A = 2 4 1 1 2 3 4 ¡1 ¡1 1 1 3 5; B = 2 4 1 1 2 6 3 4 ¡1 5 ¡1 1 1 2 3 5: 28. A = 2 4 1 1 ¡3 1 2 ¡5 ¡1 ¡3 7 3 5; B = 2 4 1 1 ¡3 ¡1 1 2 ¡5 ¡2 ¡1 ¡3 7 3 3 5: 1.1. INTRODUCTION TO MATRICES AND SYSTEMS OF LINEAR EQUATIONS 3 29. A = 2 4 1 1 1 2 3 1 1 ¡1 3 3 5; B = 2 4 1 1 1 1 2 3 1 2 1 ¡1 3 2 3 5: 30. Elementary operations on equations: E2 ¡2E1 : Reduced system of equations: 2x1 +3x2 = 6¡7x 2 = ¡5 : Elementary row operations: R2 ¡2R1 : Reduced augmented matrix: ? 2 3 6 0 ¡7 ¡5 ? : 31. Elementary operations on equations: E2 ¡E1; E3 +2E1 : Reduced system of equations: x1 +2x2 ¡x3 = 1 ¡x2 +3x3 = 1 5x2 ¡2x3 = 6 : Elementary row operations: R2 ¡R1; R3 +2R1 : Reduced augmented matrix: 2 4 1 2 ¡1 1 0 ¡1 3 1 0 5 ¡2 6 3 5: 32. Elementary operations on equations: E1 $ E2; E3 ¡2E1 : Reduced system of equations: x1 ¡x2 +2x3 = 1 x2 + x3 = 4 3x2 ¡5x3 = 4 : Elementary row operations: R1 $ R2; R3 ¡2R1 : Reduced augmented matrix: 2 4 1 ¡1 2 1 0 1 1 4 0 3 ¡5 4 3 5: 33. Elementary operations on equations: E2 ¡E1; E3 ¡3E1 : Reduced system of equations: x1 + x2 = 9 ¡2x2 = ¡2 ¡2x2 = ¡21 : Elementary row operations: R2 ¡R1; R3 ¡3R1 : Reduced augmented matrix: 2 4 1 1 9 0 ¡2 ¡2 0 ¡2 ¡21 3 5: 4 CHAPTER 1. MATRICES AND SYSTEMS OF EQUATIONS 34. Elementary operations on equations: E2 + E1; E3 +2E1 : Reduced system of equations: x1 + x2 + x3 ¡x4 = 1 2x2 = 4 3x2 +3x3 ¡3x4 = 4 : Elementary row operations: R2 + R1; R3 +2R1 : Reduced augmented matrix: 2 4 1 1 1 ¡1 1 0 2 0 0 4 0 3 3 ¡3 4 3 5: 35. Elementary operations on equations: E2 $ E1; E3 + E1 : Reduced system of equations: x1 +2x2 ¡x3 + x4 = 1 x2 + x3 ¡x4 = 3 3x2 +6x3 = 1 : Elementary row operations: R2 $ R1; R3 + R1 : Reduced augmented matrix: 2 4 1 2 ¡1 1 1 0 1 1 ¡1 3 0 3 6 0 1 3 5: 36. Elementary operations on equations: E2 ¡E1; E3 ¡3E1 : Reduced system of equations: x1 + x2 = 0 ¡2x2 = 0 ¡2x2 = 0 : Elementary row operations: R2 ¡R1; R3 ¡3R1 : Reduced augmented matrix: 2 4 1 1 0 0 ¡2 0 0 ¡2 0 3 5: 37. (b) In each case, the graph of the resulting equation is a line. 38. Now if a11 = 0 we easily obtain the equivalent system a21x1 + a22x2 = b2 a12x2 = b1 Thus we may suppose that a11 6= 0. Then : a11x1 + a12x2 = b1 a21x1 + a22x2 = b2 ? E 2 ¡(a21=a11)E1 =) 1.1. INTRODUCTION TO MATRICES AND SYSTEMS OF LINEAR EQUATIONS 5 a11x1 + a12x2 = b1 ((¡a21=a11)a12 + a22)x2 = (¡a21=a11)b1 + b2 ? a 11E2 =) a11x1 + a12x2 = b1 (a11a22 ¡a12a21)x2 = ¡a21b1 + a11b2 Each of a11 and (a11a22 ¡a12a21) is non-zero. 39. Let A= ? a 11x1 + a12x2 = b1 a21x1 + a22x2 = b2 and let B = ? a 11x1 + a12x2 = b1 ca21x1 + ca22x2 = cb2 Suppose that x1 = s1; x2 = s2 is a solution to A . Then a11s1 + a12s2 = b1; and a21s1 + a22s2 = b2: But this means that ca21s1 + ca22s2 = cb2 and so x1 = s1; x2 = s2 is also a solutiontoB. Nowsupposethat x1 = t1; x2 = t2 isasolutiontoB . Then a11t1+a12t2 = b1 and ca21t1 + ca22t2 = cb2 . Since c 6= 0 , a21x1 + a22x2 = b2 . 40. Let A= ? a 11x1 + a12x2 = b1 a21x1 + a22x2 = b2 and let B = ? a 11x1 + a12x2 = b1 (a21 + ca11)x1 +(a22 + ca12)x2 = b2 + cb1 : Let x1 = s1 and x2 = s2 beasolutiontoA. Then a11s1+a12s2 = b1 and a21s1+a22s2 = b2 so a11s1+a12s2 = b1 and(a21+ca11)s1+(a22+ca12)s2 = b2+cb1 asrequired. Nowif x1 = t1 and x2 = t2 isasolutiontoB then a11t1+a12t2 = b1 and(a21+ca11)t1+(a22+ca12)t2 = b2+cb1, so a11t1 + a12t2 = b1 and a21t1 + a12t2 = b2 as required. 41. The proof is very similar to that of 45 and 46. 42. By adding the two equations we obtain: 2x21 ¡ 2x1 = 4. Then x1 = 2 or x1 = ¡1 and substituting these values in the second equation we flnd that there are three solutions: x1 =¡1; x2 = 0 ; x1 = 2; x2 =p3; ; x1 = 2; x2 =¡p3. 6 CHAPTER 1. MATRICES AND SYSTEMS OF EQUATIONS 1.2 Echelon Form and Gauss-Jordan Elimination 1. The matrix is in echelon form. The row operation R2 ¡ 2R1 transforms the matrix to reduced echelon form ? 1 0 0 1 ? : 2. Echelon form. R2 ¡2R1 yields reduced row echelon form ? 1 0 ¡7 0 1 3 ? : 3. Not in echelon form. (1=2)R1; R2 ¡4R1; (¡1=5)R2 yields echelon form ? 1 3=2 1=2 0 1 2=5 ? : 4. Not in echelon form. R1 $ R2 yields echelon form ? 1 2 3 0 1 1 ? : 5. Not in echelon form. R1 $ R2; (1=2)R1; (1=2)R2 yields the echelon form ? 1 0 1=2 2 0 0 1 3=2 ? : 6. Not in echelon form. (1=2)R1 yields the echelon form ? 1 0 3=2 1=2 0 0 1 2 ? : 7. Not in echelon form. R2 ¡ 4R3; R1 ¡ 2R3; R1 ¡ 3R2 yields the reduced echelon form2 4 1 0 0 5 0 1 0 ¡2 0 0 1 1 3 5: 8. Not in echelon form. (1=2)R1; (¡1=3)R3 yields the echelon form 2 4 1 ¡1=2 3=2 0 1 1 0 0 1 3 5: 9. Not in echelon form. (1=2)R2 yields the echelon form 2 4 1 2 ¡1 ¡2 0 1 ¡1 ¡3=2 0 0 0 1 3 5: 10. Not in echelon form ¡R1; (1=2)R2 yields the echelon form 2 4 1 ¡4 3 ¡4 ¡6 0 1 1=2 ¡3=2 ¡3=2 0 0 0 1 2 3 5: 11. x1 = 0; x2 = 0: 12. The system is inconsistent. 13. x1 =¡2+5x3; x2 = 1¡3x3; x3 is arbitrary. 14. x1 = 1¡2x3; x2 = 0: 1.2. ECHELON FORM AND GAUSS-JORDAN ELIMINATION 7 15. x1 = 0; x2 = 0; x3 = 0: 16. x1 = 0; x2 = 0; x3 = 0: 17. x1 = x3 = x4 = 0; x2 is arbitrary. 18. The system is inconsistent. 19. The system is inconsistent. 20. x1 = 3x4 ¡5x5 ¡2; x2 = x4 + x5 ¡2; x3 =¡2x4 ¡x5 +2; x4 and x5 are arbitrary. 21. x1 =¡1¡(1=2)x2 +(1=2)x4; x3 = 1¡x4; x2 and x4 arbitrary, x5 = 0. 22. x1 = (5+3x2)=2; x2 arbitrary. 23. The system is inconsistent. 24. x1 = x3; x2 =¡3+2x3; x3 arbitrary. 25. x1 = 2¡x2; x2 arbitrary. 26. x1 = 10+ x2; x2 arbitrary, x3 =¡6: 27. x1 = 2¡x2 + x3; x2 and x3 arbitrary. 28. x1 = 2x3; x2 = 1; x3 arbitrary. 29. x1 = 3¡2x3; x2 =¡2+3x3; x3 arbitrary. 30. x1 =¡3x4 ¡6x5; x2 = 1+3x4 +7x5; x3 =¡2x4 ¡5x5; x4 and x5 arbitrary. 31. x1 = 3¡(7x4¡16x5)=2; x2 = (x4 +2x5)=2; x3 =¡2+(5x4¡12x5)=2; x4 and x5 arbitrary. 32. x1 = 2; x2 =¡1: 33. The system is inconsistent. 34. x1 = 1¡2x2; x2 arbitrary. 35. The system is inconsistent. 36. x1 +2x2 = ¡3ax 1 ¡2x2 = 5 ? E 1 + E2 =) x 1 +2x2 = ¡3 (a +1)x1 = 2 Hence if a =¡1 there is no solution. 37. x1 +3x2 = 42x 1 +6x2 = a ? E 2 ¡2E1 =) x 1 +3x2 = 4 0 = a¡8 Thus, if a 6= 8 there is no solution. 8 CHAPTER 1. MATRICES AND SYSTEMS OF EQUATIONS 38. 2x1 +4x2 = a3x 1 +6x2 = 5 ? E 2 ¡(3=2)E1 =) 2x 1 +4x2 = a 0 = 5¡(3=2)a Thus, if a 6= 10=3 there is no solution. 39. 3x1 + ax2 = 3ax 1 +3x2 = 5 ? E 2 ¡(a=3)E1 =) 3x 1 + ax2 = 3 (a2=3¡3)x2 = 5¡a Thus, if a =§3 there is no solution. 40. x1 + ax2 = 6ax 1 +2ax2 = 4 ? E 2 ¡aE1 =) x 1 + ax2 = 6 (2a¡a2)x2 = 4¡6a 41. cosfi = 1=2 and sinfl = 1=2, so fi = ?=3 or fi = 5?=3 and fl = ?=6 or fl = 5?=6. 42. cos2 fi = 3=4 and sin2 fl = 1=2. The choices for fi are ?=6, 5?=6, 7?=6, and 11?=6. The choices for fl are ?=4, 3?=4, 5?=4, and 7?=4. 43. x1 = 1¡2x3; x2 = 2+ x3; x3 arbitrary. (a) x3 = 1=2: (b) In order for x1 ? 0; x2 ? 0; we must have ¡2 ? x3 ? 1=2; for a given x1 and x2; y = ¡6¡7x3; so the minimum value is y = 8 at x3 =¡2: (c) The minimum value is 20. 44. ? 1 d c b ? ? R 2 ¡cR1 =) ? 1 d 0 b¡cd ? 8< : R1 ¡(d=(b¡cd))R2 ( recall b¡cd 6= 0 =) 9 = ; ? 1 0 0 b¡cd ? ? 1=(b¡cd)R 2 =) ? 1 0 0 1 ? : 45. ? 1 x x 0 1 x ? ; ? 1 x x 0 0 1 ? ; ? 1 x x 0 0 0 ? ; ? 0 1 x 0 0 1 ? ? 0 1 x 0 0 0 ? ; ? 0 0 1 0 0 0 ? ; ? 0 0 0 0 0 0 ? : 46. (a) 2 4 1 x 0 1 0 0 3 5 ; 2 4 1 x 0 0 0 0 3 5 ; 2 4 0 1 0 0 0 0 3 5 ; 2 4 0 0 0 0 0 0 3 5: (b) 2 4 1 x x 0 1 x 0 0 1 3 5 ; 2 4 1 x x 0 1 x 0 0 0 3 5 ; 2 4 1 x x 0 0 1 0 0 0 3 5 ; 2 4 1 x x 0 0 0 0 0 0 3 5; 2 4 0 1 x 0 0 1 0 0 0 3 5; 2 4 0 1 x 0 0 0 0 0 0 3 5; 2 4 0 0 1 0 0 0 0 0 0 3 5; 2 4 0 0 0 0 0 0 0 0 0 3 5: 1.2. ECHELON FORM AND GAUSS-JORDAN ELIMINATION 9 (c) 2 4 1 x x x 0 1 x x 0 0 1 x 3 5; 2 4 1 x x x 0 1 x x 0 0 0 1 3 5; 2 4 1 x x x 0 1 x x 0 0 0 0 3 5; 2 4 1 x x x 0 0 1 x 0 0 0 1 3 5; 2 4 1 x x x 0 0 1 x 0 0 0 0 3 5; 2 4 1 x x x 0 0 0 1 0 0 0 0 3 5: 2 4 1 x x x 0 0 0 0 0 0 0 0 3 5; 2 4 0 1 x x 0 0 1 x 0 0 0 1 3 5; 2 4 0 1 x x 0 0 0 1 0 0 0 0 3 5; 2 4 0 1 x x 0 0 0 0 0 0 0 0 3 5; 2 4 0 0 1 x 0 0 0 1 0 0 0 0 3 5; 2 4 0 0 1 x 0 0 0 0 0 0 0 0 3 5; 2 4 0 0 0 1 0 0 0 0 0 0 0 0 3 5; 2 4 0 0 0 0 0 0 0 0 0 0 0 0 3 5: 47. ? 1 2 2 3 ? ? 2R 2 =) ? 1 2 4 6 ? ? R 2 ¡R1 =) ? 1 2 3 4 ? : 48. ? 1 4 3 7 ? ? R 2 ¡3R1 =) ? 1 4 0 ¡5 ? ? R 1 +(2=5)R2 =) ? 1 2 0 ¡5 ? ? (3=5)R 2 =) ? 1 2 0 ¡3 ? ? R 2 +2R1 =) ? 1 2 2 1 ? : 49. 100x1 +10x2 + x3 = 15(x1 + x2 + x3) 100x3 +10x2 + x1 = 100x1 +10x2 + x3 +396 x3 = x1 + x2 +1 x1 = 1, x2 = 3, and x3 = 5, so N = 135. 50. a¡b + c = 6 a + b + c = 4 4a +2b + c = 9 a = 2, b =¡1, c = 3. So y = 2x2 ¡x +3. 51. Let x1, x2, x3 be the amounts initially held by players one, two and three, respectively. Also assume that player one loses the flrst game, player two loses the second game, and player three loses the third game. Then after three games, the amount of money held by each player is given by the following table 10 CHAPTER 1. MATRICES AND SYSTEMS OF EQUATIONS Player Amount of money 1 4x1 ¡4x2 ¡4x3 = 24 2 ¡2x1 +6x2 ¡2x3 = 24 3 ¡x1 ¡x2 +7x3 = 24 Solving yields x1 = 39, x2 = 21, and x3 = 12. 52. The resulting system of equations is x1 + x2 + x3 = 34 x1 + x2 = 7 x2 + x3 = 22 The solution is x1 = 12, x2 =¡5, x3 = 27. 53. If x1 is the number of adults, x2 the number of students, and x3 the number of children, then x1 +x2 +x3 = 79, 6x1 +3x2 +(1=2)x3 = 207, and for j = 1;2;3, xj is an integer such that 0? xj ?79. Following is a list of possiblities Number of Adults 0 5 10 15 20 25 30 Number of Students 67 56 45 34 23 12 1 Number of Children 12 18 24 30 36 42 48 54. The resulting system of equations is a + b + c + d = 5 b +2c +3d = 5 a +2b +4c +8d = 17 b +4c +12d = 21: The solution is a = 3, b = 1, c =¡1, d = 2. So p(x) = 3+ x¡x2 +2x3. 55. By (7), 1+2+3+¢¢¢+ n = a1n + a2n2. Setting n = 1 and n = 2 gives a1 + a2 = 1 2a1 +4a2 = 3 The solution is a1 = a2 = 1=2, so 1+2+3+ : : : + n = n(n +1)=2. 56. By (7), 12 +22 +32 +¢¢¢+ n2 = a1n + a2n2 + a3n3. Setting n = 1, n = 2, n = 3, gives a1 + a2 + a3 = 1 2a1 +4a2 +8a3 = 5 3a1 +9a2 +27a3 = 14 Thesolutionis a1 = 1=6, a2 = 1=2and a3 = 1=3,so12+22+32+: : :+n2 = n(n+1)(2n+1)=6. 1.3. CONSISTENT SYSTEMS OF LINEAR EQUATIONS 11 57. The system of equations obtained from (7) is a1 + a2 + a3 + a4 + a5 = 1 2a1 +4a2 +8a3 +16a4 +32a5 = 17 3a1 +9a2 +27a3 +81a4 +242a5 = 98 4a1 +16a2 +64a3 +256a4 +1024a5 = 354 5a1 +25a2 +125a3 +625a4 +3125a5 = 979 The solution is a1 = ¡1=30, a2 = 0, a3 = 1=3, a4 = 1=2, a5 = 1=5. Therefore, 14 +24 + 34 +¢¢¢+ n4 = n(n +1)(2n +1)(3n2 +3n¡1)=30. 58. 15 +25 +35 +¢¢¢+ n5 = n2(n +1)2(2n2 +2n¡1)=12. 1.3 Consistent Systems of Linear Equations 1. The augmented matrix reduces to 2 66 4 1 1 0 5=6 0 0 1 2=3 0 0 0 0 0 0 0 0 3 77 5: n = 3, r = 2, x2 is independent. 2. The augmented matrix reduces to 2 4 1 0 ¡3=2 0 1 2 0 0 0 3 5: n = 2, r = 2. 3. The augmented matrix reduces to 2 4 1 0 4 0 13=2 0 1 ¡1 0 ¡3=2 0 0 0 1 1=2 3 5: n = 4, r = 3, x3 is independent. 4. The augmented matrix reduces to 2 66 4 1 2 3 0 1=3 0 0 0 1 1=3 0 0 0 0 0 0 0 0 0 0 3 77 5: n = 4, r = 2, x2 and x3 are independent. 5. n = 2 and r ? 2 so r = 0; n¡r = 2; r = 1; n¡r = 1; r = 2; n¡r = 0: There could be a unique solution. 6. n = 4 and r ? 3 so r = 0, n¡r = 4; r = 1, n¡r = 3; r = 2, n¡r = 2; r = 3, n¡r = 1. By the corollary to Theorem 3, there are inflnitely many solutions. 7. Inflnitely many solutions. 12 CHAPTER 1. MATRICES AND SYSTEMS OF EQUATIONS 8. Inflnitely many solutions. 9. Inflnitely many solutions, a unique solution or no solution. 10. Inflnitely many solutions, a unique solution, or no solution. 11. A unique solution or inflnitely many solutions. 12. Inflnitely many solutions or a unique solution. 13. Inflnitely many solutions. 14. Inflnitely many solutions. 15. Inflnitely many solutions or a unique solution. 16. Inflnitely many solutions or a unique solution. 17. Inflnitely many solutions. 18. Inflnitely many solutions. 19. There are nontrivial solutions. 20. There are nontrivial solutions. 21. There is only the trivial solution. 22. There is only the trivial solution. 23. If a =¡1 then when we reduce the augmented matrix we obtain a row of zeroes and hence inflnitely many nontrivial solutions. 24. (a) Reduced row echelon form of the augmented matrix is 2 4 1 0 2 ¡2b1 +3b2 0 1 ¡1 b1 ¡b2 0 0 0 b3 ¡b1 ¡2b2 3 5. Hence, if b3 ¡ b1 ¡2b2 6= 0 then the system is inconsistent. Therefore, the system of equations is consistent if and only if b3 ¡b1 ¡2b2 = 0. (b) (i) The system is consistent. For example, a solution is x1 = ¡1, x2 = 1 and x3 = 1. (ii)Thesystemisinconsistentbypart(a). (iii)Thesystemisconsistent. Forexample, a solution is x1 = 1, x2 = 0 and x3 = 1. 25. (a) B = 2 66 4 ? x x 0 ? x 0 0 ? 0 0 0 3 77 5: 1.3. CONSISTENT SYSTEMS OF LINEAR EQUATIONS 13 (b) In the third row of the matrix of 25(a) for B, we need 0¢x1+0¢x2 =? and, in general, this can?t be. 26. The resulting system of equations is 3a + b + c = 07a +2b + c = 0 : The general solution is a = c, b = -4c. Thus x¡4y +1 = 0 is an equation for the line. 27. The resulting system of equations is 2a +8b + c = 04a + b + c = 0 : The general solution is a = (-7/30)c, b = (-1/15)c. Thus ¡7x¡2y +30 = 0 is an equation for the line. 28. The resulting system of equations is 16a¡4d + f = 0 4a +4b +4c¡2d¡2e + f = 0 9c +3e + f = 0 a + b + c + d + e + f = 0 16a +4d + f = 0 : The general solution is: a = (-1/16)f, b = (-71/144)f, c = (1/18)f, d = 0, e = (-1/2)f. An equation is 9x2 +71xy ¡8y2 +72y ¡144 = 0 : 29. The resulting system of equations is 16a¡4b + c¡4d + e + f = 0 a¡2b +4c¡d +2e + f = 0 9a +6b +4c +3d +2e + f = 0 25a +5b + c +5d + e + f = 0 49a¡7b + c +7d¡e + f = 0 : The general solution is: a = (-3/113)f, b = (3/113)f, c = (1/113)f, d = 0, e = (-54/113)f. An equation is ¡3x2 +3xy + y2 ¡54y +113 = 0 : 30. Using equation (4), the given points result in a system of 9 equations in 10 unknowns, with the solution: a = (¡15=16)j; b = (¡1=16)j; c = (7=8)j; d = (15=16)j; e = (¡15=16)j; f = (1=8)j; g = (15=8)j; h = (¡15=16)j; i = (¡15=8)j: An equation is: ¡15x2 ¡y2 +14z2 +15xy ¡15xz +2yz +30x¡15y ¡15z +16 = 0 : 31. Omitted 32. The resulting system of equations is: 2a + b + c + d = 0 5a +2b + c + d = 0 13a +3b +2c + d = 0 : 14 CHAPTER 1. MATRICES AND SYSTEMS OF EQUATIONS The general solution is: a = (1=6)d; b = (¡1=2)d; c = (¡5=6)d: Thus, x2 + y2 ¡3x¡5y +6 = 0 ; is an equation for the circle. 33. The resulting system of equations is: 25a +4b +3c + d = 0 5a + b +2c + d = 0 4a +2b + d = 0 : The general solution is: a = (7=50)d; b = (¡39=50)d; c = (¡23=50)d: Thus, 7x2 +7y2 ¡39x¡23y +50 = 0 ; is an equation for the circle. 1.4 Applications 1. (a) x1 + x4 = 1200 x1 + x2 = 1000 x3 + x4 = 600 x2 + x4 = 400 The solution is x1 = 1200¡x4, x2 =¡200+ x4, x3 = 600¡x4. (b) x1 = 1100, x2 =¡100, x3 = 500. (c) 200? x4 ?600 so 600? x1 ?1000 2. (a) x1 = 1200 x1 + x2 = 1000 x3 + x4 = 1000 x2 + x3 = 800 The solution is x1 = 1200¡x4, x2 =¡200+ x4, x3 = 1000¡x4: (b) x1 = 1100, x2 =¡100, x3 = 900. (c) 200? x4 ?1000 so 200? x1 ?1000. 3. x2 = 800, x3 = 400, x4 = 200. 4. x2 = 400, x3 = 700, x4 = 300, x5 = 500, x6 = 100. 5. 4I1 + 3I2 = 2, 3I2 + 4I3 = 4, and I1 + I3 = I2. Therefore, I1 = 1=20, I2 = 3=5, and I3 = 11=20. 6. I1+I2 = 7, I1+2I3 = 3, and I1+I3 = I2. Therefore, I1 = 18=5, I2 = 17=5, and I3 =¡1=5. 7. 5=7;20=7;15=7 8. 7=4;15=8;¡13=8;1=8;1=8;27=8. 1.5. MATRIX OPERATIONS 15 9. (a) x1 ¡x4 = a1 ¡a2 x1 ¡x2 = ¡b1 + b2 ¡x3 + x4 = d1 ¡d2 x2 ¡x3 = ¡c1 + c2 10. Let I1; I2; : : : ; I5 be the currents owing through R1; R2; : : : ; R5, respectively. If I5 = 0 then I1 = I2, I3 = I4, I1R1 ¡I3R3 = 0, and I2R2 ¡I4R4 = 0. It follows that either all currents are zero or R1R4 = R2R3. 1.5 Matrix Operations 1. (a) ? 2 0 2 6 ? ; (b) ? 0 4 2 4 ? ; (c) ? 0 ¡6 6 18 ? ; (d) ? ¡6 8 4 6 ? : 2. (a) ? ¡2 2 2 4 ? ; (b) ? 6 3 3 9 ? ; (c) ? ¡2 7 3 5 ? ; (d) ? ¡2 3 1 1 ? : 3. ? ¡2 ¡2 0 0 ? : 4. ? ¡2 1 0 ¡1 ? : 5. ? ¡1 ¡1 0 0 ? : 6. ? 2 4 ¡2 ¡6 ? : 7. (a) ? 3 ¡3 ? ; (b) ? 3 4 ? ; (c) ? 0 0 ? : 8. (a) ? 3 1 ? ; (b) ? ¡11 18 ? ; (c) ? ¡5 24 ? : 9. (a) ? 2 1 ? ; (b) ? 0 1 ? ; (c) ? 17 14 ? : 10. (a) ? ¡4 13 ? ; (b) ? ¡15 0 ? ; (c) ? 3 ¡6 ? : 11. (a) ? 2 3 ? ; (b) ? 20 16 ? : 16 CHAPTER 1. MATRICES AND SYSTEMS OF EQUATIONS 12. (a) ? 0 2 ? ; (b) ? ¡33 ¡17 ? : 13. a1 = 11=3; a2 =¡(4=3) . 14. a1 = 0; a2 =¡2. 15. a1 =¡2; a2 = 0. 16. a1 = 4=11; a2 = 14=11 . 17. The equation has no solution. 18. The equation has no solution. 19. a1 = 4; a2 =¡(3=2): 20. a1 = 9=11; a2 =¡(17=11) . 21. w1 = ? 0 1 ? ; w2 = ? 1 3 ? ; AB = ? 1 1 3 8 ? ; (AB)r= ? 1 3 ? 22. w1 = ? ¡13 ¡1 ? ; w2 = ? ¡27 ¡16 ? ; Q = ? ¡3 7 1 6 ? ; Qs= ? ¡27 ¡16 ? : 23. w1 = ? ¡2 1 ? ; w2 = ? ¡1 1 ? ; w3 = ? ¡1 2 ? ; Q = ? ¡1 4 2 17 ? ; Q r= ? ¡1 2 ? : 24. w1 = ? 2 1 ? ; w2 = ? ¡1 3 ? ; w3 = ? ¡3 8 ? : Q = ? ¡3 ¡4 8 19 ? ; Q r= ? ¡3 8 ? : 25. ? ¡4 6 2 12 ? : 26. ? 3 ¡1 15 30 ? : 27. ? 4 12 4 10 ? : 28. ? 0 0 0 0 ? : 1.5. MATRIX OPERATIONS 17 29. ? 0 0 0 0 ? : 30. ? 0 0 0 0 ? : 31. AB = ? 5 16 5 18 ? ; BA = ? 4 11 6 19 ? : 32. 2 66 4 50 11 16 10 3 ¡2 28 4 3 77 5: 33. Au= ? 11 13 ? ; vA = [8;22]: 34. uv= ? 2 4 6 12 ? ; vu= 14: 35. vBu = 66 . 36. Bu= ? 7 13 ? : 37. CA = 2 66 4 5 10 8 12 15 20 8 17 3 77 5: 38. CB = 2 66 4 3 8 4 8 7 12 5 14 3 77 5: 39. C(B)u) = 2 66 4 27 28 43 47 3 77 5: 40. (AB)u= ? 53 59 ? ; A(Bu) = ? 53 59 ? : 41. (BA)u= ? 37 63 ? ; B(Au) = ? 37 63 ? : 18 CHAPTER 1. MATRICES AND SYSTEMS OF EQUATIONS 42. 2 66 4 x1 x2 x3 x4 3 77 5 = 2 66 4 x3 +2x4 ¡2x3 ¡3x4 x3 x4 3 77 5 = x3 2 66 4 1 ¡2 1 0 3 77 5 + x4 2 66 4 2 ¡3 0 1 3 77 5: 43. 2 4 x1 x2 x3 3 5 = 2 4 ¡2+ x3 3¡2x3 x3 3 5 = 2 4 ¡2 3 0 3 5 + x3 2 4 1 ¡2 1 3 5: 44 2 66 4 x1 x2 x3 x4 3 77 5 = 2 66 4 ¡1+ x3 1¡2x3 x3 1 3 77 5 = 2 66 4 ¡1 1 0 1 3 77 5 + x3 2 66 4 1 ¡2 1 0 3 77 5: 45. 2 66 66 4 x1 x2 x3 x4 x5 3 77 77 5 = 2 66 66 4 x3 + x5 ¡2x3 ¡x5 x3 ¡x5 x5 3 77 77 5 = x3 2 66 66 4 1 ¡2 1 0 0 3 77 77 5 + x5 2 66 66 4 1 ¡1 0 ¡1 1 3 77 77 5 : 46. 2 66 66 4 x1 x2 x3 x4 x5 3 77 77 5 = 2 66 66 4 1+ x3 +2x4 +3x5 ¡2x3 ¡3x4 ¡4x5 x3 x4 x5 3 77 77 5 = 2 66 66 4 1 0 0 0 0 3 77 77 5 + x3 2 66 66 4 1 ¡2 1 0 0 3 77 77 5 + x4 2 66 66 4 2 ¡3 0 1 0 3 77 77 5 + x5 2 66 66 4 3 ¡4 0 0 1 3 77 77 5 : 47. 2 66 66 4 x1 x2 x3 x4 x5 3 77 77 5 = 2 66 66 4 x3 +2x4 +3x5 ¡2x3 ¡3x4 ¡4x5 x3 x4 x5 3 77 77 5 = x3 2 66 66 4 1 ¡2 1 0 0 3 77 77 5 + x4 2 66 66 4 2 ¡3 0 1 0 3 77 77 5 + x5 2 66 66 4 3 ¡4 0 0 1 3 77 77 5 : 48. 2 66 66 66 4 x1 x2 x3 x4 x5 x6 3 77 77 77 5 = 2 66 66 4 x3 + x5 +2x6 ¡2x3 ¡x5 ¡2x6 x3 ¡x5 ¡x6 x6 3 77 77 5 = x3 2 66 66 66 4 1 ¡2 1 0 0 0 3 77 77 77 5 + x5 2 66 66 66 4 1 ¡1 0 ¡1 1 0 3 77 77 77 5 + x6 2 66 66 66 4 2 ¡2 0 ¡1 0 1 3 77 77 77 5 : 1.5. MATRIX OPERATIONS 19 49. 2 66 66 4 x1 x2 x3 x4 x5 3 77 77 5 = 2 66 66 4 x2 +2x4 x2 ¡2x4 x4 0 3 77 77 5 = x2 2 66 66 4 1 1 0 0 0 3 77 77 5 + x4 2 66 66 4 2 0 ¡2 1 0 3 77 77 5 : 50. A(Bu) has 8 multiplications while (AB)u has 12 multiplications. 51. C(A(B u)) has 12 multiplications, (CA)(Bu) has 16 multiplications, [C(AB)](u) has 20 multiplications, and C[(AB)u] has 16 multiplications. 52. (a) A1 = ? 2 1 ? ; A2 = ? 3 4 ? ; D1 = 2 66 4 2 2 1 1 3 77 5; D2 = 2 66 4 1 0 ¡1 3 3 77 5; D3 = 2 66 4 3 0 1 1 3 77 5; D4 = 2 66 4 6 4 ¡1 2 3 77 5: (b) A1 is in R2;D1 is in R4: (c) AB1 = ? 5 5 ? ; AB2 = ? 16 18 ? ; AB = ? 5 16 5 18 ? : 53. (a) AB is a 2 x 4 matrix, BA is not deflned. (b) AB is not deflned, BA is not deflned. (c) AB is not deflned. BA is a 6 x 7 matrix. (d) AB is a 2 x 2 matrix, BA is a 3 x 3 matrix. (e) AB is a 3 x 1 matrix, BA is not deflned. (f) A(BC) and (AB)C are 2 x 4 matrices. (g) AB is a 4 x 4 matrix. BA is a 1 x 1 matrix. 54. (AB)(CD) is a 2 x 2 matrix, A(B(CD)) and ((AB)C)D are 2 x 2 matrices. 55. A2 = AA provided A is a square matrix. 56. Since b 6= 0 is arbitrary in B, the equation has inflnitely many solutions. 57. (a) Px = 2 4 135;000 120;000 45;000 3 5 is the state vector after one year and P 2x = 2 4 126;000 132;000 42;000 3 5 is the state vectore after two years. 20 CHAPTER 1. MATRICES AND SYSTEMS OF EQUATIONS (b) Pnx 58. (a) Setting AB = BA yields the system of equations 3b ¡2c = 0, 2a +3b ¡2d = 0, and 3a+3c¡3d = 0. The solution is a =¡c+d and b = 2c=3, so B = ? ¡c + d 2c=3 c d ? . (b) B = ? ¡2 2 3 1 ? and C = ? 0 0 1 1 ? are possible choices for B and C. 59. Let A be an (m £ n) matrix and B be a (p £ r) matrix. Since AB is deflned, n = p and AB is an (m £ r) matrix. But AB is a square matrix, so m = r. Thus, B is an (n £ m) matrix, so BA is deflned and is an (n£n) matrix. 60. Let B = [B1;B2; : : : ;Bs]. Then AB = [AB1; AB2; : : : ; ABs]. (a) If Bj = then the jth column of AB is ABj = . (b) If Bi = Bj then ABi = ABj. 61. (a) (i) A = ? 2 ¡1 1 1 ? ; x= ? x 1 x2 ? ; b= ? 3 3 ? : (ii) A = 2 4 1 ¡3 1 1 ¡2 1 0 1 ¡1 3 5; x= 2 4 x1 x2 x3 3 5; b= 2 4 1 2 ¡1 3 5: (b) (i) x1 ? 2 1 ? + x2 ? ¡1 1 ? = ? 3 3 ? : (ii) x1 2 4 1 1 0 3 5+ x2 2 4 ¡3 ¡2 1 3 5+ x3 2 4 1 1 ¡1 3 5 = 2 4 1 2 ¡1 3 5: (c) (i) x1 = 2; x2 = 1; 2A1 +A2 = b: (ii) x1 = 2; x2 = 1; x3 = 2; 2A1 +A2 +2A3 = b: 62. ? 1 1 2 1 2 3 ? ? R 2 ¡R1 =) ? 1 1 2 0 1 1 ? : Thus x1 = 1 and x2 = 1 . 63. (a) We solve each of the systems (i) Ax= ? 1 0 ? ; (ii) ) Ax= ? 0 1 ? : (i) x= ? 2 ¡1 ? ; (ii)x= ? ¡1 1 ? : 1.6. ALGEBRAIC PROPERTIES OF MATRIX OPERATIONS 21 (b) B = ? 2 ¡1 ¡1 1 ? andAB = I = BA: 64. The ith component of Ax is the Pnj=1 aijxj: Now the ith components of x1A1 ; x2A2 ; : : : ; xnAn are x1ai1; x2ai2 : : : ; xnain; respectively. Thus the ith component of x1A1 +x2A2 +¢¢¢+ xnAn is Pnj=1 aijxj as required. 65. (a) B = ? ¡1 6 1 0 ? : (b) No B exists. (c) B = ? 2 2 ¡1 ¡1 ? : 66. If A = (aij) and B = (bij) then AB = 2 4 a11b11 a11b12 + a12b22 a11b13 + a12b23 + a13b33 0 a22b22 a22b23 + a23b33 0 0 a33b33 3 5: 67. Let A = (aij) and B = (bij) be upper triangular (n x n) matrices. Then the ijth entry of AB equals Pnk=1 aikbkj: Suppose i > j: If k > j then bkj = 0: If j ? k then i > k so aik = 0: Thus the ijth component of AB equals zero. 68. 2 66 66 4 x1 x2 x3 x4 x5 3 77 77 5 = 2 66 66 4 4¡3x2 ¡22x5 x2 6¡9x5 5¡x5 x5 3 77 77 5 = 2 66 66 4 4 0 6 5 0 3 77 77 5 + x2 2 66 66 4 ¡3 1 0 0 0 3 77 77 5 + x5 2 66 66 4 ¡22 0 ¡9 ¡1 1 3 77 77 5 : 69. 2 66 66 4 x1 x2 x3 x4 x5 3 77 77 5 = 2 66 66 4 5+2x4 ¡3x5 4¡3x4 ¡2x5 2¡x4 ¡x5 x4 x5 3 77 77 5 = 2 66 66 4 5 4 2 0 0 3 77 77 5 + x4 2 66 66 4 2 ¡3 ¡1 1 0 3 77 77 5 + x5 2 66 66 4 ¡3 ¡2 ¡1 0 1 3 77 77 5 : 70. 2 66 66 66 4 x1 x2 x3 x4 x5 x6 3 77 77 77 5 = 2 66 66 66 4 2¡x2 ¡2x6 x2 3¡x6 2¡2x6 3¡5x6 x6 3 77 77 77 5 = 2 66 66 66 4 2 0 3 2 3 0 3 77 77 77 5 + x2 2 66 66 66 4 ¡1 1 0 0 0 0 3 77 77 77 5 + x6 2 66 66 66 4 ¡2 0 ¡1 ¡2 ¡5 1 3 77 77 77 5 : 1.6 Algebraic Properties of Matrix Operations 1. DE = ? 8 15 11 18 ? ; EF = ? 9 9 5 5 ? ; (DE)F = D(EF) = 22 CHAPTER 1. MATRICES AND SYSTEMS OF EQUATIONS ? 23 23 29 29 ? : 2. FE = ? 5 9 5 9 ? ; ED = ? 12 27 7 14 ? ; F(ED) = (FE)D = ? 19 41 19 41 ? : 3. DE = ? 8 15 11 18 ? ; ED = ? 12 27 7 14 ? : 4. EF = ? 9 9 5 5 ? ; FE = ? 5 9 5 9 ? : 5. Fu= ? 0 0 ? ; Fv= ? 0 0 ? : 6. 3Fu= 3 ? 0 0 ? = ? 0 0 ? ; 7Fv= 7 ? 0 0 ? = ? 0 0 ? : 7. AT = ? 3 4 2 1 7 6 ? : 8. DT = ? 2 1 1 4 ? : 9. ETF = ? 5 5 9 9 ? : 10. ATC = ? 34 15 28 20 56 32 37 35 ? : 11. (F v)T = £ 0 0 ?: 12. (EF)v= ? 0 0 ? : 13. ¡6: 14. 0: 15. 36: 16. 0: 17. 2: 1.6. ALGEBRAIC PROPERTIES OF MATRIX OPERATIONS 23 18. 18: 19. p2: 20. 3p10: 21. p29: 22. 4p2: 23. 0: 24. 0: 25. 2p5 26. Let A = ? 0 1 0 0 ? and let B = ? 1 0 0 0 ? : Then (A¡B)(A + B) = ? ¡1 ¡1 0 0 ? and A2 ¡B2 = ? ¡1 0 0 0 ? : 27. Let A = ? 1 0 0 0 ? and let B = ? 1 0 0 1 ? : Then A2 = AB and A 6= B: 28. The argument depends upon the "fact" that if the product of two matrices is O then one of the factors must be O: This is not true. Let A = ? 0 1 0 0 ? : and B = ? 1 0 0 0 ? : Then A2 =O = AB and neither of A or B is O: 29. D and F are symmetric. 30. Let A = ? 1 1 1 0 ? and let B = ? 0 1 1 1 ? : Then each of A and B are symmetric and AB = ? 1 2 0 1 ? is not symmetric. 31. If each of A and B are symmetric, then a necessary and su?cient condition that AB be symmetric is that AB = BA: 32. xT Gx= £ x1 x2 ? ? 2 1 1 1 ?? x 1 x2 ? = x21 +(x1 + x2)2: This term is always greater than zero whenever x1 and x2 are not simultaneously zero. 33. xT Dx= [x2; x2] ? 2 2 1 4 ?? x 1 x2 ? = x21 +3x22 +(x1 +x2)2: This term is always greater than zero whenever x1 and x2 are not simultaneously zero. 24 CHAPTER 1. MATRICES AND SYSTEMS OF EQUATIONS 34. xT Fx= [x1; x2] ? 1 1 1 1 ?? x 1 x2 ? = (x1 +x2)2: Then xT Fx= 0 if and only if x1 +x2 = 0: 35. ? ¡3 3 3 ¡3 ? : 36. ? 0 0 0 0 ? : 37. ? ¡27 ¡9 27 9 ? : 38. ? 9 3 ¡9 ¡3 ? : 39. 2 4 ¡12 18 24 18 ¡27 ¡36 24 36 ¡48 3 5: 40. 2 4 ¡12 18 24 12 ¡18 ¡24 24 ¡36 ¡48 3 5: 41. (a) xTa = 6 means that x1 +2x2 = 6 and xTb= 2 means that 3x1 +4x2 = 2: Thus x1 =¡10; x2 = 8 and x= ? ¡10 8 ? : (b) xT(a+b) = 12 and x Ta = 2 yields 4x1 + 6x2 = 12 and x1 + 2x2 = 2: Thus x1 = 6; x2 =¡2 and x= ? 6 ¡2 ? : 42. (a) ? 1 3 0 1 ? : (b) ? 5 11 ¡3 ¡7 ? : (c) BC1 = ? 14 18 ? ; BT1 C = £ 6 8 ?; (BC1)TC2 = 132; kCB2k= 2p337: 43. (b) A5u = 25u = 32u = 2 4 32 96 64 3 5 (c) Anu = 2nu. Property 3 is required. For example, A2u = A(Au) = A(2u) = 2(Au) = 2(2u) = 22u. 1.6. ALGEBRAIC PROPERTIES OF MATRIX OPERATIONS 25 44. (a) By property (3) there exists an (m£n) matrix O such that A +O = A. (b) By property (4) there exists an (m £ n) matrix D such that C + D = O. Thus, A = A +O = A +(C + D). (c) Since matrix addition is associative (property 2), A = A +(C + D) = (A + C)+ D. Now A + C = B + C by assumption so, by substitution, A = (B + C)+ D. (d) Since matrix addition is associative, this becomes A = B +(C + D). (e) By choice of D, C + D =O, so A = B +O. (f) But B +O = B so A = B. 45. (a) Theorem 9, part(2) (b) Theorem 8, part(3) (c) Theorem 9, part(3) 46. Using Theorem 10, it can be seen that yTx = (xTy)T = 0T = 0. Thus k x ¡ y k =q (x¡y)T(x¡y) = p(xT ¡yT)(x¡y) = p xTx¡xTy¡yTx+yTy= pkxk+kyk=p 2. 47. (A + AT)T = AT +(AT)T = AT + A = A + AT. 49. (a) QT is a (n x m) matrix, QTQ is a n x n matrix and QQT is a m x m matrix. Now (QTQ)T = QT(QT)T = QTQ so QTQ is symmetric. A similar argument shows that QQT is symmetric. (b) (ABC)T = ((AB)C)T = CT(AB)T = CT(BTAT) = CTBTAT: 50. 0?kQxk2= (Qx)T(Qx) = x TQTQx: 51. Property 2. Let A = (aij); B = (bij) and C = (cij): The (ij)th com- ponent of (A + B)+ C is (aij + bij)+ cij whereas the (ij)th component of A +(B + C) is aij +(bij + cij): The two are clearly equal. Property 3. Let O denote the (m x n) matrix with all zero entries. Clearly A+O= A for every (m x n) matrix A: Property 4. If A = (aij) then set P = (¡aij): Clearly A + P =O: 52. Let A = (aij); B = (bij); C = (cij); AB = (dij); and BC = (eij): The (rs)th entry of (AB)C is Ppk=1 drkcks; where drk = Pnj=1 arjbjk: Thus the (rs)th entry of (AB)C isP p k=1( Pn j=1 arjbjk)cks =P p k=1 Pn j=1 arjbjkcks = Pn j=1 arj( Pp k=1 bjkcks) = Pn j=1 arjejs: The last sum is the (rs) th entry of A(BC) so it follows that (AB)C = A(BC): 26 CHAPTER 1. MATRICES AND SYSTEMS OF EQUATIONS 53. Property 2: If A = (aij) then the (ij)th entry of r(sA) is r(saij): Similarly the (ij)th entry of (rs)A is (rs)aij: The two are clearly equal. Property 3: Let A = (aij) and B = (bij): The (ij)th entry of r(AB) is rPnk=1 aikbkj: The (ij)th entry of (rA)B is Pnk=1(raik)bkj: Finally, the (ij)th entry of A(rB) isP n k=1 aik(rbkj): The three are equal so r(AB) = (rA)B = A(rB): 54. Property 2: Let A = (aij); B = (bij) and C = (cij): The (rs)th entry of A(B + C) isP n k=1 ark(bks +cks) = Pn k=1 arkbks + Pn k=1 arkcks: The last expression in the (rs) th entry of AB + AC so A(B + C) = AB + AC: Property 3: The (ij)th entry of (r + s)A is (r + s)aij: The (ij)th entry of rA + sA is raij + saij: The entries are equal so (r + s)A = rA + sA: Property 4: The (ij)th entry of r(A + B) is r(aij + bij): The (ij)th entry of rA + rB is raij + rbij: Since the entries are equal r(A + B) = rA + rB: 55. Property 1: Let A = (aij); B = (bij); and A+B = (cij); where cij = aij +bij: The (rs)th entry of (A + B)T is csr = asr + bsr: But asr is the (rs)th entry of AT and bsr is the (rs)th entry of BT: Thus asr + bsr is the (rs)th entry of AT + BT: Property 3: Let A = (aij); AT = (dij): and (AT)T = (eij): Thus ers = dsr = ars; that is, (AT)T = A: 56. n = 2, m = 3. 57. n = 5, m = 7. 58. n = m = 4. 59. n = 4, m = 6. 60. n = 4, m = 2. 61. n = m = 5. 1.7 Linear Independence and Nonsingular Matrices 1. x1v1 + x2v2 = has only the trivial solution so fv1 ; v2g is linearly independent. 2. Linearly dependent. v3 = 2v1: 3. Linearly dependent. v5 = 3v1: 4. x1v2 +x3v3 = has only the trivial solution so fv2 ; v3g is linearly independent. 1.7. LINEAR INDEPENDENCE AND NONSING. MATRICES 27 5. Linearly dependent. v3 = 2v1 : 6. Linearly dependent. v3 = 2v1¡2v4: 7. Linearly dependent. u4 = 4u5 : 8. x1u3 +x2u4 = has only the trivial solution. So fu3 ; u4g is linearly independent. 9. x1u1+x2u2+x3u5= hasonlythetrivialsolutionsofu1; u2; u5g islinearlyindependent. 10. Linearly dependent. u4 = 4u5 : 11. Linearly dependent. u4 = 4u5 : 12. x1u1+x2u2+x3u4= hasonlythetrivialsolutionsofu1; u2; u4g islinearlyindependent. 13. Linearly dependent. u4 = (16=5)u0 +(12=5)u1¡(4=5)u2 : 14. Linearly dependent. u4 = (16=5)u0 +(4=5)u2 +(4=5)u3: 15. Sets 5, 6, 13, and 14 are linearly dependent by inspection. 16. A is nonsingular. 17. B is singular, x1 =¡2x2: 18. C is nonsingular. 19. AB is singular, x1 =¡2x2: 20. BA is singular, 7x1 =¡10x2: 21. D is singular, x1 = x2 = 0; x3 arbitrary. 22. F is nonsingular. 23. D + F is nonsingular. 24. E is singular, x1 arbitrary, x2 = 0 = x3: 25. EF is singular, x1 arbitrary, x2 = 0 = x3: 26. DE is singular, x1 arbitrary, x2 = 0 = x3: 27. FT is nonsingular. 28. fv1 ; v2g is linearly dependent if a = 3=2: 29. a = 6: 28 CHAPTER 1. MATRICES AND SYSTEMS OF EQUATIONS 30. fv1; v2; v3g is linearly dependent if a = 1: 31. fv1; v2; v3g is linearly dependent if b(a¡2) = 4: 32. fv1; v2g is linearly dependent if 3a = b: 33. fv1; v2g is linearly dependent if c = ab: 34. x= ? 0 1=2 ? ; v1 = (1=2)A2: 35. x= ? 0 1 ? ; v3 = A2 : 36. x= ? ¡1=2 1=2 ? ; v4 = (¡1=2)C1 +(1=2)C2 : 37. x= ? 1=2 1=2 ? ; v2 = (1=2)(C1 +C2): 38. x= 2 4 ¡2=3 4=3 ¡1 3 5; u1 = (¡2=3)F1 +(4=3)F2¡F3 : 39. x= 2 4 ¡8=3 ¡2=3 3 3 5u3 = (¡8F1¡2F2 +9F3 )=3: 40. 8 ? 1 2 ? ¡3 ? 2 3 ? = ? 2 7 ? : 41. ¡11 ? 1 2 ? +7 ? 2 3 ? = ? 3 ¡1 ? : 42. 8 ? 1 2 ? ¡4 ? 2 3 ? = ? 0 4 ? : 43. 0 ? 1 2 ? +0 ? 2 3 ? = ? 0 0 ? : 44. 1 ? 1 2 ? +0 ? 2 3 ? = ? 1 2 ? : 45. ¡3 ? 1 2 ? +2 ? 2 3 ? = ? 1 0 ? : 1.8. DATA FITTING, NUMERICAL INTEGRATION 29 46. (a) Since v2 =¡2v1 the set S is linearly dependent for any value of a. (b) If a =¡3 then v3 = v1 ¡v2. 47. (a) The set S is linearly dependent for any value of a. (b) The vector v3 can be written as a linear combination of v1 and v2 for any value of a. 48. A nontrivial solution is: 1v1 +0v2 +0v3 = . 49. 0 = T = (a1v1 + a2v2 + a3v3)T(a1v1 + a2v2 + a3v3) = a21 k v1 k2 +a22 k v2 k2 +a23 k v3k2, so ai = 0; i = 1;2;3 50. If a1v1 + a2v2 + a3v3 = , where some ai 6= 0, then a1v1 + a2v2 + a3v3 +0v4 = . 51. If = a1v1 +a2(v1 +v2)+a3(v1 +v2 +v3) then = (a1 +a2 +a3)v1 +(a2 +a3)v2 +a3v3. Since fv1;v2;v3g are linearly iindependent, a1 + a2 + a3 = 0, a2 + a3 = 0, and a3 = 0. It follows that a1 = a2 = a3 = 0. 52. AB = [AB1; : : : ; ABn] = O, so ABi = for 1 ? i ? n. Since A is nonsingular, Bi = for 1? i ? n, so B =O. 53. If AB = AC then A(B ¡C) =O. By Exercise 50, B ¡C =O. Therefore, B = C. 54. Suppose, b = [b1; : : : ; bn¡1]T. If c = [b1; : : : ; bn¡1;¡1]T then Bc = b1A1+¢¢¢+bn¡1An¡1¡ Ab = Ab¡Ab = . 55. If x1 is a nontrivial vector such that Bx1 = , then ABx1 = A = . 56. By Theorem 12, A = [w1 ;w2 ] is a nonsingular matrix. By Theorem 13, Ax= b has a (unique) solution. 57. Let v be any vector such that ATv= . By Theorem 13, there exists a vector w; such that Aw= v: Then AT(Aw) = ATv= , and so wT(ATAw) = wT = 0. Then (Aw )T(Aw) = w TATAw= 0 and kAwk = 0: Thus Aw= , and since A is nonsingular, w= . Thus Aw= A = . But then v had to be and AT is nonsingular. 1.8 Data Fitting, Numerical Integration & Difierentiation 1. p(t) = (¡1=2)t2 +(9=2)t¡1: 2. p(t) = t2 ¡4t +1: 3. p(t) = 2t +3: 30 CHAPTER 1. MATRICES AND SYSTEMS OF EQUATIONS 4. p(t) = 3t +2: 5. p(t) = 2t3 ¡2t2 +3t +1: 6. p(t) = t3 + t2 +1: 7. y = 2e2x + e3x: 8. y =¡ex¡1 +2e3(x¡1): 9. y = 3e¡x +4ex + e2x: 10. y = 2ex ¡4e2x + e3x: 11. R3h0 f(t)dt ? 3h2 [f(h)+ f(2h)]: 12. Rh0 f(t)dt ? h2[f(0)+ f(h)]: 13. R3h0 f(t)dt ? 3h8 [f(0)+3f(h)+3f(2h)+ f(3h)]. 14. R4h0 ? 8h3 f(h)¡ 4h3 f(2h)+ 8h3 f(3h): 15. Rh0 f(t)dt ? h2[¡f(¡h)+3f(0)]: 16. Rh0 f(t)dt ?¡ h12f(¡h)+ 2h3 f(0)+ 5h12f(h): 17. f0(0)? ¡1h f(0)+ 1hf(h): 18. f0(0)? ¡1h f(¡h)+ 1hf(0): 19. f0(0)?¡ 32hf(0)+ 2hf(h)+ ¡12h f(2h): 20. f0(0)? ¡116h f(0)+ 3hf(h)¡ 32hf(2h)+ 13hf(3h): 21. f00(0)? 1h2[f(¡h)¡2f(0)+ f(h)]: 22. f00(0)? 1h2[f(0)¡2f(h)+ f(2h)]: 23. 2 4 1 1 1 b¡a a t b (b2 ¡a2)=2 a2 t2 b2 (b3 ¡a3)=3 3 5 2 4 1 1 1 b¡a 0 t¡a b¡a (b2 + a2 ¡2ab)=2 0 t2 ¡a2 b2 ¡a2 (b3 ¡a3 ¡3(b¡a)a2)=3 3 5 2 4 1 1 1 b¡a 0 1 2 b¡a 0 t2 ¡a2 b2 ¡a2 (b3 ¡a3 ¡3(b¡a)a2)=3 3 5 1.8. DATA FITTING, NUMERICAL INTEGRATION 31 2 4 1 1 1 b¡a 0 1 2 b¡a 0 0 (b¡a)2=2 (b¡a)3=12 3 5 2 4 1 1 1 b¡a 0 1 2 b¡a 0 0 6 b¡a 3 5 24. 2 4 1 1 1 0 a¡h a a + h 1 (a¡h)2 a2 (a + h)2 2a 3 5 2 4 1 1 1 0 0 h 2h 1 0 h(2a¡h) 4ah 2a 3 5 2 4 1 1 1 0 0 h 2h 1 0 0 2h2 h 3 5: 25. By Rolle?s Theorem there exist u1 and u2 such that t0 < u1 < t1 < u2 < t2 and p0(u1) = p0(u2) = 0: Since p0(u1) = p0(u2) = 0; u1 < u2; and p0(t) = 2at + b; it follows that b = 0 = a: Finally, p(t0) = 0 means c = 0: 26. Suppose we have seen that a nonzero polynomial of degree n¡1 can have at most n¡1 distinct real zeros. Now assume that p(t) has n+1 zeros; that is there exist real numbers t0; t1; : : : ; tn such that t0 < t1 < ¢¢¢ < tn and p(ti) = 0 for 0? i ? n: By Rolle?s Theorem there are real numbers u1; : : : ; un such that ti¡1 < ui < ti; for 1 ? i ? n; and such that p0(ui) = 0 for each i: Now p0(t) = nantn¡1 +¢¢¢+ a1 and p0(t) has n zeros. By assumption p0(t) is the zero polynomial. Thus 0 = a1 = ¢¢¢ = an: This leaves p(t) = a0; but p(t0) = 0 so a0 = 0: Therefore p(t) is the zero polynomial. 27. We must solve the system Lx= b where L = 2 66 4 0 0 0 1 0 0 1 0 1 1 1 1 3 2 1 0 3 77 5 x= 2 66 4 a b c d 3 77 5; b= 2 66 4 2 3 8 10 3 77 5: p(t) = t3 +2t2 +3t +2: 28. p(t) = t2 +2t +1: 29. p(t) = t3 + t2 +4t +3: 30. p(t) = 2t3 ¡2t +3: 31. By Rolle?s Theorem there exists a number s such that t0 < s < t1 and p0(s) = 0: Thus p0(t) has three zeros. By Exercise 25, p0(s) is the zero polynomial. It follows that p(t) is the zero polynomial. 32. The coe?cient matrix of p(t) = at3 + bt2 + ct + d must satisfy Lx= b 32 CHAPTER 1. MATRICES AND SYSTEMS OF EQUATIONS where L = 2 66 4 t30 t20 t0 1 3t20 2t0 1 0 t31 t21 t1 1 3t21 2t1 1 0 3 77 5 x= 2 66 4 a b c d 3 77 5 b= 2 66 4 y0 s0 y1 s1 3 77 5 . Suppose Lx0= , where x0= [a; b; c; d]T: If p(t) = at3 +bt2 +ct+d then p(t0) = p(t1) = 0 and p0(t0) = p0(t1) = 0: By Exercise 31 a = b = c = d = 0; that is x0 = . This proves that L is nonsingular so by Theorem 13, Lx= b has a unique solution. 33. First supppose that p(ti) = p0(ti) = 0 for 0 ? i ? n: Since p(ti¡1) = 0 = p(ti); it follows from Rolle?s Theorem that there is a real number ui such that ti¡1 < ui < ti and p0(ui) = 0: Therefore p0(t) has 2n+1 zeros, t0; t1; : : : ; tn; u1; : : : ; un: By Exercise 26, p0(t) is the zero polynomial and it follows that p(t) is the zero polynomial. Now set p(t) = P2n+1k=0 aktk and assume that p(ti) = yi and p0(ti) = si for 0 ? i ? n: These constraints yield a system of equations Lx=b,where L = 2 66 66 64 t2n+10 t2n0 : : : t0 1 (2n +1)t2n0 2nt2n¡10 : : : 1 0 ... t2n+1n t2nn : : : tn 1 (2n +1)t2nn 2nt2n¡1n : : : 1 0 3 77 77 75; x = [a2n+1; a2n; : : : ; a1; a0]T and b = [y0; s0; : : : ; yn; sn]T: Suppose Lx0 = , where x0 = [b2n+1; b2n; : : : ; b1; b0]T: If we set q(t) =P 2n+1 k=0 bkt k then it follows that q(ti) = q0(ti) = 0 for 0? i ? n: As we have shown above, this implies that b2n+1 = b2n =¢¢¢= b1 = b0 = 0: In particular x0 = and it follows that L is nonsingular. By Theorem 13 the system Lx=b has a unique solution. 34. R5h0 f(x)dx ? 5h24[1912f(0)+ 254 f(h)+ 256 f(2h)+ 256 f(3h)+ 254 f(4h)+ 1912f(h)]. 35. f0(a)? 112h[f(a¡2h)¡8f(a¡h)+8f(a + h)¡f(a +2h)] 1.9 Matrix Inverses and their Properties 5. (cf. Ex. 2) x = A¡1b = Bb = ? 1 ¡1 ¡:2 :3 ?? 6 9 ? = ? ¡3 1:5 ? : 6. (cf. Ex. 1) x = A¡1b = Bb = ? 3 ¡4 ¡5 7 ?? 5 2 ? = ? 7 ¡11 ? : 7. (cf. Ex. 3) x = B¡1b = Ab = 2 4 ¡1 ¡2 11 1 3 ¡15 0 ¡1 5 3 5 2 4 4 2 2 3 5 = 2 4 14 ¡20 8 3 5: 1.9. MATRIX INVERSES AND THEIR PROPERTIES 33 8. (cf. Ex. 4) x = B¡1b = Ab = 2 4 1 0 0 2 1 0 3 4 1 3 5 2 4 2 3 2 3 5 = 2 4 2 7 20 3 5: 9. If B is any 3 x 3 matrix, then the (1;1)th entry of AB is zero and so AB 6= I: 10. The (1;1)th entry of BA is zero. 11. Let B = (xij) be a (3 x 3) matrix and suppose that BA = I: Then the (1;1)th entry of BA must be one and the (1;2)th entry of BA be must be zero. But each of these entries equals 2x11 + x12 +3x13 and cannot simultaneously be one and zero. 12. The (1;1)th and the (2;1)th entry of AB cannot be simultaneously be zero and one. 13. ? 3 ¡1 ¡2 1 ? : 14. ? ¡7=4 3=4 3=2 ¡1=2 ? : 15. ? ¡1=3 2=3 2=3 ¡1=3 ? : 16. 2 4 0 1 3 5 5 4 1 1 1 3 5: 17. 2 4 1 0 0 ¡2 1 0 5 ¡4 1 3 5: 18. 2 4 1 11 ¡7 0 ¡7 4 0 2 ¡1 3 5: 19. 2 4 1 ¡2 0 3 ¡3 ¡1 ¡6 7 2 3 5: 20. 2 66 4 ¡35 ¡16 26 ¡17 ¡15 ¡7 11 ¡7 4 2 ¡3 2 ¡2 ¡2 2 ¡1 3 77 5: 34 CHAPTER 1. MATRICES AND SYSTEMS OF EQUATIONS 21. 2 66 4 ¡1=2 ¡2=3 ¡1=6 7=6 1 1=3 1=3 ¡4=3 0 ¡1=3 ¡1=3 1=3 ¡1=2 1 1=2 1=2 3 77 5: 22. ¡15 ? 1 ¡2 ¡1 ¡3 ? : 23. ¢ = 10 so A¡1 = 110 ? 3 2 ¡2 2 ? : 24. ¡17 ? 1 ¡3 ¡2 ¡1 ? : 25. ¢ = 0 so A¡1 does not exist. 26. A¡1 does not exist. 27. ? 6= 2; ¡2 28. ? 6= 2 29. x= A¡1 b= ? 2 ¡1 ¡3 2 ? ? 4 2 ? = ? 6 ¡8 ? : 30. ? ¡4 4 ? : 31. x= A¡1b = ? 4 ¡1 3 ¡1 ? ? 5 2 ? = ? 18 13 ? : 32. ? 17 ¡11 ? : 33. x= A¡1b= 110 ? 3 ¡1 1 3 ? ? 10 5 ? = ? 5=2 5=2 ? : 34. x= ? 6:8 ¡3:2 ? : 35. Q¡1 = C¡1A¡1 = ? ¡3 1 3 5 ? : 36. Q¡1 = A¡1C¡1 = ? ¡2 5 2 4 ? : 1.9. MATRIX INVERSES AND THEIR PROPERTIES 35 37. Q¡1 = (A¡1)T = ? 3 0 1 2 ? : 38. Q¡1 = C¡1(A¡1)T = ? ¡1 1 1 2 ? ? 3 0 1 2 ? = ? ¡2 2 5 4 ? : 39. Q¡1 = (A¡1)T(C¡1)T = ? 3 0 1 2 ? ? ¡1 1 1 2 ? = ? ¡3 3 1 5 ? : 40. Q¡1 = A¡1B = ? 5 7 4 2 ? : 41. Q¡1 = BC¡1 = ? 1 5 ¡1 4 ? : 42. Q¡1 = B = ? 1 2 2 1 ? : 43. Q¡1 = 12A¡1 = ? 3=2 1=2 0 1 ? : 44. Q¡1 = 110C¡1 = ? ¡1=10 1=10 1=10 1=5 ? : 45. Q¡1 = B(C¡1A¡1) = ? 3 11 ¡3 7 ? : 46. B = A¡1D = ? ¡4 6 7 3 ¡4 ¡4 ? ; C = EA¡1 = 2 4 8 ¡3 1 0 ¡6 3 3 5: 47. B = A¡1D = 2 4 1 10 15 12 3 3 3 5; C = EA¡1;= ? 13 12 8 2 3 5 ? : 48. a 6=¡1. 49. (AB)¡1 = B¡1A¡1 = 2 4 2 35 1 14 35 34 23 12 70 3 5 (3A)¡1 = (1=3)A¡1 = 2 4 1=3 2=3 5=3 1 1=3 2 2=3 8=3 1=3 3 5 36 CHAPTER 1. MATRICES AND SYSTEMS OF EQUATIONS (AT)¡1 = (A¡1)T = 2 4 1 3 2 2 1 8 5 6 1 3 5 50. A2 = AB +2A = A(B +2I) so A = B +2I = 2 4 4 1 ¡1 0 5 2 ¡1 4 3 3 5 51. I 52. (a) X = I and X =¡I (d) The equation (X¡I)(X+I) =O does not require that either X¡I =O or X+I =O. 53. ATA = ? u 1 u2 v1 v2 ?? u 1 v1 u2 v2 ? = ? uTu uTv uTv vTv ? : 54. AA = (I ¡uuT )(I ¡uuT ) = I 2 ¡2uuT +(uuT )(uuT ) = (I ¡2uuT +u(uTu)uT = I ¡2uuT +uuT = I ¡uuT = A: 55. I = AA¡1 = (AA)A¡1 = A(AA¡1) = AI = A: 56. Symmetry: AT = (I ¡avvT )T = I T ¡a(vvT )T = I ¡a(v T Tv T) = I ¡avvT = A: AA = (I ¡avvT )(I ¡avvT ) = I ¡2avvT +a2(vvT )(vvT ) = I ¡2avvT +a[2=vvT ](vvT vvT ) = I ¡2avvT +(2avvT ) = I: 57. Ax= Ix¡a(vvT )x= x¡av(vTx) = x¡a(vTx)v= x¡?v where ? = a(vTx): 58. kAxk= p(Ax)T(Ax) = p(xTAT)(Ax) = pxT(ATA)x =p xTIx =pxTx =kxk: 59. A(I ¡auvT ) = (I +uvT )(I ¡auvT ) = I2 +uvT¡auvT¡ a(uvT )(uvT ) = I +uvT¡auvT¡au(vTu)vT = I +uvT¡auvT (1+vTu) = I +uvT¡uvT = I: 60. AB = A(A2 ¡2A +3I) = A3 ¡2A2 +3A¡I + I = +I = I: 61. AB = A(¡1=b0)[A + b1I] = (¡1=b0)(A2 + b1A)+(¡I + I) = (¡1=b0)(A2 + b1A + b0I)+ I = +I = I: 1.9. MATRIX INVERSES AND THEIR PROPERTIES 37 62. = A2 + b1A + b0I =? 9 5 25 14 ? + ? 2b 1 b1 5b1 3b1 ? + ? b 0 0 0 b0 ? = ? 9+2b 1 + b0 5+ b1 25+5b1 14+3b1 + b0 ? : b0 = 1;b1 =¡5: (¡1=b0)[A + b1I] = ? 3 ¡1 ¡5 2 ? = A¡1 : 63. = A2 + b1A + b0I = ? 11 ¡4 ¡2 3 ? + ? ¡3b 1 2b1 b1 b1 ? + ? b 0 0 0 b0 ? = ? 11¡3b 1 + b0 ¡4+2b1 ¡2+ b1 3+ b1 + b0 ? :b0 =¡5;b1 = 2: (¡1=b0)[A + b1I] =¡(1=5) ? 1 ¡2 ¡1 ¡3 ? = A¡1 : 64. = A2 + b1A + b0I = ? 0 ¡10 10 5 ? + ? 2b 1 ¡2b1 2b1 3b1 ? + ? b 0 0 0 b0 ? = ? 2b 1 + b0 ¡10¡2b1 10+2b1 5+3b1 + b0 ? :b0 = 10;b1 =¡5: (¡1=b0)[A + b1I] = 110 ? 3 2 ¡2 2 ? = A¡1 : 65. = A2+b1A+b0I = ? 7 0 0 7 ? + ? ¡b 1 +3b1 2b1 b1 ? + ? b 0 0 0 b0 ? = ? 7¡b 1 + b0 3b1 2b1 7+ b1 + b0 ? : b0 =¡7;b1 = 0 (¡1=b0)[A + b1I] = 17 ? ¡1 3 2 1 ? = A¡1: 66. (a) Ax = b1, has solution x ? 2 4 106:395 ¡4909:194 4979:886 3 5. Ax = b2 has solution x ? 2 4 107:459 ¡4958:286 5029:685 3 5 (b) A¡1 = 2 4 25:315 21:316 ¡59:764 ¡1100:002 ¡1028:428 2780:764 1114:405 1044:910 ¡2820:572 3 5 67. (a) A = ? 1 0 0 1 ? and B = ? 0 1 1 0 ? . (b) A = ? 1 0 0 0 ? and B = ? 0 0 0 1 ? . 68. (A¡1)T = (AT)¡1 = A¡1. 69. (a) (AB = O) =) B = A¡1AB = A¡1O =O: 38 CHAPTER 1. MATRICES AND SYSTEMS OF EQUATIONS (b) B = ? 1 1 ¡1 ¡1 ? : 70. (AB = AC) =) B = IB = (A¡1A)B = A¡1(AB) = A¡1(AC) = C: 71. If either d 6= 0 or c 6= 0 then A ? d ¡c ? = and so A is singular. If c = 0 = d, then A = ? a b 0 0 ? , and Ax = ? 0 1 ? does not have a solution so A is singular. 72. (AB)¡1 = B¡1A¡1: 73. The hypothesis of Theorem 17 is not satisfled; that is, Theorem 17 assumes that A and B have inverses. 74. (a) (Bv = ) =) (A(Bv) = =) (AB)v = ) (=) v = ) so B is singular. (b) AB and B¡1 are nonsingular so A = (AB)B¡1 is nonsingular. 75. Suppose each of the systems Ax= ek is consistent and let bk be a solution. If B = [b1;b2; : : : ;bn] then AB = [Ab1;Ab2; : : : ;Abn] = [e1;e2; : : : ;en] = I: Thus B = A¡1 and A is nonsingular. 76. Clearly I¡1 = I, so by Theorem 1.5, I is non-singular. 77. Since AB is deflned, q = r and AB is a (p£s) matrix. Since BA is deflned, s = p and BA is a (r £q) matrix. But AB = BA, so p = r and q = s. 79. Suppose B and C are inverses for A. Then B = BI = B(AC) = (BA)C = IC = C. 1.10 Supplementary Exercises 1. The augmented matrix for the system reduces to ? 1 0 1 0 (a¡1)(a +2) (a¡3)(a +2) ? There are inflnitely many solutions if a = ¡2, no solution if a = 1, and a unique solution in which x2 = 0 if a = 3. 1.10. SUPPLEMENTARY EXERCISES 39 2. (a) The system is consistent if and only if ¡b1 +2b2 +b3 = 0 and in this case the solution is x1 = b2 ¡b1 ¡2x3, x2 = b2 ¡2b1 ¡3x3, x3 arbitrary. (b) (i) inconsistent (ii) x1 =¡3¡2x3, x2 =¡8¡3x2, x3 arbitrary (iii) x1 =¡4¡2x3, x2 =¡11¡3x2, x3 arbitrary (iv) inconsistent 3. (a) Reducing the matrix [A;b1;b2;b3] yields: for b1, x1 =¡48, x2 = 11, x3 = 18; for b2, x1 = 4, x2 = 2, x3 = 1; and for b3, x1 =¡3, x2 = 2, x3 = 2. (b) C = 2 4 ¡48 4 ¡3 11 2 2 18 1 2 3 5 4. Set C = [c1;c2]. Reducing the matrix [A; C] yields solution x1 = 2¡x3, x2 = 1+2x3, x3 arbitrary, for the system Ax = c1. Similarly, the system Ax = c2 has solution x1 =¡1¡x3, x2 = ¡3 + 2x3, x3 arbitrary. Therefore, if b1 = 2 4 2¡a 1+2a a 3 5 and b2 = 2 4 ¡1¡b ¡3+2b b 3 5 for arbitrary a; b, then Ab1 = c1 and Ab2 = c2, B = [b1;b2] is the desired matrix since AB = C. 5. By assumption, A5 + 3A1 + 5A4 + 7A2 + 9A3 = b. Reordering the terms yield 3A1 + 7A2 +9A3 +5A4 +A5 = b so Ax = b has solution [3;7;9;5;1]T. 6. (a) x1 = 2+ x3, x2 = 3¡3x3, x3 arbitrary. (b) x1 = x3, x2 =¡3x3, x3 nonzero. 7. (a) x = [2;1;0]T is the unique solution. (b) x = is the unique solution. 8. x = [¡1;0;1]T 9. (a) A¡1 = 2 4 ¡4 1 3 4 ¡1 ¡2 ¡3 1 1 3 5 (b) A¡1 = ? cos sin ¡sin cos ? 10. ¢ = (?¡4)(?¡1)+2 = (?¡2)(?¡3). A is singular when ¢ = 0; that is, when ? = 2 or ? = 3. When A is nonsingular A¡1 = 1(?¡2)(?¡3) ? ?¡1 1 ¡2 ?¡4 ? 40 CHAPTER 1. MATRICES AND SYSTEMS OF EQUATIONS 11. A = ? 1=2 ¡1=4 ¡5=4 3=4 ? 12. A = ? 3 4 6 8 ? and B = ? 1 2 2 2 ? 13. A99 = A; A100 = I 14. x = A¡1b = [3¡6¡1]T 15. (AB)¡1 = B¡1A¡1 = 2 4 28 ¡22 ¡17 27 ¡1 49 29 8 16 3 5 16. (3A)¡1 = (1=3)A¡1 = 2 4 2=3 1 5=3 7=3 2=3 1=3 4=3 ¡4=3 1 3 5 17. (ATB)¡1 = B¡1(A¡1)T = 2 4 15 ¡31 ¡31 36 52 47 18 21 ¡1 3 5 18. [(A¡1B¡1)¡1A¡1B]¡1 = (B¡1)2 = 2 4 70 ¡19 5 ¡39 44 21 11 8 22 3 5 1.11 Conceptual Exercises 1. False. If A = ? 1 2 2 3 ? and B = ? 1 1 1 4 ? then A and B aresymmetricbut AB = ? 3 9 5 14 ? is not symmetric. 2. True. (A + AT)T = AT +(AT)T = AT + A = A + AT. 3. True. A¡1 = A and B¡1 = B so AB¡1 = B¡1A¡1 = BA. 4. False. If A = ? 1 0 0 1 ? and B = ? 1 0 0 ¡1 ? then A and B are nonsingular, but A + B = ? 2 0 0 0 ? is singular. 5. False. The system x1 = 1 x2 = 2 x1 + x2 = 3 1.11. CONCEPTUAL EXERCISES 41 clearly has a unique solution x1 = 1, x2 = 2. 6. True. Suppose A = [A1;A2; : : : ;An]. Then for 1? j ? n, = Aej = Aj. 7. False. If fu1;u2g is linearly dependent then so is fAu1; Au2g. (cf. Exercise 12). 8. True. Since AB is deflned, n = p and AB is an (m £ q) matrix. But AB is square, so m = q. Thus BA is deflned and is an (n£n) matrix. 9. Q¡1 = RP. 10. AB = (AB)T = BTAT = BA. 11. First note that uTi uj = (uTj ui)T, since uTj ui is an (1£1) matrix. If = c1u1 +c2u2 +c3u3 then 0 = T = (c1u1 + c2u2 + c3u3)T(c1u1 + c2u2 + c3u3) = c21 ku1k2 +c22 ku2k2 +c23 k u3k2. If follows that c1 = c2 = c3 = 0. 12. Suppose c1u1+c2u2 = , where ci 6= 0 for i = 1 or i = 2. Then = A = A(c1u1+c2u2) = c1Au1 + c2Au2. 13. kAxk2= (Ax)T(Ax) = xTATAx = xTIx = xTx =kxk2. 14. A2 = AI so A = I. 15. (AB)2 = (AB)(AB) = A(BA)B = A(AB)B = A2B2 = AB. 16.(b) Since Ak¡1 6=O, there exists a vector b such that Ak¡1b 6= (cf. Exercise 6). If c = Ak¡1b then Ac = A(Ak¡1b) = Akb =Ob = . It follows that A is singular. Chapter 2 Vectors in 2-Space and 3-Space 2.1 Vectors in the Plane 1. For vector ¡¡!AB the x¡component is ¡4¡0 = 4 and the y¡component is 3¡(¡2) = 5. For vector ¡¡!CD the x¡component is 1¡5 = ¡4 and the y¡component is 4¡(¡1) = 5. The vectors are equal. 2. For vector ¡¡!AB the x¡component is 3¡(¡1) = 4 and the y¡component is ¡2¡3 = ¡5. For vector ¡¡!CD the x¡component is 1¡5 =¡4 and the y¡component is 4¡(¡1) = 5. The vectors are not equal. 3. For vector ¡¡!AB the x¡component is 0¡(¡4) = 4 and the y¡component is 1¡(¡2) = 3. For vector ¡¡!CD the x¡component is 3¡0 = 3 and the y¡component is 2¡(¡2) = 4. The vectors are not equal. 4. For vector ¡¡!AB the x¡component is ¡1¡3 = ¡4 and the y¡component is ¡1¡1 = ¡2. For vector ¡¡!CD the x¡component is ¡6¡0 =¡6 and the y¡component is 0¡3 =¡3. The vectors are not equal. 5. (a) For u: kuk= p(2¡(¡3))2 +(2¡5)2 =p25+9 = p34: For v: kvk= p(¡2¡3)2 +(7¡4)2 =p25+9 = p34: Therefore kuk=kvk. (b) Segment AB has slope: (2¡5)=(2¡(¡3)) =¡3=5. Segment CD has slope: (7¡4)=(¡2¡3) = 3=(¡5). (c) For vector ¡¡!AB the x¡component is 2¡(¡3) = 5 and the y¡component is 2¡5 =¡3. For vector ¡¡!CD the x¡component is ¡2¡3 = ¡5 and the y¡component is 7¡4 = 3. The vectors are not equal. 6. D = (4;7) 44 CHAPTER 2. VECTORS IN 2-SPACE AND 3-SPACE 7. D = (¡2;5) 8. D = (¡1;6) 9. D = (¡1;1) 11. v1 = 5, v2 = 3 12. v1 = 5, v2 =¡4 13. v1 =¡6, v2 = 5 14. v1 =¡3, v2 =¡4 15. B = (3;3) 16. B = (¡1;2) 17. A = (2;4) 18. A = (0;3) 19. (a) B = (3;2), C = (5;0) 20. (a) B = (¡2;3), C = (1;4) 21. (a) Q = (7;1) 22. (a) Q = (4;1) 23. (a) B = (¡1;4), C = (0;¡1) 24. (a) Q = (3;¡2) 25. (a) B = (3;3), C = (6;1) 26. (a) B = (0;7), C = (¡3;¡2) 27. (a) D = (6;¡3) 28. 1p5 ? ¡1 2 ? 29. 15 ? 3 4 ? 30. 1p2i+ 1p2j 2.2. VECTORS IN SPACE 45 31. 3p13i¡ 2p13j 32. B = (4;2) 33. B = (1;¡2) 34. B = (0;¡5) 35. B = (1=3;¡7) 36. u+v = ? 2 3 ? , u¡3v = ? ¡2 ¡5 ? 37. u+v = ? 2 4 ? , u¡3v = ? ¡6 8 ? 38. u+v = 2i+j, u¡3v =¡2i+5j 39. u+v = 4i+j, u¡3v =¡4i¡7j 40. Note that given u = ? a b ? with either a 6= 0 or b 6= 0, gives kuk= pa2 + b2 6= 0. Then w = 1kuku = 1kuk ? a b ? = ? a=pa2 + b2 b=pa2 + b2 ? kwk= r a2 a2 + b2 + b2 a2 + b2 = r a2 + b2 a2 + b2 = p1 = 1 2.2 Vectors in Space 1. d(P; Q) = p(0¡1)2 +(2¡2)2 +(2¡1)2 =p2 2. d(P; Q) = p(0¡1)2 +(0¡1)2 +(1¡0)2 =p3 3. d(P; Q) = p(0¡1)2 +(0¡0)2 +(1¡0)2 =p2 4. d(P; Q) = p(0¡1)2 +(0¡1)2 +(0¡1)2 =p3 5. M = (1;4;4);d(M;0) = p(0¡1)2 +(0¡4)2 +(0¡4)2 =p33 6. M = (2;1;4);d(M;0) = p(0¡2)2 +(0¡1)2 +(0¡4)2 =p21 7. B = (0;3=2;¡3=2), C = (1;3;0), D = (2;9=2;3=2) 8. plane 9. line 46 CHAPTER 2. VECTORS IN 2-SPACE AND 3-SPACE 10. line 11. plane 12. plane 17. (a) The length of the segment from P to R is p(r1 ¡p1)2 +(r2 ¡p2)2 +(r3 ¡p3)2 and the length of the segment from R to S is p(s1 ¡r1)2 +(s2 ¡r2)2 +(s3 ¡r3)2. Let a and b be the distances from P to R and R to S respectively and c be the distance from point P to S. Solving the equation a2 +b2 = c2 for c yields the desired equality. (b) This problem is worked similarly to part (a). 18. (a) v = 2 4 ¡2 3 ¡3 3 5 (b) D = (¡3;5;¡2) 19. (a) v = 2 4 3 2 ¡3 3 5 (b) D = (2;4;¡2) 20. (a) v = 2 4 ¡3 ¡2 3 3 5 (b) D = (¡4;0;4) 21. (a) v = 2 4 0 5 ¡7 3 5 (b) D = (¡1;7;¡6) 22. A = (4;0;0) 23. A = (3;2;1) 24. A = (4;1;2) 25. A = (¡2;3;¡1) 26. (a) u+2v = 2 4 9 9 14 3 5 (b) ku¡vk= 5 2.2. VECTORS IN SPACE 47 (c) w = 2 4 3=2 0 2 3 5 27. (a) u+2v = 2 4 7 7 10 3 5 (b) ku¡vk= 3 (c) w = 2 4 1 ¡1=2 1 3 5 28. (a) u+2v = 2 4 11 ¡3 4 3 5 (b) ku¡vk= p74 (c) w = 2 4 ¡4 3=2 ¡1=2 3 5 29. (a) u+2v = 2 4 ¡1 1 2 3 5 (b) ku¡vk= p150 (c) w = 2 4 7=2 ¡5 1=2 3 5 30. u = 2v = 2i+2j 31. u = 2k 32. u =¡4v = 2 4 ¡4 0 ¡4 3 5 33. u =¡53v = 2 4 5=3 ¡10=3 ¡10=3 3 5 34. u = 2 4 ¡1=2 ¡1 0 3 5 48 CHAPTER 2. VECTORS IN 2-SPACE AND 3-SPACE 35. u = 2 4 ¡2 ¡4 ¡1 3 5 2.3 The Dot Product and the Cross Product 1. ¡2 2. 0 3. 7 4. 0 5. cos( ) = 11p290 6. cos( ) = ¡3p130 7. cos( ) = 16 8. cos( ) = 1114 9. = ?6 10. = 2?3 11. = ?2 12. = 0 13. u = i+3j+4k 14. u = 0i+0j+4k 15. u = 3i+0j+4k 16. u = 12i+4j+3k or u = 12i¡4j+3k 17. u =¡1i+3j+1k 18. u = 1i+1j+2k 19. R = (33=10;11=10) 2.3. THE DOT PRODUCT AND THE CROSS PRODUCT 49 20. R = (8;2) 21. R = (¡3;¡1) 22. R = (0;0) u and q are perpendicular so w is the zero vector. 23. u1 = ? 5 5 ? , u2 = ? 2 ¡2 ? 24. u1 = ? 2 ¡2 ? , u2 = ? 4 4 ? 25. u1 = 2 4 2 4 2 3 5, u2 = 2 4 4 0 ¡4 3 5 26. u1 = 2 4 3 3 3 3 5, u2 = 2 4 ¡1 ¡2 3 3 5 27. If u = 2 4 u1 u2 u3 3 5, then u¢u = u21 + u22 + u23. Each term of this sum is a real number squared and therefore greater than or equal to zero, hence the sum is greater than or equal to zero. 28. Note that u¢v = u1v1 + u2v2 + u3v3 and v¢u = v1u1 + v2u2 + v3u3. By the commutative property of multiplication, uivi = viui for each value of i and hence u¢v = v¢u. 29. Note that u¢(cv) = 2 4 u1 u2 u3 3 5¢ 2 4 c v1 c v2 c v3 3 5 = u1 c v1 + u2 c v2 + u3 c v3 = c(u1v1 + u2v2 + u3v3) = c(u¢v) 30. u¢(v +w) = 2 4 u1 u2 u3 3 5¢ 2 4 v1 + w1 v2 + w2 v3 + w3 3 5 = u1(v1 + w1)+ u2(v2 + w2)+ u3(v3 + w3) = u1v1 + u1w1 + u2v2 + u2w2 + u3v3 + u3w3 = (u1v1 + u2v2 + u3v3)+(u1w1 + u2w2 + u3w3) = u¢v +u¢w 50 CHAPTER 2. VECTORS IN 2-SPACE AND 3-SPACE 31. With the given u and v the left-hand-side of equation (1) yields: ku¡vk2 = (u1 ¡v1)2 +(u2 ¡v2)2 = u21 ¡2u1v1 + v21 + u22 ¡2u2v2 + v22 The right-hand-side of equation (1) yields: kuk2+kvk2 ¡2kukkvkcos( ) = u21 + u22 + v21 + v22 ¡2kukkvkcos( ) Setting these expanded left and right hand sides equal to each other and then cancelling terms yields: ¡2u1v1 ¡2u2v2 =¡2kukkvkcos( ) Dividing the last equation by ¡2 gives the desired result, equation (2a). 32. 2 4 0 0 ¡11 3 5 33. 2 4 ¡2 2 8 3 5 34. 2 4 0 0 0 3 5 35. 2 4 3 ¡1 ¡5 3 5 36. w = 2 4 ¡1 ¡1 2 3 5 37. w = 2 4 ¡2 ¡5 4 3 5 38. w = 2 4 1 ¡1 ¡1 3 5 39. w = 2 4 2 ¡3 1 3 5 2.3. THE DOT PRODUCT AND THE CROSS PRODUCT 51 40. w = 2 4 ¡9 ¡9 ¡9 3 5 41. w = 2 4 1 1 ¡4 3 5 42. p41 square units 43. 4p6 square units 44. 5 p5 2 square units 45. 6 p11 2 square units 46. 22 cubic units 47. 24 cubic units 48. coplanar 49. NOT coplanar 50. Substitute the given values of x, y, and z into the left-hand-side of each of the two given equations and simplify. This will show that for the given values, the left-hand-side of each equation is zero and therefore the given values are a solution to the system. 51. By flnding the two cross products (i£i)£j = 2 4 0 0 0 3 5 and i£(i£j) = 2 4 0 ¡1 0 3 5, one can see that they are not equal. 52. (a) Expanding and simplifying the right and left hand sides of the equations, one can see that they are equal. ku£vk2 = (u2v3 ¡u3v2)2 +(u3v1 ¡u1v3)2 +(u1v2 ¡u2v1)2 = u32(v12 + v22)¡2u1u3v1v3 ¡2u2v2(u1v1 + u3v3)+ u22(v12 + v32)+ u12(v22 + v32) kuk2 kvk2 ¡(u¢v)2 = (u2v3 ¡u3v2)2 +(u3v1 ¡u1v3)2 +(u1v2 ¡u2v1)2 = u32(v12 + v22)¡2u1u3v1v3 ¡2u2v2(u1v1 + u3v3)+ u22(v12 + v32)+ u12(v22 + v32) 52 CHAPTER 2. VECTORS IN 2-SPACE AND 3-SPACE (b) Substitute kukkvkcos( ) into the equation given in 10.(a) for u¢v: ku£vk2 = kuk2 kvk2¡kuk2 kvk2 cos2 ( ) = kuk2 kvk2(1¡cos2 ( )) = kuk2 kvk2 sin2 ( ) Therefore, ku£vk=kukkvkcos( ). 53. Toseethatkukcos( )istheheightoftheparallelepiped, calculatethelengthof proj(v£w)u. Then the volume of the parallelepiped is (area of base)£(height) which is exactly the abso- lute value of the triple product given. 2.4 Lines and Planes in Space 1. x = 2+3t y = 4+2t z =¡3+4t 2. x = 1+2t y = 1 z =¡1+3t 3. As a direction vector for L we use u = ¡¡¡!P0P1 = 2 4 1¡0 2¡4 4¡1 3 5 = 2 4 1 ¡2 3 3 5. Therefore, one set of parametric equations for L are x = t, y = 4¡2t, z = 1+3t. 4. As a direction vector for L we use u = ¡¡¡!P0P1 = 2 4 6¡5 6¡1 4¡(¡3) 3 5 = 2 4 1 5 7 3 5. Therefore, one set of parametric equations for L are x = 5+ t, y = 1+5t, z =¡3+7t. 5. The direction vector for the flrst line is u = 2 4 2 ¡1 3 3 5. The direction vector for the second line is v = 2 4 ¡4 2 6 3 5 = ¡2u. Since the two direction vectors are scalar multiples of each other, the lines are parallel. 6. The direction vector for the flrst line is u = 2 4 6 4 3 3 5. The direction vector for the second line is v = 2 4 3 2 3 3 5. Since there is no real number k such that u = kv, the two lines are NOT 2.4. LINES AND PLANES IN SPACE 53 parallel. 7. The direction vector for the flrst line is u = 2 4 ¡2 3 ¡2 3 5. The direction vector for the second line is v = 2 4 2 ¡3 4 3 5. Since there is no real number k such that u = kv, the two lines are NOT parallel. 8. The direction vector for the flrst line is u = 2 4 ¡1 2 ¡3 3 5. The direction vector for the second line is v = 2 4 3 ¡6 9 3 5 = ¡3u. Since the two direction vectors are scalar multiples of each other, the lines are parallel. 9. The normal vector to the plane is n = 2 4 3 4 ¡1 3 5. Therefore, the equation of the line is x = 1+3t, y = 2+4t, z = 1¡t. 10. The normal vector to the plane is n = 2 4 1 ¡1 2 3 5. Therefore, the equation of the line is x = 2+ t, y = 0¡t, z =¡3+2t. 11. By substituting the parametric equations of the line into the equation for the plane, we flnd that t =¡1. Thus P = (¡1;4;1). 12. When substituting the parametric equations of the line into the equation for the plane, we flnd a contradiction. Therefore, the line does not intersect the plane at any point. It can also be noted that the normal vector to the plane and the direction vector of the line are perpendicular so that the line is at least parallel to the plane. Then since the point P0 is not in the plane, the line is not in the plane either. 13. By substituting the parametric equations of the line into the equation for the plane, we flnd that t =¡4. Thus P = (¡8;¡13;36). 14. By substituting the parametric equations of the line into the equation for the plane, we flnd that t =¡2. Thus P = (¡3;5;¡3). 15. 6x + y ¡z = 16 54 CHAPTER 2. VECTORS IN 2-SPACE AND 3-SPACE 16. x¡2y +3z = 9 17. A normal vector to the plane can be found by taking the cross product of the vectors ¡¡!PQ = 2 4 1 1 2 3 5 and ¡!PR = 2 4 0 4 1 3 5. n =¡¡!PQ£¡!PR =¡7i¡j+4k. Therefore, the equation of the plane is ¡7x¡y +4z = 5. 18. A normal vector to the plane can be found by taking the cross product of the vectors ¡¡!PQ = 2 4 1 8 ¡5 3 5 and ¡!PR = 2 4 2 1 2 3 5. n =¡¡!PQ£¡!PR = 21i¡12j¡15k. Therefore, the equation of the plane is 21x¡12y ¡15z =¡12 or equivalently 7x¡4y ¡5z =¡4. 19. A normal vector to the plane can be found by taking the cross product of the vectors ¡¡!PQ = 2 4 ¡2 ¡1 1 3 5 and ¡!PR = 2 4 3 0 2 3 5. n =¡¡!PQ£¡!PR =¡2i+7j+3k. Therefore, the equation of the plane is ¡2x +7y +3z = 17. 20. A normal vector to the plane can be found by taking the cross product of the vectors ¡¡!PQ = 2 4 ¡1 1 1 3 5 and ¡!PR = 2 4 1 0 1 3 5. n =¡¡!PQ£¡!PR = 1i+2j¡k. Therefore, the equation of the plane is x +2y ¡z = 6. 21. v = 2 4 2=3 1=3 ¡2=3 3 5 22. v = 2 4 2=3 ¡2=3 1=3 3 5 23. x +2y ¡2z = 17 2.5. SUPPLEMENTARY EXERCISES 55 24. x + y +3z = 11 25. x = 4¡t y = 5+ t z = t 26. x =¡3+ t y = 2¡t z = t 2.5 Supplementary Exercises 1. Solving the system of equations c1u+c2v = x for c1 and c2 we flnd that c1 = 3 and c2 =¡2. Thus x1 = 3u = ? 15 6 ? and x2 =¡2v = ? ¡14 ¡2 ? . 2. The island dock is at (5;5p3). Assuming that the boat travels t miles west and t miles north to get to the buoy, the buoy is at (5¡ t;5p3+ t). The distance from the mainland dock to the buoy is q (5¡t)2 +(5p3+ t)2 = q 2t2 +(10p3¡2)t +100 miles. 3. R = (13;¡1) 4. kak2 = a¢a = (2u+3v)¢(2u+3v) = 4(u¢u)+12(u¢v)+9(v¢v) = 4kuk2 +12(0)+9kvk2 = 4+9 = 13 Since kak2 = 13 then kak= p13. 5. kak2 = a¢a = (u+v +w)¢(u+v +w) = (u¢u)+2(u¢v)+2(u¢w)+2(v¢w)+(v¢v)+(w¢w) = 22 +2(0)+2(0)+2(0)+12 +22 = 4+1+4 = 9 Since kak2 = 9 then kak= 3. 6. (u¡v)¢(u+v) = (u¢u)¡(v¢v) = 4¡9 =¡5 Therefore (u¡v)¢(u+v) =¡5. 7. Note that since u¢v = 0 then u and v are perpendicular. So the angle between u and v, , is ?=2. Therefore, ku£vk=kukkvksin( ) = 2¢3¢1 = 6. 56 CHAPTER 2. VECTORS IN 2-SPACE AND 3-SPACE 8. Note that since u and v are perpendicular then ku¡vk=ku+vk= p13. Then k(u¡v)£(u+v)k2 = ku¡vk2 ku+vk2 sin2 ( ) = 169(1¡sin2 ( ) But cos( ) = (u¡v)¢(u+v)ku¡vkku+vk and from Exercise 6 we have that (u¡v¢u+v) =¡5. Therefore, k(u¡v)£(u+v)k2 = 169(1¡(¡5=13)2) = 169¡25 = 144. Thus k(u¡v)£ (u+v)k= 12. 9. Unit vectors in the xz¡plane are of the form v = 2 4 x 0 y 3 5 where px2 + z2 = 1. For v to be perpendicular to 2 4 3 ¡2 4 3 5 requires that 2 4 x 0 y 3 5¢ 2 4 3 ¡2 4 3 5 = 0 or 3x +4y = 0. Hence, v is of the form v = k 2 4 ¡4=3 0 1 3 5 for some number k. Since v is a unit vector then k = §3=5. Therefore v = 2 4 ¡4=5 0 3=5 3 5 or v = 2 4 4=5 0 ¡3=5 3 5. 10. Letting Pa = (a;0;0), Pb = (0; b;0) and Pc = (0;0; c) then ¡¡¡!PaPb = 2 4 ¡a b 0 3 5 and ¡¡¡!P aPc = 2 4 ¡a 0 c 3 5. Thus a normal vector to the plane is n =¡¡¡!PaPb £¡¡¡!PaPc = (bc)i+(ac)j+(ab)k. Then the equation of the plane is (ab)x+(ac)y +(ab)z = abc. Dividing through by abc the equation of the plane can be rewritten as 1 a ¶ x + 1 b ¶ y + 1 c ¶ z = 1. 11. Letting P1 = (1;1;2), P2 = (2;3;9), P3 = (¡2;1;¡1) and P4 = (1;2;5), then the equation oftheplanedeterminedby P1, P2 and P3 is¡x¡3y+z =¡2. Since P4 satisflestheequation of this plane, i. e. ¡1(1)¡3(2)+1(5) =¡2 then P4 is also in the plane. Therefore, all four points lie in the plane ¡x¡3y + z =¡2. 12. Consider a circle of radius r in the xy¡plane centered at the origin. Letting A = (0; r), B = (0;¡r) and C = (x; y) where x2 + y2 = r2, then the vector ¡!AC = ? x y ¡r ? and the 2.6. CONCEPTUAL EXERCISES 57 vector ¡¡!BC = ? x y + r ? . Since ¡!AC¢¡¡!BC = x2 +(y+r)(y¡r) = (x2 +y2)¡r2 = r2¡r2 = 0 then the vectors ¡!AC and ¡¡!BC are orthogonal and therefore, the triangle 4ABC is a right triangle. 13. Let the midpoints of the four sides of the quadrilateral ABCD be P1 = (a1 + b12 ; a2 + b22 ), P1 = (b1 + c12 ; b2 + c22 ), P1 = (c1 + d12 ; c2 + d22 ), and P1 = (d1 + a12 ; d2 + a22 ). Then note that the line segments P1P3 and P2P4 share the same midpoint, namely (a1 + b1 + c1 + d14 ; a2 + b2 + c2 + d24 ). 2.6 Conceptual Exercises 1. False. For example, if u = 2i +4j and v = ¡2i + j then u¢v = 0 while both u and v are nonzero. 2. False. For example, if u = 2i+j+2k and v = 3u = 6i+3j+6k then u£v = 0 while both u and v are nonzero. 3. Let u = 2 4 u1 u2 u3 3 5 and v = 2 4 v1 v2 v3 3 5. Then ku+vk2+ku¡vk2 = u21 +2u1v1 + v21 + u22 +2u2v2 + v22 + u23 +2u3v3 + v23 +u21 ¡2u1v1 + v21 + u22 ¡2u2v2 + v22 + u23 ¡2u3v3 + v23 = 2(u21 + u22 + u23)+2(v21 + v22 + v23) = 2kuk2 +2kvk2 4. The vectors u and v form the sides of a parallelogram while the vectors u+v and u¡v form the diagonals of the parallelogram. Thus, the parallelogram law states that the sum of the squares of the sides of a parallelogram equals the sum of the squares of the diagonals. In otherwords, theparallelogramlawistheapplicationofthelawofcosinestoaparallelogram. 5. For example, u = i+j+k, v = 2i¡j+k, and w = i¡2j¡k. Then u£(v£w) =¡6i+6j and (u£v)£w =¡7i¡j¡5k. Thus u£(v£w)6= (u£v)£w. 6. xi = Ai ¢bA i ¢Ai for i = 1;2;3. 7. It is equivalent to show that the columns of A form a linearly independent set (see Section 1.7). That is, if c1A1 + c2A2 + c3A3 = 0 then the only solution is c1 = c2 = c3 = 0. But due to Exercise 6. we have that ci = Ai ¢0A i ¢Ai = 0. Therefore, the columns of A are linearly independent and so A is nonsingular. 58 CHAPTER 2. VECTORS IN 2-SPACE AND 3-SPACE 8. Due to Exercise 7. the matrix A is nonsingular. Thus the solution of Ax = b can be written as x = A¡1b. From Exercise 6., we have that xi = Ai¢b (since Ai¢Ai = 1 for each i). Then let A = 2 4 a11 a12 a13 a21 a22 a23 a31 a32 a33 3 5 so Ai = 2 4 ai1 ai2 ai3 3 5. Then xi = Ai ¢ b = A1ib1 + A2ib2 + A3ib3. This leads to the following system of equations for x1, x2 and x3: x1 = A11b1 + A21b2 + A31b3 x2 = A12b1 + A22b2 + A32b3 x3 = A13b1 + A23b2 + A33b3 In matrix form these equations are written as x = 2 4 a11 a21 a31 a12 a22 a32 a13 a23 a33 3 5b = ATb. Since we already know that x = A¡1b then it must be true that A¡1 = AT. 9. It is equivalent to show that kAxk2 =kxk2. Recall that the dot product of u and v can be expressed as (u¢v) = uTv. Then kAxk2 = (Ax)¢(Ax) = (Ax)T(Ax) = xTATAx = xTx = x¢x =kxk2 10. (Au)¢(Av) = (Au)T(Av) = uTATAv = uTv = u¢v. 11. To show that the angle 1 between the vectors u and v equals the angle 2 between the vectors Au and Av consider that cos( 1) = u¢vkukkvk = Au¢AvkAukkAvk = cos( 2). Since cos( 1) = cos( 2) it follows that 1 = 2. 12. (u¡v)¢(u+v) = (u¢u)¡(v¢v) =kuk2¡kvk2 = 0 since kuk=kvk. Therefore u and v are orthogonal. Chapter 3 The Vector Space Rn 3.1 Introduction 12. Geometrically, W consists of the points in the plane that lie on the line with equation x + y = 1. 13. Geometrically, W consists of the points in the plane that lie on the line with equation x =¡3y. 14. Geometrically, W consists of the points in the plane that lie on the y¡axis. 15. Geometrically, W consists of the points in the plane that have coordinates (x; y) satisfying x + y ?0. 16. Geometrically, W consists of the points in the plane that lie on the line passing through the origin and the point (1;3). [Let t = 0 and t = 1; respectively. If (x; y) is a point on the line then x = t and y = 3t, so the equation for the line is y = 3x. 17. Geometrically, W consists of the points in the plane that lie on the circle with equation x2 + y2 = 4. 18. Geometrically, W consists of the points in three-space that lie on positive x¡axis. 19. Geometrically, W consists of the points in three-space that lie on the plane with equation x + y +2z = 0. 20. Geometrically, W consists of the points in three-space that lie on the line which passes through the origin and the point with coordinates (2;0;1). The line can be represented by the equations x = 2r; y = 0; z = r: 21. Geometrically, W consists of the points in three-space that are on or above the xy¡plane and that lie on the sphere with equation x2 + y2 + z2 = 1. 60 CHAPTER 3. THE VECTOR SPACE RN 22. W =fx: x= ? a b ? ; a¡2b = 1g. 23. W =fx: x= ? a 0 ? ; a any real number g. 24. W =fx: x= ? a b ? ; b > 0g. 25. W =fx: x= ? a 2 ? ; a any real number g. 26. W =fx: x= ? a b ? ; b = a2g. 27. W =fx: x= 2 4 x1 x2 x3 3 5; x1 + x2 ¡2x3 = 0g. 28. W =fx: x= t 2 4 2 ¡3 1 3 5; t any real numberg. 29. W =fx: x= 2 4 0 x2 x3 3 5; x2; x3 any real number g. 30. W =fx: x= 2 4 x1 2 x3 3 5; x1; x3 any real number g. 3.2 Vector Space Properties of Rn 1. Clearly is in W . Suppose u and v are in W , where u = ? u 1 u2 ? and v = ? v 1 v2 ? . Then u1 = 2u2 and v1 = 2v2 . If a is any scalar then u+v = ? u 1 + v1 u2 + v2 ? and au = ? au 1 au2 ? : But u1 +v1 = 2u2 +2v2 = 2(u2 +v2) and au1 = a(2u2) = 2(au2); so u+v and au are in W: By Theorem 2, W is a subspace of R2: Geometrically, W consists of the points on the line with equation x = 2y: 3.2. VECTOR SPACE PROPERTIES OF RN 61 2. W isnotasubspaceof R2: Noteforexamplethat isnotin W: Alsoif u = ? 2 0 ? andv = ? 0 ¡2 ? ; then u and v are in W; but u+v is not in W: 3. W is not a subspace of R2 since, for example, u = ? 1 1 ? and v = ? 1 ¡1 ? are in W whereas u+v is not in W: Note that W satisfles properties (s1) and (s3) of Theorem 2. 4. W is not a subspace of R2: For example, u = ? 1 1 ? is in W; but p2u is not in W: Note that W satisfles properties (s1) and (s2) of Theorem 2. 5. Clearly, = ? 0 0 ? is in W: If u = ? u 1 u2 ? and v = ? v 1 v2 ? are in W; then u1 = v1 = 0: Thus u+v = ? 0 u2 + v2 ? is in W: Likewise, if a is any scalar, au = ? 0 au2 ? is in W: By Theorem 2, W is a subspace of R2: Geometrically, W consists of the points on the y¡axis . 6. If u = ? u 1 u2 ? is in W; then j u1 j+j u2 j= 0 so u1 = u2 = 0: Thus W =f g and W is a subspace of R2: 7. W satisfles none of the properties (s1) - (s3) of Theorem 2, so W is not a subspace of R2: Clearly, = ? 0 0 ? is not in W: Also, u = ? 1 0 ? and v = ? 0 1 ? are in W whereas u+v is not in W: Finally, if a 6= 1 then av is not in W: 8. W is not a subspace of R2: For example u = ? 1 0 ? and v = ? 0 1 ? are in W whereas u+v is not in W: Note that W satisfles the properties (s1) and (s3) of Theorem 2. 9. Clearly = 2 4 0 0 0 3 5 is in W: If u = 2 4 u1 u2 u3 3 5 and v = 2 4 v1 v2 v3 3 5 are in W and a is any scalar, then u3 = 2u1 ¡ u2 and v3 = 2v1 ¡ v2: Now u3 + v3 = (2u1 ¡ u2)+(2v1 ¡ v2) = 2(u1 + v1)¡(u2 + v2) and au3 = a(2u1 ¡u2) = 2(au1)¡ au2: Therefore u+v and au are in W: By Theorem 2, W is a subspace of R3: Geometrically, W consists of the points on the plane with equation 2x¡y ¡z = 0: 62 CHAPTER 3. THE VECTOR SPACE RN 10. Clearly = 2 4 0 0 0 3 5 is in W: If u = 2 4 u1 u2 u3 3 5 and v = 2 4 v1 v2 v3 3 5 are in W; it follows that u2 +v2 = (u3 +v3)+(u1 +v1) and for any scalar a; au2 = au3 +au1: Thus u+v and au are in W: By Theorem 2, W is a subspace of R3: Geometrically, W consists of the points in the plane with equation x¡y + z = 0: 11. W is not a subspace of R3: For example, u = 2 4 1 1 1 3 5 and v = 2 4 1 0 0 3 5 are in W but u+v is not in W: Also if a 6= 1 and a 6= 0 then au is not in W: 12. Clearly = 2 4 0 0 0 3 5 is in W: If u = 2 4 u1 u2 u3 3 5 and v = 2 4 v1 v2 v3 3 5 are in W then u1 = 2u3 and v1 = 2v3: Therefore u1 + v1 = 2(u3 + v3) and, for any scalar a; au1 = 2au3: Thus u+v and au are in W and by Theorem 2, W is a subspace of R3: Geometrically, W consists of the points on the plane with equation x¡2z = 0: 13. W is not a subspace of R3: For example u = 2 4 1 0 0 3 5 and v = 2 4 2 2 0 3 5 are in W but u+v is not in W: Also, if a 6= 1 and a 6= 0 then au is not in W: 14. Clearly is in W: Vectors u and v in W can be written in the form u = 2 4 u1 0 u3 3 5 and v = 2 4 v1 0 v3 3 5: Thus u+v = 2 4 u1 + v1 0 u3 + v3 3 5 is in W: Likewise, for any scalar a; au = 2 4 au1 0 au3 3 5 is in W: Geometrically, W consists of the points in the xz¡ plane . 15. Clearly is in W: If u and v are vectors in W then u and v can be expressed in the form u = 2 4 2a ¡a a 3 5 and v = 2 4 2b ¡b b 3 5: Then u+v = 2 4 2(a + b) ¡(a + b) a + b 3 5: Similarly, for any scalar c; cu = 2 4 2ca ¡ca ca 3 5: By Theorem 2, W is a subspace of R3: Geometrically W consists of the points on the line with parametric equations x = 2t; y =¡t; z = t: 3.2. VECTOR SPACE PROPERTIES OF RN 63 16. Clearly is in W: Elements u and v in W can be expressed in the form u = 2 4 a 2a 2a 3 5 and v = 2 4 b 2b 2b 3 5: Therefore u+v = 2 4 a + b 2(a + b) 2(a + b) 3 5 is in W and, for any scalar c; cu = 2 4 ca 2ca 2ca 3 5 is in W: By Theorem 2, W is a subspace of R3: Geometrically, W consists of the points on the line with parametric equations x = t; y = 2t; z = 2t: 17. Clearly is in W: Moreover, any two elements u and v in W can be written in the form u = 2 4 a 0 0 3 5 and v = 2 4 b 0 0 3 5: Therefore u+v = 2 4 a + b 0 0 3 5 is in W and for any scalar c; cu = 2 4 ca 0 0 3 5 is in W?. By Theorem 2, W is a subspace of R3: Geometrically, W consists of the points on the x¡ axis. 18. Clearly aT = 0 so is in W?. If u and v are in W then aTu= 0 and aTv= 0: It follows that aT(u+v) = aT u+aT v= 0; so u+v is in W: If c is a scalar, then aT (cu ) = caTu= 0 and hence cu is in W: This proves that W is a subspace of R3: 19. Thevector u = 2 4 u1 u2 u3 3 5 isin W ifandonlyif0 = aTu= u1+2u2+u3: Thusgeometrically W consists of the points in R3 which lie in the plane x +2y +3z = 0: 20. The vector u = 2 4 u1 u2 u3 3 5 is in W if and only if 0 = aTu= u1: Thus geometrically W consists of thepoints in R3 which lie on the yz- plane: 21. Clearly, a T = bT = 0; so is in W: Let aTu= bTu= aTv= bTv= 0: It then follows that aT(u+v)= 0 and bT(u+v)= 0: Therefore u+v is in W: Likewise, aT(cu) = c(aTu ) = 0 and bT (cu) = c(bTu) = 0 for any scalar c: Therefore cu is in W: Thus W is a subspace of R3: 64 CHAPTER 3. THE VECTOR SPACE RN 22. The vector u = 2 4 u1 u2 u3 3 5 is in W if and only if 0 = aTu= u1 ¡ u2 +2u3 and 0 = bTu = 2u1 ¡ u2 +3u3: Thus W is the set of points in R3 formed by the intersecting planes x¡y+2z = 0 and 2x¡y+3z = 0: The parametric equations for the line are x =¡t; y = t; z = t: 23. The vector u = 2 4 u1 u2 u3 3 5 is in W if and only if 0 = aTu= u1 +2u2 +2u3 and 0 = bTu = u1 + 3u2: Thus W is the set of points on the line formed by the intersecting planes x + 2y + 2z = 0 and x + 3y = 0: Solving yields x = ¡6z and y = 2z so the line has parametric equations x =¡6t; y = 2t; z = t: 24. The vector u = 2 4 u1 u2 u3 3 5 is in W if and only if 0 = aTu= u1 + u2 + u3 and 0 = bTu = 2u1 +2u2 +2u3: Clearly the latter condition is redundant so W consists of the points on the plane x + y + z = 0: 25. The vector u = 2 4 u1 u2 u3 3 5 is in W if and only if 0 = aTu = u1 ¡ u3 and 0 = bTu =¡2u1 +2u3: Clearly the latter condition is redundant so W consists of the points in the plane x¡z = 0: 27. Property (m1) is not satisfled. For example 3(2x) = 3 ? 4x 1 4x2 ? = ? 24x 1 24x2 ? where 6x = ? 12x 1 12x2 ? . Also, (m4) is not satisfled since 1x = ? 2x 1 2x2 ? 6= x. 28. Property (c2) is not satisfled. For example, if x = ? 1 1 ? and a = ¡1 then x is in W but ax is not in W. This also illustrates that (a4) is not satisfled. 29. The set of points on the line can be expressed as the set W = ft 2 4 a b c 3 5 : t any real number g. Taking t = 0 we see that is in W: If u= r 2 4 a b c 3 5 and v = s 2 4 a b c 3 5 then u+v= (r + s) 2 4 a b c 3 5 is in W: Likewise, if k is any scalar then ku = kr 2 4 a b c 3 5 is in W: Therefore W is a subspace of R3: 3.3. EXAMPLES OF SUBSPACES 65 30. Since is in both U and V; = + is in U + V: Suppose x and y are in U + V and write x= u1+v1 ; y= u2+v2 where u1 ;u2 are in U and v1 ; v2 are in V: Then x +y = (u1+u2)+(v1+v2) is in U +V: If a is a scalar then ax = au1+av1 is in U +V: It follows that U + V is a subspace of Rn: 31. Clearly is in U \V: Suppose x and y are in U \V: Then x and y are in U and since U is a subspace, x+y is in U: Also for any scalar a; ax is in U: Similarly, x+y and ax are in V: Therefore x +y and ax are in U \ V: It follows that U \ V is a subspace of Rn: 32. The vector u = 2 4 1 ¡1 0 3 5 is in U and the vector v = 2 4 0 1 1 3 5 is in V: Thus u and v are in U [V but u+v is in neither U nor V: 33. (a) Clearly is in U [V: Suppose x is in U [V and let a be a scalar. If x is in U; then ax is also in U: Similarly, if x is in V; then ax is in V: In either case, ax is in U [V: (b) Assume that u+v is in U: Since ¡u is in U and U is closed under addition we see that v= (u+v)+(¡u) is in U: This contradicts the assumption that v is not in U: Similarly, u+v is not in V: 34. Since W is non-empty, W contains a vector x: By (s3) the vector 0x= is in W: By Theorem 2, W is a subspace of Rn: 3.3 Examples of Subspaces 1. By deflnition Sp(S) = ft ? 1 ¡1 ? : t any real number g: Thus if x = ? x 1 x2 ? is in R2 , then x is in Sp(S) if and only if x1 + x2 = 0. In particular Sp(S) is the line with equation x + y = 0: 2. By deflnition Sp(S) = ft ? 2 3 ? : t any real number g: If x = ? x 1 x2 ? is in R2 , then x is in Sp(S) if and only if 3x1 +2x2 = 0: In particular Sp(S) is the line with equation 3x +2y = 0: 3. Sp(S) = ft ? 0 0 ? : t any real number g = feg:Sp(S) is the point (0;0): 4. Sp(S) = fx in R2 : x = k1a + k2b for scalars k1 and k2 g: For an arbitrary vector x in R2; x = ? x 1 x2 ? ; x = k1a+ k2b if k1 = 3x1 + 2x2 and k2 = ¡x1 ¡x2 : It follows that Sp(S) = R2 : 66 CHAPTER 3. THE VECTOR SPACE RN 5. Sp(S) = fx in R2 : x = k1a + k2d for scalars k1 and k2g: For an arbitrary vector x in R2; x = ? x 1 x2 ? ; the equation k1a + k2d = x has augmented matrix ? 1 1 x 1 ¡1 0 x2 ? : This matrix reduces to? 1 1 x1 0 1 x1 + x2 ? and backsolving yields k1 = ¡x2 ; and k2 = x1 + x2 : It follows that Sp(S) = R2 : 6. If x = ? x 1 x2 ? the equation k1a + k2c = x has augmented matrix ? 1 ¡2 x 1 ¡1 2 x2 ? : This matrix reduces to ? 1 ¡2 x 1 0 0 x1 + x2 ? ; so the equation has a solution if and only if x1 + x2 = 0: It follows that Sp(S) =fx: x1 + x2 = 0g?; that is, Sp(S) is the line with equation x + y = 0: 7. Sp(S) = fx in R2 : x = k1b + k2e for scalars k1 and k2g: But k1b + k2e = k1b so Sp(S) = Sp(fbg): It follows that Sp(S) =ft ? 2 ¡3 ? : t any real numberg. If x = ? x 1 x2 ? is in R2; x is in Sp(S) if and only if 3x1 +2x2 = 0: Thus, Sp(S) =fx : 3x1 +2x2 = 0g; so Sp(S) is the line with equation 3x +2y = 0: 8. If x = ? x 1 x2 ? the equation k1a+ k2b+ k3d = x has augmented matrix ? 1 2 1 x 1 ¡1 ¡3 0 x2 ? : This matrix reduces to ? 1 2 1 x 1 0 ¡1 1 x1 + x2 ? : It follows that the system is consistent for arbitrary x , so Sp(S) = R2 : 9. Sp(S) = fx in R2 : x = k1b + k2c + k3d for scalars k1 ; k2 ; k3g: With x = ? x 1 x2 ? ; the equation k1b+ k2c+ k3d = x has augmented matrix? 2 ¡2 1 x1 ¡3 2 0 x2 ? : The reduction reveals that the system is consistent for every x , so Sp(S) = R2 : 10. Since k1a+ k2b+ k3e = k1a+ k2b , Sp(S) = Spfa;bg= R2: 11. Since k1a+k2c+k3e = k1a+k2c; Sp(S) = Spfa;cg: But Spfa;cg=fx : x1 +x2 = 0g: Thus Sp(S) is the line with equation x + y = 0: (cf. Exercise 6). 12. Sp(S) =fx : x= tv; t a scalarg so Sp(S) is the line through (0;0;0) and (1;2;0): The parametric equations for the line are x = t; y = 2t; z = 0: Equivalently, if x= [x1; x2; x3]T; then Sp(S) = fx : ¡2x1 + x2 = 0 and x3 = 0g: Thus Sp(S) the line formed by the intersecting planes ¡2x + y = 0 and z = 0: 3.3. EXAMPLES OF SUBSPACES 67 13. Sp(S) =fx in R3 : x= tw forsomescalar t g.ThereforeSp(S) isthelinethrough (0;0;0) and (0;¡1;1): The parametric equations are x = 0; y = ¡t; z = t: If x= [x1 ;x2 ;x3 ]T then tw= x has augmented matrix 2 4 0 x1 ¡1 x2 1 x3 3 5 which reduces to 2 4 1 x3 0 x2 + x3 0 x1 3 5: The system is consistent if x2 +x3 = 0 and x1 = 0 so we have Sp(S) =fx : x2 +x3 = 0 and x1 = 0g: Therefore Sp(S) is the line formed by the the intersecting planes y + z = 0 and x = 0: 14. For x = [x1 ;x2 ;x3 ]T the equation k1v + k2w = x is consistent if and only if ¡2x1 + x2 + x3 = 0: Thus Sp(S) = fx : ¡2x1 + x2 + x3 = 0g and is the plane with equation ¡2x + y + z = 0: 15. Sp(S) =fu in R3 : u = k1v+k2xg: The equation k1v+k2x = u has augmented matrix2 4 1 1 u1 2 1 u2 0 ¡1 u3 3 5 , which reduces to 2 4 1 1 u1 0 1 2u1 ¡u2 0 0 2u1 ¡u2 + u3 3 5: Asolutionexistsifandonlyif2u1¡u2+u3 = 0 , soSp(S) =fu : 2u1 ¡u2 + u3 = 0g: Sp(S) is the plane with equation 2x¡y + z = 0: 16. For arbitrary u in R3 the equation k1v+k2w+k3x = u has a solution, so Sp(S) = R3 : 17. Sp(S) = fu in R3 : u = k1w + k2x + k3zg: For u = [u1 ;u2 ;u3]T the equation k1w + k2x+ k3z = u has augmented matrix 2 4 0 1 1 u1 ¡1 1 0 u2 1 ¡1 2 u3 3 5: This matrix reduces to 2 4 1 ¡1 2 u3 0 1 1 u1 0 0 2 u2 + u3 3 5: Since the system is consistent for every u in R3, Sp(S) = R3 : 18. For u = [u1 ;u2;u3]T the system of equations k1v + k2w + k3z = u is consistent if and only if ¡2u1 + u2 + u3 = 0: Therefore Sp(S) = fu : ¡2u1 + u2 + u3 = 0g and Sp(S) is the plane with equation ¡2x + y + z = 0: 19. The matrix 2 4 0 1 ¡2 u1 ¡1 1 ¡2 u2 1 ¡1 2 u3 3 5 reduces to 68 CHAPTER 3. THE VECTOR SPACE RN 2 4 1 ¡1 2 u3 0 1 ¡2 u1 0 0 0 u2 + u3 3 5 so the system of equations k1w+k2x+k3y = u is consistent if and only if u2 + u3 = 0: Therefore Sp(S) = fu : u2 + u3 = 0g and Sp(S) is the plane with equation y + z = 0: 20. By Exercise 14, Sp(S) = fx : ¡2x1 + x2 + x3 = 0g: Then the vectors given in (a), (c) and (d) are in Sp(S): Moreover when the system of equations k1v+k2w = x is consistent, the unique solution is k1 = x1 ; andk2 = x3 : Thus in (a) [1;1;1]T = v + w; in (c) [1;2;0]T = v; and in (d), [2;3;1]T = 2v +w: 21. By Exercise 15, Sp(S) = fu in R3 : 2u1 ¡u2 + u3 = 0g: Thus the vectors given in (b), (c), and (e) are in Sp(S): From the calculations done in Exercise 15, it follows that when the system of equations k1v+k2x = u is consistent, the unique solution is k1 =¡u1 +u2 and k2 = 2u1 ¡ u2 : Thus in (b), [1;1;¡1]T = x; in (c), [1;2;0]T = v; and in (e), [¡1;2;4]T = 3v ¡4x: 22. The vectors a;c; and e are in N(A): 23. The vectors d and e are in N(A) since by direct calculation, Ad = and Ae = . 24. The vectors v;w; and z are in N(A): 25. The vectors x and y are in N(A) since, by direct calculation Ax = and Ay = . 26. The homogeneous system Ax = has solution x1 = 2x2 , where x2 is arbitrary. Thus N(A)=f x in R2 : x1 ¡2x2 = 0g: If b= ? b 1 b2 ? , then Ax=b is consistent if and only if 3b1 +b2 = 0 , so R(A) =fb in R2 : 3b1 +b2 = 0g: 27. The matrix [A j b] is row equivalent to the matrix :? ¡1 3 b1 0 0 2b1 + b2 ? : It follows that the homogeneous system Ax = has solution x1 = 3x2 whereas the system Ax = b is consistent if and only if 2b1 + b2 = 0: Therefore N(A) = fx : x1 ¡3x2 = 0g and R(A) = fb : 2b1 + b2 = 0g: 28. N(A) =f g and R(A) = R2 : 29. The matrix [A j b] reduces to ? 1 1 b 1 0 3 ¡2b1 + b3 ? : Setting b= and solving yields x1 = x2 = 0 , so N(A)=f g. Since the system Ax= b is consistent for every b in R2;R(A) = R2 : 3.3. EXAMPLES OF SUBSPACES 69 30. N(A) =fx inR3 : x1 =¡3x3 and x2 =¡x3g and R(A) = R2 : 31. The matrix [A j b] reduces to ? 1 2 1 b 1 0 0 1 ¡3b1 + b2 ? : Setting b = and backsolving yields x1 =¡2x2 ; x3 = 0 asthesolutionto Ax= . ThusN(A)=fx in R3 : x1 +2x2 = 0 and x3 = 0g: Since Ax=b is consistent for arbitrary b in R2, R(A)=R2 : 32. The homogeneous system Ax = has only the trivial solution so N(A) = f g. The system Ax = b is consistent precisely when 3b1 ¡ 2b2 + b3 = 0 so R(A) = fb in R3 : 3b1 ¡2b2 + b3 = 0g: 33. The matrix [A j b] reduces to 2 4 0 1 b1 0 0 ¡2b1 + b2 0 0 ¡3b1 + b3 3 5: Setting b= yields x2 = 0; x1 arbitrary as the solution to Ax = . ThusN(A)=fx in R2 : x2 = 0g: The system Ax = b is consistent if and only if¡2b1+b2 = 0 and¡3b1+b3 = 0: Therefore R(A) = fb in R3 : b2 = 2b1 and b3 = 3b1g: 34. N(A) =fx in R3 : x1 =¡7x3 and x2 =¡3x3 g andR(A) =fb in R3 : 3b1¡2b2+b3 = 0g: 35. The matrix [A j b] reduces to 2 4 1 2 3 b1 0 1 ¡2 ¡b1 + b2 0 0 0 ¡4b1 +2b2 + b3 3 5: Setting b= and backsolving the reduced system yields N(A) =fx in R3 : x1 =¡7x3 and x2 = 2x3g: The system Ax = b is consistent if and only if ¡4b1 +2b2 + b3 = 0 so R(A) =fb in R3 :¡4b1 +2b2 + b3 = 0g: 36. N(A)=f g and R(A)= R3: 37. The matrix [A j b] reduces to 2 4 1 2 1 b1 0 1 2 ¡2b1 + b2 0 0 1 b1 ¡b2 + b3 3 5: Setting b= and solving yields N(A) =f g. The system Ax = b is consistent for all b so R(A) = R3: 38. (a) The vectors b in (i), (iv), and (vi) are in R(A): (b) For (i), x = [1;0]T is one choice; for (iv), x = [¡2;0]T is one choice; for (vi), x = [0;0]T is one choice. (c) For (i), b= A1 ; for (iv), b=¡2A1 , for (vi), b= 0A1 +0A2 : 39. (a) From the description of R(A) obtained in Exercise 27 it follows that the vectors b in (ii),(v), and (vi) are in R(A): 70 CHAPTER 3. THE VECTOR SPACE RN (b) When the system of equations Ax = b is consistent, the calculations done in Exercise 27 show that the solution is given by x1 =¡b1 +3x2 , where x2 is arbitrary. Thus for (ii), x= [1;0]T is one choice; for (v), x= [0;1]T is one choice; for (vi), x = [0;0]T is one choice. (c) If Ax = b, where x= [x1; x2]T , then b = x1A1 +x2A2: Therefore for (ii), b = A1 ?; for (v), b = A2; for (vi), b = 0A1 +0A2: 40. (a) The vectors b in (ii), (iii), (iv), and (vi) are in R(A): (b) For (ii), x= [¡1;¡1;0]T is one choice; for (iii), x= [2;¡1;0]T is one choice; for (iv), x= [2;1;0]T is one choice; for (vi), x= [0;0;0]T is one choice. (c) For (ii), b=¡A1¡A2 ; for (iii), b= 2A1¡A2 ; for (iv), x= 2A1 +A2 ; for (vi), x= 0A1 +0A2 +0A3 : 41. (a) From the description of R(A) obtained in Exercise 35, the vectors b in (i), (iii), (v), and (vi), are in R(A): (b) When the system Ax= b is consistent, the solution is given by x1 = 3b1¡2b2¡7x3 and x2 = ¡b1 + b2 +2x3 , where x3 is arbitrary. Thus for (i), x= [¡1;1;0]T is one choice; for (iii), x= [¡2;3;0]T is one choice for (v), x= [¡2;1;0]T is one choice; for (vi), x= [0;0;0]T is one choice. (c). If Ax= b, where x= [x1; x2; x3]T , then b = x1A1 + x2A2 + x3A3: Thus it follows from (b) that for (i), b= ¡A1 +A2?; for (iii), b= ¡2A1 +3A2?; for (v), b=¡2A1 +A2?; for (vi), b= 0A1 +0A2 +0A3 , 42. A = 2 4 2 ¡3 1 ¡1 4 ¡2 2 1 4 3 5 43. A = [ 3 ¡4 2 ] 44. A = [v;w;x] = 2 4 1 0 1 2 ¡1 1 0 1 ¡1 3 5 45. A = [w;x;z] = 2 4 0 1 1 ¡1 1 0 1 ¡1 2 3 5 46. Let A be the (3 x 3) matrix whose columns are the vectors given in S: Then AT reduces to BT = 2 4 1 0 ¡1 0 2 3 0 0 0 3 5: The nonzero columns of B; w1= [1;0;¡1]T; and w2= [0;2;3]T, form a basis for Sp(S): 3.3. EXAMPLES OF SUBSPACES 71 47. Let A be the (3 x 3) matrix whose columns are the vectors given in S. Then AT reduces to BT = 2 4 ¡2 1 3 0 3 2 0 0 0 3 5: The nonzero columns of B; w1 = [¡2;1;3]T and w2 = [0;3;2]T form a basis for Sp(S): 48. w1 = [1;0;1]T and w2 = [0;1;1]T: 49. Let A be the (3 x 4) matrix whose columns are the vectors given in S: Then AT reduces to BT = 2 66 4 1 2 2 0 3 1 0 0 0 0 0 0 3 77 5: The nonzero columns of B; w1= [1;2;2]T and w2= [0;3;1]T form a basis for Sp(S): 50. (a) R(I) = Rn and N(I)=f g. (b) R(O)=f g and N(O)= Rn: (c) R(A) = Rn and N(A)=f g. 51. Let x be in N(A)\N(B): Then Ax= and Bx= . Therefore (A + B)x= Ax+Bx = + ?= . It follows that x is in N(A + B): 52. (a) If Bx= then (AB)x= A(Bx) = A ?= . (b) Suppose b = (AB)x for some vector x in Rn: Then y = Bx is in Rr and b = A(Bx) = Ay: 53. Let m and n denote the zero vectors in Rm and Rn respectively. Then n is in W and m = A n: Therefore m is in V: Suppose u and v are in V: Then there exist vectors x and z in W such that u= Ax and v= Az: Since x+z is in W, u+v= Ax+Az = A(x+z) is in V: If a is a scalar then ax is in W so au= a(Ax) = A(ax) is in V: Thus V is a subspace of Rn: 54. If A has row vectors a1 ; : : : ;ak ; : : : ;am then B has row vectors a1 ; : : : ;cak ; : : : ;am : Note that d1a1 +¢¢¢+ dkak +¢¢¢+ dmam = d1a1 +¢¢¢+ (dk=c)cak +¢¢¢+ dmam : It follows that : Spfa1 ; : : : ;ak ; : : : ;amg= Spfa1 ; : : : ;cak ; : : : ;amg: 72 CHAPTER 3. THE VECTOR SPACE RN 3.4 Bases for Subspaces 1. Backsolving the given system yields x1 = x3 ¡x4; and x2 = x4: Thus 2 66 4 x1 x2 x3 x4 3 77 5 = 2 66 4 x3 ¡x4 x4 x3 x4 3 77 5 = x3 2 66 4 1 0 1 0 3 77 5+ x4 2 66 4 ¡1 1 0 1 3 77 5: As in Example 5,f[1;0;1;0]T;[¡1;1;0;1]Tg is a basis for W: 2. Backsolving yields x1 =¡x3 ¡2x4 and x2 = 2x3 + x4: It follows that f[¡1;2;1;0]T;[¡2;1;0;1]Tg is a basis for W: 3. Writing x1 = x2 ¡x3 +3x4 we have 2 66 4 x1 x2 x3 x4 3 77 5 = 2 66 4 x2 ¡x3 +3x4 x2 x3 x4 3 77 5 = x2 2 66 4 1 1 0 0 3 77 5+ x3 2 66 4 ¡1 0 1 0 3 77 5+ x4 2 66 4 3 0 0 1 3 77 5: Thus f[1;1;0;0]T;[¡1;0;1;0]T;[3;0;0;1]Tg is the desired basis. 4. Writing x1 = x2 ¡ x3 and noting that x1; x3 and x4 are unconstrained variables, we obain f[1;1;0;0]T;[¡1;0;1;0]T;[0;0;0;1]Tg as the desired basis. 5. Since x1 =¡x2 we have 2 66 4 x1 x2 x3 x4 3 77 5 = 2 66 4 ¡x2 x2 x3 x4 3 77 5 = x2 2 66 4 ¡1 1 0 0 3 77 5+ x3 2 66 4 0 0 1 0 3 77 5+ x4 2 66 4 0 0 0 1 3 77 5: It follows that f[¡1;1;0;0]T;[0;0;1;0]T;[0;0;0;1]Tg is a basis for W. 6. Backsolving yields x1 = 2x4; x2 = 2x4; x3 = x4: Thus f[2;2;1;1]Tg is a basis for W: 7. Backsolving yields x1 =¡2x3 ¡x4 and x2 =¡x3: Thus 2 66 4 x1 x2 x3 x4 3 77 5 = 2 66 4 ¡2x3 ¡x4 ¡x3 x3 x4 3 77 5 = x3 2 66 4 ¡2 ¡1 1 0 3 77 5+ x4 2 66 4 ¡1 0 0 1 3 77 5: Therefore f[¡2;¡1;1;0]T;[¡1;0;0;1]Tg is a basis for W: 3.4. BASES FOR SUBSPACES 73 8. Backsolving yields x1 =¡x4 and x2 =¡x3: Therefore the set f[¡1;0;0;1]T;[0;¡1;1;0]Tg is a basis for W. 9. Let fw1;w2g be the basis found in Exercise 1. (a) x = 2w1 + w2 (b) x is not in W. (c) x =¡3w2 (d) x = 2w1. 10. Let fw1;w2g be the basis found in Exercise 2. (a) x = w1 +w2 (b) x = 2w1 ¡w2 (c) x is not in W. (d) x =¡2w2. 11. (a) B = 2 4 1 2 3 ¡1 0 ¡1 ¡1 1 0 0 0 0 3 5. (b) Backsolving the reduced system Bx = yields the solution x1 = ¡x3 ¡ x4; x2 = ¡x3 + x4 for the homogeneous system Ax = . Thus x = [x1; x2; x3; x4]T is in N(A) if and only if 2 66 4 x1 x2 x3 x4 3 77 5 = 2 66 4 ¡x3 ¡x4 ¡x3 + x4 x3 x4 3 77 5 = x3 2 66 4 ¡1 ¡1 1 0 3 77 5+ x4 2 66 4 ¡1 1 0 1 3 77 5 It follows that f[¡1;¡1;1;0]T, [¡1;1;0;1]Tg is a basis for N(A). (c) It follows from (b) that x1A1+x2A2+x3A3+x4A4= if and only if x1 =¡x3¡x4 and x2 = ¡x3 + x4. Since x3 and x4 are unconstrained variables fA1;A2g is a basis for R(A). Setting x3 = 1 and x4 = 0 yields x1 =¡1 and x2 =¡1 so ¡A1¡A2 +A3 = . Therefore A3 = A1 +A2 : Similarly, setting x3 = 0 and x4 = 1 yields A4 = A1¡A2 : (d) The nonzero rows of B form a basis for the row space of A ; that is f[1;2;3;¡1];[0;¡1;¡1;1]g is the desired basis. 12. (a) B = 2 4 1 1 2 0 1 1 0 0 0 3 5: (b) The system Ax= has solution x1 =¡x3 and x2 =¡x3: Thereforef[¡1;¡1;1]Tg is a basis for N(A): (c) fA1 ;A2g is a basis for R(A) and A3 = A1 +A2 : (d) f[1;1;2];[0;1;1]g is a basis for the row space of A : 13. (a) B = 2 66 4 1 2 1 0 0 1 1 ¡1 0 0 0 0 0 0 0 0 3 77 5: 74 CHAPTER 3. THE VECTOR SPACE RN (b) The homogeneous system Ax= has solution x1 = x3 ¡2x4; x2 =¡x3 + x4: Thus x= [x1; x2; x3; x4]T is in N(A) if and only if 2 66 4 x1 x2 x3 x4 3 77 5 = 2 66 4 x3 ¡2x4 ¡x3 + x4 x3 x4 3 77 5 = x3 2 66 4 1 ¡1 1 0 3 77 5+ x4 2 66 4 ¡2 1 0 1 3 77 5: The set f[1;¡1;1;0]T;[¡2;1;0;1]Tg is a basis for N(A). (c) It follows from (b) that in the equation x1A1 +x2A2 +x3A3 +x4A4 = ; x3 and x4 are unconstrained variables. Therefore fA1 ;A2 g is a basis for R(A): Furthermore A1¡A2+A3= , so A3=¡A1+A2 : Likewise ¡2A1+A2+A4= , so A4= 2A1 ¡A2 : (d) The nonzero rows of B; [1;2;1;0];[0;1;1;¡1]; form a basis for the row space of A: 14. (a) B = 2 4 2 2 0 0 ¡1 1 0 0 1 3 5: (b) The system Ax= has only the trivial solution so N(A)=f g. (c) It follows from (b) that the columns of A are linearly independent so fA1 ;A2 ;A3g is a basis for R(A): (d) The set f[2;2;0];[0;¡1;1];[0;0;1]g is a basis for the row space of A: 15. (a) B = 2 4 1 2 1 0 0 ¡1 0 0 0 3 5: (b) The system Ax= has solution x1 =¡2x2; x3 = 0: Thus N(A)=fx: x= [¡2x2; x2;0]Tg and f[¡2;1;0]Tg is a basis for N(A): (c) In the equation x1A1 +x2A2 +x3A3 = , x2 is an unconstrained variable, so fA1 ;A3g is a basis for R(A): Furthermore, ¡2A1 +A2 = , so A2 = 2A1 : (d) f[1;2;1];[0;0;¡1]g is a basis for the row space of A: 16. (a) B = 2 4 2 1 2 0 1 ¡1 0 0 0 3 5: (b) x= [x1; x2; x3]T is in N(A) if and only if x1 = (¡3=2)x3 and x2 = x3: Therefore f[¡3=2;1;1]Tg is a basis for N(A): 3.4. BASES FOR SUBSPACES 75 (c) fA1 ;A2g is a basis for R(A) and A3 = (3=2)A1¡A2 : (d) f[2;1;2];[0;1;¡1]g is a basis for the row space of A: 17. The matrix AT is row equivalent to BT = 2 66 4 1 3 1 0 ¡1 ¡1 0 0 0 0 0 0 3 77 5: The desired basis is f[1;3;1]T; [0;¡1;¡1]Tg, formed by taking the nonzero columns of B: 18. The matrix AT is row equivalent to BT = 2 4 1 1 2 0 0 1 0 0 0 3 5: The desired basis is f[1;1;2]T;[0;0;1]Tg, formed by taking the nonzero columns of B: 19. The matrix AT is row equivalent to BT = 2 66 4 1 2 2 0 0 1 ¡2 1 0 0 0 0 0 0 0 0 3 77 5 so f[1;2;2;0]T;[0;1;¡2;1]Tg is a basis for R(A): 20. The matrix AT is row equivalent to BT = 2 4 2 2 2 0 ¡1 1 0 0 1 3 5 so f[2;2;2]T;[0;¡1;1]T;[0;0;1]Tg is a basis for R(A): 21. (a) For the given vectors u1 and u2 the equation x1u1 +x2u2 = has solution x1 = ¡2x2 where x2 is an unconstrained variable. Therefore fu1 g is a basis for Sp(S), where u1 = [1;2]T: (b) If A = [u1;u2] then AT is row equivalent to BT = ? 1 2 0 0 ? . Therefore f[1;2]Tg is a basis for Sp(S). 22. (a) For the given vectors u1 ;u2 and u3 the equation x1u1 +x2u2 +x3u3 = has solution x1 = (¡1=3)x3 and x2 = (¡4=3)x3; where x3 is arbitrary. Thus fu1 ;u2 g is a basis for Sp(S), where u1 = [1;2]T and u2 = [2;1]T. (b) If A = [u1 ;u2 ;u3 ] then AT is row equivalent to BT =2 4 1 2 0 ¡3 0 0 3 5: Therefore f[1;2]T;[0;¡3]Tg is a basis for Sp(S): 23. (a) For the given vectors u1 ;u2 ;u3 ;u4 the equation x1u1 +x2u2 +x3u3 +x4u4 = has solution x1 = ¡x3 ¡3x4; x2 = ¡x3 + x4: Since x3 and x4 are unconstrained variables, fu1 ;u2g is a basis for Sp(S); where u1 = [1;2;1]T and u2 = [2;5;0]T: 76 CHAPTER 3. THE VECTOR SPACE RN (b) If A = [u1 ;u2 ;u3 ;u4 ] then AT is row equivalent to BT = 2 66 4 1 2 1 0 1 ¡2 0 0 0 0 0 0 3 77 5: Therefore f[1;2;1]T;[0;1;¡2]Tg is a basis for Sp(S): 24. (a) For the given vectors u1 ;u2 ;u3 ;u4 , in the equation x1u1 +x2u2 +x3u3 +x4u4 = , x4 is an unconstrained variable. The desired basis is fu1 ;u2 ;u3g; where u1 = [1;2;¡1;3]T;u2 = [¡2;1;2;¡1]T; and u3 = [¡1;¡1;1;¡3]T: (b) If A = [u1 ;u2 ;u3 ;u4 ]; then AT reduces to BT =2 66 4 1 2 ¡1 3 0 1 0 0 0 0 0 5 0 0 0 0 3 77 5: Therefore f[1;2;¡1;3]T;[0;1;0;0]T; [0;0;0;5]Tg is a basis for Sp(S): 25. (a) Let A denote the given matrix. The homogeneous system Ax= has solution x1 = 0; x2 is arbitrary, x3 = 0: Thus f[0;1;0]Tg is a basis for N(A): (b) Let A denote the given matrix. The system Ax= has solution x1 = ¡x2 , where x2 and x3 are arbitrary. Thus f[¡1;1;0]T; [0;0;1]Tg is a basis for N(A): (c) The system Ax= has solution x1 = ¡x2; x3 = 0; where x2 is arbitrary. The set f[¡1;1;0]Tg is a basis for N(A): 26. (a) f[1;1]T;[0;1]Tg: (b) f[1;1]Tg: (c) f[1;1]T;[0;1]Tg: 27. The equation x1v1 +x2v2 +x3v3 = has solution x1 =¡2x3; x2 =¡3x3; x3 arbitrary. In particular, x1 =¡2; x2 =¡3; x3 = 1 is a nontrivial solution and the set S is linearly dependent. Moreover, from ¡2v1 ¡3v2 +v3 = we obtain v3 = 2v1 +3v2 : If v is in Sp(S) then v= a1v1 +a2v2 +a3v3 = (a1 +2a3)v1 +(a2 +3a3)v2 ; so v is in Spfv1;v2g: It follows that Spfv1 ;v2 ;v3g= Spfv1 ;v2g. 28. The subsets fv1;v2g, fv1;v3g, fv2;v3g are bases for R2. 29. The subsets are fv1;v2;v3g, fv1;v3;v4g, and fv1;v2;v4g. Note that v4 = 3v2 ¡v3. 30. By Theorem 12 of Section 1.8, the matrix V = [v1 ;v2 ;v3 ] is nonsingular. Thus, by Theorem 13 of Section 1.8, the system of equations Ax= b has a solution for each b in R3; that is each vector b in R3 can be written in the form x1v1+x2v2+x3v3= b: This shows that R3 = Sp(B) so B is a basis for R3: 3.5. DIMENSION 77 31. Set V = [v1 ;v2 ;v3]: By assumption the system V x= b has a solution for every b in R3 . By Theorem 13 of Section 1.8, V is a nonsingular matrix. Therefore, by Theorem 12 of Section 1.8, the set fv1 ;v2 ;v3g is linearly independent. 32. The set S is linearly independent so, by Exercise 30, S is a basis for R3: 33. The set S is linearly dependent so S is not a basis for R3: 34. The set S is linearly dependent so S is not a basis for R3: 35. If u= [u1; u2; u3]T then u is in Sp(S) if and only if 4u1 ¡2u2 +u3 = 0: In particular, Sp(S)6= R3 and S is not a basis for R3. 36. A vector w= [w1; w2; w3]T is in Spfv1;v2g if and only if w1 + w3 = 0: In particular w= [0;0;1]T is not a linear combination of v1 and v2 : 37. (a) By Theorem 11 of Section 1.8, any set of three or more vectors in R2 is linearly dependent and is not a basis for R2: (b) Suppose fvg is a basis for R2: Then e1 = a1v and e2 = a2v for some nonzero scalars a1 and a2: But then a2e1 ¡a1e2 = , contradicting the fact that fe1 ;e2 g is a linearly independent set. We conclude that fvg is not a basis for R2: It follows that every basis for R2 contains exactly two vectors. 38. If vT= [x1; x2; : : : ; xn] then the constraints vTui = 0;1? i? p; yield a homogeneous system of p equations in the unknowns x1; x2; : : : ; xn. By Theorem 4 of Section 1.4 the system has nontrivial solutions. Suppose v= a1u1+a2u2+¢¢¢+apup: Thenkvk2 = vTv= vT(a1u1+a2u2+¢¢¢+apup) = a1v Tu1 +a2v Tu2 +¢¢¢+ apv Tup = 0; contradicting that v is a nonzero vector. 39. By Theorem 11 of Section 1.8, any set of n + 1 or more vectors in Rn is linearly dependent so it is not a basis for Rn: By Exercise 38, any set of less than n vectors cannot span Rn: Therefore a basis for Rn must contain exactly n vectors. 3.5 Dimension 1. S contains only one vector and dim(R2) = 2, so by property 2 of Theorem 9, S does not span R2: 2. S does not span R2 by property 2 of Theorem 9 3. Since S contains three vectors and dim(R2) = 2, S is linearly dependent by property 1 of Theorem 9. 78 CHAPTER 3. THE VECTOR SPACE RN 4. S is linearly dependent by property 1 of Theorem 9. 5. Since u4 = , S is a linearly dependent set; for example 0u1 +au4 = for any nonzero scalar a. Also S does not span R2 since Spfu1;u4g= Spfu1g: 6. S is linearly dependent since, for example, 3u1¡u2 = . 7. S contains two vectors and dim(R3) = 3 so by property 2 of Theorem 9, S does not span R3: 8. S does not span R3 by property 2 of Theorem 9. 9. Since S contains four vectors and dim(R3) = 3; S is linearly dependent by property 1 of Theorem 9. 10. It is easily checked that S is a linearly independent set. Therefore, by property 3 of Theorem 9, S is a basis for R2: 11. It is easily checked that S is a linearly independent set. Since S contains two vectors and dim(R2) = 2 it follows from property 3 of Theorem 9 that S is a basis for R2: 12. The set S is linearly independent so, by property 3 of Theorem 9, S is a basis for R3: 13. It is easily shown by direct calculation that S is a linearly dependent set. Therefore S is not a basis for R3: 14. The set S is linearly independent so, by property 3 of Theorem 9, S is a basis for R3: 15. If we write x1 = 2x2 ¡x3 +x4 then the procedure described in Example 5 of Section 2.4 yields a basis f[2;1;0;0]T;[¡1;0;1;0]T; [1;0;0;1]Tg for W: It follows that dim(W) = 3: 16. dim(W) = 3: 17. Following the procedure used in Example 5 of Section 2.4, we obtain a basis f[1;¡1;0;0]T;[2;0;¡1;0]Tg for W: In particular dim(W) = 2: 18. dim(W) = 2: 19. The set f[¡1;3;2;1]Tg is a basis for W, so dim(W) = 1: 20. dim(W) = 1: 21. The homogeneous system Ax= has solution x1 =¡2x2: Therefore f[¡2;1]Tg is a basis for N(A) and nullity(A) = 1: Since 2 = rank(A) + nullity(A); it follows that rank(A) = 1: 3.5. DIMENSION 79 22. The set f[2;1;1]Tg is a basis for N(A): Therefore nullity(A) = 1 and rank(A) = 2: 23. Thehomogeneoussystem Ax= hassolution x1 =¡5x3; x2 =¡2x3: Thusf[¡5;¡2;1]Tg is a basis for N(A) and nullity(A) = 1: Since 3 = rank(A) + nullity(A); it follows that rank(A) = 2: 24. The set f[2;¡1;1;0]Tg is a basis for N(A): Therefore nullity(A) = 1 and rank(A) = 3: 25. AT reduces to BT = 2 4 1 ¡1 1 0 2 3 0 0 0 3 5: It follows that f[1;¡1;1]T; [0;2;3]Tg isabasisforR(A): Consequentlyrank(A) = 2: Since3 = rank(A)+nullity(A); it follows that nullity(A) = 1: 26. The matrix AT reduces to BT = 2 66 4 1 2 2 0 2 ¡1 0 0 0 0 0 0 3 77 5: Therefore f[1;2;2]T;[0;2;¡1]Tg is a basis for R(A), rank(A) = 2 and nullity(A) = 2: 27. (a) Following the methods of Example 7 in Section 2.4, let A = 2 4 1 ¡1 1 2 1 ¡2 0 ¡1 ¡2 3 ¡1 0 3 5: Then AT reduces to BT =2 66 4 1 1 ¡2 0 ¡1 1 0 0 1 0 0 0 3 77 5: It follows that f [1;1;¡2]T;[0;¡1;1]T; [0;0;1]T g is a basis for W . In particular dim(W) = 3: (b) Following the procedure in (a), we obtain a basis f[1;2;¡1;1]T; [0;1;¡1;1]T;[0;0;¡1;4]Tg for W: In particular, dim(W) = 3: 28. W =fx in R4 : x1 +2x2 ¡3x3 ¡x4 = 0g: It follows that dim(W) = 3: 29. The constraints aTx= 0;bTx= 0 and cTx= 0 yield the homogeneous system of equations x1 ¡ x2 = 0; x1 ¡ x3 = 0; and x2 ¡ x3 = 0: Solving we obtain x1 = x3 and x2 = x3 where x3 and x4 are arbitrary. Thus f[1;1;1;0]T;[0;0;0;1]Tg is a basis for W and dim(W) = 2: 30. Following the procedure described in the hint, suppose we have obtained a linearly in- dependent subset Sk = fw1 ; : : : ;wk g of W: If Sk spans W we are done. If not there exists a vector wk+1 in W such that wk+1 is not in Sp(Sk): Suppose a1w1 +¢¢¢+ akwk +ak+1wk+1 = . Now ak+1 = 0 since otherwise we could solve for wk+1 ; contradicting that wk+1 is not in Sp(Sk) . Since Sk is linearly independent, it follows 80 CHAPTER 3. THE VECTOR SPACE RN that ai = 0; 1 ? i ? k: This shows that the set Sk+1 = fw1 ; : : : ;wk ;wk+1 g is linearly independent. A linearly independent subset of Rn contains at most n vectors, so the process must eventually stop. That is, there is a linearly independent subset Sm = fw1 ; : : : ;wmg such that Sm spans W: Thus Sm is a basis for W: 31. Suppose x= a1u1 +a2u2 +¢¢¢+ apup and x= b1u1 +b2u2 +¢¢¢+ bpup : Then =x¡x = (a1 ¡ b1)u1 +(a2 ¡ b2)u2 +¢¢¢ + (ap ¡ bp)up: Since fu1 ;u2 ; : : : ;up g is linearly independent, a1¡b1 = 0; a2¡b2 = 0; : : : ; ap¡bp = 0: Therefore a1 = b1; a2 = b2; : : : ; ap = bp: 32. Let B =fu1 ; : : : ;umg be a basis for U: Then B is a linearly independent subset of V so by property 1 of Theorem 9, m ?dim(V ) Moreover if m = dim(V ) then by property 3 of Theorem 9, B is also a basis for V: It follows that V = W: 33. (a) rank(A)?3 and nullity(A)?0. (b) rank(A)?3 and nullity(A)?1. (c) rank(A)?4 and nullity(A)?0. 34. Use Theorem 9, part(1). The columns of A are vectors in R3. 35. Use Theorem 9, part(1). The rows of A, when transposed, are vectors in R3. 36. Since n = rank(A)+nullity(A), it follows that rank(A)? n: Further, R(A) is a subset of Rm so, by Exercise 32, rank(A)? m: 37. Clearly 2 = rank(A)?rank([A j b]): By Exercise 36, rank([A j b])?2: Therefore rank([A j b]) = 2 = rank(A) and, by Theorem 11, the system Ax= b is consistent. 38. 4 = rank(A)+nullity(A), so 3 = rank(A) ? rank([A j b]): By Exercise 36, rank([A j b ])?3: Therefore rank(A) = 3 = rank([A j b]) and, by Theorem 11, the system Ax= b is consistent. 39. The matrix A is, by deflnition, nonsingular if and only if N(A) = f g. 40. If x is in N(B) then (AB)x = A(Bx) = A = , so x is also in N(AB). Conversely, if x is in N(AB) then = (AB)x = A(Bx). Since A is nonsingular, Bx = and x is in N(B). 41. Suppose c1w1 +¢¢¢ + cpwp = where ci 6= 0: Then wi = a1w1 +¢¢¢ + ai¡1wi¡1 +ai+1wi+1 +¢¢¢+ apwp ; where aj = ¡cj=ci: If w is any vector in W then w= b1w1 +¢¢¢+ bpwp for some scalars b1; : : : ; bp: Substituting for wi yields w= (b1 + bia1)w1 +¢¢¢+(bi¡1 + biai¡1)wi¡1 +(bi+1 + biai+1)wi+1 +¢¢¢+(bp + biap)wp : It follows that W = Spfw1 ; : : : ;wi¡1 ;wi+1 ; : : : ;wp g: By Theorem 8 any set of p vectors in W is linearly dependent. This contradicts the assumption that dim(W) = p: We conclude that ci = 0 for each i so S is a linearly independent set. 3.6. ORTHOGONAL BASES FOR SUBSPACES 81 42. See the proof given in Exercise 30. 3.6 Orthogonal Bases for Subspaces 1. u1Tu2 = 1(¡1) + 1(0) + 1(1) = 0;u1Tu3 = 1(¡1) + 1(2) + 1(¡1) = 0;u2Tu3 = ¡1(¡1) + 0(2)+1(¡1) = 0: 2. u1Tu2 = u1Tu3 = u2Tu3 = 0: 3. u1Tu2 = 1(2)+1(0)+2(¡1) = 0;u1Tu3 = 1(1)+1(¡5)+2(2) = 0;u2Tu3 = 2(1)+0(¡5)+ (¡1)2 = 0. 4. u1Tu2 = u1Tu3 = u2Tu3 = 0: 5. 0 = u1Tu3 = a+b+c and 0 = u2Tu3 = 2a+2b¡4c: Solving yields a =¡b; b arbitrary, and c = 0. 6. a = (¡1=2)c; b = (5=2)c; c arbitrary. 7. 0 = u1Tu2 = ¡3 + a;0 = u1Tu3 = 4 + b + c;0 = u2Tu3 = ¡8 ¡ b + ac: Solving yields a = 3; b =¡5; c = 1: 8. 0 = u1Tu2 = 2a + 2; 0 = u1Tu3 = 2b + 3 ¡ c; 0 = u2Tu3 = ab + 3 ¡ c: Solving yields a =¡1; ; b = 0; c = 3: 9. v= a1u1 +a2u2 +a3u3 where a1 = (u1Tv)=(u1Tu1) = 2=3; a2 = (u2Tv)=(u2Tu2 ) =¡1=2; a3 = (u3Tv)=(u3Tu3 ) = 1=6: 10. v= u1 +u2 : 11. v= a1u1 +a2u2 +a3u3 where a1 = (u1Tv)=(u1Tu1 ) = 9=3 = 3, a2 = (u2Tv)=(u2Tu2 ) = 0, a3 = (u3Tv)=(u3Tu3 ) = 0: 12. v= (4=3)u1 +(1=3)u3 : 13. Denote the given vectors by, w1 ; w2 ; w3 ; respectively. Then u1 = w1 = [0;0;1;0]T: u2 = w2¡c1u1 ; where c1 = (u1Tw2 )=(u1Tu1 ) = 2: Then u2 = [1;1;0;1]T; u3 = w3¡b1u1¡b2u2 where b1 = (u1Tw3 )=(u1Tu1 ) = 1 and b2 = (u2Tw3 )=(u2Tu2 ) = 2=3: Therefore u3 = [1=3;¡2=3;0;1=3]T: 14. u1 = [1;0;1;2]T; u2 = [1;1;¡1;0]T; u3 = [1=2;¡1;¡1=2;0]T: 15. Denote the given vectors by w1,w2, w3, respectively. Then u1 = w1= [1;1;0]T, u2= w2 ¡c1u1, where c1 = (u1Tw2 )=(u1Tu1 ) = 2=2 = 1. Thus u2 = [¡1;1;1]T. u3 = w3 ¡b1u1 ¡b2u2 , where b1 = (u1Tw3 )=(u1Tu1 ) = 2=2 = 1 and b2 = (u2Tw3 )=(u2Tu2 ) = 6=3 = 2: Therefore u3 = [2;¡2;4]T: 82 CHAPTER 3. THE VECTOR SPACE RN 16. u1 = [0;1;2]T; u2 = [3;4;¡2]Tu3 = [10;¡6;3]T: 17. Denote the given system by w1, w2, w3, respectively. Then u1 = w1 = [0;1;0;1]T, u2= w2 ¡c1u1, where c1 = (u1Tw2)=(u1Tu1) = 2=2 = 1. Thus u2 = [1;1;0;¡1]T, u3 = w3 ¡b1 u1 ¡ b2u2, where b1 = (u1Tw3 )=(u1Tu1 ) = 2=2 = 1 and b2 = (u2Tw3 )=(u2Tu2 ) = 2=3. Therefore u3 = [¡2=3;1=3;1;¡1=3]T. 18. u1 = [1;1;0;2]T;u2 = [¡1;1;1;0]T;u3 = [¡1=2;¡1=6;¡1=3;1=3]T: 19. If A denotes the given matrix then the homogeneous system Ax = has solution x1 = ¡3x3 ¡ x4; x2 = ¡x3 ¡ 3x4; where x3 and x4 are arbitrary. It follows that f[¡3;¡1;1;0]T;[¡1;¡3;0;1]Tg is a basis for N(A) and fA1 ;A2g is a basis for R(A) , where A1= [1;2;1]T and A2= [¡2;1;¡1]T: TheGram-Schmidtprocessyieldsorthogonal bases f[¡3;¡1;1;0]T, [7=11;¡27=11;¡6=11;1]Tg and f[1;2;1]T, [¡11=6;8=6;¡5=6]Tg for N(A) and R(A) respectively. 20. A basis for N(A) is f[¡1;¡3;1;0;0]T, [¡2;¡3;0;1;0]T, [¡3;¡2;0;0;1]Tg. The Gram-Schmidt process yields the orthogonal basis f[¡1;¡3;1;0;0]T, [¡1;0;¡1;1;0]T, [¡13=11;5=11;2=11;¡1;1]Tg. A basis for R(A) is f[1;¡1;2]T, [3;2;¡1]Tg and the Gram-Schmidt process yields the or- thogonal basis f[1;¡1;2]T, [19=6;11=6;¡4=6]Tg. 21. By Theorem 13 an orthogonal set of nonzero vectors is linearly independent. By property 1 from Theorem 9, Section 2.5, any set of four or more vectors in R3 is linearly dependent. 22. It follows from Theorem 13 of Section 2.6 and Theorem 12 of Section 1.8 that A is nonsingular. Note that ATA is the (3 x 3) matrix [cij] where cij = uiTuj : Since S is orthogonal cij = uiTuj = 0 if i 6= j and cii = uiTui =k ui k2 for i = 1;2;3: For the vectors given in Exercise 1, ATA = 2 4 3 0 0 0 2 0 0 0 6 3 5: 23. Since v is in W; 0 = vTv= (kvk)2: Therefore v= . 24. Suppose y6= . For any scalar c;0?k x¡cyk2= (x¡cy)T (x¡cy ) = xTx¡cxTy¡cyTx+c2yTy = (kxk)2 ¡2cxTy +c2(kyk)2: For c = xTy =yTy this implies that 0 ?kxk2 ¡(xTy)2= kyk2: It follows that j xTyj?kxk kyk: If y= then j xTyj= kxk kyk= 0: 25. kx+yk2 = (x+y)T(x+y) =kxk2 +2(xTy)+kyk2 ? kxk2 +2j xTyj+kyk2 ?kxk2 +2kxk kyk+kyk2 = (kxk+kyk)2: 26. Note that kyk=kx+(y¡x)k ?kxk+ky¡xk so ¡(kxk¡kyk) =kyk¡kxk?ky¡xk=kx¡yk: Similarly, kxk= 3.7. LINEAR TRANSFORMATIONS FROM RN TO RM 83 k(x¡y)+yk?kx¡yk+ kyk: Therefore, kxk¡ kyk?kx¡yk: It follows that jkxk¡kykj?kx¡yk: 27. If W = Spfwigp¡1i=1 then fuigp¡1i=1 is an orthogonal basis for W?. Sincefwigpi=1 is linearly dependent, wp is in W . Therefore up is in W and upTui = 0 for 1? i ? p¡1: It follows that upTw= 0 for every vector w in W . By Exercise 23, up = . 28. kvk2 = (a1u1 +¢¢¢+ apup )T(a1u1 +¢¢¢+ apup ) =P 1?i;j?p aiajui Tuj = Pp i=1 a 2iuiTui = Pp i=1 a 21 since B is an orthonormal basis. 3.7 Linear Transformations from Rn to Rm 1. (a) T ? 0 0 ?¶ = ? 0 0 ? : (b) T ? 1 1 ?¶ = ? ¡1 0 ? : (c) T ? 2 1 ?¶ = ? 1 ¡1 ? : (d) T ? ¡1 0 ?¶ = ? ¡2 1 ? : 2. (a) T ? 2 2 ?¶ = ? 0 0 ? : (b) T ? 3 1 ?¶ = ? 2 ¡6 ? : (c) T ? 2 0 ?¶ = ? 2 ¡6 ? : (d) T ? 0 0 ?¶ = ? 0 0 ? : 3. (a), (b), and (d) are in the null space of T. 4. T(x) = b requires 2x1 ¡3x2 = 2 and ¡x1 + x2 = ¡2. Solving yields x1 = 4; x2 = 2; so x= [4;2]T. 5. If b = [b1; b2]T then T(x ) = b requires that 2x1 ¡3x2 = b1 and ¡x1 + x2 = b2: Solving yields x1 =¡b1 ¡3b2 and x2 =¡b1 ¡2b2; that is x= [¡b1 ¡3b2;¡b1 ¡2b2]T: 6. T(x ) = b if and only if Ax = b . Solving yields x1 = ¡2 + x2; x2 arbitrary. For example, if x= [¡2;0]T then T(x) = b. 7. The system of equations Ax= b is easily seen to be inconsistent. 84 CHAPTER 3. THE VECTOR SPACE RN 8. Let u= [u1; u2]T and v= [v1; v2]T. Then F(u+v) = ? 2(u 1 + v1)¡(u2 + v2) (u1 + v1)+3(u2 + v2) ? = ? 2u 1 ¡u2 u1 +3u2 ? + ? 2v 1 ¡v2 v1 +3v2 ? = F(u)+F(v): Also F(au) = ? 2au 1 ¡au2 au1 +3au2 ? = a ? 2u 1 ¡u2 u1 +3u2 ? = aF(u): This shows that F is a linear transformation. 9. F is a linear transformation. 10. F is not a linear transformation. For example if u= [u1; u2]T and v= [v1; v2]T then F(u+v) = ? (u 1 + v1)+(u2 + v2) 1 ? whereas F(u)+F(v) = ? (u 1 + v1)+(u2 + v2) 2 ? : Likewise, F(au) = ? au 1 + au2 1 ? whereas aF(u) = ? au 1 + au2 a ? : 11. F is not a linear transformation. For example F ? 1 2 ? + ? 2 1 ?¶ = F ? 3 3 ?¶ = ? 9 9 ? whereas F ? 1 2 ?¶ + F ? 2 1 ?¶ = ? 1 2 ? + ? 4 2 ? = ? 5 4 ? : 12. If u= [u1; u2; u3]T and v= [v1; v2; v3]T then F(u+v) =? (u1 + v1)¡(u2 + v2)+(u3 + v3) ¡(u1 + v1)+3(u2 + v2)¡2(u3 + v3) ? = ? u 1 ¡u2 + u3 ¡u1 +3u2 ¡2u3 ? + ? v 1 ¡v2 + v3 ¡v1 +3v2 ¡2v3 ? = F(u)+ F(v): For any scalar a; F(au) = ? au 1 ¡au2 + au3 ¡au1 +3au2 ¡2au3 ? = a ? u 1 ¡u2 + u3 ¡u1 +3u2 ¡2u3 ? = a F(u). Thus F is a linear transformation. 13. F is a linear transformation. 14. If u= [u1; u2]T and v= [v1; v2]T then F(u+v) = 2 4 (u1 + v1)¡(u2 + v2) ¡(u1 + v1)+(u2 + v2) u2 + v2 3 5 = 2 4 u1 ¡u2 ¡u1 + u2 u2 3 5+ 2 4 v1 ¡v2 ¡v1 + v2 v2 3 5 = 3.7. LINEAR TRANSFORMATIONS FROM RN TO RM 85 F(u)+ F(v): Similarly, F(au) = 2 4 au1 ¡au2 ¡au1 + au2 au2 3 5 = a 2 4 u1 ¡u2 ¡u1 + u2 u2 3 5 = aF(u): It follows that F is a linear transformation. 15. F is a linear transformation. 16. Let u = [u1; u2]T and v = [v1; v2]T: Then F(u+v ) = 2(u1 + v1) + 3(u2 + v2) = (2u1 +3u2)+(2v1 +3v2) = F(u)+ F(v): Likewise F(au) = 2au1 +3au2 = a(2u1 +3u2) = a F(u): This means that F is a linear transfor- mation. 17. F is not a linear transformation. For example note that F(¡e1 ) = 1 whereas ¡F(e1 ) =¡1: 18. The set fe1g is an orthonormal basis for W and T(v) = (vTe1 )e1 = [a;0;0]T. 19. (a) T ? 1 1 ?¶ = T(e1 +e2 ) = T(e1 )+ T(e2 ) = u1 +u2 = [3;1;¡1]T. (b) T ? 2 ¡1 ?¶ = T(2e1¡e2 ) = 2T(e1 )¡T(e2 ) = 2u1¡u2 = [0;¡1;¡2]T. (c) T ? 3 2 ?¶ = T(3e1 +2e2 ) = 3T(e1 )+2T(e2 ) = 3u1 +2u2 = [7;2;¡3]T. 20. (a) T ? 1 1 ?¶ = T(2v1¡v2 ) = 2u1¡u2 = [¡3;3]T. (b) T ? 2 ¡1 ?¶ = T(v1¡2v2 ) = u1¡2u2 = [¡6;0]T: (c) T ? 3 2 ?¶ = T(5v1¡3v2 ) = 5u1¡3u2 = [¡9;7]T: 21. Let u1 = [1;1]T and u2 = [1;¡1]T: If x= [x1; x2]T then x= [(x1 + x2)=2]u1 +[(x1 ¡x2)=2]u2 : Thus T(x) = [(x1 + x2)=2] ? 2 ¡1 ? +[(x1 ¡x2)=2] ? 0 3 ? = ? x 1 + x2 x1 ¡2x2 ? : 22. T(x) = 2 4 (x1 + x2)=2 2x1 (3x1 ¡x2)=2 3 5: 86 CHAPTER 3. THE VECTOR SPACE RN 23. Let u1 = [1;0;1]T; u2 = [0;¡1;1]T; u3 = [1;¡1;0]T: If x= [x1; x2; x3]T then x= c1u1 +c2u2 +c3u3 ; where c1 = (x1 + x2 + x3)=2; c2 = (¡x1 ¡x2 + x3)=2 and c3 = (x1 ¡x2 ¡x3)=2: Therefore T(x) = c1[0;1]T + c2[1;0]T + c3[0;0]T; that is, T 0 @ 2 4 x1 x2 x3 3 5 1 A = ? (¡x 1 ¡x2 + x3)=2 (x1 + x2 + x3)=2 ? : 24. T 0 @ 2 4 x1 x2 x3 3 5 1 A = 2 4 ¡x1 ¡x2 + x3 ¡x1 ¡x2 x1 3 5: 25. A = [T(e1 ); T(e2 )] = ? 1 3 2 1 ? : The homogeneous system of equations Ax = has only the trivial solution so N(T) = N(A) = f g and nullity(T) = 0: Since rank(T) = 2¡nullity(T) = 2; it follows that R(T) = R2 : 26. A = 2 4 1 ¡1 1 1 0 1 3 5;N(T) =N(A)=f g;R(T) =R(A)= f[b1; b2; b3]T : b1 ¡b2 +2b3 = 0g;nullity(T) = 0;rank(T) = 2: 27. A = [T(e1 ); T(e2 )] = [3;2];N(T) = fx : 3x1 + 2x2 = 0g;R(T) = R1;rank (T) = 1;nullity (T) = 1. 28. A = 2 4 1 1 0 0 0 1 0 1 0 3 5;N(T) =N(A) =f g;R(T) =R(A) = R3; rank(T) = 3;nullity(T) = 0: 29. A = [T(e1 ); T(e2 ); T(e3 )] = ? 1 ¡1 0 0 1 ¡1 ? ;N(T) =N(A) = fx: x1 = x3; x2 = x3g;R(T) =R(A) = R2;rank(T) = 2; nullity(T) = 1: 30. A = [2;¡1;4];N(T) =fx: 2x1 ¡x2 +4x3 = 0g;R(T) = R1; rank(T) = 1;nullity(T) = 2: 31. For any x and y in R; f(x + y) = a(x + y) = ax + ay = f(x)+ f(y): If b is any real number then f(bx) = a(bx) = b(ax) = bf(x): Therefore f is a linear transformation. 3.7. LINEAR TRANSFORMATIONS FROM RN TO RM 87 32. Since T is a linear transformation and x can be viewed as a scalar, T(x) = T(x ¢1) = xT(1) = xa = ax for each x in R. 33. T ? x 1 x2 ?¶ = ? x 1 ¡x2 ? 34. T ? x 1 x2 ?¶ = ? x 2 x1 ? 35. For vectors u and v in V;[F + G](u+v) = F(u+v)+ G(u+v) = [F(u)+ F(v)]+ [G(u)+ G(v)] = [F(u)+ G(u)]+[F(v)+ G(v)] = [F + G](u)+[F + G](v): Similarly, [F + G](au) = F(au)+ G(au) = aF(u)+ aG(u) = a[F(u)+ G(u)] = a[F + G](u) for every scalar a. This proves that F + G is a linear transformation. 36. (a) (F + G) 0 @ 2 4 x1 x2 x3 3 5 1 A = ? x 1 + x2 +3x3 2x1 +5x2 ¡2x3 ? . (b) A = ? 2 ¡3 1 4 2 ¡5 ? , B = ? ¡1 4 2 ¡2 3 3 ? , C = ? 1 1 3 ¡2 3 3 ? 37. [aT](u+v) = a[T(u+v)] = a[T(u)+ T(v)] = aT(u)+ aT(v) = [aT](u)+[aT](v): Also [aT](bu) = a(T(bu)) = a(bT(u)) = b(aT(u)) = b[aT](u): This proves that aT is a linear transformation. 38. (a) [3T] 0 @ 2 4 x1 x2 x3 3 5 1 A = ? 3x 1 ¡3x2 3x2 ¡3x3 ? (b) A = ? 1 ¡1 0 0 1 ¡1 ? , B = ? 3 ¡3 0 0 3 ¡3 ? 39. For u1 and u2 in U;[G?F](u1 +u2 ) = G(F(u1 +u2 )) = G(F(u1 )+ F(u2 )) = G(F(u1 ))+ G(F(u2 )) = [G?F](u1 )+ [G?F](u2 ): If u is in U and a is any scalar then [G?F](au) = G(F(au)) = G(aF(u)) = aG(F(u)) = a[G?F](u): Thus G?F is a linear transformation. 40. (a) [G?F] 0 @ 2 4 x1 x2 x3 3 5 1 A = 2 4 ¡5x1 ¡8x2 ¡6x3 x1 +16x2 ¡10x3 3x1 +3x2 +5x3 3 5 88 CHAPTER 3. THE VECTOR SPACE RN (b) A = ? ¡1 2 ¡4 2 5 1 ? , B = 2 4 1 ¡2 3 2 ¡1 1 3 5, 2 4 ¡5 ¡8 ¡6 1 16 ¡10 3 3 5 3 5 41. Write B = [B1 ;B2 ; : : : ;Bn ]: Then T(ei ) = Bei = Bi ; so A = [T(e1 ); T(e2 ); : : : ; T(en )] = B: 42. [G?F](x) = G(F(x)) = G(Ax) = B(Ax) = BAx : By Exercise 41, BA is the matrix for G?F. 43. A = [e1 ;e2 ; : : : ;en ] = I; the (n x n) identity matrix. 44. A = [T(e1 ); T(e2 ); : : : ; T(en )] = [ae1 ;ae2 ; : : : ;aen ]: Thus A = 2 66 64 a 0 ¢¢¢ 0 0 a ¢¢¢ 0 ... ... ... 0 0 ¢¢¢ a 3 77 75: 45. (a) A = ? 0 ¡1 1 0 ? : (b) A = ? 1=2 ¡p3=2 p3=2 1=2 ? : (c) A = ? ¡1=2 ¡p3=2 p3=2 ¡1=2 ? : 46. (a) = 0 so A = ? 1 0 0 ¡1 ? ;T(e1 ) = e1 ;T(e2 ) =¡e2 : (b) = ? so A = ? ¡1 0 0 1 ? ;T(e1 ) =¡e1 ;T(e2 ) = e2 : (c) = ?=2 so A = ? 0 1 1 0 ? ;T(e1 ) = e2 ;T(e2 ) = e1 : (d) = 2?=3 so A = ? ¡1=2 p3=2 p3=2 1=2 ? ;T(e1 ) = [¡1=2;p3=2]T; T(e2 ) = [p3=2;1=2]T: 47. Set u1 = T(e1 ) and u2 = T(e2 ): By assumption ku1k=ku2k= 1 and u1Tu2 = 0: Moreover T(v) = au1 +bu2 so k T(v)k2= (au1 +bu2 )T(au1 +bu2 ) = a2 + b2 =kvk2: Thus k T(v)k= k vk and T is orthogonal. 3.8. LEAST-SQUARES SOLUTIONS TO INCONSISTENT SYSTEMS 89 48. Set u1 = T(e1 ) and u2 = T(e2 ): Then A = [u1 ;u2 ] and ATA = ? u 1 Tu1 u1 Tu2 u2 Tu1 u2 Tu2 ? : It then follows from Theorem 16 that ATA = I. 49. (a) ATA = ? A 1 TA1 A1 TA2 A2 TA1 A2 TA2 ? so it follows that A1 TA1 = A2 TA2 = 1 whereas A1 TA2 = A2 TA1 = 0: Thus fA1 ;A2g is an orthonormal set. (b) Now k T(e1 )k=k A1k= 1;k T(e2 )k=k A2k= 1 and T(e1 ) is perpendicular to T(e2 ): By Theorem 16, T is orthogonal. 3.8 Least-Squares Solutions to Inconsistent Systems 1. ATA = ? 3 4 4 14 ? and ATb= ? 1 6 ? : The system of equations ATAx= ATb has unique solution x?= ? ¡5=13 7=13 ? : 2. ATA = 2 4 6 11 23 11 22 44 23 44 90 3 5 and ATb = 2 4 1 5 7 3 5: The system of equations ATAx = ATb has solution x?= 2 4 ¡3 19=11 0 3 5+ x3 2 4 ¡2 ¡1 1 3 5; where x3 is arbitrary. 3. ATAx = 2 4 11 16 17 16 30 18 17 18 33 3 5 and ATb= 2 4 10 17 13 3 5: The system of equations ATAx= ATb has solution x? = 2 4 28=74 27=74 0 3 5+ x3 2 4 ¡3 1 1 3 5 where x3 is arbitrary. 4. ATA = 2 4 15 24 3 24 39 3 3 3 6 3 5 and ATb= 2 4 0 1 ¡3 3 5: The system of equations ATAx= ATb has solution x?= 2 4 ¡8=3 5=3 0 3 5+ x3 2 4 ¡5 3 1 3 5 where x3 is arbitrary. 90 CHAPTER 3. THE VECTOR SPACE RN 5. ATA = ? 14 28 28 56 ? and ATb= ? 52 104 ? : Thesystemofequations ATAx= ATb hassolution x?= ? 26=7 0 ? + x2 ? ¡2 1 ? ; where x2 is arbitrary. 6. ATA = 2 4 11 1 1 1 1 1 1 1 1 3 5 and ATb = 2 4 21 1 1 3 5: The system of equations ATAx = ATb has solution x?= 2 4 2 ¡1 0 3 5+ x3 2 4 0 ¡1 1 3 5 where x3 is arbitrary. 7. We must obtain the least-squares solution to Ax= b where A = 2 66 4 ¡1 1 0 1 1 1 2 1 3 77 5;x= ? m c ? ; and b= 2 66 4 0 1 2 4 3 77 5: ATA = ? 6 2 2 4 ? and ATb= ? 10 7 ? : The system of equations ATAx= ATb has solution x?= ? 1:3 1:1 ? : Therefore y = (1:3)t +1:1 is the least-squares linear flt. 8. y = (¡19=35)t +31=35 is the least-squares linear flt. 9. We must obtain the least-squares solution to Ax= b where A = 2 66 4 ¡1 1 0 1 1 1 2 1 3 77 5;x= ? m c ? ; and b= 2 66 4 ¡1 1 2 3 3 77 5: In this case ATA = ? 6 2 2 4 ? and ATb= ? 9 5 ? : The system of equations ATAx= ATb has solution x?= ? 13=10 3=5 ? so y = 2t +1 is the least-squares linear flt. 10. y = 4t¡3=2 is the least-squares linear flt. 11. Wemustobtaintheleast-squaressolutionto Ax= b where A = 2 66 4 1 ¡2 4 1 ¡1 1 1 1 1 1 2 4 3 77 5;x= 2 4 a0 a1 a2 3 5; 3.8. LEAST-SQUARES SOLUTIONS TO INCONSISTENT SYSTEMS 91 and b = 2 66 4 2 0 1 2 3 77 5: In this case ATA = 2 4 4 0 10 0 10 0 10 0 34 3 5 and ATb = 2 4 5 1 17 3 5: The system of equations ATAx = ATb has solution x? = [0;1=10;1=2]T so y = a0 + a1t + a2t2 = (1=10)t +(1=2)t2 is the least-squares quadratic flt. 12. y =¡1=20¡(1=20)t +(1=4)t2 is the least-squares quadratic flt. 13. Wemustobtaintheleast-squaressolutionto Ax= b where A = 2 66 4 1 ¡2 4 1 ¡1 1 1 0 0 1 1 1 3 77 5;x= 2 4 a0 a1 a2 3 5; and b= 2 66 4 ¡3 ¡1 0 3 3 77 5: Inthiscase ATA = 2 4 4 ¡2 6 ¡2 6 ¡8 6 ¡8 18 3 5and ATb= 2 4 ¡1 10 ¡10 3 5: Thesystemof equations ATAx= ATb hassolution x?= [9=20;43=20;1=4]T so y = 9=20+(43=20)t+(1=4)t is the least-squares quadratic flt. 14. y = 31=55¡(4=55)t +(12=11)t2 is the least-squares quadratic flt. 15. Note that Ax¡b= [f(t1)¡y1; : : : ; f(tm)¡ym]T so, by deflnition, kAx¡bk2= Pmi=1[f(ti)¡ yi]2: 16. (a) A = 2 66 4 1 ¡1 2 1 3 ¡1 4 1 3 77 5;x= ? a 1 a2 ? ;b= 2 66 4 0 2 4 5 3 77 5: (b) ATA = ? 30 2 2 4 ? and ATb = ? 36 3 ? : The system of equations ATAx = ATb has solution x? = [69=58;9=58]T: Therefore Q(a1; a2) is minimized if f(t) = (69=58)pt + (9=58)cos?t: 17. Suppose that Ax = , where x = [a0; a1; : : : ; an]T: If p(t) = a0 + a1t + ¢¢¢ + antn then p(ti) = 0 for 0 ? i ? m; that is, p(t) has m + 1 roots and m + 1 > n: It follows that a0 = a1 =¢¢¢= an = 0: Thus nullity(A) = 0 and, consequently, rank(A) = n: 18. The matrix reduces to: 2 66 4 1 0 2 0 1 1 0 0 0 0 0 0 3 77 5: So, rank A = 2. 92 CHAPTER 3. THE VECTOR SPACE RN 3.9 Fitting Data and Least Squares Solutions 1. If u1 = [2;1;0]T and u2 = [¡1;0;1]T then fu1 ;u2g is a basis for W. For A = [u1 ;u2 ] the system of equations ATAx= ATv is given by 5x1 ¡2x2 = 4; ¡2x1 +2x2 = 5: Solving we obtain x1 = 3 and x2 = 11=2. Thus w?= 3u1 +(11=2)u2 = [1=2;3;11=2]T. 2. The system of equations ATAx = ATv has solution x1 = x2 = 2: Thus w? = 2u1 +2u2 = [2;2;2]T: 3. The basisfu1 ;u2gis given in Exercise 1. The system ATAx= ATv is given by 5x1¡2x2 = 3; ¡2x1 +2x2 = 0: The solution is x1 = x2 = 1; so w?= u1 +u2 = [1;1;1]T = v: 4. R(B) has basisfu1;u2gwhere u1= [1;1;0]T and u2= [2;1;1]T: If A = [u1;u2] the system ATAx= ATv is given by 2x1 +3x2 = 2; 3x1 +6x2 = 9: The solution is x1 = ¡5; x2 = 4: Thus w?=¡5u1 +4u2 = [3;¡1;4]T. 5. R(B)hasbasisfu1;u2gwhere u1= [1;1;0]T and u2= [2;1;1]T: If A = [u1;u2]; thesystem ATAx= ATv is given by 2x1 +3x2 = 6; 3x1 +6x2 = 12: Solving yields x1 = 0; x2 = 2; so w?= 2u2 = [4;2;2]T: 6. The system of equations ATAx = ATv has solution x1 = ¡3; x2 = 3: Thus w? = ¡3u1 +3u2 = [3;0;3]T = v: 7. R(B) has basis fu1 ;u2 g where u1 = [1;¡1;1]T and u2 = [2;0;1]T: If A = [u1 ;u2 ], the system ATAx= ATv is given by 3x1 +3x2 = 6; 3x1 +5x2 = 8: Solving yields x1 = x2 = 1; so w?= u1 +u2 = [3;¡1;2]T. 8. The system of equations ATAx= ATv has solution x1 = ¡1; x2 = 2 so w? = ¡u1 +2u2 = [3;1;1]T. 9. W has basis fug where u= [0;¡1;1]T: If A = [u] then the system ATAx= ATv is given by 2x =¡2: Thus x =¡1 and w?=¡u= [0;1;¡1]T: 10. w?=¡2u= [0;2;¡2]T: 11. An orthogonal basis for W is the set fu1 ;u2g where u1 = [2;1;0]T and u2 = [¡1=5;2=5;1]T: The vector w? is given by w? = a1u1 +a2u2 where a1 = u1Tv =u1Tu1 = 4=5 and a2 = u2Tv=u2Tu2 = 11=2: Thus w?= [1=2;3;11=2]T: 12. w?= a1u1 +a2u2 where u1 and u2 are given in Exercise 11, a1 = u1Tv=u1Tu1 = 6=5; and a2 = u2Tv=u2Tu2 = 2. Thus w?= [2;2;2]T. 13. If u1 = [1;1;0]T and u2 = [1=2;¡1=2;1]T then fu1 ;u2g is an orthogonal basis for W. The vector w? is given by w?= a1u1 +a2u2 where a1 = u1Tv=u1Tu1 = 1 and a2 = u2Tv=u2Tu2 = 4: Thus w?= [3;¡1;4]T: 3.10. SUPPLEMENTARY EXERCISES 93 14. w? = a1u1 +a2u2 where u1 and u2 are given in Exercise 13, a1 = u1Tv =u1Tu1 = 3 and a2 = u2Tv=u2Tu2 = 2: Thus w?= [4;2;2]T. 15. If u1 = [1;¡1;1]T and u2 = [1;1;0]T then fu1 ;u2 g is an orthogonal basis for W. The vector w? is given by w?= a1u1 +a2u2 where a1 = u1Tv=u1Tu1 = 2 and a2 = u2Tv=u2Tu2 = 1: Therefore w?= [3;¡1;2]T. 16. w? = a1u1 +a2u2 where u1 and u2 are given in Exercise 15, a1 = u1Tv =u1Tu1 = 1 and a2 = u2Tv=u2Tu2 = 2: Therefore w?= [3;1;1]T. 3.10 Supplementary Exercises 1. Clearly = ? 0 0 ? is in W. Suppose x = ? x 1 x2 ? is in W. Then x1x2 = 0. If a is any scalar then ax = ? ax 1 ax2 ? and (ax1)(ax2) = a2x1x2 = 0, so ax is in W. Now u = ? 1 0 ? and v = ? 0 1 ? are in W, but u + v = ? 1 1 ? is not in W. Therefore, W does not satisfy (S2). 2. Clearly = [0;0]T is in W. Let u and v be in W, where u = ? u 1 u2 ? and v = ? v 1 v2 ? . Then u1 ? 0, u2 ? 0, v1 ? 0, and v2 ? 0. If follows that u1 + v1 ? 0 and u2 + v2 ? 0. Thus u+v = ? u 1 + v1 u2 + v2 ? is in W. If u = ? 1 1 ? and a =¡1 then u is in W but au is not in W. Therefore W does not satisfy (S3). 3. (a) Ax = 3x if and only if (A¡3I)x = . Thus, W is the null space of the matrix A¡3I. (b) Thesystemofequations(A¡3I)x = hassolution x1 =¡x2+x3, x2 and x3 arbitrary. Therefore 8 < : 2 4 ¡1 1 0 3 5; 2 4 1 0 1 3 5 9 = ; is a basis for W. 4. Let S = fu1;u2g, T = fv1;v2;v3g, and b = [b1; b2; b3]T: Reducing the matrices [u1;u2;b] and [v1;v2;v3;b] yields 2 4 1 0 2b2 ¡b1 0 1 b1 ¡b2 0 0 ¡5b1 +7b2 + b3 3 5and 2 4 1 0 3 b1 0 1 2 b2 0 0 0 ¡5b1 +7b2 + b3 3 5, re- spectively. Thus Sp(S) = Sp(T) = fb : ¡5b1 + 7b2 + b3 = 0g. Alternatively, reducing the matrices [u1; u2]T and [v1; v2; v3]T gives ? 1 0 5 0 1 ¡7 ? and 2 4 1 0 5 0 1 ¡7 0 0 0 3 5, respectively. Therefore, f[1;0;5]T, [0;1;¡7]Tg is a basis for both Sp(S) and Sp(T). 94 CHAPTER 3. THE VECTOR SPACE RN 5. (a) A reduces to 2 4 1 ¡1 0 7 0 0 1 ¡2 0 0 0 0 3 5 so rank(A) = 2 and nullity(A) = 2. (b) f[1;¡1;0;7];[0;0;1;¡2]g. (c) f[1;2;1]T;[2;5;0]Tg. (d) f[1;2;1];[2;5;0]g is a basis for the row space of AT and f[1;¡1;0;7]T;[0;0;1;¡2]Tg is a basis for the column space of AT. (e) The homogenous system of equations Ax = has solution x1 = x2 ¡7x4, x3 = 2x4, x2 and x4 arbitrary.A basis for N(A) is f[1;1;0;0]T;[¡7;0;2;1]Tg. 6. (a) The matrix A = [v1;v2;v3] reduces to 2 4 1 0 1 0 1 2 0 0 0 3 5. If follows that fv1;v2g is a basis for Sp(S). (b) AT reduces to 2 4 3 0 1 0 3 ¡2 0 0 0 3 5 so f[3;0;1]T;[0;3;¡2]Tg is a basis for Sp(S). (c) Let b = [b1; b2; b3]T. The matrix [A;b] reduces to 2 4 1 0 1 (2b1 ¡b2)=3 0 1 2 (b1 + b2)=3 0 0 0 (¡b1 +2b2 ¡3b3)=3 3 5. Therefore, Sp(S) =fb :¡b1 +2b2 ¡3b3 = 0g. If follows that f[2;1;0]T;[3;0;1]Tg is a basis for Sp(S). 7. The matrix A is row equivalent to the (m£n) martrix 2 66 66 64 1 1 : : : 1 1 0 1 : : : n¡2 n¡1 0 0 : : : 0 0 ... ... ... ... 0 0 : : : 0 0 3 77 77 75 Thus, rank(A) = 2, nullity(A) = n¡2, and f[1;1; : : : ;1;1];[0;1; : : :; n¡2; n¡1]g is a basis for the row space of A. 8. (a) 1 (b) 1 (c) 0 9. (a) 1 (b) 2 (c) 2 10. T(e1) = T(x1)¡2T(x2) = ? ¡3 3 ? and T(e2) = 2T(x1)+ T(x2) = ? 4 1 ? so ? ¡3 4 3 1 ? is the matrix of T. 3.10. SUPPLEMENTARY EXERCISES 95 11. (a) x1 = 2 4 ¡5¡8u 2+3u u 3 5 and x2 = 2 4 3¡8v ¡1+3v v 3 5, where u and v are arbitrary. (b) x3 = 2 4 ¡8w 3w w 3 5 for any nonzero w. (c) Taking u = v = 0 and w = 1 in (a) and (b) gives B =fx1;x2;x3g = f[¡5;2;0]T;[3;¡1;0]T;[¡8;3;1]Tg. The set B is linearly independent, so is a basis for R3. (d) e1 = x1 +2x2 so T(e1) = T(x1)+2T(x2) = ? 1 0 ? +2 ? 0 1 ? = ? 1 2 ? . e2 = 3x1 +5x2 so T(e2) = 3T(x1) + 5T(x2) = 3 ? 1 0 ? + 5 ? 0 1 ? = ? 3 5 ? . e3 = ¡x1 + x2 + x3 so T(e3) =¡T(x1)+T(x2)+T(x3) =¡ ? 1 0 ? + ? 0 1 ? + ? 0 0 ? = ? ¡1 1 ? . If follows that A = ? 1 3 ¡1 2 5 1 ? . 12. f[1;0;2;0;¡3;1];[0;1;¡1;0;2;2];[0;0;0;1;¡1;¡2]gisa basis for the row space of A. There- fore rank(A) = 3 and nullity(A) = 3. 13. b = [a; b; c; d]T is in R(T) if and only if ¡16a ¡7b +9c + d = 0. Therefore, w1, w3, and w4 are in R(T). 14. For w1, x1 = 2 66 66 66 4 1¡2u +3v ¡w u¡2v ¡2w u v +2w v w 3 77 77 77 5 ; for w3, x3 = 2 66 66 66 4 4¡2u +3v ¡w 7+ u¡2v ¡2w u ¡3+ v +2w v w 3 77 77 77 5 ; for w4, x4 = 2 66 66 66 4 1¡2u +3v ¡w 1+ u¡2v ¡2w u v +2w v w 3 77 77 77 5 In each case, u, v, and w are arbitrary. 15. See the solution to Exercise 14. T(xi) = wi if and only if Axi = wi. 16. (a) In the solutions to Exercise 14, take u = v = w = 0. This gives w1 = Ax1 = A1 w3 = Ax3 = 4A1 +7A2 ¡3A4, and w4 = Ax4 = A1 +A2. (b) fA1;A2;A4g 96 CHAPTER 3. THE VECTOR SPACE RN (c) The homogenous system of equations, Ax = , has solution x1 = ¡2x3 + 3x5 ¡ x6, x2 = x3 ¡2x5 ¡2x6, x4 = x5 +2x6, x3; x5; x6 arbitrary. Setting x3 = 1, x5 = x6 = 0 yields x1 = ¡2, x2 = 1, and x4 = 0. If follows that ¡2A1 + A2 + A3 = , so A3 = 2A1 ¡A2. Similarly, A5 =¡3A1 +2A2 ¡A4 and A6 = A1 +2A2 ¡2A4. (d) b is in the column space of A since, by the condition established in the solution to Exercise 13, b is in R(T) . The system of equations Ax = b has solution x1 = 0, x2 =¡5, x3 = 0, x4 = 2, x5 = 0, x6 = 0, so b =¡5A2 +2A4. (e) Ax = 2A1+3A2+A3¡A4+A5+A6. Substitutingfor A3, A5, and A6 theexpressions obtained in (c) gives Ax = 2A1 +6A2 ¡4A4. 17. (a) If b = [a; b; c; d]T then R(T) = fb : ¡16a ¡ 7b + 9c + d = 0g (cf. the solution to Exercise 13.). The set fu1;u2;u3g is the basis for R(T), where u1 = [1;0;0;16]T, u2 = [0;1;0;7]T, and u3 = [0;0;1;¡9]T. Moreover, if b is in R(T) then b = [a; b; c;16a +7b¡9c]T = au1 + bu2 + cu3. (b) b = u1 +2u2 +3u3. 18. (a) If v1 = [¡2;1;1;0;0;0]T, v2 = [3;¡2;0;1;1;0]T, and v3 = [¡1;¡2;0;2;0;1]T, then fv1;v2;v3g is a basis for N (T) (cf. the solution to Exercise 16(c)). Moreover,if x = [x1; x2; x3; x4; x5; x6]T is in N (T) then x = [¡2x3 +3x5 ¡x6; x3 ¡2x5 ¡2x6; x3; x5 +2x6; x5; x6]T, where x3, x5, x6 are arb- itrary. Thus, x = x3v1 + x5v2 + x6v3. (b) x = v1 +2v2 ¡2v3. 3.11 Conceptual Exercises 1. False. In R2 let W =f[a; a]T : a arbitrary g. Then e1 +e2 is in W but neither e1 nor e2 is in W. 2. True. Since ax is in W, a¡1(ax) = x is in W. 3. False. f g is a linearly dependent subset of Rn. 4. True. cf. Theorem 9(1). 5. True. cf. Theorem 9(2). 6. False. Consider S =f[0;0]T;[1;0]T;[3;0]Tg in R2. 7. False. Consider S1 =f[1;1]Tg and f[1;0]T;[0;1]Tg in R2. 8. False. The sets f[1;0]T;[0;1]Tg and f[2;0]T;[1;1]Tg are both bases for R2. 9. False. A basis for W must contain exactly k vectors but if W 6= f g then W contains inflnitely many vectors. 3.11. CONCEPTUAL EXERCISES 97 10. False. Let B = f[1;0]T;[0;1]Tg. Then B is a basis for R2 but no subset of B is a basis for W =f[a; a]T : a arbitrary g. 11. True. A basis for W must also be a basis for Rn. 12. False. B1 = f[1;1;0]T;[1;0;1]Tg is a basis for W1 = f[b + c; b; c]T : b; c arbitrary g and B2 =f[2;1;0]T;[¡1;0;1]Tg is a basis for W2 =f[2b¡c; b; c]T : b; c arbitrary g. B1\B2 =; but W1 \W2 =f[3c;2c; c]T : c arbitrary g. 13. No. is not in V . 14. B need not be a subset of W. Chapter 4 The Eigenvalue Problems 4.1 Introduction 1. The matrix A ¡ ?I = ? 1¡? 0 2 3¡? ? is singular if and only if 0 = (1¡ ?)(3¡ ?): Thus ? = 1 and ? = 3 are eigenvalues for A. The eigenvectors corresponding to ? = 1 (A ¡ I)x= . Solving yields x1 = ¡x2; x2 arbitrary. Therefore any vector of the form x = a ? ¡1 1 ? ; a 6= 0; is an eigenvector for ? = 1: Similarly the eigenvectors corresponding to ? = 3 are the nontrivial solutions to (A¡3I)x= . Solving yields x1 = 0; x2 arbitrary, so any vector of the form x= a ? 0 1 ? ; a 6= 0; is an eigenvector for ? = 3. 2. The matrix A¡?I = ? 2¡? 1 0 ¡1¡? ? is singular if and only if 0 = (2¡?)(¡1¡?): Therefore A has eigenvalues ? = 2 and ? =¡1. For ? = 2 the corresponding eigenvectors are x= a ? 1 0 ? ; a 6= 0: For ? =¡1 the corresponding eigenvectors are x= a ? ¡1 3 ? ; a 6= 0: 3. The matrix A¡?I = ? 2¡? ¡1 ¡1 2¡? ? is singular if and only if 0 = (2¡?)(2¡?)¡1 = ?2 ¡4? +3 = (? ¡1)(? ¡3): Therefore A has eigenvalues ? = 1 and ? = 3?. Solving (A¡I)x= yields x1 = x2; x2 arbitrary, so any vector of the form x= a ? 1 1 ? ; a 6= 0; is an eigenvector corresponding to ? = 1?. Solving (A¡3I)x= yields x1 = ¡x2; x2 arbitrary, so any vector of the form x= a ? ¡1 1 ? ; a 6= 0; is an eigenvector for ? = 3. 4. ? = 2;x= a ? ¡2 1 ? ; a 6= 0; ? = 3;x= a ? ¡1 1 ? ; a 6= 0: 100 CHAPTER 4. THE EIGENVALUE PROBLEMS 5. The matrix A¡?I = ? 2¡? 1 1 2¡? ? is singular if and only if 0 = (2¡?)(2¡?)¡1 = ?2 ¡4? +3 = (? ¡1)(? ¡3): Therefore A has eigenvalues ? = 1 and ? = 3 . Solving (A¡I)x= yields x1 =¡x2; x2 arbitrary, so any vector of the form x= a ? ¡1 1 ? ; a 6= 0; is an eigenvector for ? = 1. Solving (A ¡3I)x= yields x1 = x2; x2 arbitrary, so any vector x= a ? 1 1 ? ; a 6= 0; is an eigenvector for ? = 3: 6. ? = 2;x= a ? 1 1 ? ; a 6= 0: 7. The matrix A¡?I = ? 1¡? 0 2 1¡? ? is singular if and only if 0 = (1¡?)2; so ? = 1 is the only eigenvalue for A: Solving (A¡I)x= yields x1 = 0; x2 arbitrary, so any vector x= a ? 0 1 ? ; a 6= 0; is an eigenvector for ? = 1. 8. ? = 2;x= a ? 1 0 ? ; a 6= 0. 9. The matrix A¡?I = ? 2¡? 2 3 3¡? ? is singular if and only if 0 = (2¡?)(3¡?)¡6 = ?2 ¡5? = ?(?¡5). Therefore A has eigenvalues ? = 0 and ? = 5. Solving Ax= yields x1 =¡x2; x2 arbitrary, so x= a ? ¡1 1 ? ; a 6= 0; is an eigenvector for ? = 0: Solving (A¡5I)x= yields x1 = (2=3)x2; x2 arbitrary, so x= a ? 2 3 ? ; a 6= 0; is an eigenvector for ? = 5. 10. ? = 0;x= a ? ¡2 1 ? ; a 6= 0;? = 9;x= a ? 1 4 ? ; a 6= 0: 11. The matrix A¡?I = ? 1¡? ¡1 1 3¡? ? is singular if and only if 0 = (1¡?)(3¡?)+1 = ?2 ¡4? +4 = (?¡2)2: Therefore ? = 2 is the only eigenvalue for A: Solving (A¡2I)x = yields x1 =¡x2; so x= a ? ¡1 1 ? ; a 6= 0 is an eigenvector for ? = 2: 12. ? = 3;x= a ? ¡1 1 ? ; a 6= 0: 4.2. DETERMINANTS AND THE EIGENVALUE PROBLEM 101 13. The matrix A¡?I = ? ¡2¡? ¡1 5 2¡? ? is singular if and only if 0 = (¡2¡?)(2¡?)+5 = ?2 +1: Solving yields ? =§i: 14. The matrix A¡?I is singular if and only if ?2 +1 = 0?. Solving yields ? =§i. 15. The matrix A¡?I = ? 2¡? ¡1 1 2¡? ? is singular if and only if 0 = (2¡?)(2¡?)+1 = ?2 ¡4? +5: Solving we obtain ? = 2§i: 16. A¡?I is singular if and only if ?2 ¡2? +2 = 0. Solving yields ? = 1§i: 17. The matrix A¡?I = ? a¡? b b d¡? ? is singular if and only if 0 = (a¡?)(d¡?)¡b2 = ?2 ¡(a + d)? +(ad¡b2): Note that (a + d)2 ¡4(ad¡b2) = (a¡d)2 +4b2 ? 0, so the equation has real roots. 18. Thematrix A¡?I issingularifandonlyif ?2¡2a?+(a2+b2) = 0: Since(2a)2¡4(a2+b2) = ¡4b2 < 0 the equation has no real roots. 19. Let A = ? a b c d ? : The matrix AT ¡ ?I = ? a¡? c b d¡? ? is singular if and only if 0 = (a¡?)(d¡?)¡bc = ?2 ¡(a + d)?+ (ad¡bc): Therefore the eigenvalues of AT are roots of (5)?, so coincide with the eigenvalues of A. 4.2 Determinants and the Eigenvalue Problem 1. M11 = 2 4 1 3 ¡1 2 4 1 2 0 ¡2 3 5: A11 = det(M11) = flfl flfl 4 10 ¡2 flfl flfl¡3 flfl flfl 2 12 ¡2 flfl flfl¡ flfl flfl 2 42 0 flfl flfl = 18: 2. M21 = 2 4 ¡1 3 1 2 4 1 2 0 ¡2 3 5: A21 =¡det(M21) =¡18: 3. M31 = 2 4 ¡1 3 1 1 3 ¡1 2 0 ¡2 3 5: A31 = det(M31) = ¡ flfl flfl 3 ¡10 ¡2 flfl flfl¡3 flfl flfl 1 ¡12 ¡2 flfl flfl+ flfl flfl 1 32 0 flfl flfl = 0: 102 CHAPTER 4. THE EIGENVALUE PROBLEMS 4. M41 = 2 4 ¡1 3 1 1 3 ¡1 2 4 1 3 5: A41 =¡det(M41) = 18: 5. M34 = 2 4 2 ¡1 3 4 1 3 2 2 0 3 5: A34 =¡det(M34) = ¡2 flfl flfl 1 32 0 flfl flfl¡ flfl flfl 4 32 0 flfl flfl¡3 flfl flfl 4 12 2 flfl flfl = 0: 6. M43 = 2 4 2 ¡1 1 4 1 ¡1 6 2 1 3 5: A43 =¡det(M43) =¡18: 7. det(A) = 2A11 +4A21 +6A31 +2A41 = 2(18)+4(¡18)+6(0)+2(18) = 0: 8. det(A) = 5;A is nonsingular. 9. det(A) = 0;A is singular 10. det(A) = 0;A is singular. 11. det(A) =¡1;A is nonsingular. 12. det(A) = 0;A is singular. 13. det(A) = 2 flfl flfl ¡2 11 ¡1 flfl flfl+3 flfl flfl ¡1 13 ¡1 flfl flfl+2 flfl flfl ¡1 ¡23 1 flfl flfl = 6; A is nonsingular. 14. det(A) = 0; A is singular. 15. det(A) = 2 flfl flfl 3 21 4 flfl flfl= 20; A is nonsingular. 16. By Theorem 4, det(A) = 2(1)(2) = 4; A is nonsingular. 17. Expansion along the flrst column of A yields det(A) = flfl flfl flfl 3 0 0 4 1 2 3 1 4 flfl flfl flfl: Now expansion along the flrst row gives det(A) = 3 flfl flfl 1 21 4 flfl flfl = 6: A is nonsingular. 18. det(A) = 1;A is nonsingular. 4.2. DETERMINANTS AND THE EIGENVALUE PROBLEM 103 19. Expansion along the flrst column in successive steps yields det(A) =¡3 flfl flfl flfl 0 0 2 0 3 1 2 1 2 flfl flfl flfl = (¡3)(2) flfl flfl 0 23 1 flfl flfl = (¡3)(2)(¡6) = 36: A is nonsingular. 20. (a) The described algorithm yields a11a22a33 + a12a23a31 + a13a21a32 ¡ a31a22a13 ¡ a32a23a11 ¡ a33a21a12 which equals det(A). (b) Note that for a (4 £ 4) matrix, A = (aij), the deflnition of det(A) yields a sum of products with 24 terms, whereas the \basketweave algorithm" yields an expression with only eight summands. For A = 2 66 4 1 1 1 1 1 2 2 2 1 2 3 3 1 2 3 4 3 77 5, the basket weave algorithm gives 7, but det(A) = 1. 21. Det(A) = 4x¡2y ¡2, so A is singular when 4x¡2y ¡2 = 0, that is when y = 2x¡1. 22. Det(A) = (x¡2)(y +1), so A is singular if either x = 2 or y =¡1. 23. For n = 2; flfl flfl d 11 d flfl flfl = d2 ¡ 1 = (d ¡ 1)(d + 1): For n = 3; flfl flfl flfl d 1 1 1 d 1 1 1 d flfl flfl flfl = d flfl flfl d 11 d flfl flfl¡ flfl flfl 1 11 d flfl flfl+ flfl flfl 1 d1 1 flfl flfl = d(d¡1)(d +1)¡(d¡1)+(1¡d) = (d¡1)2(d +2): For n = 4; flfl flfl flfl flfl d 1 1 1 1 d 1 1 1 1 d 1 1 1 1 d flfl flfl flfl flfl = d flfl flfl flfl d 1 1 1 d 1 1 1 d flfl flfl flfl¡ flfl flfl flfl 1 1 1 1 d 1 1 1 d flfl flfl flfl+ flfl flfl flfl 1 d 1 1 1 1 1 1 d flfl flfl flfl¡ flfl flfl flfl 1 d 1 1 1 d 1 1 1 flfl flfl flfl = d(d¡1)2(d +2)¡3(d¡1)2 = (d¡1)3(d +3): 24. (a) If A is singular then det(A) = 0 so det(AB) = det(A)det(B) = 0: Therefore AB is singular. Similarly if B is singular then so is AB. (b) If AB is singular then 0 = det(AB) = det(A)det(B): Therefore either det(A) = 0 or det(B) = 0; that is either A or B is singular. 25. 1 = det(I) = det(AA¡1) = det(A)det(A¡1): Therefore det(A¡1) = 1=det(A): 26. det(AB) = det(A)det(B) = det(B)det(A) = det(BA): 27. Det(ABA¡1) = det(A)det(B)=det(A) = det(B) = 5. 104 CHAPTER 4. THE EIGENVALUE PROBLEMS 28. Det(A2B) = [det(A)]2 det(B) = 325 = 45. 29. Det(A¡1B¡1A2) = [det(A)]2=[det(A)det(B)] = det(A)=det(B) = 3=5. 30. Det(AB¡1A¡1B) = [det(A)=det(B)][det(B)=det(A)] = 1. 31. (a) H(n) = n!=2. (b) n = 2, 3 secs; n = 5, 3 min; n = 10, 63 days. 32. If U = [uij] and V = [vij] then, by Theorem 4, det(U) = u11u22¢¢¢unn and det(V ) = v11v22¢¢¢vnn: By Exercise 59, Section 1.6, UV is an upper triangular matrix. Moreover UV is the (n x n) matrix [aij] where a11 = u11v11; : : : ; ann = unnvnn: It follow that det(UV ) = (u11v11)(u22v22)¢¢¢(unnvnn) = det(U)det(V ): 33. Suppose that V is a lower triangular matrix with diagonal entries t1; t2; : : : ; tn: Then V T is an upper triangular matrix with the same diagonal entries so det(V ) = det(V T) = t1t2¢¢¢tn: 34. For n = 2;det(T) = flfl flfl t11 t120 t 22 flfl flfl = t11t22: Assume that if T is a (k x k) matrix then det(T) = t11t22¢¢¢tkk:?. Now assume that T is a [(k+1) x (k+1)] matrix. Thus T = 2 66 64 t11 t12 ¢¢¢ t1;k+1 0 t22 ¢¢¢ t2;k+1 ... ... 0 0 : : : tk+1;k+1 3 77 75: Expansion along the flrst column yields det(T) = t11 flfl flfl flfl fl t22 ¢¢¢ t2;k+1 ... ... 0 ¢¢¢ tk+1;k+1 flfl flfl flfl fl : Since the result holds for (k x k) matrices, we have det(T) = t11t22¢¢¢tk+1;k+1: It follows by induction that for any integer n ?2;det(T) = t11t22¢¢¢tnn: 4.3 Elementary Operations and Determinants 1. det(A) = flfl flfl flfl 1 2 1 3 0 2 ¡1 1 3 flfl flfl flfl ? R 2 ¡3R1 R3 + R1 = flfl flfl flfl 1 2 1 0 ¡6 ¡1 0 3 4 flfl flfl flfl = flfl flfl ¡6 ¡13 4 flfl flfl =¡21: 2. det(A) = 20: 4.3. ELEMENTARY OPERATIONS AND DETERMINANTS 105 3. det(A) = flfl flfl flfl 3 6 9 2 0 2 1 2 0 flfl flfl flfl = (3)(2) flfl flfl flfl 1 2 3 1 0 1 1 2 0 flfl flfl flfl ? R 2 ¡R1 R3 ¡R1 = (6) flfl flfl flfl 1 2 3 0 ¡2 ¡2 0 0 ¡3 flfl flfl flfl = 36: 4. det(A) =¡24: 5. det(A) = flfl flfl flfl 2 4 ¡3 3 2 5 2 3 4 flfl flfl flfl = (1=2) flfl flfl flfl 2 4 ¡3 6 4 10 2 3 4 flfl flfl flfl ? R 2 ¡3R1 R3 ¡R1 = (1=2) flfl flfl flfl 2 4 ¡3 0 ¡8 19 0 ¡1 7 flfl flfl flfl = (2)(1=2) flfl flfl ¡8 19¡1 7 flfl flfl =¡37: 6. det(A) =¡21: 7. flfl flfl flfl flfl 1 0 0 0 2 0 0 3 1 1 0 1 1 4 2 2 flfl flfl flfl flfl fC3 $ C4g = (¡1) flfl flfl flfl flfl 1 0 0 0 2 0 3 0 1 1 1 0 1 4 2 2 flfl flfl flfl flfl fC2 $ C3g = flfl flfl flfl flfl 1 0 0 0 2 3 0 0 1 1 1 0 1 2 4 2 flfl flfl flfl flfl = (1)(3)(1)(2) = 6: 8. flfl flfl flfl flfl 0 0 3 1 2 1 0 1 0 0 0 2 0 2 2 1 flfl flfl flfl flfl = (¡1) flfl flfl flfl flfl 2 1 0 1 0 2 2 1 0 0 3 1 0 0 0 2 flfl flfl flfl flfl =¡24: 9. flfl flfl flfl flfl 0 0 2 0 0 0 1 3 0 4 1 3 2 1 5 6 flfl flfl flfl flfl ? C 1 $ C4 C2 $ C3 = flfl flfl flfl flfl 0 2 0 0 3 1 0 0 3 1 4 0 6 5 1 2 flfl flfl flfl flfl fC1 $ C2g = (¡1) flfl flfl flfl flfl 2 0 0 0 1 3 0 0 1 3 4 0 5 6 1 2 flfl flfl flfl flfl = (¡1)(2)(3)(4)(2) =¡48: 106 CHAPTER 4. THE EIGENVALUE PROBLEMS 10. flfl flfl flfl flfl 0 0 1 0 1 2 1 3 0 0 0 5 0 3 1 2 flfl flfl flfl flfl = (¡1) flfl flfl flfl flfl 1 2 1 3 0 3 1 2 0 0 1 0 0 0 0 5 flfl flfl flfl flfl =¡15: 11. flfl flfl flfl flfl 0 0 1 0 0 2 6 3 2 4 1 5 0 0 0 4 flfl flfl flfl flfl fR1 $ R3g = (¡1) flfl flfl flfl flfl 2 4 1 5 0 2 6 3 0 0 1 0 0 0 0 4 flfl flfl flfl flfl = (¡1)(2)(2)(1)(4) =¡16: 12. flfl flfl flfl flfl 0 1 0 0 0 2 0 3 2 1 0 6 3 2 2 4 flfl flfl flfl flfl = (¡1) flfl flfl flfl flfl 1 0 0 0 2 3 0 0 1 6 2 0 2 4 3 2 flfl flfl flfl flfl =¡12: 13. det(B) = 3det(A) = 6: 14. det(B) = det(A) = 2: 15. det(B) =¡det(A) =¡2: 16. det(B) = det(A) = 2: 17. det(B) =¡2det(A) =¡4: 18. det(B) = det(A) = 2: 19. flfl flfl flfl flfl 2 4 2 6 1 3 2 1 2 1 2 3 1 2 1 1 flfl flfl flfl flfl fR1 ¡2R4g = flfl flfl flfl flfl 0 0 0 4 1 3 2 1 2 1 2 3 1 2 1 1 flfl flfl flfl flfl = (¡4) flfl flfl flfl 1 3 2 2 1 2 1 2 1 flfl flfl flfl ? R 2 ¡2R1 R3 ¡R1 = (¡4) flfl flfl flfl 1 3 2 0 ¡5 ¡2 0 ¡1 ¡1 flfl flfl flfl = (¡4) flfl flfl ¡5 ¡2¡1 ¡1 flfl flfl =¡12: 20. ¡5: 21. flfl flfl flfl flfl 0 4 1 3 0 2 2 1 1 3 1 2 2 2 1 4 flfl flfl flfl flfl fR4 ¡2R3g = flfl flfl flfl flfl 0 4 1 3 0 2 2 1 1 3 1 2 0 ¡4 ¡1 0 flfl flfl flfl flfl = 4.3. ELEMENTARY OPERATIONS AND DETERMINANTS 107 flfl flfl flfl 4 1 3 2 2 1 ¡4 ¡1 0 flfl flfl flfl fR1 + R3g = flfl flfl flfl 0 0 3 2 2 1 ¡4 ¡1 0 flfl flfl flfl = (3) flfl flfl 2 2¡4 ¡1 flfl flfl = 18: 22. 4. 23. flfl flfl flfl 1 a a2 1 b b2 1 c c2 flfl flfl flfl ? R 2 ¡R1 R3 ¡R1 = flfl flfl flfl 1 a a2 0 b¡a b2 ¡a2 0 c¡a c2 ¡a2 flfl flfl flfl = (b¡a)(c¡a) flfl flfl flfl 1 a a2 0 1 b + a 0 1 c + a flfl flfl flfl fR3 ¡R2g = (b¡a)(c¡a) flfl flfl flfl 1 a a2 0 1 b + a 0 0 c¡b flfl flfl flfl = (b¡a)(c¡a)(c¡b) flfl flfl flfl 1 a a2 0 1 b + a 0 0 1 flfl flfl flfl = (b¡a)(c¡a)(c¡b): 24. (b¡a)(c¡a)(c¡b)(d¡a)(d¡b)(d¡c): 25. Write A = 2 66 64 A1 A2 ... An 3 77 75 where Ai = [ai1; ai2; : : : ; ain] is the ith row of A. Then cA = 2 66 64 cA1 cA2 ... cAn 3 77 75 so, by Theorem 7, det(cA) = cndet(A): 26. Suppose the ith and jth rows of A are identical and let B denote the matrix obtained by interchanging these two rows. By Theorem 6 det(B) = ¡det(A): But B = A so det(A) = det(B). It follows that det(A) =¡det(A); so det(A) = 0: 27. A = ? 1 0 0 0 ? and B = ? 0 0 0 1 ? is one possibility. 28. By Theorem 5, det(A) = det(AT): But AT = ¡A so by Exercise 25, det(AT) = (¡1)ndet(A): Therefore det(A) = (¡1)ndet(A): In particular if n is odd then det(A) = ¡det(A): It follows that det(A) = 0 and hence A is singular. 108 CHAPTER 4. THE EIGENVALUE PROBLEMS 4.4 Eigenvalues and the Characteristic Polynomial 1. p(t) = (1¡t)(3¡t): The eigenvalues are ? = 1 and ? = 3; each with algebraic multiplicity 1. 2. p(t) = (2 ¡ t)(¡1 ¡ t): The eigenvalues are ? = 2 and ? = ¡1 each with algebraic multiplicity 1. 3. p(t) = flfl flfl 2¡t ¡1¡1 2¡t flfl flfl = (2¡t)(2¡t)¡1 = t2 ¡4t +3 = (t¡1)(t¡3): The eigenvalues are ? = 1 and ? = 3; each with algebraic multiplicity 1. 4. p(t) = (t¡1)2: The only eigenvalue is ? = 1 and it has algebraic multiplicity 2. 5. p(t) = flfl flfl 1¡t ¡11 3¡t flfl flfl = (1¡t)(3¡t)+1 = t2 ¡4t +4 = (t¡2)2: The only eigenvalue is ? = 2 and it has algebraic multiplicity 2. 6. p(t) = t(t¡5): The eigenvalues are ? = 0 and ? = 5; each with algebraic multiplicity 1. 7. p(t) = flfl flfl flfl ¡6¡t ¡1 2 3 2¡t 0 ¡14 ¡2 5¡t flfl flfl flfl =¡t3 +t2 +t¡1 =¡(t¡1)2(t+1): The eigenvalues are ? = 1 with algebraic multiplicity 2 and ? =¡1 with algebraic multiplicity 1. 8. p(t) = ¡t(t +1)2: The eigenvalues are ? = 0 with algebraic multiplicity 1, and ? = ¡1 with algebraic multiplicity 2. 9. p(t) = flfl flfl flfl 3¡t ¡1 ¡1 ¡12 ¡t 5 4 ¡2 ¡1¡t flfl flfl flfl =¡t3 +2t2 + t¡2 = ¡(t¡2)(t¡1)(t +1): The eigenvalues are ? = 2; ? = 1; and ? = ¡1 each with algebraic multiplicity 1. 10. p(t) =¡(t¡1)3:?The only eigenvalue is ? = 1 and it has algebraic multiplicity 3. 11. p(t) = flfl flfl flfl 2¡t 4 4 0 1¡t ¡1 0 1 3¡t flfl flfl flfl = (2¡t) flfl flfl 1¡t ¡11 3¡t flfl flfl = (2¡t)(t2¡4t+4) =¡(t¡2)3: The only eigenvalue is ? = 2 and it has algebraic multiplicity 3. 12. p(t) = (t ¡5)2(t + 1)(t ¡15): The eigenvalues are ? = 5 with algebraic multiplicity 2, ? =¡1 with algebraic multiplicity 1, and ? = 15 with algebraic multiplicity 1. 4.4. EIGENVALUES AND THE CHARACTERISTIC POLYNOMIAL 109 13. p(t) = flfl flfl flfl flfl 5¡t 4 1 1 4 5¡t 1 1 1 1 4¡t 2 1 1 2 4¡t flfl flfl flfl flfl = t4¡18t3+97t2¡180t+100 = (t¡1)(t¡2)(t¡5)(t¡ 10): The eigenvalues are ? = 1; ? = 2; ? = 5; ? = 10; each with algebraic multiplicity 1. 14. p(t) = (t¡2)3(t+2): The eigenvalues are ? = 2 with algebraic multiplicity 3, and ? =¡2 with algebraic multiplicity 1. 15. Let x be an eigenvector corresponding to ?: Thus x6= and Ax= ?x : Multiplication by A¡1 yields x= A¡1(?x) = ?A¡1x: Since A is nonsingular ? 6= 0 (cf. Theorem 13). Thus multiplication by ?¡1 gives A¡1x= ?¡1x: 16. If Ax= ?x then (A + fiI)x= Ax+fiIx= ?x+fix= (? + fi)x: 17. Let x be an eigenvector corresponding to ? and suppose Akx= ?kx for some integer k ?2: Then Ak+1x= A(Akx) = A(?kx) = ?k(Ax) = ?k(?x) = ?k+1x : It follows by the principle of induction that Anx = ?nx for each positive integer n; n ?2: 18. (a) q(H)x= (H3 ¡2H2 ¡H +2I)x= H3x¡2H2x¡Hx+2Ix= ?3x¡2?2x¡?x+2x= (?3 ¡2?2 ¡? +2)x= q(?)x: (b) The eigenvalues for q(A) are q(1) = 0 and q(¡1) = 0: The eigenvalues for q(B) are q(0) = 2 and q(¡1) = 0: 19. q(C) = C3 ¡2C2 ¡C +2I = 2 4 35 ¡3 ¡15 ¡44 2 19 68 ¡6 ¡29 3 5¡ 2 2 4 17 ¡1 ¡7 ¡16 2 7 32 ¡2 ¡13 3 5¡ 2 4 3 ¡1 ¡1 ¡12 0 5 4 ¡2 ¡1 3 5+ 2 4 2 0 0 0 2 0 0 0 2 3 5 =O: 20. p(t) = t2 ¡4t +3. p(A) = A2 ¡4A +3I =O. 21. p(t) = t2 ¡2t +1. p(A) = A2 ¡2A + I =O. 22. p(t) =¡t3 +2t2 + t¡2. p(A) =¡A3 +2A2 + A¡2I =O. 23. p(t) = t4 ¡18t3 +97t2 ¡180t +100. p(A) = A4 ¡18A3 +97A2 ¡180A +100I =O. 24. (a) Suppose B = [B1 ;B2 ;B3 ]: Then = Be1 = B1 ; = Be2 = B2 ; and = Be3 = B3 : Thus B =O: (b) Since Aui = ?iui for i = 1;2;3; it follows, as in Exercise 18a, that p(A)ui= p(?i)ui = (0)ui = . By property 3 of Theorem 9 in Section 2.5, fu1 ;u2 ;u3 g is a basis for 110 CHAPTER 4. THE EIGENVALUE PROBLEMS R3: Therefore every vector x in R3 can be expressed in the form x = a1u1 +a2u2 +a3u3 : It follows that p(A)x= a1p(A)u1 +a2p(A)u2 +a3p(A)u3 = . By part (a), p(A) =O: 25. p(A) = A2 ¡(a + d)A +(ad¡bc)I = ? a2 + bc ab + bd ca + dc cb + d2 ? ¡(a + d) ? a b c d ? +(ad¡bc) ? 1 0 0 1 ? = ? 0 0 0 0 ? : 26. Expansion yields p(t) =¡[t3¡(a+b+c)t2+(ab+ac+bc)t¡abc]: Similarly the properties of matrix multiplication imply that ¡(A¡aI)(A¡bI)(A¡cI) =¡[A3 ¡(a + b + c)A2 + (ab + ac + bc)A¡abcI] = p(A): Therefore p(A) =¡(A¡aI)(A¡bI)(A¡cI) =2 4 0 d f 0 b¡a e 0 0 c¡a 3 5 2 4 a¡b d f 0 0 e 0 0 c¡b 3 5 2 4 a¡c d f 0 b¡c e 0 0 0 3 5 =O: 27. (a) For n = 2, det(A¡tI) = flfl flfl ¡a1 ¡t ¡a01 ¡t ? = t2 + a1t + a0 = q(t): For n = 3, det(A ¡ tI) = flfl flfl flfl ¡a2 ¡t ¡a1 ¡a0 1 ¡t 0 0 1 ¡t flfl flfl flfl: Expanding along the third column and applying the case n = 2 yields det(A¡tI) =¡a0 flfl flfl 1 ¡t0 1 flfl flfl¡t flfl flfl ¡a2 ¡t ¡a11 ¡t ? = ¡t(t2 + a2t + a1)¡a0 =¡q(t): (b) A = 2 66 4 ¡3 1 ¡2 2 1 0 0 0 0 1 0 0 0 0 1 0 3 77 5: The characteristic polynomial for A is det(A ¡ tI) = flfl flfl flfl flfl ¡3¡t 1 ¡2 2 1 ¡t 0 0 0 1 ¡t 0 0 0 1 ¡t flfl flfl flfl flfl = q(t): (c) For some integer k ?2; assume that if A = 2 66 66 64 ¡ak¡1 ¡ak¡2 ¢¢¢ ¡a1 ¡a0 1 0 ¢¢¢ 0 0 0 1 ¢¢¢ 0 0 ... 0 0 ¢¢¢ 1 0 3 77 77 75 4.4. EIGENVALUES AND THE CHARACTERISTIC POLYNOMIAL 111 then det(A ¡ tI) = (¡1)k(tk + ak¡1tk¡1 +¢¢¢+ a1t + a0): If A is the [(k + 1) x (k +1)] companion matrix then det(A¡tI) = flfl flfl flfl flfl flfl fl ¡ak ¡t ¡ak¡1 ¢¢¢ ¡a1 ¡a0 1 ¡t ¢¢¢ 0 0 0 1 ¢¢¢ 0 0 ... ... 0 0 ¢¢¢ 1 ¡t flfl flfl flfl flfl flfl fl : Expanding along the (k +1)st column and using the case n = k yields det(A¡tI) = (¡1)k+1a0 flfl flfl flfl flfl fl 1 ¡t ¢¢¢ 0 0 1 ¢¢¢ 0 ... ... 0 0 ¢¢¢ 1 flfl flfl flfl flfl fl ¡t flfl flfl flfl flfl fl ¡ak ¡t ¡ak¡1 ¢¢¢ ¡a1 1 ¡t ¢¢¢ 0 ... 0 0 ¢¢¢ ¡t flfl flfl flfl flfl fl = (¡1)k+1a0 ¡ t(¡1)k(tk + aktk¡1 +¢¢¢+ a2t + a1) = (¡1)k+1q(t): By mathematical induction det(A¡tI) = (¡1)nq(t) for all n; n ?2: 28. x0 = [1;1;1]T;x1 = [1;¡7;1]T;x2 = [9;¡7;17]T;x3 = [17;¡23;33]T; x4 = [41;¡39;81]T;x5 = [81;¡87;161]T:fl0 =¡5=3? ¡1:667; fl1 = 75=51?1:471; fl2 = 875=419?2:088; fl3 = 4267=1907 ?2:238; fl4 = 19755=9763?2:023: 29. Notethat xj= Ajx0 sobyproperty(a)ofTheorem11,xj= c1?j1u1+c2?j2u2+¢¢¢+cn?jnun : Set aj = ?j=?1;1 ? j ? n: Then a1 = 1 whereas jajj< 1 for 2 ? j ? n: In particular limk!1 akj = 0 if j 6= 1: It follows that xkTxk+1 = Pcicj?ki ?k+1j uiTuj 1? i; j ? n = ?2k+11 (c21u1Tu1 +rk); where rk = Pcicjaki ak+1j uiTuj 1? i; j ? n (i; j)6= (1;1): In particular limk!1 rk = 0: Similarly xkTxk = ?2k1 (c1u1Tu1 +tk) where limk!1 tk = 0: Therefore limk!1 flk = ?1: 30. Let A and AT have characteristic polynomials p(t) and q(t); respectively. Note that (A¡tI)T = AT ¡tI: It follows that p(t) = det(A¡tI) = det(A¡tI)T = det(AT ¡tI) = q(t): 31. Suppose p(t) = t2 +a1t+a0: Then p(0) = a0 = det(A¡0I) = 4 and p(1) = 1+a1 +a0 = det(A¡I) = 1: Thus we have a0 = 4 a0 + a1 = 0 : Solving yields a0 = 4; a1 =¡4: Therefore p(t) = t2 ¡4t +4: 112 CHAPTER 4. THE EIGENVALUE PROBLEMS 32. p(0) = a0 = det(A ¡ 0I) = 0 and p(1) = 1 + a1 + a0 = det(A ¡ I) = ¡4: Therefore a0 = 0; a1 =¡4; and p(t) = t2 ¡5t: 33. Suppose p(t) = ¡t3 + a2t2 + a1t + a0: Then p(¡1) = 1+ a2 ¡ a1 + a0 = det(A + I) = 0; p(0) = a0 = det(A¡0I) =¡1; and p(1) =¡1+a2 +a1 +a0 = det(A¡I) = 0: It follows that a0 =¡1; a1 = 1; and a2 = 1: Therefore p(t) =¡t3 + t2 + t¡1: 34. p(¡1) = 1+a2¡a1+a0 = det(A+I) = 0; p(0) = a0 = det(A¡0I) = 0; and p(1) =¡1+a2+ a1 + a0 = det(A¡I) =¡4: Therefore a0 = 0; a1 =¡1; a2 =¡2; and p(t) =¡t3 ¡2t2 ¡t: 4.5 Eigenvalues and Eigenvectors 1. (A¡3I)x= is the system ¡x1 ¡x2 = 0 ¡x1 ¡x2 = 0 : The solution is x1 = ¡x2; x2 arbitrary, so E? consists of the vectors of the form x2 ? ¡1 1 ? : Thus f[¡1;1]Tg is a basis for E?. The eigenvalue ? = 3 has algebraic and geometric multiplicity 1. 2. The system (A¡I)x= has solution x1 = x2; x2 arbitrary. f[1;1]Tg is a basis for E?. The eigenvalue ? = 1 has algebraic and geometric multiplicity 1. 3. (B ¡2I)x= is the system ¡x1 ¡x2 = 0 x1 + x2 = 0 : Thesolutionis x1 =¡x2; x2 arbitrary, so E? consistsofthevectorsoftheform x2 ? ¡1 1 ? : Thus f[¡1;1]Tg is a basis for E?: The eigenvalue ? = 2 has algebraic multiplicity 2 and geometric multiplicity 1. 4. The system (C ¡I)x= has solution x1 = (1=2)x3; x2 = (¡3=2)x3; x3 arbitrary. The set f[1;¡3;2]Tg is a basis for E?: The eigenvalue ? = 1 has algebraic multiplicity 2 and geometric multiplicity 1. 5. (C + I)x= is the system ¡5x1 ¡ x2 + 2x3 = 0 3x1 + 3x2 = 0 ¡14x1 ¡ 2x2 + 6x3 = 0 : 4.5. EIGENVALUES AND EIGENVECTORS 113 The solution is x1 = (1=2)x3; x2 = (¡1=2)x3; x3 arbitrary, so E? consists of vectors of the form a 2 4 1 ¡1 2 3 5 where a is an arbitrary scalar. Thus f[1;¡1;2]Tg is a basis for E?: The eigenvalue ? =¡1 has algebraic and geometric multiplicity 1. 6. The system (D ¡I)x= has solution x1 = (1=2)x2 ¡(3=8)x3; x2 and x3 arbitrary. The set f[1;2;0]T;[¡3;0;8]Tg is a basis for E?: The eigenvalue ? = 1 has algebraic multiplicity 3 and geometric multiplicity 2. 7. (E + I)x= is the system 7x1 + 4x2 + 4x3 + x4 = 0 4x1 + 7x2 + x3 + 4x4 = 0 4x1 + x2 + 7x3 + 4x4 = 0 x1 + 4x2 + 4x3 + 7x4 = 0 : The solution is x1 = x4; x2 =¡x4; x3 =¡x4; x4 arbitary, so E? consists of vectors of the form x4 2 66 4 1 ¡1 ¡1 1 3 77 5: Thus f[1;¡1;¡1;1]Tg is a basis for E?: The eigenvalue ? =¡1 has geometric and alge- braic multiplicity 1. 8. The system (E ¡5I)x= has solution x1 =¡x4; x2 =¡x3; x3 and x4 arbitrary. The set f[¡1;0;0;1]T;[0;¡1;1;0]Tg is a basis for E?: The eigenvalue ? = 5 has algebraic and geometric multiplicity 2. 9. (E ¡15I)x= is the system ¡9x1 + 4x2 + 4x3 + x4 = 0 4x1 ¡ 9x2 + x3 + 4x4 = 0 4x1 + x2 ¡ 9x3 + 4x4 = 0 x1 + 4x2 + 4x3 ¡ 9x4 = 0 : The solution is x1 = x2 = x3 = x4; x4 arbitrary so E? con- sists of vectors of the form x4 2 66 4 1 1 1 1 3 77 5: Thus f[1;1;1;1]Tg is a basis for E?: The eigenvalue ? = 15 has algebraic and geometric multi- plicity 1. 114 CHAPTER 4. THE EIGENVALUE PROBLEMS 10. The system (F +2I)x= has solution x1 = x2 = x3 = x4; x4 arbi- trary. The set f[1;1;1;1]Tg is a basis for E?: The eigenvalue ? =¡2 has algebraic and geometric multiplicity 1. 11. (F ¡2I)x= is the system ¡x1 ¡ x2 ¡ x3 ¡ x4 = 0 ¡x1 ¡ x2 ¡ x3 ¡ x4 = 0 ¡x1 ¡ x2 ¡ x3 ¡ x4 = 0 ¡x1 ¡ x2 ¡ x3 ¡ x4 = 0 : The solution is x1 = ¡x2 ¡ x3 ¡ x4; x2; x3; x4 arbitrary so E? consists of vectors of the form x2 2 66 4 ¡1 1 0 0 3 77 5+x3 2 66 4 ¡1 0 1 0 3 77 5+x4 2 66 4 ¡1 0 0 1 3 77 5: Thus f[¡1;1;0;0]T;[¡1;0;1;0]T;[¡1;0;0;1]Tg is a basis for E?: The eigenvalue ? = 2 has algebraic and geometric multiplicity 3. 12. The characteristic equation is p(t) = ¡(t ¡1)2(t ¡2) = 0 so the eigenvalues are ? = 1 and ? = 2: The eigenvectors for ? = 1 are the nonzero vectors of the form 2 4 x1 x3 x3 3 5 = x1 2 4 1 0 0 3 5 + x3 2 4 0 1 1 3 5: The eigenvectors for ? = 2 are the nonzero vectors of the form 2 4 x2 x2 0 3 5 = x2 2 4 1 1 0 3 5: The matrix is not defective. 13. The characteristic equation for the given matrix A is 0 = det(A¡tI) = flfl flfl flfl 2¡t 1 2 0 3¡t 2 0 0 2¡t flfl flfl flfl =¡(t¡2)2(t¡3): The eigenvalues are ? = 2 and ? = 3: The system (A¡2I)x= is given by x2 +2x3 = 0 x2 +2x3 = 0 : In the solution x1 is arbitrary, x2 =¡2x3; and x3 is arbitrary. The eigenvectors for ? = 2 are the nonzero vectors of the form 2 4 x1 ¡2x3 x3 3 5 4.5. EIGENVALUES AND EIGENVECTORS 115 = x1 2 4 1 0 0 3 5+ x3 2 4 0 ¡2 1 3 5. Therefore ? = 2 has algebraic and geometric multiplicity 2. The system (A¡3I)x= is given by ¡x1 + x2 +2x3 = 0 2x3 = 0 ¡x3 = 0 : The solution is x1 = x2; x2 arbitrary, x3 = 0: The eigenvectors for ? = 3 are the nonzero vectors of the form 2 4 x2 x2 0 3 5 = x2 2 4 1 1 0 3 5: Therefore ? = 3 has algebraic and geometric multiplicity 1. The matrix is not defective. 14. The characteristic polynomial is p(t) = ¡(t ¡1)3 so ? = 1 is the only eigenvalue. The eigenvectors for ? = 1 are the nonzero vectors of the form 2 4 x1 0 0 3 5 = x1 2 4 1 0 0 3 5: Thus ? = 1 has algebraic multiplicity 3 and geometric multiplicity 1. The matrix is defective. 15. The given matrix A has characteristic equation 0 = det(A ¡ tI) = ¡(t ¡ 2)2(t ¡ 1) so the eigenvalues are ? = 1 and ? = 2: The system of equations (A ¡ I)x = has solution x1 = ¡3x3; x2 = ¡x3; x3 arbitrary so the eigenvectors for ? = 1 are the nonzero vectors of the form x= [¡3x3;¡x3; x3]T: For the system (A ¡2I)x= , x1 and x2 are arbitrary and x3 = 0: The eigenvectors for ? = 2 are the nonzero vectors of the form x = x1 2 4 1 0 0 3 5+ x2 2 4 0 1 0 3 5: The matrix is not defective. 16. The characteristic polynomial is p(t) =¡(t¡3)2(t+4) so the eigenvalues are ? = 3 and ? =¡4: The eigenvectors for ? = 3 are the nonzero vectors of the form x= [5x3;3x3; x3]T: The eigenvectors for ? =¡4 are the nonzero vectors of the form x= [¡2x3;(2=3)x3; x3]T: Since ? = 3 hasalgebraicmultiplicity2andgeometricmultiplicity1,thematrixisdefective. 17. The given matrix A has characteristic polynomial p(t) = ¡(t +1)(t ¡1)(t ¡2) so the eigenvalues for A are ? = ¡1; ? = 1; ? = 2: The system of equations (A+I)x= has solution x1 = (1=2)x3; x2 = x3; x3 arbitrary so the eigenvectors for ? = ¡1 are the nonzero vectors of the form x = [(1=2)x3; x3; x3]T: The system of equations (A¡I)x= has solution x1 =¡3x2; x2 arbitrary,x3 =¡7x2 so the eigenvectors for ? = 1 are the nonzero vectors of the form x= [¡3x2; x2;¡7x2]T: The system of equations (A¡2I)x= has solution x1 = (1=2)x3; x2 = (¡1=2)x3 so the eigenvectors for ? = 2 are the nonzero vectors of the form x= [(1=2)x3;(¡1=2)x3; x3]T: The matrix is not defective. 116 CHAPTER 4. THE EIGENVALUE PROBLEMS 18. u1 = [1;1]T is an eigenvector for the eigenvalue ? = 2 and u2 = [2;5]T is an eigenvector for the eigenvalue ? = ¡1: Moreover x= ¡6u1 +3u2 so A10x= ¡6(2)10u1 +3(¡1)10u2 = [¡6138;¡6129]T: 19. The characteristic polynomial for A is p(t) = ¡(t ¡1)2(t ¡2) so the eigenvalues for A are ? = 1 and ? = 2: The vectors u1 = [1;0;0]T and u2 = [0;1;2]T are the eigenvectors for ? = 1 and u3 = [1;2;3]T is an eigenvector for ? = 2: Moreover x= u1 +2u2 +u3 so A10x= (1)10u1 +2(1)10u2 +(2)10u3 = [1025;2050;3076]T: 20. Since ? is an eigenvalue for H;nullity(H ¡?I) ? 1: It follows that rank(H ¡?I) ? 3: But a; b and c? are nonzero so the flrst three columns of H ¡?I are linearly independent. Therefore rank(H ¡?I)?3: Thus rank(H ¡?I) = 3 and, hence, nullity(H ¡?I) = 1: This proves that ? has geometric multiplicity 1. 21. P = P¡1P2 = P¡1P = I: 22. Suppose Px= ?x;x6= . Then ?2x= P2x= Px= ?x so (?2 ¡?)x= . Since x6= , 0 = ?2 ¡? = ?(?¡1): Therefore either ? = 0 or ? = 1: 23. P2 = (uuT)(uuT) = u(uTu)uT = uuT = P: 24. (I ¡ Q)2 = I2 ¡ IQ ¡ QI + Q2 = I ¡ Q: Also (I ¡2Q)2 = I2 ¡2IQ ¡2QI +4Q2 = I so (I ¡2Q)¡1 = I ¡2Q: 25. P2 = (uuT + vvT)(uuT + vvT) = u(uTu)uT +u(uTv)vT + v(vTu)uT +v(vTv)vT = uuT +vvT = P: 26. P(au+bv) = aPu+bPv= a(uuT +vvT)u+b(uuT +vvT)v= au(uTu)+ av(vTu)+ bu(uTv)+ bv(vTv) = au+bv: 27. (a) A has eigenvalues ? = 1 and ? = 3 with corresponding eigenvectors u1 = [1=p2;1=p2]T and u2 = [¡1=p2;1=p2]T; respectively, where ku1k=ku2k = 1: It is easily checked that u1u1T +3u2u2T = A: (b) A has eigenvalues ? =¡1 and ? = 3 with corresponding eigenvectors u1 = [¡1=p2;1=p2]T and u2 = [1=p2;1=p2]T; respec- tively, where ku1k=ku2k= 1: It is easily checked that ¡u1u1T +3u2u2T = A: (c) A has eigenvalues ? = 4 and ? = ¡1 with corresponding eigenvectors u1 = [2=p5;1=p5]T and u2 = [1=p5;¡2=p5]T; respec- tively, where ku1k=ku2k= 1: It is easily checked that 4u1u1T ¡u2u2T = A. 28. fl(uTv) = uT(flv) = u T(Av) = (Av)Tu= vTATu= vT(Au) = vT(?u) = ?(vTu) = ?(uTv): Since fl 6= ? it follows that uTv= 0: 4.6. COMPLEX EIGENVALUES AND EIGENVECTORS 117 29. Set B = A¡C: For each i;1? i ? n; Bui = Aui¡Cui = Aui¡?1u1(u1Tui )¡¢¢¢¡?iui (uiTui )¡¢¢¢¡ ?nun (unTui ) = ?iui ¡?iui = . If x is in Rn then x may be written in the form x= a1u1 +¢¢¢+ anun ; so Bx= a1Bu1 +¢¢¢+ anBun = . In particular Bej = for 1 ? j ? n: If B = [B1 ; : : : ;Bn ] then Bej = Bj ; so Bj = for 1 ? j ? n: This shows that A¡C =O so A = C: 4.6 Complex Eigenvalues and Eigenvectors 1. u = 3+2i: 2. z = 1¡i: 3. u + v = 7¡3i: 4. z + w = 3¡2i: 5. u + u = 6: 6. s¡s = 4i: 7. vv = 17: 8. uv = 10¡11i: 9. s2 ¡w =¡5+5i: 10. z2w = 2+4i: 11. uw2 = 17¡6i: 12. s(u2 + v) = 31+7i: 13. u=v = (uv)=(vv) = 10=17¡(11=17)i: 14. v=u2 = 8=169+(53=169)i: 15. s=z = (sz)=(zz) = 3=2+(1=2)i: 16. (w + v)=u = 22=13+(6=13)i: 17. w + iz = 1: 18. s¡iw = 0: 118 CHAPTER 4. THE EIGENVALUE PROBLEMS 19. For the given matrix A the characteristic polynomial is p(t) = t2 ¡ 8t + 20 and the eigenvalues are ? = 4+2i and ? = 4¡2i: The system of equations (A¡(4+2i)I)x= is given by (2¡2i)x1 + 8x2 = 0 ¡x1 + (¡2¡2i)x2 = 0 : The solution is x1 = (¡2¡2i)x2; x2 arbitrary. Thus the eigenvectors for ? = 4+2i are the nonzero vectors of the form x= [(¡2¡2i)x2; x2]T: By Theorem 16, x is an eigenvector corresponding to ?. 20. The characteristic polynomial is p(t) = t2 +4 so the eigenvalues are ? = 2i and ? =¡2i: The eigenvectors for ? = 2i are the nonzero vectors of the form x= [(¡1¡i)x2; x2]T: By Theorem 16, x is an eigenvector for ? =¡2i: 21. The given matrix A has characteristic polynomial p(t) = t2 + 1 so the eigenvalues are ? = i and ? =¡i: The system (A¡iI)x= is given by (¡2¡i)x1 ¡ x2 = 0 5x1 + (2¡i)x2 = 0 : The solution is x1 = ((¡2+ i)=5)x2; so the eigenvectors for ? = i are the nonzero vectors of the form x= [((¡2+ i)=5)x2; x2]T: By Theorem 16, x is an eigenvector for ? =¡i: 22. The characteristic polynomial is p(t) = ¡(t ¡ 2)(t2 ¡ 4t + 5): The eigenvalues are ?1 = 2; ?2 = 2 + i; and ?2 = 2 ¡ i. The eigenvectors for ?1 = 2 have the form x = [0;¡x3; x3]T; x3 6= 0: The eigenvectors for ?2 = 2+ i have the form x= [x3;(¡2=5¡ (1=5)i)x3; x3]T; x3 6= 0 and x is an eigenvector for ?2 = 2¡i: 23. The given matrix A has characteristic polynomial p(t) = ¡(t¡2)(t2 ¡4t +13) so the eigenvalues for A are ?1 = 2; ?2 = 2+3i; and ?2 = 2¡3i: The system (A¡2I)x= is given by ¡x1 ¡ 4x2 ¡ x3 = 0 3x1 + 3x3 = 0 x1 + x2 + x3 = 0 : The solution is x1 =¡x3; x2 = 0; x3 arbitrary so the eigenvectors corresponding to ?1 = 2 are of the form x= [¡x3;0; x3]T: The system [A¡(2+3i)I ]x= is given by (¡1¡3i)x1 ¡ 4x2 ¡ x3 = 0 3x1 ¡ 3ix2 + 3x3 = 0 x1 + x2 + (1¡3i)x3 = 0 : 4.6. COMPLEX EIGENVALUES AND EIGENVECTORS 119 The solution is x1 = (¡5=2+(3=2)i)x3 and x2 = (3=2+(3=2)i)x3 so the eigenvectors for ?2 = 2 + 3i are the nonzero vectors of the form x = [(¡5=2 + (3=2)i)x3;(3=2+(3=2)i)x3; x3]T: By Theorem 16, x is an eigenvector for ?2 = 2¡3i: 24. The characteristic polynomial is p(t) = (t2 ¡2t +26)(t2 ¡2t +5) so the eigenvalues are ?1 = 1 + 5i; ?1 = 1¡5i; ?2 = 1 + 2i; ?2 = 1¡2i: Eigenvectors for ?1 are the nonzero vectors of the form x= [ix2; x2;0;0]T and x is an eigenvector for ?1: Eigenvectors for ?2 are the nonzero vectors of the form x= [0;0; ix4; x4]T and x is an eigenvector for ?2: 25. x = 2¡i; y = 3¡2i: 26. x = i; y = 2: 27. xTx= (1¡i)(1+ i)+2(2) = 6 so kxk= p6: 28. xTx= (3¡i)(3+ i)+(2+ i)(2¡i) = 10+5 = 15 so kxk= p15: 29. xTx= (1+2i)(1¡2i)+(¡i)(i)+(3¡i)(3+ i) = 5+1+10 = 16: Thus kxk= p16 = 4: 30. xTx= (¡2i)(2i)+(1+ i)(1¡i)+3(3) = 4+2+9 = 15 so kxk= p15: 31. ?1 = ¡1:4937 + 1:2616i, x1 = 2 4 0:5835¡0:1460i 0:1650¡0:4762i ¡0:4369+0:4397i 3 5 ; ?2 = ¡1:4937 ¡ 1:2616i, x2 = 2 4 0:5835+0:1460i 0:1650+0:4762i ¡0:4369¡0:4397i 3 5; ?3 = 10:9873, x3 = 2 4 ¡0:4486 ¡0:7312 ¡0:5139 3 5. In each case, the eigenvectors are chosen to have length 1. 32. ?1 = ¡3:6884 + 2:8416i, x1 = 2 4 ¡0:0558¡0:6977i ¡0:4571+0:1436i 0:3948+0:3532i 3 5; ?2 = ¡3:6884 ¡ 2:8416i, x2 = 2 4 ¡0:0558+0:6977i ¡0:4571¡0:1436i 0:3948¡0:3532i 3 5; ?3 = 13:3769, x3 = 2 4 ¡0:4184 ¡0:7889 ¡0:4501 3 5. For i = 1;2;3;kxik= 1. 33. ?1 = 1:1857+2:6885i, x1 = 2 66 4 ¡0:0781¡0:6033i ¡0:3495+0:5754i 0:1199¡0:1125i 0:1963+0:3334i 3 77 5; ?2 = 1:1857¡2:6885i, x2 = 2 66 4 ¡0:0781+0:6033i ¡0:3495¡0:5754i 0:1199+0:1125i 0:1963¡0:3334i 3 77 5; 120 CHAPTER 4. THE EIGENVALUE PROBLEMS ?3 = 16:8037, x3 = 2 66 4 ¡0:5484 ¡0:0550 ¡0:1746 ¡0:8160 3 77 5; ?4 = 4:8249, x4 = 2 66 4 0:7046 ¡0:6728 ¡0:2027 ¡0:0995 3 77 5. For i = 1;2;3;4;kxik= 1. 34. ?1 = 0:2617+2:0076i, x1 = 2 66 4 ¡0:2742+0:1318i ¡0:6251¡0:1139i 0:1594+0:2495i 0:5871¡0:2672i 3 77 5; ?2 = 0:2617¡2:0076i, x2 = 2 66 4 ¡0:2742¡0:1318i ¡0:6251+0:1139i 0:1594¡0:2495i 0:5871+0:2672i 3 77 5; ?3 = 16:6911, x3 = 2 66 4 0:5848 0:5644 0:1660 0:5585 3 77 5; ?4 = 3:7856, x4 = 2 66 4 ¡0:4955 0:6192 0:2253 ¡0:5659 3 77 5 35. Let z = a + bi and w = c + di. (a) z + w = (a + c)+(b + d)i so z + w = (a + c)¡(b + d)i = z + w: (b) zw = (ac ¡ bd) + (ad + bc)i so zw = (ac ¡ bd) ¡ (ad + bc)i: Therefore z w = (a¡bi)(c¡di) = (ac¡bd)¡(ad + bc)i = zw: (c) z + z = (a + bi)+(a¡bi) = 2a: (d) z ¡z = (a + bi)¡(a¡bi) = 2bi: (e) zz = (a + bi)(a¡bi) = a2 ¡b2i2 = a2 + b2: 36. If A = [aij] and B = [bij] then AB is the (n x p) matrix [cij] where cij = Pnk=1 aikbkj: Likewise A B is the (n x p) matrix [dij] where dij = Pnk=1 aikbkj = Pnk=1 aikbkj =P n k=1 aikbkj = cij: Thus A B = AB: If A is a real matrix and x is an (n x 1) vector then Ax = Ax = Ax: 37. (AB)? = (AB)T = (A B)T = BTAT = B?A?: 4.7. SIMILARITY TRANSFORMATIONS & DIAGONALIZATION 121 38. (a) First note that for vectors x and y;y?x = y?x?? = (x?y)? = x?y: Now suppose that Ax = ?x where x 6= . Then ? kxk 2 = ?xTx = ?x?x = x?(?x) = x?(Ax) = (Ax)?x = (x?A?)x = x?(Ax) = x?(?x) = ?x?x = ?kxk2 = ? kxk2 : Since kxk26= 0; ? = ? and ? is a real number. (b) A? is the (n x n) matrix [bij] where bij = aji: Since A = A?; aij = aji for 1? i; j ? n: In particular, aii = aii for 1? i ? n; so aii is a real number. 39. (a) Since p(r) = 0 wehave0 = 0 = p(r) = a0 + a1r +¢¢¢+ anrn = a0+a1 r+¢¢¢+anrn = a0 + a1r +¢¢¢+ anrn = p(r): (b) Write p(t) = c(t¡r1)(t¡r2)(t¡r3) where c is a real number and r1; r2; r3 are the (not necessarily distinct) roots of p(t). If all three roots are real numbers then there is nothing to prove, so assume that r1 = a + bi; b 6= 0: By (a) we may also assume that r2 = r1 = a¡bi: Thus p(t) = c(t¡r1)(t¡r2)(t¡r3) = c[t2¡2at+(a2+b2)](t¡r3): Since the coe?cients of p(t) are real numbers, it follows that r3 is a real number. (c) Thecharacteristicpolynomial p(t) = det(A¡tI) hasdegreethreeandrealcoe?cients. By (b) p(t) has a real root so A has at least one real eigenvalue. 40. For any vector x;k Ax k2= (Ax)TAx = (Ax)T(Ax) = xTATAx = xTx =k x k 2: In particular if Ax = ?x;x 6= , then kxk2 = k?xk2= ?xT(?x) = ??(xTx) = ?? kxk2: Thus ?? = 1: 41. Let ? be an eigenvalue for A (note that ? is real by Theorem 17) and suppose x in Rn is a corresponding eigenvector. Then x6= and 0 < x TAx= x T(?x) = ?xTx = ? kxk2: Since kxk2 > 0 it follows that ? > 0: 42. cf. Exercise 40. 4.7 Similarity Transformations and Diagonalization 1. A has eigenvalues ? = 1 and ? = 3 with corresponding eigenvectors u1 = [1;1]T and u2 = [¡1;1]T; respectively. If S = [u1 ;u2 ] then S¡1AS = D where D = ? 1 0 0 3 ? : Now D5 = ? 1 0 0 243 ? so A5 = SD5S¡1 = ? 122 ¡121 ¡121 122 ? : 2. For S = ? 1 ¡1 1 1 ? ; S¡1AS = ? 0 0 0 2 ? : A5 = S ? 0 0 0 25 ? S¡1 = ? 16 ¡16 ¡16 16 ? : 122 CHAPTER 4. THE EIGENVALUE PROBLEMS 3. A has only one eigenvalue, ? = ¡1. The corresponding eigenvectors are the nonzero vectors of the form x= [x2; x2]T: Since we cannot flnd a linearly independent set fu1 ;u2g of eigenvectors for A; A is not diagonalizable. 4. A has eigenvalue ? = 1 with associated eigenvectors of the form x= [x1;0]T; x1 6= 0: A is not diagonalizable, 5. A has eigenvalues ? = 1 and ? = 2 with corresponding eigenvectors u1 = [1;¡10]T and u2 = [0;1]T; respectively. If S = [u1 ;u2 ] then S¡1AS = D where D = ? 1 0 0 2 ? : Thus A5 = SD5S¡1 = S ? 1 0 0 32 ? S¡1 = ? 1 0 310 32 ? : 6. For S = ? 1 7 0 2 ? ; S¡1AS = ? ¡1 0 0 1 ? : Thus A5 = S ? (¡1)5 0 0 (1)5 ? S¡1 = ? ¡1 7 0 1 ? = A: 7. A has eigenvalue ? = 1 with algebraic multiplicity 3. The eigenvectors for ? = 1 have the form x = [x2 + 2x3 ;x2 ;x3]T: In particular we cannot obtain 3 linearly independent eigenvectors so A is not diagonalizable. 8. A has eigenvalue ? = 1 with algebraic multiplicity 3 and geometric multiplicity 1, so A is not diagonalizable. 9. A has eigenvalues 1;2; and ¡1 with corresponding eigenvectors u1 = [¡3;1;¡7]T;u2 = [¡1;1;¡2]T;u3 = [1;2;2]T; respectively. If S = [u1 ;u2 ;u3 ] then S¡1 = 2 4 2 0 ¡1 16=3 1=3 7=3 5=3 1=3 ¡2=3 3 5 and S¡1AS = D; where D = 2 4 1 0 0 0 2 0 0 0 ¡1 3 5: A5 = SD5S¡1 = 2 4 163 ¡11 ¡71 ¡172 10 75 324 ¡22 ¡141 3 5: 10. A has eigenvalues ? = 1 and ? = 2. For ? = 1; u1 = [1;0;0]T and u2 = [0;1;1]T are linearly independent eigenvectors. ? = 2 has corresponding eigenvector u3 = [1;1;0]T: If S = [u1 ;u2 ;u3 ] then S¡1AS = 2 4 1 0 0 0 1 0 0 0 2 3 5 = D: Moreover A5 = SD5S¡1 = 2 4 1 31 ¡31 0 32 ¡31 0 0 1 3 5: 4.7. SIMILARITY TRANSFORMATIONS & DIAGONALIZATION 123 11. A has eigenvalue ? = 1 with algebraic multiplicity 2 and geometric multiplicity 1 so A is not diagonalizable. 12. If S = 2 4 1 0 3 0 ¡1 4 0 1 0 3 5 then S¡1AS = 2 4 1 0 0 0 1 0 0 0 5 3 5: A5 = S 2 4 1 0 0 0 1 0 0 0 55 3 5S¡1 = 2 4 1 2343 2343 0 3125 3124 0 0 1 3 5: 13. q1Tq2 = 0 and q1Tq1 = q2Tq2 = 1 so Q is orthogonal. 14. Q is orthogonal. 15. q1Tq1 = 5 so Q is not orthogonal. 16. q1Tq1 = 13 so Q is not orthogonal. 17. 0 = q1Tq2 = q1Tq3 = q2Tq3 and 1 = q1Tq1 = q2Tq2 = q3Tq3 so Q is orthogonal. 18. q1Tq1 = 6 so Q is not orthogonal. 19. If Q isorthogonalthen2fi2 = 1;6fl2 = 1; a2+b2+c2 = 1; fia+fic = 0; and fla+2flb¡flc = 0: This implies that fi = 1=p2; fl = 1=p6; a = ¡c and b = c; where c = §1=p3: Thus we one choice for Q is Q = 2 4 1=p2 1=p6 ¡1=p3 0 2=p6 1=p3 1=p2 ¡1=p6 1=p3 3 5: 20. If Q is orthogonal then fi = 1=p3; fl = 1=p14; a = (¡5=4)c; and b = (1=4)c where c =§4=p42: 21. Q = 2 4 ¡0:8807 0:4332 0:1918 0:1849 0:6870 ¡:7028 0:4361 0:5835 0:6851 3 5 T = 2 4 0:5048 2:9498 ¡1:4966 0:0 8:3443 0:2429 0:0 0:0 ¡2:8491 3 5 22. Q = 2 4 0:5892 ¡0:7155 ¡0:3754 0:7516 0:6559 ¡0:0703 ¡0:2965 0:2407 ¡0:9242 3 5 T = 2 4 ¡0:5223 ¡7:6134 ¡8:2480 0:0 ¡5:6251 0:0735 0:0 0:0 7:1474 3 5 124 CHAPTER 4. THE EIGENVALUE PROBLEMS 23. Q = 2 66 4 ¡0:5276 ¡0:5463 ¡0:6406 0:1130 ¡0:4235 ¡0:4440 0:6477 ¡0:4516 ¡0:3315 0:0292 0:3989 0:8545 ¡0:6576 0:7096 ¡0:1044 ¡0:2306 3 77 5 T = 2 66 4 19:2422 1:4541 ¡3:5019 0:0983 0:0 ¡3:8383 ¡3:4646 ¡0:5055 0:0 0:0 4:8409 3:5148 0:0 0:0 0:0 0:7552 3 77 5 24. Q = 2 66 4 0:5118 ¡0:5287 0:6725 ¡0:0794 0:6815 ¡0:2030 ¡0:6454 0:2788 0:3517 0:3331 ¡0:1083 ¡0:8681 0:3871 0:7539 0:3457 0:4029 3 77 5 T = 2 66 4 19:1640 5:4524 1:6686 ¡1:9529 0:0 2:9467 ¡2:5265 ¡2:3201 0:0 0:0 ¡3:8881 0:9528 0:0 0:0 0:0 ¡2:2226 3 77 5 25. (a) (S¡1AS)2 = (S¡1AS)(S¡1AS) = S¡1A(SS¡1)AS = S¡1A2S: (S¡1AS)3 = (S¡1AS)2(S¡1AS) = (S¡1A2S)(S¡1AS) = S¡1A2(SS¡1)AS = S¡1A3S: (b) Suppose (S¡1AS)n = S¡1AnS for some integer n ? 1: Then (S¡1AS)n+1 = (S¡1AS)n(S¡1AS) = (S¡1AnS)(S¡1AS) = S¡1An(SS¡1)AS = S¡1An+1S: By in- duction (S¡1AS)k = S¡1AkS for any positive integer k. 26. Suppose S¡1AS = D; where D is a diagonal matrix, and suppose W¡1AW = B: If T = W¡1S then T is invertible and T¡1BT = S¡1WBW¡1S = S¡1AS = D: Therefore B is diagonalizable. 27. Suppose that S¡1AS = B: (a) S¡1(A + fiI)S = S¡1AS + S¡1(fiI)S = B + fiS¡1IS = B + fiI: (b) Set Q = (S¡1)T = (ST)¡1: Then Q¡1ATQ = STAT(S¡1)T = (S¡1AS)T = BT: (c) A product of nonsingular matrices is nonsingular so if A is nonsingular then so is B: Moreover B¡1 = (S¡1AS)¡1 = S¡1A¡1(S¡1)¡1 = S¡1A¡1S: Therefore B¡1 is similar to A¡1: 28. For (b) if x;y are in Rn then (Qx)T(Qy) = x TQTQy= xTIy = xTy: For (c) let t = det(Q): Recall that from Theorem 5 of Section 3.3 that det(QT) = det(Q): Since QTQ = I; Theorem 2 of Section 3.2 gives 1 = det(I) = det(QTQ) = det(QT)det(Q) = t2: But t2 = 1 implies that t =§1: 4.7. SIMILARITY TRANSFORMATIONS & DIAGONALIZATION 125 29. First note that QT = (I ¡2uuT)T = I T ¡2u TTu T = I ¡2uu T = Q; so Q is symmetric. Thus QTQ = QQ = (I ¡2uuT )(I ¡2uuT ) = I 2 ¡2uu TI ¡2IuuT +4u(uTu)u T = I: Thus Q is orthogonal. Moreover Qu = (I ¡2uuT)(u) = Iu¡2u(uTu) = u¡2u =¡u; so u s an eigenvector corresponding to the eigenvalue ? =¡1. 30. (AB)T(AB) = BT(ATA)B = BTB = I: 31. y= [b=pa2 + b2;¡a=pa2 + b2]T is one choice. ¡y is another choice. 32. QTQ = ? uTu uTv vTu vTv ? = ? 1 0 0 1 ? so Q is orthogonal. Now QTAQ is the product ? uT vT ? A[u;v] = ? uT vT ? [Au; Av] = ? uT vT ? [?u; Av] = ? ?uTu u TAv ?vTu v TAv ? = ? ? u TAv 0 v TAv ? : 33. If u= [1=p2;¡1=p2]T then Au= 2u and uTu= 1: If v= [1=p2;1=p2]T then uTv= 0 and vTv= 1: If Q = [u;v] then QTAQ =? 2 ¡2 0 2 ? : 34. If u= [1=p5;2=p5]T then Au= u and uTu= 1: If v= [2=p5;¡1=p5]T then uTv = 0 and vTv = 1: If Q = [u ;v ] then QTAQ = ? 1 8 0 2 ? : Similarly if u= [2=p13;3=p13]T and v= [3=p13;¡2=p13]T then Au = 2u ;uTu = vTv = 1; and uTv = 0: If Q = [u ;v ] then QTAQ = ? 2 8 0 1 ? : 35. If u= [1=p2;1=p2]T then Au= u and uTu= 1: If v= [1=p2;¡1=p2]T then uTv= 0 and vTv= 1: If Q = [u;v] then QTAQ = ? 1 0 0 3 ? : 36. If Q = ? ¡1=p2 1=p2 1=p2 1=p2 ? then QTAQ = ? 0 1 0 5 ? : Also if Q = ? 2=p13 3=p13 3=p13 ¡2=p13 ? then QTAQ = ? 5 1 0 0 ? : 126 CHAPTER 4. THE EIGENVALUE PROBLEMS 37. Note that ARj is the jth column of AR and RTi is the ith row of RT: Thus RTi ARj is the ijth entry of RTAR: 38. It is an consequence of Exercise 37 that QTAQ = 2 4 uTAu uTAv uTAw vTAu vTAv vTAw wTAu wTAv wTAw 3 5: But Au= ?u;uTu= 1;vTu= 0 = wTu; so QTAQ has the desired form. 39. The matrix B has characteristic polynomial p(t) = det(B ¡tI): Expansion along the flrst column gives p(t) = (?¡t)det(A1 ¡tI) = (?¡t)q(t) where q(t) is the characteristic polynomial for A1: Since every root of q(t) is also a root of p(t); each eigenvalue for A1 is also an eigenvalue for B: Since Q is an orthogonal matrix QT = Q¡1 and B is similar to A: Thus A and B have the same eigenvalues. In particular B has only real eigenvalues. It follows that A1 has only real eigenvalues. 40. (a) It is straightforward to show that RT = 2 64 1 j 0 0 ¡¡ + ¡¡ ¡¡ 0 j ST0 j 3 75 and that RTR = 2 64 1 j 0 0 ¡¡ + ¡¡ ¡¡ 0 j STS0 j 3 75 = I: (b) RTQTAQR = RTBR and RTBR has the form 2 64 ? j a b ¡¡ + ¡¡ ¡¡ 0 j STA1S0 j 3 75 = 2 66 4 ? j a b ¡¡ + ¡¡ ¡¡ 0 j T1 0 j 3 77 5: Since T1 is upper triangular so is RTBR: 4.7. SIMILARITY TRANSFORMATIONS & DIAGONALIZATION 127 41. Assume that Theorem 22 is true for any [(k-1) x (k-1)] matrix with only real eigenvalues. Now let A be a (k x k) matrix with only real eigenvalues and suppose Au= ?u where uTu= 1: By the Gram-Schmidt process there is an orthonormal basis fu1 ;u2 ; : : : ;ukg for Rk such that u1 = u: The matrix Q = [u1 ;u2 ; : : : ;uk ] is orthogonnal and QTAQ = 2 66 66 64 ? j u1TAu2 ¢¢¢ u1TAuk ¡¡ + ¡¡¡¡ ¡¡¡¡ ¡¡¡¡ 0 j ... j A 1 0 j 3 77 77 75; where A1 is a [(k-1) x (k-1)] matrix. If B = QTAQ then B has characteristic polynomial p(t) = det(B ¡tI): Expanding along the flrst column yields p(t) = (?¡t)det(A1 ¡tI) = (? ¡ t)q(t) where q(t) is the characteristic polynomial for A1: Thus every eigenvalue for A1 is also an eigenvalue for B: But B is similar to A (since QT = Q¡1?) so B has only real eigenvalues and, therefore A1 has only real eigenvalues. By assumption there exists a [(k-1) x (k-1)] orthogonal matrix S such that STA1S = T1 where T1 is upper triangular. If R is the (k x k) matrix R = 2 66 66 64 1 j 0 ¢¢¢ 0 ¡¡ + ¡¡ ¡¡¡¡ ¡¡ 0 j ... j S 0 j 3 77 77 75 then R is orthogonal and P = QR is orthogonal. Furthermore PTAP = RTQTAQR = RTBR and RTBR has the form 2 66 66 64 ? j c2 ¢¢¢ ck ¡¡ + ¡¡ ¡¡¡¡ ¡¡ 0 j ... j STA 1S 0 j 3 77 77 75 = 2 66 66 64 ? j c2 ¢¢¢ ck ¡¡ + ¡¡ ¡¡¡¡ ¡¡ 0 j ... j T 1 0 j 3 77 77 75: 128 CHAPTER 4. THE EIGENVALUE PROBLEMS Since T1 is upper triangular so is RTBR: The theorem now follows by induction. 42. Au1= (2¡n)u1 while A(e1¡ei) = Ae1¡Aei= A1¡Ai= [2;0; : : : ;0;¡2;0; : : : ;0]T = 2(e1 ¡ei ); for 2? i ? n: 43. Since A is symmetric we have fl(uTv) = uT(flv) = uTAv= uTATv= (Au)Tv= (?u)Tv= ?uTv: Since fl 6= ?;uTv= 0: 4.8 Applications 1. x1 = ? 4 2 ? ; x2 = ? 2 4 ? ; x3 = ? 4 2 ? ; x4 = ? 2 4 ? : 2. x1 = x2 = x3 = x4 = ? 12 12 ? : 3. x1 = ? 80 112 ? ; x2 = ? 68 124 ? ; x3 = ? 65 127 ? ; x4 = ? 64:25 127:75 ? : 4. x1 = ? 5 ¡1 ? ; x2 = ? 11 ¡7 ? ; x3 = ? 29 ¡25 ? ; x4 = ? 83 ¡79 ? : 5. x1 = ? 7 1 ? ; x2 = ? 11 8 ? ; x3 = ? 43 19 ? ; x4 = ? 119 62 ? : 6. x1 = ? 6 8 ? ; x2 = ? 26 48 ? ; x3 = ? 126 248 ? ; x4 = ? 626 1248 ? : 7. A has eigenvalues ?1 = 1 and ?2 = ¡1 with corresponding eigenvectors u1 = [1;1]T and u2 = [¡1;1]T; respectively. x0 = 3u1 +u2 so xk = 3(1)ku1 +(¡1)ku2 = [3+(¡1)k+1;3+ (¡1)k]T: In particular x4 = [2;4]T = x10 : The sequence fxk g has no limit but kxk k =p20 for all k. 8. A has eigenvalues ?1 = 1 and ?2 = 0 with corresponding eigenvectors u1 = [1;1]T and u2 = [¡1;1]T; respectively. x0 = 12u1 ¡4u2 so xk = 12u1 : In particular x4 = x10 = [12;12]T: The limit of the sequence fxkg is x? = [12;12]T: 9. A has eigenvalues ?1 = 1 and ?2 = 0:25 with corresponding eigenvectors u1 = [1;2]T and u2 = [¡1;1]T; respectively. x0 = 64u1 ¡64u2 so xk = 64(1)ku1 ¡64(0:25)ku2 = [64+64(0:25)k;128¡64(0:25)k]T: In particular x4 = [64:25;127:75]T and x10 = [64:00006;127:99994]T: The sequence fxkg converges to x?= [64;128]T: 4.8. APPLICATIONS 129 10. A has eigenvalues ?1 = 3 and ?2 = 1 with corresponding eigenvectors u1 = [¡1;1]T and u2 = [1;1]T; respectively. x0 = ¡u1 +2u2 ; so xk = ¡3ku1 +2u2 = [2 + 3k;2¡3k]T: In particular x4 = [83;¡79]T and x10 = [59051;¡59047]T: The sequence fxkg has no limit and limk!1 kxkk=1: 11. A has eigenvalues ?1 = 3 and ?2 = ¡1 with corresponding eigenvectors u1 = [2;1]T and u2 = [¡2;1]T; respectively. x0 = (3=4)u1 +(5=4)u2 so xk = (3=4)(3)ku1 +(5=4)(¡1)ku2 = (1=4)[2(3)k+1 ¡10(¡1)k;3k+1 +5(¡1)k]T: In particular x4 = [119;62]T and x10 = [88571;44288]T: The sequence fxk g has no limit and limk!1 kxkk =1: 12. A has eigenvalues ?1 = 5 and ?2 = 1 with corresponding eigenvectors u1 = [1;2]T and u2 = [1;¡2]T; respectively. x0 = u1 +u2 so xk = 5ku1 +u2 = [5k + 1;2(5)k ¡2]T: In particular, x4= [626;1248]T and x10= [390626;781248]T: The sequence fxkg has no limit and limk!1 kxkk=1: 13. A has eigenvalues ?1 = 2; ?2 = 1; ?3 = ¡1 with corresponding eigenvectors u1 = [1;¡1;2]T;u2 = [3;¡1;7]T; and u3 = [1;2;2]T; re- spectively. x0 = 2u1 +2u2¡5u3 ; so xk = 2k+1u1 +2u2 +5(¡1)k+1u3; thus : xk = 2 4 2k+1 +6+5(¡1)k+1 ¡2k+1 ¡2+10(¡1)k+1 2k+2 +14+10(¡1)k+1 3 5: In particular x4 = [33;¡44;68]T and x10 = [2049;¡2060;4100]T: The sequence fxkg has no limit and limk!1 kxkk=1: 14. A has eigenvalues ?1 = 4; ?2 = 1; ?3 = ¡1 with corresponding eigenvectors u1 = [14;11;43]T;u2 = [1;1;2]T; and u3 = [1;¡1;2]T; respectively. x0 =¡0:2u1 +3:5u2 +0:3u3 so xk = ¡0:2(4)ku1 +3:5u2 +0:3(¡1)ku3 ; that is xk = 2 4 (¡2:8)4k +3:5+0:3(¡1)k (¡2:2)4k +3:5¡0:3(¡1)k (¡8:6)4k +7+0:6(¡1)k 3 5: In particular x4 = [¡713;¡560;¡2195:2]T and x10 = 130 CHAPTER 4. THE EIGENVALUE PROBLEMS [¡2936009;¡2306864;¡9017746]T: The sequence fxk g has no limit and limk!1 kxk k =1: 15. If x(t) = ? u(t) v(t) ? then x0(t) = Ax(t) for A = ? 5 ¡6 3 ¡4 ? . Eigenvalues and corresponding eigenvectors for A are ?1 = 2, u1 = ? 2 1 ? and ?2 = ¡1, u2 = ? 1 1 ? . Setting x(t) = ae2tu1 + be¡tu2 and x(0) = ? 4 1 ? yields a = 3 and b = ¡2. Thus, x(t) = 3e2tu1 ¡2e¡tu2. Equivalently, u(t) = 6e2t ¡2e¡t and v(t) = 3e2t ¡2e¡t. 16. If x(t) = ? u(t) v(t) ? then x0(t) = Ax(t) for A = ? 1 2 2 1 ? . Eigenvalues and corresponding eigenvectors for A are ?1 = 3, u1 = ? 1 1 ? and ?2 = ¡1, u2 = ? ¡1 1 ? . Setting x(t) = ae3tu1 + be¡tu2 and x(0) = ? 1 5 ? yields a = 3 and b = 2. Thus, x(t) = 3e3tu1 +2e¡tu2. Equivalently, u(t) = 3e3t ¡2e¡t and v(t) = 3e3t +2e¡t. 17. The matrix A = 2 4 1 1 1 0 3 3 ¡2 1 1 3 5 has eigenvalues ?1 = 0, ?2 = 2, ?3 = 3. Corresponding eigenvectors are u1 = [0;¡1;1]T, u2 = [¡2;¡3;1]T, u3 = [1;2;0]T, respectively. The solution is x(t) = 2u1¡e2tu2+e3tu3. Equivalently, u(t) = 2e2t+e3t, v(t) =¡2+3e2t+2e3t, and w(t) = 2¡e2t. 18. The matrix A = 2 4 ¡2 2 ¡3 2 1 ¡6 ¡1 ¡2 0 3 5 has eigenvalues ?1 = 5 and ?2 = ¡3, where ?2 has algebraicmultiplicity2. Aneigenvectorfor ?1 is u1 = [¡1;¡2;1]T. Further, u2 = [¡2;1;0]T and u3 = [3;0;1]T are linearly independent eigenvectors for ?2. The solution is x(t) = e5tu1 + e¡3tu2 +2e¡3tu3. Equivalently, u(t) = ¡e5t +4e¡3t, v(t) = ¡2e5t + e¡3t, w(t) = e5t +2e¡3t. 19. (a) The eigenvectors corresponding to ? = 1 have the form x= [a;0]; a 6= 0: In particular ? = 1 has algebraic multiplicity 2 and geometric multiplicity 1. Thus A is defective. (b) For k = 0 we have x0 = [1;1]T = [2(0)+1;1]T: Suppose xm = [2m +1;1]T for some integer m ?0. Then xm+1 = Axm = [2m +3;1]T = [2(m +1)+1;1]T: It follows that xk = [2k +1;1]T for each k ?0: 20. fi =¡0:5: For each k ?0;xk = x0 so limk!1xk = [1;1]T: 4.8. APPLICATIONS 131 21. For fi =¡0:18 A has eigenvalues ?1 = 1 and ?2 =¡0:18 with corresponding eigenvectors u1 = [3;10]T and u2 = [10;¡6]T; respectively. Moreover x0 = (16=118)u1 +(7=118)u2 : Therefore xk = (16=118)u1 +(7=118)(¡0:18)ku2 : It follows that limx!1xk = (16=118)u1 : 22. Note that u1= Au0= Av0= v1: Suppose we have shown that um= vm for some m ?1: Then um+1 = Aum = Avm = vm+1 : It follows by induction that uk = vk for all k. 23. Bw is the vector [c1; c2; : : : ; cn]T where ci = bi1 + bi2 +¢¢¢+ bin = 1: Therefore Bw = w and ? = 1 is an eigenvalue for B with corresponding eigenvector w: 24. Suppose u= [u1; u2; : : : ; un]T and choose i so that juij= max1?i?nfju1 j; : : : ;jun jg: Set fi = 1=ui and v = fiu : Then v = [v1; v2; : : : ; vn]T where vi = 1 and jvjj? 1 for 1 ? j ? n: Note also that Bv= v : Equating the ith component yields bi1v1+¢¢¢+binvn = vi = 1: But bi1+¢¢¢+bin = 1 so it follows that v1 =¢¢¢= vn = 1: Thus v= w= fiu; so u= fi¡1w: 25. Suppose Bu = ?u;u 6= . As in Exercise 24, deflne v = [v1; : : : ; vn]T such that Bv = ?v ; vi = 1 for some i; and jvjj? 1 for 1 ? j ? n: The ith component of ?v is ? whereas the ith component of Bv is bi1v1 +¢¢¢+ binvn: Therefore j?j=jbi1v1 +¢¢¢+ binvnj? bi1 jv1j+¢¢¢+ bin jvnj? bi1 +¢¢¢+ bin = 1: 26. The matrix AT is a stochastic matrix. Moreover A and AT have the same eigenvalues so we may apply Exercises 23 and 25. 27. If au+bv= then = A = A(au+bv) = aAu+bAv= a?u+ b(?v+u) = ?(au+bv)+ bu= bu : Since u6= it follows that b = 0: Thus au= and a = 0: This proves that fu;vg is a linearly independent set. 28. The formula holds for k = 1 by Exercise 27. Suppose Amv= ?mv+m?m¡1u for some m ? 1: Then Am+1v = A(Amv ) = A(?mv +m?m¡1u ) = ?mAv +m?m¡1Au = ?m(?v +u)+ m?m¡1?u= ?m+1v+(m +1)?mu. By induction the given formula holds for every integer k ?1: 29. (a) The vector u= [1;0]T is an eigenvector for ? = 1 and v= [0;1=2]T satisfles the equation (A¡I)v= u: (b) x0 = u+2v: (c) Akx0 = Ak(u+2v) = Aku+2Akv= u+2(v+ku) = (2k +1)u+2v: (d) It follows immediately from (c) that Akx0 = [2k +1;1]T: 132 CHAPTER 4. THE EIGENVALUE PROBLEMS 4.9 Supplementary Exercises 1. Det(A) = x2 ¡9 so A is singular when x =§3. 2. x ?¡1=4. 3. [1;1]T is an eigenvector for the eigenvalue ? = 2. 4. (a) det(A¡1B2) = (detB)2=detA = 81=2. (b) det(3A) = 33 det(A) = 54. (c) det(AB2A¡1) = (detB)2 = 81. 5. x 6= 0. 6. A2 = ¡3A + I so A2u = ¡3Au+u = [¡5;0]T. A3 = ¡3A2 + A so A3u = ¡3A2u+ Au = [17;1]T. 7. Suppose Ax = . Then A2x =¡3Ax+x (cf. Exercise 6), so = x. 8. I = A2 +3A = A(A +3I) so A¡1 = A +3I. Thus A¡1u = Au+3u = [5;10]T. 9. A2 =¡3A + I; A3 = 10A¡3I; A4 =¡33A +10I; A5 = 109A¡33I. 10. ?1 =¡1, ?2 =¡2. 11. ?1 = 2, ?2 = 3. 12. x = 8, ? =¡6. 13. x = 0, y =¡1. 14. x = 2, y = 3. 4.10 Conceptual Exercises 1. False. A = ? ¡1 0 0 1 ? . 2. True. If Ax = ?x then A¡1x = (1=?)x. 3. True. Det(A4) = (detA)4. 4. False. A = ? 1 1 0 1 ? . 5. True. kAxk2 = (Ax)T(Ax) = xTATAx = xTx =kxk2. 4.10. CONCEPTUAL EXERCISES 133 6. True. Det(S¡1AS ¡tI) = det[S¡1(A¡tI)S] = det(A¡tI). 7. False. A = ? 1 0 0 2 ? and B = ? 2 0 0 1 ? . 8. False. A = ? 1 0 0 0 ? . 9. (a) Qx (b) QTu. 10. If Ax = ?x then A3x = ?3x. But A3 =O. 11. If P¡1AP = I then A = PP¡1 = I. 12. A¡1(AB)A = BA. 13. Suppose S¡1AS = B. Then S¡1A2S = (S¡1AS)(S¡1AS) = B2. Similarly, S¡1A3S = (S¡1A2S)(S¡1AS) = B2B = B3 and S¡1A4S = (S¡1A3S)(S¡1AS) = B3B = B4. 14. (a) Yes. AT = (I ¡2uuT)T = IT ¡2uuT = I ¡2uuT = A. (b) Yes. AAT = A2 = (I ¡2uuT)2 = I ¡4uuT +4u(uTu)uT = I. (c) Au = (I ¡2uuT)u = u¡2u(uTu) =¡u. (d) Aw = (I ¡2uuT)w = w¡2u(uTw) = w. (e) ? = 1 has geometric multiplicity n¡1 and ? =¡1 has geometric multiplicity 1. Chapter 5 Vector Spaces and Linear Transformations 5.1 Introduction (No exercises) 5.2 Vector Spaces 1. u¡2v= ? 0 ¡7 5 ¡11 ¡3 ¡12 ? ;u¡(2v¡3w) = ? 12 ¡22 38 ¡50 ¡6 ¡15 ? ; ¡2u¡v+3w= ? 7 ¡21 28 ¡42 ¡7 ¡14 ? : 2. u¡2v=¡x2 ¡4x; u¡(2v¡3w) =¡x2 +2x +3; ¡2u¡v+3w=¡3x2 +4x +8: 3. u¡2v= ex¡2sinx; u¡(2v¡3w) = ex¡2sinx+3px2 +1; ¡2u¡v+3w=¡2ex¡sinx+ 3px2 +1; 4. For u;v; and w in Exercise 2 we may take c1 = c3; c2 =¡c3; c3 arbitrary. For example, c1 = 1; c2 = ¡1; c3 = 1 is one choice. For u;v; and w in Exercise 1, c1u+c2v+c3w= if and only if c1 = c2 = c3 = 0: 5. Note that c1u+c2v+c3w= (c1 +c2)x2 +(2c2 +2c3)x+(¡2c1 ¡c2 +c3): Thus c1u+c2v +c3w= x2 +6x +1 if and only if c1 + c2 = 1 2c2 + 2c3 = 0 ¡2c1 ¡ c2 + c3 = 0: Solving yields c1 =¡2+ c3; c2 = 3¡c3; c3 arbitrary. One choice is 136 CHAPTER 5. VECTOR SPACES AND LINEAR TRANSFORMATIONS c1 =¡2; c2 = 3; c3 = 0 and a direct calculation shows that ¡2u+ 3v= x2 +6x +1: Similarly c1u+c2v+c3w= x2 if and only if c1 + c2 = 1 2c2 + 2c3 = 6 ¡2c1 ¡ c2 + c3 = 1: The system is easily seen to be inconsistent. 6. S is a vector space. 7. S is not a vector space. None of properties (c1), (c2), (a3), and (a4) of Deflnition 1 is satisfled. For example v= [1;0;0;0]T and w= [0;0;0;1]T are in S but u+w is not in S. 8. P is a vector space. 9. P is not a vector space. Properties (c1), (c2), and (a3) of Deflnition 1 fail to hold in P: For example p(x) = 1+2x2 and q(x) = x¡2x2 are in P but p(x)+ q(x) is not in P: 10. P is a vector space. 11. P is not a vector space (cf. Exercise 9). 12. S is a vector space. 13. S is a vector space. 14. S is not a vector space. For example the (3 x 4) zero matrix is not in S: 15. S is not a vector space. Properties (c1), (c2), and (a3) of Deflnition 1 fail to hold in S. 16. S is not a vector space. For example if A = [aij] is a nonzero matrix in S then p2A is not in S: Thus property (c2) of Deflnition 1 fails to hold. Note that S satisfles the remaining properties of Deflni- tion 1. 17. Let A = ? 1 0 0 1 ? and let B = ? 0 1 1 0 ? : Then A and B are in Q but A+B is not in Q: Also 0A; the (2 x 2) zero matrix, is not in Q: 18. Q is not a vector space. For example, if A = ? 1 0 0 0 ? and B = ? 0 0 0 1 ? then A and B are in Q but A + B is nonsingular. 5.2. VECTOR SPACES 137 19. The set of all (2 x 2) matrices is a vector space so axioms (a1), (a2), (m1), (m2), (m3), and (m4) are satisfled by any subset. Now let A and B be (2 x 2) symmetric matrices; that is A = AT and B = BT: Therefore (A + B)T = AT + BT = A + B; so A + B is symmetric. This verifles that property (c1) holds. For any scalar c; (cA)T = cAT = cA; so cA is symmetric and (c2) is satisfled. Clearly the (2 x 2) zero matrix is symmetric so (a3) holds. Finally if A is symmetric then so is ¡A; so (a4) holds. Therefore Q is a vector space. 20. Suppose u, v, and w are vectors in a vector space V such that u+ v=u+w. By property (a4) of Deflnition 1, V contains a vector -u such that u+(-u)= . By property (a1),(-u)+u= . Applying properties (a1), (a3), and (a2) yields v=v+ = +v= [(-u)+u]+ v= (-u)+(u+ v)= (-bfu)+(u+w)= [(-u)+u]+w= +w= w+ =w. Similarly, v+u=w+u implies that v=w. 21. Let u and w be inverses for v : Thus u + v = v +u = and w + v = v +w = . Therefore u= u+ = u+(v+w) = (u+ v)+w= +w= w: 22. Note that 0v+ = 0v= (0+0)v= 0v+0v: By the cancellation laws, = 0v: 23. If a = 0 then we are done, so suppose that a 6= 0: Then v= 1v= (a¡1a)v= a¡1(av) = a¡1 = . 24. We show as illustrations that properties (a2) and (m1) hold. Thus +( + ) = + = = + = ( + )+ , so (a2) holds. If a and b are scalars then a(b ) = a = = (ab) so (m1) is satisfled. 25. F is a vector space. Since F is a subset of C[¡1;1] and C[¡1;1] is a vector space, properties (a1), (a2), (m1), (m2), (m3), and (m4) hold in F: Now let g(x); h(x) be in F; that is g(x) and h(x) are continuous, g(¡1) = g(1) and h(¡1) = h(1): It follows that (g + h)(x) = g(x)+ h(x) is continuous and (g + h)(¡1) = g(¡1)+ h(¡1) = g(1)+ h(1) = (g + h)(1): Therefore (g + h)(x) is in F and property (c1) holds. If a is a scalar then (ag)(x) = ag(x) is continuous and (ag)(¡1) = ag(¡1) = ag(1) = (ag)(1); so (ag)(x) is in F: This verifles that (c2) holds. The zero vector in C[¡1;1] is the function deflned by (x) = 0 for all x; ¡1 ? x ? 1: In particular (¡1) = 0 = (1) so (x) is in F: Thus (x) is also the zero vector for F and (a3) is satisfled. Property (a4) is an immediate consequence of (c2) since ¡g(x) = (¡1)g(x) for g(x) in C[¡1;1]: Since F satisfles the properties of Deflnition 1, F is a vector space. 26. F is a vector space. 27. F is not a vector space. For example set f(x) = 2x ¡1 and g(x) = 2x2 ¡1: Then f(x) and g(x) are in F whereas f(x)+ g(x) = 2x2 +2x¡2 is not. 138 CHAPTER 5. VECTOR SPACES AND LINEAR TRANSFORMATIONS 28. F is a vector space. 29. F is a vector space. As in Exercise 25, it su?ces to check that properties (c1), (c2), (a3), and (a4) of Deflnition 1 are satisfled. To check (c1) for example, let f(x) and g(x) be in F . Then R1¡1[f(x)+ g(x)]dx = R1¡1 f(x)dx +R1¡1 g(x)dx = 0+0 = 0; so f(x)+ g(x) is in F . 30. If f(x) and g(x) are in C2[a; b] then (f +g)(x) = f(x)+g(x) is continuous on [a; b];(f + g)0(x) = f0(x)+g0(x) is continuous on [a; b]; and (f +g)00(x) = f00(x)+g00(x) is continuous on [a; b]: Similarly if c is a scalar then the functions (cf)(x) = cf(x);(cf)0(x) = cf0(x); and (cf)00(x) = cf00(x) are all continuous on [a; b]: It follows that C2[a; b] is a vector space. 31. (a) F is a vector space. Since F is a subset of C2[¡1;1] and C2[¡1;1] is a vector space by Exercise 30, properties (a1), (a2), (m1), (m2), m(3), and (m4)holdin F?. Nowlet g(x); h(x) bein F . Thus g00(x)+g(x) = 0 and h00(x)+h(x) = 0: It follows that (g+h)00(x)+(g+h)(x) = [g00(x)+g(x)]+[h00(x)+h(x)] = 0+0 = 0 for ¡1 ? x ? 1. Therefore (g + h)(x) is in F and property (c1) is satisfled. If a is any scalar then (ag)00(x)+(ag)(x) = a[g00(x)+ g(x)] = a0 = 0 for ¡1? x ?1: Thus (ag)(x) is in F and property (c2) is satisfled. The zero vector in C2[¡1;1] is the function deflned by (x) = 0;¡1 ? x ? 1: In particular 00(x)+ (x) = 0+0 = 0 for ¡1 ? x ? 1 so (x) is in F: Therefore (x) is the zero vector in F and (a3) is satisfled. Property (a4) follows from (c2) since ¡f(x) = (¡1)f(x) for every f(x) in C2[¡1;1]: (b) F is not a vector space. For example suppose that g(x) and h(x) are in F ; that is, assume that g00(x)+g(x) = x2 and h00(x)+h(x) = x2: Then (g+h)00(x)+(g+h)(x) = [g00(x)+ g(x)]+[h00(x)+ h(x)] = 2x2: Therefore (g + h)(x) is not in F . 32. Let p(x) and q(x) be in P: Then we may write p(x) = a0 + a1x + ¢¢¢ + anxn and q(x) = b0 + b1x +¢¢¢+ bnxn where for 0 ? i ? n; ai and bi are real numbers. (By using zero coe?cients as necessary we may list the same terms for both p(x) and q(x):) Thus p(x) + q(x) = (a0 + b0) + (a1 + b1)x + ¢¢¢ + (an + bn)xn and for any scalar c; cp(x) = ca0 + ca1x +¢¢¢+ canxn: It is now straightforward to verify that P is a vector space. 33. The proof that F(R) is a vector space requires checking all ten properties of Deflnition 1. We illustrate by verifying properties (a2), (a3) and (m1). If f(x); g(x); h(x) are in F(R) then [f +(g + h)](x) = f(x)+[g + h](x) = f(x)+[g(x)+ h(x)] = [f(x)+ g(x)]+ h(x) = [f + g](x) + h(x) = [(f + g)) + h](x): Thus property (a2) holds. Deflne : R ! R by (x) = 0 for all x in R. If f(x) is inF(R) then (f+ )(x) = f(x)+ (x) = f(x); f+ = f so is the zero of F(R): Therefore property (a3) is satisfled. To check (m1) let a and b be scalars and suppose f(x) is in F(R): Then [a(bf)](x) = a[(bf)(x)] = a[b(f(x))] = ab(f(x)) = [(ab)f](x): 5.3. SUBSPACES 139 34. (a) We will show that the given operations satisfy (a3) and (a4) of Deflnition 1. Note that z= [¡1;1]T is in V and u+z= u for every element u in V: Thus z is the zero of V: If u= [u1; u2]T then w= [¡u1 ¡2;¡u2 +2] is in V and u+w= z: Thus w is an additive inverse for u: (b) Note that 2(e1 +e2 ) = [4;0]T whereas 2e1 +2e2 = [3;1]T: Simil- arly, e1 +e1 = [3;¡1]T whereas (1+1)e1 = 2e1 = [2;0]T: 35. To check (m2) as an illustration, note that a(u+ v) = = + = au+av for any u;v in V and scalar a: If u6= then 1u= and (m4) fails. 36. The zero of V is the vector z= [0;1]T: If u= [u1; u2] is in V then w= [¡u1;1=u2] is in V and u+w= z: Thus w is the inverse of u: To show that (m2) holds let u= [u1; u2]T and v= [v1; v2]T be in V and let a be a scalar. Then a(u+v) = a ? u 1 + v1 u2v2 ? = ? a(u 1 + v1) (u2v2)a ? = ? au 1 + av1 ua2va2 ? = ? au 1 ua2 ? + ? av 1 va1 ? = au+ av: 5.3 Subspaces 1. W is not a subspace of V: None of the properties (s1), (s2), and (s3) of Theorem 2 is satisfled. For example, if A = ? 1 0 0 0 0 0 ? and B = ? 0 0 1 0 0 0 ? then A and B are in W but A + B is not in W: 2. W is a subspace of V: 3. W is a subspace of V: Clearly the (2 x 3) zero matrix is in W: Let A = [aij] and B = [bij] be in W: Therefore a11¡a12 = 0; a12 +a13 = 0; a23 = 0; b11¡b12 = 0; b12 +b13 = 0; b23 = 0: Now A + B is the (2 x 3) matrix A + B = [cij]; where cij = aij + bij: Thus c11 ¡ c12 = (a11 +b11)¡(a12 +b12) = (a11 ¡a12)+(b11 ¡b12) = 0:Similarly, c12 +c13 = 0 and c23 = 0: This shows that A + B is in W: If k is a scalar then kA is the (2 x 3) matrix kA = [dij] where dij = kaij: Consequently d11 ¡d12 = ka11 ¡ka12 = k(a11 ¡a12) = k0 = 0: Likewise d12 + d13 = 0 and d23 = 0: Therefore kA is in W: It follows from Theorem 2 that W is a subspace of V: 4. W is not a subspace of V: For Example if A = ? 1 0 0 0 0 0 ? and B = ? 0 1 1 0 0 0 ? then A and B are in W but A + B is not in W: Note that properties (s1) and (s3) of Theorem 2 are satisfled. 140 CHAPTER 5. VECTOR SPACES AND LINEAR TRANSFORMATIONS 5. W is a subspace of P2: If (x) = 0 + 0x + 0x2 is the zero polynomial then clearly (0)+ (2) = 0; so (x) is in W: Suppose g(x) and h(x) are in W?; that is g(0)+g(2) = 0 and h(0) + h(2) = 0: Then (g + h)(0) = +(g + h)(2) = [g(0) + h(0)] + [g(2) + h(2)] = [g(0)+g(2)]+[h(0)+h(2)] = 0+0 = 0 and it follows that g(x)+h(x) is in W: If c is a scalar then (cg)(0)+(cg)(2) = c[g(0)+g(2)] = 0 and hence (cg)(x) is in W: By Theorem 2, W is a subspace of P2: 6. W is a subspace of P2: 7. W is not a subspace of P2 . For example if p(x) = x2 + x¡2 and q(x) = x2 ¡9 then p(x) and q(x) are in W but p(x)+q(x) = 2x2 +x¡11 is not in W: Note that properties (s1) and s(3) of Theorem 2 are satisfled. 8. W is a subspace of P2 . 9. F is a subspace of C[¡1;1]: First recall that the zero of C[¡1;1] is the function deflned by (x) = 0 for ¡1 ? x ? 1: Since (¡1) = 0 = ¡ (1); (x) is in F: Now assume that g(x) and h(x) are in F: Thus g(¡1) =¡g(1) and h(¡1) =¡h(1): It follows that (g + h)(¡1) = g(¡1)+ h(¡1) =¡g(1)¡h(1) =¡(g + h)(1): Therefore (g + h)(x) is in F: If c is a scalar then (cg)(¡1) = c(g(¡1)) = c(¡g(1)) = ¡(cg)(1) so (cg)(x) is in F: by Theorem 2, F is a subspace of C[¡1;1]: 10. F is not a subspace of C[¡1;1]: If f(x) is a nonzero function in F and c < 0 then cf(x) is not in F . Note that properties (s1) and (s2) of Theorem 2 are satisfled. 11. F is not a subspace of C[¡1;1]: None of the properties (s1), s(2), (s3) of Theorem 2 is satisfled. For example if g(x) and h(x) are in F then (g + h)(¡1) = g(¡1) + h(¡1) = ¡2+(¡2) =¡4 so (g + h)(x) is not in F . 12. F is a subspace of C[¡1;1]: 13. F is a subspace of C2[¡1;1]: If (x) is the zero function then 00(x) = (x) = 0 for ¡1 ? x ? 1: In particular 00(0) = 0 so (x) is in F: Let g(x) and h(x) be in F; that is g00(0) = 0 = h00(0): Therefore (g + h)00(0) = g 00(0)+ h00(0) = 0; so (g + h)(x) is in F . If c is a scalar then (cg)00(0) = cg00(0) = 0 and (cg)(x) is in F . By Theorem 2, F is a subspace of C2[¡1;1]: 14. F is a subspace of C2[¡1;¡]: 15. F is not a subspace of C2[¡1;1]: None of the properties (s1), (s2), s(3) of Theorem 2 is satisfled. For example suppose g(x) and h(x) are in F . Then g00(x)+ g(x) = sinx and h00(x)+h(x) = sinx for¡1? x ?1: But (g+h)00(x)+(g+h)(x) = 2sinx for¡1? x ?1: Therefore (g + h)(x) is not in F . 5.3. SUBSPACES 141 16. F is a subspace of C2[¡1;1]: 17. Note that c1p1(x)+c2p2(x)+c3p3(x) = (c1 +2c2 +3c3)+(2c1 +5c2 +8c3)x+(c1 ¡2c3)x2: Therefore c1p1(x) + c2p2(x) + c3p3(x) = ¡1 ¡ 3x + 3x2 requires that c1 + 2c2 + 3c3 = ¡1;2c1 +5c2 +8c3 = ¡3; and c1 ¡2c3 = 3: Solving we obtain c1 = ¡1; c2 = 3; c3 = ¡2 and it is easily verifled that p(x) =¡p1(x)+3p2(x)¡2p3(x): 18. p(x) = P4i=1 cipi(x) if and only if c1 =¡1¡2c3; c2 = 2+3c3; c3 is arbitrary, and c4 =¡3: For example p(x) =¡p1(x)+2p2(x)¡3p4(x): 19. From the matrix equation A = P4i=1 ciBi we obtain the system of equations c1 + c2 ¡ c3 + c4 = ¡2 c2 ¡ 3c3 + 2c4 = ¡4 2c1 + 4c3 ¡ c4 = 1 c1 + 2c2 ¡ 4c3 + c4 = 0 : The solution is c1 = ¡1¡2c3; c2 = 2+3c3; c3 arbitrary, and c4 = ¡3: Taking c3 = 0 we see that A =¡B1 +2B2 ¡3B4: 20. ex = sinhx +coshx: 21. cos2x = (¡1)sin2 x +(1)cos2 x: 22. ? ? 1 1 0 ¡1 ? ; ? 0 0 1 0 ? : 23. Let p(x) = a0+a1x+a2x2+a3x3 bein W: Theconstraints p(1) = p(¡1) and p(2) = p(¡2) imply that a0+a1+a2+a3 = a0¡a1+a2¡a3 and a0+2a1+4a2+8a3 = a0¡2a1+4a2¡8a3: This forces a1 = a3 = 0 while a0 and a2 are arbitrary. Thus f1; x2g is a spanning set for W: 24. f¡3+2x + x2;2¡3x + x3g: 25. For Exercise 2, a matrix A = [aij] is in W if and only if A has the form A =? a12 ¡2a13 a12 a13 a21 a22 a23 ? ; where a12; a13; a21; a22; a23 are arbitrary. Therefore W = Sp ? ? 1 1 0 0 0 0 ? ; ? ¡2 0 1 0 0 0 ? ; ? 0 0 0 1 0 0 ? ; ? 0 0 0 0 1 0 ? ; ? 0 0 0 0 0 1 ? : 142 CHAPTER 5. VECTOR SPACES AND LINEAR TRANSFORMATIONS For Exercise 3, A = [aij] is in W if and only if A =? a11 a11 ¡a11 a21 a22 0 ? ; where a11;a21; and a22 are arbitrary. Therefore W = Sp ? ? 1 1 ¡1 0 0 0 ? ; ? 0 0 0 1 0 0 ? ; ? 0 0 0 0 1 0 ? : For Exercise 5 let p(x) = a0 + a1x + a2x2: The condition p(0)+ p(2) = 0 implies that 2a0 +2a1 +4a2 = 0: Therefore p(x) = (¡a1 ¡2a2)+ a1x + a2x2 where a1 and a2 are arbitrary. It follows that W = Spf¡1+x;¡2+x2g: For Exercise 6, W = Spf 1;¡4x + x2g: For Exercise 8, W = Spf x;1¡x2g: 26. That W is a subspace follows from Theorem 2 and from the properties of the transpose given in Theorem 10 of Section 1.6. W = Sp 8 < : 2 4 1 0 0 0 0 0 0 0 0 3 5; 2 4 0 0 0 0 1 0 0 0 0 3 5; 2 4 0 0 0 0 0 0 0 0 1 3 5 9 = ;; 2 4 0 1 0 1 0 0 0 0 0 3 5; 2 4 0 0 1 0 0 0 1 0 0 3 5; 2 4 0 0 0 0 0 1 0 1 0 3 5 9 = ;: 27. It is straightforward to show that tr(A + B) = tr(A)+ tr(B) and tr(cA) = ctr(A): It then follows easily from Theorem 2 that W is a subspace of V: If A is in W then A has the form A =2 4 ¡a22 ¡a33 a12 a13 a21 a22 a23 a31 a32 a33 3 5: It follows that W = SpfB1;B2;E12;E13;E21;E23;E31;E32g where B1 = 2 4 ¡1 0 0 0 1 0 0 0 0 3 5 and B2 = 2 4 ¡1 0 0 0 0 0 0 0 1 3 5: 28. BT = [(1=2)(A + AT)]T = (1=2)(AT + ATT) = (1=2)(A + AT) = B; so B is symmetric. Similarly CT = [(1=2)(A¡AT)]T = (1=2)(AT ¡A) =¡C and C is skew-symmetric. 29. For any (n x n) matrix A; A = B +C where B and C are the matrices given in Exercise 28. 5.3. SUBSPACES 143 30. The vector space of all (3 x 3) matrices is spanned by the set fE11; E22; E23; A1; A2; A3; B1; B2; B3g where A1 = 2 4 0 1 0 1 0 0 0 0 0 3 5; A2 = 2 4 0 0 1 0 0 0 1 0 0 3 5; A3 = 2 4 0 0 0 0 0 1 0 1 0 3 5; B1 = 2 4 0 1 0 ¡1 0 0 0 0 0 3 5; B2 = 2 4 0 0 1 0 0 1 ¡1 0 0 3 5; B3 = 2 4 0 0 0 0 0 1 0 ¡1 0 3 5: If A = [aij] is a (3 x 3) matrix then A = a11E11 +a22E22 +a33E33 +(1=2)(a12 +a21)A1 +(1=2)(a13 +a31)A2 +(1=2)(a23 +a32)A3 + (1=2)(a12 ¡a21)B1 +(1=2)(a13 ¡a31)B2 +(1=2)(a23 ¡a32)B3: 31. (a) A is in W if and only if A has the form A = 2 4 0 a12 a13 0 0 a23 0 0 0 3 5; where a12; a13; a23 are arbitrary. Thus W = SpfE12;E13;E23g: (b) A is in W if and only if A has form A =2 4 ¡a22 ¡a33 ¡a23 a13 0 a22 a23 0 0 a33 3 5: Therefore W = Sp 8 < : 2 4 ¡1 0 0 0 1 0 0 0 0 3 5; 2 4 ¡1 0 0 0 0 0 0 0 1 3 5; 2 4 0 ¡1 0 0 0 1 0 0 0 3 5; 2 4 0 0 1 0 0 0 0 0 0 3 5 9 = ;: (c) A is in W if and only if A has the form A = 2 4 a11 a11 a13 0 a22 a13 0 0 a22 3 5:: Therefore W = Sp 8 < : 2 4 1 1 0 0 0 0 0 0 0 3 5; 2 4 0 0 1 0 0 1 0 0 0 3 5; 2 4 0 0 0 0 1 0 0 0 1 3 5 9 = ;: (d) A is in W if and only if A has the form A =2 4 a11 ¡a23 a13 0 a11 a23 0 0 a11 3 5: Therefore W = Sp 8 < : 2 4 1 0 0 0 1 0 0 0 1 3 5; 144 CHAPTER 5. VECTOR SPACES AND LINEAR TRANSFORMATIONS 2 4 0 ¡1 0 0 0 1 0 0 0 3 5; 2 4 0 0 1 0 0 0 0 0 0 3 5 9 = ;: 32. p(x) = (a0 ¡a1 +a2)(1)+(a1 ¡2a2)(x+1)+a2(x+1)2: In particular q(x) = 4(1)¡5(x+ 1)+2(x +1)2 and r(x) = 6(1)¡5(x +1)+(x +1)2: 33. The equation A = P4i=1 xiBi implies that x1 + 2x2 ¡ x3 + x4 = a x2 + 3x3 + x4 = b x1 + x2 ¡ 3x3 ¡ 2x4 = c ¡2x1 ¡ 2x2 + 6x3 + 5x4 = d : Solvingweobtain x1 =¡6a+5b+37c+15d; x2 = 3a¡2b¡17c¡7d; x3 =¡a+b+5c+2d; x4 = 2c + d: Therefore C =¡12B1 +6B2 ¡B3 ¡B4 and D = 8B1 ¡3B2 + B3 + B4: 5.4 Linear Independence, Bases, and Coordinates 1. If A is in W then A = ? ¡b¡c¡d b c d ? = b ? ¡1 1 0 0 ? + c ? ¡1 0 1 0 ? + d ? ¡1 0 0 1 ? : The set ? ? ¡1 1 0 0 ? ; ? ¡1 0 1 0 ? ; ? ¡1 0 0 1 ? is a basis for W. 2. The set ? ? ¡1 2 ¡3 1 ? is a basis for W. 3. The set fE12; E21; E22g is a basis for W: 4. The set ? ? 1 1 0 2 ? ; ? 0 ¡1 1 1 ? is a basis for W: 5. For p(x) in W; p(x) = a0 + a1x +(a0 ¡2a1)x2 = a0(1+ x2)+ a1(x¡2x2): The set f1+ x2; x¡2x2g is a basis for W: 6. The set f3¡x + x2g is a basis for W: 7. The set fx; x2g is a basis for W: 8. The set f1¡2x + x2g is a basis for W: 5.4. LINEAR INDEPENDENCE, BASES, AND COORDINATES 145 9. Let p(x) = P4i=0 aixi be in V: The given constraints are as follows: p(0) = 0 : a0 = 0 p0(1) = 0 : a1 +2a2 +3a3 +4a4 = 0 p00(¡1) = 0 : 2a2 ¡6a3 +12a4 = 0 : Solving yields a0 = 0; a1 = ¡9a3 + 8a4; a2 = 3a3 ¡ 6a4; a3 and a4 arbitrary. Thus f¡9x +3x2 + x3;8x¡6x2 + x4g is a basis for V: 10. The set ? ? 1 0 0 0 ? ; ? 0 0 0 1 ? ; ? 0 1 1 0 ? is a basis for the subspace of (2 x 2) symmetric matrices. 11. If A = [aij] is a (2 x 2) matrix then A = a11E11 +a12E12 +a21E21 +a22E22 so B spans V: It is easy to see that B is a linearly independent set, so B is a basis for W: 12. (a) [1;¡1;1]T; (b) [¡1;4;1]T (c) [5;2;0]T: 13. (a) [2;¡1;3;2]T (b) [1;0;¡1;1]T (c) [2;3;0;0]T: 14. Set p(x) = a0 + a1x +¢¢¢+ anxn and assume p(x) = (x): Then p(n)(x) = (x) so n!an = 0: It follows that an = 0: Suppose we have seen that am+1 = ¢¢¢ = an = 0 where 0? m < n: Then p(m)(x) = (x) so m!am = 0: Thus am = 0: By continuing the process we see that ai = 0;0? i ? n: Therefore the set f1; x; : : : ; xng is linearly independent. 15. The given matrices have coordinate vectors u1 = [2;1;2;1]T; u2 = [3;0;0;2]T;u3 = [1;1;2;1]T; respectively. The equation x1u1 +x2u2 +x3u3 = has onlythetrivialsolutionsofu1;u2;u3g isalinearlyindependentsubsetof R4: Byproperty (2) of Theorem 5, the set fA1; A2; A3g is a linearly independent subset of the vector space of (2 x 2) matrices. 16. The set is linearly dependent. For example ¡2A1 ¡A2 + A3 =O: 17. The given matrices have coordinate vectors u1 = [2;2;1;3]T; u2 = [1;4;0;5]T;u3 = [4;10;1;13]T; respectively. The set fu1 ;u2 ;u3g is linearly dependent in R4: For example ¡u1¡2u2 +u3 = . It follows that fA1; A2; A3g is linearly dependent; indeed ¡A1 ¡2A2 + A3 =O: 18. The set is linearly independent. 19. The polynomials p1(x); p2(x); p3(x); have coordinate vectors u1 = [¡1;2;1;0]T;u2 = [2;¡5;1;0]T;u3 = [0;¡1;3;0]T; respectively. The set fu1 ;u2 ;u3 g is linearly dependent; for example ¡2u1 ¡u2 +u3 = . It follows that the set fp1(x); p2(x); p3(x)g is linearly dependent; indeed ¡2p1(x)¡p2(x)+ p3(x) = 0: 146 CHAPTER 5. VECTOR SPACES AND LINEAR TRANSFORMATIONS 20. The set is linearly dependent; for example ¡p1(x)¡p2(x)¡p3(x)+ p4(x) = 0: 21. The given polynomials have coordinate vectors u1 = [1;0;0;1]T; u2 = [1;0;1;0]T;u3 = [1;1;0;0]T;u4 = [1;0;0;0]T; respectively. Sincefu1;u2;u3;u4g is a linearly independent subset of R4; the given set of polynomials is a linearly independent subset of P3: 22. A basis for Sp(S) is f1+2x + x2; x¡2x2g: 23. The given polynomials have coordinate vectors u1 = [1;2;1]T; u2 = [2;5;0]T;u3 = [3;7;1]T;u4 = [1;1;3]T; respectively. In the equation x1u1 +x2u2 +x3u3 +x4u4 = , x3 and x4 are arbitrary. It follows that fu1 ;u2g is a basis for Spfu1 ;u2 ;u3 ;u4g: Therefore fp1(x); p2(x)g is a basis for Sp(S). 24. The set ? ? 1 2 ¡1 3 ? ; ? 0 1 0 0 ? ; ? 0 0 0 1 ? is a basis for Sp(S). 25. The given matrices have coordinate vectors u1 = [1;2;¡1;3]T; u2 = [¡2;1;2;¡1]T;u3 = [¡1;¡1;1;¡3]T;u4 = [¡2;2;2;0]T; respectively. In the equation x1u1 +x2u2 +x3u3 +x4u4 = , x4 is arbitrary so fu1 ;u2 ;u3 g is a basis for Spfu1 ;u2 ;u3 ;u4 g. It follows that fA1; A2; A3g is a basis for Sp(S). 26. The coordinate vectors u1 = [¡1;1;2]T;u2 = [0;1;3]T;u3 = [1;2;8]T form a linearly independent set in R3: Since dim(R3) = 3 the set fu1 ;u2 ;u3 g is a basis for R3: By the Corollary to Theorem 5, Q is a basis for P2: 27. p(x) =¡4p1(x)+11p2(x)¡3p3(x) so [p(x)]Q = [¡4;11;¡3]T: 28. [p(x)]Q = [¡2a0 ¡3a1 + a2;4a0 +10a1 ¡3a2;¡a0 ¡3a1 + a2]T: 29. The coordinate vectors for the given matrices are u1 = [1;0;0;0]T; u2 = [1;¡1;0;0]T;u3 = [0;2;1;0]T;u4 = [¡3;0;2;1]T: It is easily verifled that fu1 ;u2 ;u3 ;u4g is a basis for R4: It follows from the Corollary to Theorem 5 that Q is a basis for V: 30. [A]Q = [9;¡5;¡1;¡1]T: 31. A = (a + b ¡2c +7d)A1 +(¡b +2c ¡4d)A2 +(c ¡2d)A3 + dA4 so [A]Q = [a + b ¡2c + 7d;¡b +2c¡4d; c¡2d; d]T: 32. The suggested constraints yield the following system of equations: p(¡1) = 0 : a0 ¡a1 + a2 = 0 p(0) = 0 : a0 = 0 p(1) = 0 : a0 + a1 + a2 = 0 Solving we obtain a0 = a1 = a2 = 0 as the unique solution. 5.5. DIMENSION 147 33. Let f(x) = c1 sinx + c2 cosx and suppose f(x) = (x): Then f(0) = c2 = 0 and f(?=2) = c1 = 0: It follows that fsinx;cosxg is a linearly independent set. 34. The conditions h(0) = h0(0) = h00(0) = h000(0) = 0 yield the system of equations c1 + c2 + c3 + c4 = 0 c1 + 2c2 + 3c3 + 4c4 = 0 c1 + 4c2 + 9c3 + 16c4 = 0 c1 + 8c2 + 27c3 + 64c4 = 0 The system has only the trivial solution so B is a linearly independent set. Since V = Sp(B) by deflnition, B is a basis for V: 35. Note that [g1(x)]B = [1;0;0;¡1]T = u1 ;[g2(x)]B = [0;1;1;0]T = u2 ; and [g3(x)]B = [¡1;0;1;1]T = u3: It is easy to verify that fu1;u2;u3g is a linearly independent subset of R4: By Theorem 5 the set fg1(x); g2(x); g3(x)g is linearly independent in V: 36. It follows from the note that w6= . Suppose that a1v1 +¢¢¢+ amvm +bw= . If b 6= 0 then we can solve for w ; contradicting the assumption that w is not in Sp(Q). Thus b = 0: This leaves a1v1 +¢¢¢ + amvm = . But the set Q is linearly independent so a1 =¢¢¢= am = 0: This proves that Q[fwg is linearly independent. 37. Suppose a1v1 +¢¢¢+anvn = , where ai 6= 0: Then vi = b1v1 +¢¢¢+bi¡1vi¡1 +bi+1vi+1 +¢¢¢+ bnvn ; where bj =¡aj=ai; for 1? j ? n; j 6= i: 38. A set of two vectors is linearly dependent if and only if one of the vectors is a scalar multiple of the other. The sets given in (a) and (b) are linearly independent whereas the sets given in (c), (d), and (e) are linearly dependent. 5.5 Dimension 1. (a) We show that V1 is a subspace. The proof that V2 is a subspace is similar. Clearly the (3 x 3) zero matrix is lower-triangular so it is in V1: Now let A and B be in V1; A = 2 4 a11 0 0 a21 a22 0 a31 a32 a33 3 5 and B = 2 4 b11 0 0 b21 b22 0 b31 b32 b33 3 5: ThenA + B = 2 4 a11 + b11 0 0 a21 + b21 a22 + b22 0 a31 + b31 a32 + b32 a33 + b33 3 5; so A + B is in V1: If c is any scalar then 148 CHAPTER 5. VECTOR SPACES AND LINEAR TRANSFORMATIONS cA = 2 4 ca11 0 0 ca21 ca22 0 ca31 ca32 ca33 3 5; so cA is in V1: By Theorem 2 of Section 4.3, V is a subspace of V1: (b) The set fE11; E21; E22; E31; E32; E33g is a basis for V1 and fE11; E12; E13; E22; E23; E33g is a basis for V2: (c) dim(V ) = 9;dim(V1) = 6;dim(V2) = 6: 2. Since V1 and V2 are subspaces of V , is in V1 \V2: Suppose u and v are vectors in V1 \ V2: Then u and v are in V1 and V1 is a vector space. Therefore u+ v is in V1 and for any scalar a; au is in V1: Similarly, u+ v and au are in V2: This shows that u+ v and au are in V1 \V2 so by Theorem 2 of Section 4.3, V1 \V2 is a subspace of V: Set V1 =f[a;0]T : a any real number g and let V2 =f[0; b]T : b any real number g: Then V1 and V2 are subspaces of R2 but V1 [ V2 is not a subspace. For example u= [1;0]T and v= [0;1]T are in V1 [V2 but u+ v is not in V1 [V2: 3. Let A = [aij] be in V1 \ V2: Since A is in V1; aij = 0 for i < j: Likewise A is in V2 so aij = 0 for i > j: Thus aij = 0 if i 6= j and V1 \ V2 is the set of (3 x 3) diagonal matrices. dim(V1 \V2) = 3: 4. dim(W) = 6: 5. The set 8 < : 2 4 0 1 0 ¡1 0 0 0 0 0 3 5; 2 4 0 0 1 0 0 0 ¡1 0 0 3 5; 2 4 0 0 0 0 0 1 0 ¡1 0 3 5 9 = ; is a basis for W so dim(W) = 3: 6. dim(W) = 2: 7. p(x) = P4i=0 aixi is in W if and only if a0 = 4a4; a2 =¡5a4; a1; a3; a4 arbitrary. A basis for W is fx; x3;4¡5x2 + x4g so dim(W) = 3: 8. The set S does not span V by property (1) of Theorem 9. 9. S contains 4 elements and dim(P2) = 3: By property (1) of Theorem 8, S is linearly dependent. 10. The set S is a basis for V by property (2) of Theorem 8. 5.5. DIMENSION 149 11. S contains only two vectors and dim(V ) = 4: By property (1) of Theorem 9, S does not span V: 12. The set S is a basis for V by property (2) of Theorem 8. 13. The set S contains 5 elements whereas dim(V ) = 4: By property (1) of Theorem 8, S is a linearly dependent set. 14. dim(W) = 2: 15. (a) First note that since V is not a subset of an already familiar vector space, we must check all the properties of Deflnition 1 given in Section 4.2. The closure properties (c1) and (c2) are evident. As illustrations we will check (a2), (a3), and (m2). Let x=fxig1i=1;y=fyig1i=1; and z=fzig1i=1 be in V: Then x+(y+z) =fxi+(yi+ zi)g1i=1 =f(xi + yi)+ zig1i=1 = (x+y)+ z: Therefore (a2) is satisfled. If = f ig1i=1; where i = 0 for each i then is the zero for V and property (a3) holds. To check (m2) let a be a scalar. Then a(x+y ) =fa(xi + yi)g1i=1 =faxi + ayig1i=1 =faxig1i=1 +fayig1i=1 = ax+ay: (b) If x= a1s1 +¢¢¢+ ansn then x is the sequence fxig1i=1 where xi = ai;1 ? i ? n; and xi = 0 for i > n: In particular if = a1s1 +¢¢¢+ ansn (where is the zero sequence described in (a)) then a1 =¢¢¢= an = 0: It follows that fs1 ;s2 ; : : : ;sng is a linearly independent subset of V: Since n is arbitrary, it follows from property (1) of Theorem 8 that V has inflnite dimension. 16. Apply property (1) of Theorem 8. 17. Suppose dim(V ) = n and let w1 be in W; w16= . If fw1g spans W then it is a basis. If not then by Exercise 36, Section 4.4, there is a vector w2 in W such that fw1 ;w2g is a linearly independent set. In general suppose we have constructed a linearly independent subset Sk =fw1 ;w2 ; : : : ;wkg of W: If W = Sp(Sk) then Sk is a basis and we are done. If Sk does not span W then, by Exercise 36, Section 4.4, there is a vector wk+1 in W such that Sk+1 =fw1 ; : : : ;wk ; wk+1 g is linearly independent. This process must stop since, by property (1) of Theorem 8, any set of n+1 vectors in V is linearly dependent. Therefore there exists an integer m; 1 ? m ? n; and a linearly independent subset Sm = fw1 ; : : : ;wm g of W such that W = Sp(Sm): Thus Sm is a basis for W and dim(W) = m ? n: 18. The set T = f[u1 ]B; : : : ;[uk ]Bg contains k vectors in Rp; where k ? p +1: Thus T is a linearly dependent subset of Rp: By Theorem 5, fu1 ; : : :uk g is a linearly dependent subset of V: 19. Since B is a linearly independent subset of V containing p vectors and Q is a basis of V containing m vectors, Theorem 6 implies that p ? m: Reversing the roles of B and Q gives m ? p: Therefore 150 CHAPTER 5. VECTOR SPACES AND LINEAR TRANSFORMATIONS m = p: 20. Property (1) of Theorem 8 is a direct consequence of Theorem 6. To prove property (2) let B be a basis for V and let S =fu1; : : : ;upg be a linearly independent subset of V: By property (2) of Theorem 5 the set T =f[u1]B; : : : ;[up]Bg is a linearly independent subset of Rp: Thus T is a basis for Rp: If v is in V then [v]B is in Sp(T). By property 1 of Theorem 5, v is in Sp(S). It follows that S spans V; so S is a basis for V: 21. Note that [w]B = [d1; : : : ; dn]T and [w]C = [c1; : : : ; cn]T: Thus A[w]C = c1[u1 ]B +¢¢¢+ cn[un ]B = [c1u1 +¢¢¢+ cnun ]B = [w]B: (a) A = 2 4 1 ¡1 1 0 1 ¡2 0 0 1 3 5: (b) [p(x)]C = [8;4;1]T and [p(x)]B = A[p(x)]C = [5;2;1]T: 22. (a) p(x) =¡4+(x+1)+(x+1)2: (b) p(x) = 15¡9(x+1)+2(x+1)2: (c) p(x) = 4¡(x+1)2: (d) p(x) =¡10+(x +1): 23. A¡1 = 2 4 1 1 1 0 1 2 0 0 1 3 5: (a) p(x) = 6+11x +7x2: (b) p(x) = 4+2x¡x2: (c) p(x) = 5+ x: (d) p(x) = 8¡2x¡x2: 24. A = 2 66 4 1 0 0 0 0 1 1 1 0 0 1 3 0 0 0 1 3 77 5: (a) p(x) =¡9+4x + x(x¡1)+ x(x¡1)(x¡2): (b) p(x) =¡2+8x + x(x¡1): (c) p(x) = 1+ x +3x(x¡1)+ x(x¡1)(x¡2): (d) p(x) = 3+5x +5x(x¡1)+ x(x¡1)(x¡2): 5.6 Inner-products 1. (1) = 4x21 + x22 ?0 and = 0 if and only if x1 = x2 = 0: (2) = 4x1y1 + x2y2 = 4y1x1 + y2x2 = : (3) = 4ax1y1 + ax2y2 = a(4x1y1 + x2y2) = a : (4) Let z= [z1; z2]T: Then = 4x1(y1 + z1)+ x2(y2 + z2) = (4x1y1 + x2y2)+ (4x1z1 + x2z2) =+ : 5.6. INNER-PRODUCTS 151 2. (i) = Pni=1 aix2i ?0 with equality if and only if xi = 0 for each i: (2) = Pni=1 aixiyi = Pni=1 aiyixi = : (3) = Pni=1 aiaxiyi = aPni=1 aixiyi = a : (4) = Pni=1 aixi(yi + zi) = Pni=1 aixiyi +Pni=1 aixizi = + : 3. (1) is immediate since A is positive deflnite. (2) = xTAy = (xTAy)T = yTATxTT = yTAx = : (3) = (ax)TAy= a[xTAy] = a : (4) = xTA(y +z) = xTAy +xTAz =+ : 4. xTAx= x21 +2x1x2 +2x22 = (x1 + x2)2 + x22: 5. (1) = a20 + a21 + a22 ?0 with equality if and only if ai = 0 for 0? i ?2: (2) = a0b0 + a1b1 + a2b2 = b0a0 + b1a1 + b2a2 = : (3) = aa0b0 + aa1b1 + aa2b2 = a(a0b0 + a1b1 + a2b2) = a : (4) Let r(x) = c0 +c1x+c2x2: Then = a0(b0 +c0)+a1(b1 +c1)+a2(b2 +c2) = (a0b0 + a1b1 + a2b2)+(a0c0 + a1c1 + a2c2) = + : 6. Clearly < p; p >= p(0)2 + p(1)2 + p(2)2 ? 0: Suppose 0 =< p; p >= a20 + (a0 + a1 + a2)2 +(a0 +2a1 +4a2)2: Then 0 = a0 = a0 + a1 + a2 = a0 +2a1 +4a2: It follows that a0 = a1 = a2 = 0: The remaining properties of Deflnition 7 are straightforward to verify. 7. (1) = a211 + a212 + a221 + a222 ?0 with equality if and only if A =O: (2) = a11b11+a12b12+a21b21+a22b22 = b11a11+b12a12+b21a21+b22a22 = : (3) = aa11b11+aa12b12+aa21b21+aa22b22 = a(a11b11+a12b12+a21b21+a22b22) = a : (4) Let C = [cij]: Then < A; B + C >= a11(b11 + c11)+ a12(b12 + c12)+ a21(b21 + c21)+ a22(b22 +c22) = (a11b11 +a12b12 +a21b21 +a22b22)+(a11c11 +a12c12 +a21c21 +a22c22) =< A; B> + : 8. =¡2;kxk= 2p2;kyk= 1;kx¡yk= p13: 152 CHAPTER 5. VECTOR SPACES AND LINEAR TRANSFORMATIONS 9. = [1;¡2] ? 1 1 1 2 ?? 0 1 ? =¡3;kxk2 == xTAx= 5 so kxk= p5;kyk= p2;x¡y= [1;¡3]T and kx¡yk2 = (x¡y)TA(x¡y) = 13: Thus kx¡yk= p13: 10. =¡1;kpk= p6;kqk= p6;kp¡qk= p14: 11. < p; q >= (¡1)1 + (2)2 + 7(7) = 52;kpk 2 =< p; p >= (¡1)2 + 22 + 72 = 54 so k p k= 3p6;kqk2== 12 +22 +72 = 54 so kqk= 3p6;kp ¡ qk2=

= 22 so kp¡qk= 2: 12. With the inner product deflned in Exercise 5, <1; x>=<1; x2 >= = 0 so f1; x; x2g is an orthogonal set. With the inner product deflned in Exercise 6, <1; x>= 3 so f1; x; x2g is not an orthogonal set. 13. For = xTy the graph of S is the circle with equation x2 + y2 = 1: For = 4x1y1 + x2y2 the graph of S is the ellipse with equation 4x2 + y2 = 1: 14. u1 = [1;0]T;u2 = [¡1;1]T: 15. v= a1u1 +a2u2 where a1 = = = 7 and a2 = = = 4: 16. p0(x) = 1; p1(x) = x¡1; p2(x) = x2 ¡2x +1=3: 17. q(x) = a0p0(x)+ a1p1(x)+ a2p2(x) where a0 = = = ¡5=3; a1 = = =¡5; and a2 = = =¡4: 18. For every scalar c; if p(x) = 2cx¡3cx2 + cx3 then = 0: 19. p0(x) = 1; p1(x) = x¡a0p0 where a0 === 0: Thus p1(x) = x: p2(x) = x2 ¡b0p0 ¡b1p1 where b0 = == 2 and b1 === 0: Thus p2(x) = x2 ¡2: p3(x) = x3 ¡c0p0 ¡c1p1 ¡c2p2 where c0 = == 0; c1 === 17=5; and c2 === 0: Therefore p3(x) = x3 ¡(17=5)x: p4(x) = x4 ¡d0p0 ¡d1p1 ¡d2p2 ¡d3p3 where d0 = = = 34=5; d1 = = = 0; d2 = < x4; p2 > = < p2; p2 >= 31=7; d3 =< x4; p3 > = < p3; p3 >= 0: Therefore p4(x) = x4 ¡(31=7)x2 +72=35: 5.6. INNER-PRODUCTS 153 20. == + : Therefore = 0: 21. By assumption kuk2 == 0: Therefore u= . 22. kavk= p = pa2 =jajkvk: 23. Suppose a1v1 +a2v2 +¢¢¢+ akvk = . For each i;1? i ? k; 0 = == Pkj=1 aj = ai = ai kvik2: Thus ai = 0 and the set is linearly independent. 24. Suppose u= Pnj=1 ajvj: Then == Pn j=1 aj = ai : Therefore ai = = : 25. From Examples 4 and 5,p0(x) = 1; p1(x) = x¡(1=2); p2(x) = x2 ¡x +(1=6); = 1; < p1; p1 >= 1=12; and < p2; p2 >= 1=180: Moreover < x3; p0 >= 1=4; < x3; p1 >= 3=40; and = 1=120: By Theorem 13,p?(x) = (1=4)p0(x)+ (9=10)p1(x)+(3=2)p2(x) = (3=2)x2 ¡(3=5)x +(1=20): 26. The required constants are < p0; x4 >= 1=5, < p1; x4 >= 1=15, < p2; x4 >= 1=105, and < p3; x4 >= 1=1400. The remaining constants have already been calculatedinExamples4, 5, and7. Iffollowsthat p4(x) = x4¡2x3+(9=7)x2¡(2=7)x+1=70. 27. With p3(x) as determined in Example 7 and with the calculations done in Example 6 we obtain p?(x)?0:841471p0(x)¡ 0:467544p1(x)¡0:430920p2(x)+0:07882p3(x): 28. T0(x) = 1; T1(x) = x; T2(x) = 2x2 ¡1; T3(x) = 4x3 ¡3x: 29. (a) Clearly T0(cos ) = 1 = cos(0 ) and T1(cos ) = cos = cos(1 ): Suppose we have seen that Tk(cos ) = cos(k ) for 0? k ? n; where n ?1: Then Tn+1(cos ) = 2cos Tn(cos )¡ Tn¡1(cos ) = 2cos cos(n ) ¡ cos(n¡1) = cos(n +1) [ since cos(fi + fl) = 2cosficosfl ¡cos(fi¡fl)]: (b) = (2=?)R1¡1[Ti(x)Tj(x)=p1¡x2]dx = ¡(2=?)R0? cos(i )cos(j )d = 0 if i 6= j: (c) T0(x) = 1 has degree zero and T1(x) = x has degree one. Suppose Tk(x) has degree k for 0 ? k ? n; where n ? 1: Thus Tn(x) = anxn + ¢¢¢ + a1x + a0 and Tn¡1(x) = bn¡1xn¡1 + ¢¢¢ + b1x + b0; where an 6= 0: Using (R) we obtain Tn+1(x) = 2anxn+1 +¢¢¢+(2a0 ¡b1)x¡b0: In particular, Tn+1(x) has degree n+1: It follows by induction that Tk(x) is a polynomial of degree k for each integer k ?0: 154 CHAPTER 5. VECTOR SPACES AND LINEAR TRANSFORMATIONS (d) T2(x) = 2x2 ¡1; T3(x) = 4x3 ¡3x; T4(x) = 8x4 ¡8x2 +1; T5(x) = 16x6 ¡20x3 +5x: 30. It follows from Exercise 29 that fT0(x); T1(x); : : : ; Tn(x)g is an orthogonal basis for Pn: Moreover < T0; T0 >= 2 whereas < Tj; Tj >= 1 for j ? 1: The given formula is now an immediate consequence of Theorem 13. 32. By property (2) of Deflnition 7, < ei;ej >=< ej;ei > for unit vectors, e1;e2; : : : ;en in R2. But < ei;ej >= eTi Aej = aij, whereas < ej;ei >= eTj Aei = aji. It follows that aij = aji, so A is symmetric. If x is a nonzero vector in Rn then xTAx =< x;x >> 0 by property (1) of Deflnition 7. Therefore, A is positive deflnite. 5.7 Linear Transformations 1. Let A = ? 1 0 0 0 ? and B = ? 0 0 0 1 ? : Then T(A + B) = det(A + B) = det(I) = 1 whereas T(A)+ T(B) = det(A)+ det(B) = 0: Therefore T is not a linear transformation. 2. T is a linear transformation. 3. T is a linear transformation. If A and B are (2 x 2) matrices it is straightforward to see that tr(A + B) = tr(A)+ tr(B); thus T(A + B) = T(A)+ T(B): Likewise if c is a scalar, tr(cA) = c tr(A) so T(cA) = cT(A): 4. T is not a linear transformation. For example if A = ? 1 0 0 0 ? and B = ? 0 1 0 0 ? then T(A) = T(B) = 0 whereas T(A + B) = 1: 5. Let f and g be in C[¡1;1] and let c be a scalar. Then T(f + g) = (f + g)0(0) = f0(0)+ g0(0) = T(f)+ T(g); and T(cf) = (cf)0(0) = cf0(0) = cT(f): Therefore T is a linear transformation. 6. T is a linear transformation. 7. T is not a linear transformation. For example T(1+ (x)) = T(1) = 2+ x + x2 whereas T(1)+ T( (x)) = (2+ x + x2)+(1+ x + x2) = 3+2x +2x2: 8. T is a linear transformation. 9. (a) T(p) = 3T(1)¡2T(x)+4T(x2) = 3(1+ x2)¡2(x2 ¡x3)+ 4(2+ x3) = 11+ x2 +6x3: 5.7. LINEAR TRANSFORMATIONS 155 (b) T(a0+a1x+a2x2) = a0T(1)+a1T(x)+a2T(x2) = a0(1+x2)+a1(x2¡x3)+a2(2+x3) = (a0 +2a2)+(a0 + a1)x2 +(¡a1 + a2)x3: 10. p(x) = (¡5)1+3(x +1)+(x2 +2x +1) so T(p(x)) =¡5x4+ 3(x3 ¡2x)+ x =¡5x4 +3x3 ¡5x: Similarly, q(x) = (¡3)1+ 7(x +1)+(x2 +2x +1) so T(q(x)) =¡3x4 +7x3 ¡13x: 11. (a) T(A) =¡2T(E11)+2T(E12)+3T(E21)+4T(E22) = 8+14x¡9x2: (b) T ? a b c d ?¶ = aT(E11)+ bT(E12)+ cT(E21)+ dT(E22) = (a + b +2d)+(¡a + b +2c + d)x +(b¡c¡2d)x2: 12. (a) Let A = [aij] and B = [bij] be (2 x 2) matrices. Then T(A + B) = T([aij + bij]) = ? (a 11 + b11)+2(a22 + b22) (a12 + b12)¡(a21 + b21) ? = ? a 11 +2a22 a12 ¡a21 ? + ? b 11 +2b22 b12 ¡b21 ? = T(A)+ T(B): If c is a scalar then T(cA) = T([caij]) = ? ca 11 + ca22 ca12 ¡ca21 ? = c ? a 11 + a22 a12 ¡a21 ? = cT(A): Therefore T is a linear transformation. (b) N(T)= ?? a b c d ? : a =¡2d; b = c; canddarbitrary : (c) ?? ¡2 0 0 1 ? ; ? 0 1 1 0 ? is a basis for N(T): (d) nullity(T) = 2 and rank(T) = 2: (e) is a subspace of R2 and 2 = dim(R2) = rank(T) = dim(R(T)): Therefore R(T) = R2: (f) T(A) = v where A is any matrix of the form A = ? x¡2d y + c c d ? ; where c and d are arbitrary. For example, A = ? x y 0 0 ? is one choice for A: 13. (a) By property 1 of Theorem 15, R(T)= Sp fT(1); T(x); T(x2); T(x3); T(x4)g =Spf0;0;2;6x;12x2g= Spf2;6x;12x2g: It follows that rank(T) = 3: Since R(T) P2 and dim(P2) = 3; we have R(T)=P2: (b) By property 3 of Theorem 15, nullity(T) = dim(P4)¡rank(T) = 5 ¡3 = 2: Since nullity(T) > 0; T is not one to one. 156 CHAPTER 5. VECTOR SPACES AND LINEAR TRANSFORMATIONS (c) We wish to determine q(x) = b0 + b1x + b2x2 + b3x3 + b4x4 in P4 such that a0 + a1x + a2x2 = T(q) = 2b2 +6b3x +12b4x2: Equating coe?cients gives b2 = a0=2; b3 = a1=6; b4 = a2=12; b0 and b1 arbitrary. In particular,q(x) = (a0=2)x2+(a1=6)x3+(a2=12)x4 is one choice. 14. R(T)= Spf1¡x+2x2 +3x3;¡1+3x¡3x2¡x3;2¡2x+5x2 +7x3;¡1+3x¡x2 +2x3;1¡ x+x2 +2x3g: Utilizing the spanning set we obtain the basis f1¡x+2x2 +3x3;2x¡x2 + 2x3; x2 + x3; x3g for R(T): In particular, rank(T) = 4: Thus nullity(T) = 1 and T is not one to one. 15. N(T)=fp(x) = a0 + a1x + a2x2 : a0 +2a1 +4a2 = 0g: It follows that nullity(T) = 2: Consequently rank(T) = 1 and R(T)= R1: 16. N(T)=ff in C[0;1] : R10 f(t)dt = 0g: For any a in R1; T(2ax) = R10 2atdt = a; so R(T) = R1: 17. (a) Let u, v be vectors in V and let c be a scalar. Then I(u+v) = u+v= I(u)+ I(v) and I(cu) = cu= cI(u): Therefore I is a linear transformation. (b) The vector v is in N(I) if and only if = I(v) = v: Thus N(I) =f g: For each v in V; I(v) = v so R(I) = V : 18. (a) Let u1;u2 be in U and let a be a scalar. Then T(u1 +u2) = V = V + V = T(u1)+T(u2) and T(au1) = V = a V = aT(u1 ): This proves that T is a linear transformation. (b) N(T)= U and R(T)=f Vg: 19. Recall that 5 = dim(P4) = rank(T)+nullity(T): Moreover R(T) P2 so rank(T)?3: The possibilities are: rank(T) 3 2 1 0 nullity(T) 2 3 4 5: Since nullity(T)?2; T cannot be one to one. 20. By property 3 of Theorem 15, dim(U) = rank(T) + nullity(T): But R(T) V so rank(T)?dim(V ) < dim(U): It follows that nullity(T) > 0 and, hence, T cannot be one to one. 5.7. LINEAR TRANSFORMATIONS 157 21. Recall that 3 = dim(R3) = rank(T)+nullity(T): Moreover N(T) R3 so nullity(T)?3: The possibilities are: rank(T) 3 2 1 0 nullity(T) 0 1 2 3: Since dim(P3) = 4 and rank(T) < 4;R(T)=P3 is not a possibility. 22. By property 3 of Theorem 15, dim(U) = rank(T)+nullity(T): In particular, rank(T)? dim(U) < dim(V ) so R(T)= V is not possible. 23. It follows from property 1 of Theorem 14 that V is in R(T): Sup- pose that v1 and v2 are in R(T); thus there exist vectors u1 and u2 in U such that T(u1) = v1 and T(u2) = v2: Therefore v1 +v2 = T(u1 )+ T(u2 ) = T(u1 +u2 ) so v1 +v2 is in R(T): If a is a scalar then av1 = aT(u1 ) = T(au1 ); and av1 is in R(T): This proves that R(T) is a subspace of V: 24. Suppose T is one to one and let u be in N(T): It follows from property 1 of Theorem 14 that T(u) = V = T( U): Since T is one to one, u= U so N(T) =f Ug: 25. (a) If rank(T) = p then, in the notation of Theorem 15, C = fT(u1); : : : ; T(up)g is a basis for R(T) [ cf. property 2 of Theorem 9 in Section 4.5 ]. In particular the set C is linearly independent. By property 2 of Theorem 15, T is one to one so, by property 4 of Theorem 14, nullity(T) = 0: (b) If rank(T) = 0 then R(T)= f Vg and T is the zero linear transformation deflned by T(u) = V for all u in U: Thus N(T)= U and nullity(T) = dim(U) = p: 26. T is one to one if and only if N(T)= f g: Thus T is one to one if and only if Ax= has only the trivial solution, that is, if and only if A is nonsingular. 27. (a) Let A and B be (2 x 2) matrices. Then T(A + B) = (A + B)T = AT + BT = T(A)+T(B): If c is a scalar then T(cA) = (cA)T = cAT = cT(A): This proves that T is a linear transformation. (b) If A isinN(T) then T(A) = AT =O: Itfollowsthat A =O andthatnullity(T) = 0: Therefore rank(T) = 4: Consequently T is one to one and R(T)= V: (c) Let B be in V and set C = BT: Then T(C) = CT = BTT = B: Therefore R(T) = V: 158 CHAPTER 5. VECTOR SPACES AND LINEAR TRANSFORMATIONS 5.8 Operations with Linear Transformations 1. (S +T)(p) = S(p)+T(p) = p0(0)+(x+2)p(x): In particular, (S +T)(x) = 1+(x+2)x = x2 +2x +1 and (S + T)(x2) = 0+(x +2)x2 = x3 +2x2: 2. (2T)(p) = 2[T(p)] = 2(x +2)p(x) = (2x +4)p(x): Therefore (2T)(x) = 2x2 +4x: 3. (H?T)(p) = H(T(p)) = H((x+2)p(x)) = [(x+2)p(x)]0+2p(0) = (x+2)p0(x)+p(x)+2p(0): The domain for H ?T is P3 and (H ?T)(x) = 2x +2: 4. (T ?H)(p) = (x +2)[p0(x)+ p(0)];T ?H has domain P4; (T ?H)(x) = x +2: 5. (a) If p(x) = P3i=0 aixi then T(p) = 2a0+(a0+2a1)x+(a1+2a2)x2+(a2+2a3)x3+a3x4: In particular T(p) = (x) if and only if p(x) = (x): Therefore T is one to one. Now rank(T) = dim(P3)¡nullity(T) = 4: Since R(T) P4 and dim(P4) = 5; R(T) 6=P4; that is, T is not onto. (b) It is easy to verify that T(p) = x is impossible. Therefore T¡1(x) is not deflned. 6. (a) R(H) = Sp fH(1); H(x); H(x2); H(x3); H(x4)g= Spf 1;1;2x;3x2;4x3g= Spf 1;2x;3x2;4x3g: It follows that rank(H) = 4 and nullity(H) = 1: Therefore H is onto but not one to one. (b) Note that H((1=2)x2) = x = H((1=2)x2 +x¡1): Therefore H¡1(x) is not uniquely determined. 7. Let p(x) = aex + be2x + ce3x be in V: Then T(p(x)) = p0(x) = aex + 2be2x + 3ce3x: Since B = fex; e2x; e3xg is a linearly independent set, it follows that T(p(x)) = (x) if and only if a = b = c = 0: Thus N(T)= f (x)g and T is one to one. The set B is a basis for V so dim(V ) = 3: Thus rank(T) = dim(V )¡nullity(T) = 3: It follows that T is onto. Therefore T is invertible. Moreover T¡1(ex) = ex; T¡1(e2x) = (1=2)e2x; and T¡1(e3x) = (1=3)e3x: This implies that T¡1(aex +be2x +ce3x) = aex +(b=2)e2x +(c=3)e3x: 8. T¡1(sinx) =¡cosx; T¡1(cosx) = sinx; and T¡1(e¡x) =¡e¡x: Therefore T¡1(asinx + bcosx + ce¡x) =¡acosx + bsinx¡ce¡x: 9. If A is in N(T) then T(A) = AT =O: Therefore A =O; so N(T)=fOg: It follows that T is one to one. Further rank(T) = dim(V )¡nullity(T) = 4 so T is onto. Therefore T is invertible. In fact T¡1 = T since (AT)T = A: 10. T¡1(A) = QAQ¡1: 11. (a) Since dim(V ) = 4; V is isomorphic to R4 by Theorem 17. 5.8. OPERATIONS WITH LINEAR TRANSFORMATIONS 159 (b) Since dim(P3) = 4 = dim(V); V andP3 are isomorphic by the corollary to Theorem 17. (c) It is easily shown that T : V !P3 deflned by T ? a b c d ?¶ = a + bx + cx2 + dx3 is an isomorphism. 12. (a) Note that dim(U) = 3: (b) dim(U) = 3 = dim(P2): (c) Deflne T : U !P2 by T ? a b b c ?¶ = a + bx + cx2: 13. If u and w are in U then S(u+w) = T1(u+w)+ T2(u+w) = T1(u)+ T1(w)+ T2(u)+ T2(w) = T1(u)+ T2(u)+ T1(w)+ T2(w) = S(u)+S(w): If c is a scalar then S(cu) = T1(cu)+ T2(cu) = cT1(u)+ cT2(u) = c[T1(u)+ T2(u)] = cS(u): This proves that S is a linear transformation. 14. Let u and w be in U:[aT](u+w) = a[T(u+w)] = a[T(u)+ T(w)] = a[T(u)]+ a[T(w)] = [aT](u)+[aT](w): If c is a scalar then [aT](cu) = a[T(cu)] = a[cT(u)] = c[aT(u)] = c[aT](u): There- fore aT is a linear transformation. 15. Suppose that T¡1(v) = u: Then T(u) = v and T(cu) = cT(u) = cv: Therefore T¡1(cv) = cu= cT¡1(v): 16. By formula 1, (T¡1)¡1(u) = v where v is chosen so that T¡1(v) = u: But T¡1(v) = u precisely when T(u) = v: Therefore (T¡1)¡1 = T: 17. Let u be in U and set T(u) = v: Then T¡1(v) = u so (T¡1 ?T)(u) = T¡1(T(u)) = T¡1(v) = u: It follows that T¡1 ?T = IU: Likewise (T ?T¡1)(v) = T(T¡1(v)) = T(u) = v; so T ?T¡1 = IV : 18. (a) Suppose S and T are both one to one and let u be in N(T ?S): Thus W = (T ?S)(u) = T(S(u)): Since T is one to one it follows that S(u) = V : But S is also one to one so u= U: 160 CHAPTER 5. VECTOR SPACES AND LINEAR TRANSFORMATIONS (b) Suppose that S and T are onto and let w be in W: By assumption there exists v in V such that T(v) = w: Likewise, S is onto so there exists u in U such that S(u) = v: Therefore (T?S)(u) = T(S(u)) = T(v) = w: It follows that T ?S is onto. (c) S and T are both one to one and both onto so by (a) and (b) T ? S is one to one and onto. Therefore T ?S is invertible. To see that (T ?S)¡1 = S¡1?T¡1; let w be in W. Since T ?S is onto there exists u in U such that (T ?S)(u) = w: Therefore (T ?S)¡1(w) = u: Now set v= S(u): Then T(v) = w so T¡1(w) = v and S¡1(v) = u: Therefore (S¡1 ?T¡1)(w) = S¡1(T¡1(w)) = S¡1(v) = u= (T ?S)¡1(w): 19. Let S : U ! V be an isomorphism and let T : V ! W be an isomorphism. By Exercise 18, T ?S : U ! W is an isomorphism. 20. (a) Since T is one to one, nullity(T) = 0: Therefore rank(T) = n¡nullity(T) = n so R(T)= V: Since T is onto, T is invertible. (b) By assumption rank(T) = n: Thus nullity(T) = n¡rank(T) = 0 and T is one to one. Therefore T is invertible. 21. It is easy to show that T(p(x)) = (x) if and only if p(x) = (x): Thus N(T)=f (x)g and T is one to one. Clearly there exists no polynomial p(x) in P such that T(p(x)) = 1: Therefore T is not onto. This does not contradict Exercise 20(a) since P has inflnite dimension. 22. Let q(x) = b0+b1x+¢¢¢+bnxn beinP andset p(x) = b0x+(b1=2)x2+¢¢¢+[bn=(n+1)]xn+1: Then S(p) = p0(x) = q(x) so S is onto. Note that N(T) is the set of all constant polynomials. In particular, N(T)6=f (x)g so T is not one to one. This does not contradict Exercise 20(b) since P has inflnite dimension. 23. If u is in N(S) then S(u) = V : Therefore (T ?S)(u) = T(S(u)) = T( V ) = W and u isinN(T ?S): If T?S isonetoonethenN(T ?S) =f Ug: Therefore N(S) =f Ug and S is one to one. 24. If w is in R(T ?S) then there exists u in U such that w= (T ?S)(u): Set v= S(u): Then v is in V and T(v) = T(S(u)) = (T ?S)(u) = w: This shows that w is in R(T): If T ?S is onto we have R(T ?S) = V and R(T ?S) R(T) V: It follows that R(T)= V and T is onto. 25. Assume that T ?S is invertible. Then T ?S is one to one so, by Exercise 23, S is one to one. Exercise 20(a) now implies that S is invertible. Since T ?S is also onto, Exercise 24 implies that T is onto. Exercise 20(b) now implies that T is invertible. 5.9. MATRIX REPRESENTATIONS FOR LINEAR TRANSFORMATIONS 161 26. Deflne S : Rn ! Rp by S(x ) = Bx and deflne T : Rp ! Rm by T(y ) = Ay : Then T ? S : Rn ! Rm is deflned by (T ? S)(x ) = ABx: Therefore nullity(B) = nullity(S);nullity(AB) = nullity(T ?S);rank(A) = rank(T); and rank(AB) = rank(T ?S): Now apply Exercises 23 and 24. 27 It follows from Exercise 20(a) that T is invertible if and only if T is one to one. Now apply Exercise 26 of Section 4.7. 28. Deflne S : Rn ! Rnby S(x ) = B(x ) and deflne T : Rn ! Rn by T(x) = Ax. Then T ? S : Rn ! Rn is deflned by (T ? S)(x) = ABx. If AB is nonsingular then T ? S is invertible by Exercise 27. By Exercise 25, both T and S are invertible. Applying Exercise 27 again, we see that A and B are nonsingular. 29. To prove that L(U; V ) is a vector space requires checking all ten properties of Deflnition 1 in Section 4.2. We shall verify only properties (c1), (c2), (a2), (a3), (a4), and (m2). If S and T are in L(U; V ) and c is a scalar then S + T and cT are in L(U; V ) by Exercises 13 and 14. Thus properties (c1) and (c2) hold. Now let R be in L(U; V ): To show that R+(S+T) = (R+S)+T wemustshowthateachofthetransformationshasthesameaction on any vector u in U: But addition is associative in V so [R +(S + T)](u) = R(u)+(S + T)(u) = R(u)+(S(u)+T(u)) =(R(u)+S(u))+T(u) = (R+S)(u)+T(u) = [(R+S)+T](u). Recall that the zero linear transformation T0 : U ! V is deflned by T0(u) = V for every u in U: Thus (T + T0)(u) = T(u)+ T0(u) = T(u)+ V = T(u): It follows that T + T0 = T; so T0 is the zero of L(U; V ): For T in L(U; V ) it is easily seen that T + (¡1)T = T0; so (¡1)T =¡T and property (a4) of Deflnition 1 is satisfled, Tocheck(m2)let a beascalarandlet S and T bein L(U; V ): Forany u in U [a(S+T)](u) = a[(S + T)(u)] = a[S(u)+ T(u)] = aS(u)+ aT(u) = [aS](u)+[aT](u) = [aS + aT](u): It follows that a(S + T) = aS + aT: 5.9 Matrix Representations for Linear Transformations 1. S(1) = 0; S(x) = 1; S(x2) = 0; and S(x3) = 0: Thus [S(1)]C = [S(x2)]C = [S(x3)]C = [0;0;0;0;0]T; while [S(x)]C = [1;0;0;0;0]T: The matrix for S is 2 66 66 4 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 77 77 5 : 162 CHAPTER 5. VECTOR SPACES AND LINEAR TRANSFORMATIONS 2. The matrix for T is 2 66 66 4 2 0 0 0 1 2 0 0 0 1 2 0 0 0 1 2 0 0 0 1 3 77 77 5 : 3. (a) (S + T)(1) = 2+ x;(S + T)(x) = 1+2x + x2;(S + T)(x2) = 2x2 + x3;(S + T)(x3) = 2x3 +x4: Therefore [(S +T)(1)]C = [2;1;0;0;0]T;[(S +T)(x)]C = [1;2;1;0;0]T;[(S + T)](x2)]C = [0;0;2;1;0]T; and [(S + T)(x3)]C = [0;0;0;2;1]T: The matrix for S + T is the matrix 2 66 66 4 2 1 0 0 1 2 0 0 0 1 2 0 0 0 1 2 0 0 0 1 3 77 77 5 : (b) By Theorem 19 the matrix for S + T is the sum of matrices for S and T: This is easily verifled. 4. The matrix for 2T is 2 66 66 4 4 0 0 0 2 4 0 0 0 2 4 0 0 0 2 4 0 0 0 2 3 77 77 5 : 5. H(1) = 1; H(x) = 1; H(x2) = 2x; H(x3) = 3x2; and H(x4) = 4x3: Therefore [H(1)]B = [H(x)]B = [1;0;0;0]T;[H(x2)]B = [0;2;0;0]T;[H(x3)]B = [0;0;3;0]T; and [H(x4)]B = [0;0;0;4]T: The matrix for H is the matrix 2 66 4 1 1 0 0 0 0 0 2 0 0 0 0 0 3 0 0 0 0 0 4 3 77 5: 6. (a) The matrix for H ?T is the matrix 2 66 4 3 2 0 0 0 2 4 0 0 0 3 6 0 0 0 4 3 77 5: (b) Denote by D; E; and F the matrices in Exercises 5,2, and 6(a), respectively. By Theorem 20, F = DE and it is easily verifled that this is the case. 7. (a) (T ? H)(1) = 2 + x;(T ? H)(x) = 2 + x;(T ? H)(x2) = 4x + 2x2;(T ? H)(x3) = 6x2 +3x3;(T ?H)(x4) = 8x3 +4x4: Therefore [(T ?H)(1)]C = [(T ?H)(x)]C = [2;1;0;0;0]T;[(T ?H)(x2)]C = [0;4;2;0;0]T;[(T ?H)(x3)]C = [0;0;6;3;0]T; and 5.9. MATRIX REPRESENTATIONS FOR LINEAR TRANSFORMATIONS 163 [(T ?H)(x4)]C = [0;0;0;8;4]T: Thus the matrix for T ?H is the matrix 2 66 66 4 2 2 0 0 0 1 1 4 0 0 0 0 2 6 0 0 0 0 3 8 0 0 0 0 4 3 77 77 5 : (b) Let D, E, and F denote the matrices for T, H, and T ?H, respectively (cf. Exercises 2, 5, and 7(a)). By Theorem 20, F = DE and it is easily verifled that this the case. 8. (a) [p]B = [a0; a1; a2; a3]T; S(p)=a1 so [S(p)]C = [a1;0;0;0;0]T: 9. (a) [p]B = [a0; a1; a2; a3]T;T(p) = 2a0 +(a0 +2a1)x +(a1 +2a2)x2 +(a2 +2a3)x3 + a3x4 so [T(p)]C = [2a0; a0 +2a1; a1 +2a2; a2 +2a3; a3]T: 10. [q]C = [a0; a1; a3; a4]T;H(q) = (a0 + a1) + 2a2x + 3a3x2 + 4a4x3 so [H(q)]B = [a0 + a1;2a2;3a3;4a4]T: It is easily seen that N[q]C = [H(q)]B: 11. (a) Q = 2 4 1 0 0 0 2 0 0 0 3 3 5: (b) P = 2 4 1 0 0 0 1=2 0 0 0 1=3 3 5: (c) Clearly P = Q¡1: 12. (a) Q = 2 4 0 ¡1 0 1 0 0 0 0 ¡1 3 5: (b) P = 2 4 0 1 0 ¡1 0 0 0 0 ¡1 3 5: (c) Clearly P = Q¡1: 13. (a) T(E11) = E11; T(E12) = E21; T(E21) = E12; and T(E22) = E22: Therefore Q =2 66 4 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 3 77 5: (b) If A = [aij] is a (2 x 2) matrix then [A]B = [a11; a12; a21; a22]T whereas [AT]B = [a11; a21; a12; a22]T: Clearly Q[A]B = [AT]B: 164 CHAPTER 5. VECTOR SPACES AND LINEAR TRANSFORMATIONS 14. 2 66 4 3 0 0 0 3 0 0 ¡1 3 0 0 0 3 77 5: 15. S(x+1) = 3+3x¡x2; S(x+2) = 6+3x¡x2; and S(x2) = 3x2: Therefore [S(x+1)]C = [3;3;¡1;0]T;[S(x +2)]C = [6;3;¡1;0]T; and [S(x2)]C = [0;0;3;0]T and the matrix representation for S is 2 66 4 3 6 0 3 3 0 ¡1 ¡1 3 0 0 0 3 77 5: 16. 2 66 4 1 0 0 0 1 0 0 0 1 0 0 0 3 77 5: 17. T(1) = 2 4 1 0 0 3 5; T(x) = 2 4 0 3 1 3 5; and T(x2) = 2 4 0 6 4 3 5; so the matrix for T is given by 2 4 1 0 0 0 3 6 0 1 4 3 5: 18. 2 4 0 ¡2 ¡2 ¡1 1 4 1 2 2 3 5: 19. T(v1) = 0v1+1v2+0v3+0v4; T(v2) = 0v1+0v2+1v3+0v4; T(v3) = 1v1+1v2+0v3+0v4; and T(v4) = 1v1 +0v2 +0v3 +3v4; so the matrix of T is 2 66 4 0 0 1 1 1 0 1 0 0 1 0 0 0 0 0 3 3 77 5: 20. T(e1 ) = Ae1 = ? 1 3 ? ; T(e2 ) = Ae2 = ? 2 0 ? and T(e3 ) = Ae3 = ? 1 4 ? : Therefore the matrix for T is the given matrix A: 21. T(1) = ¡4 + 3x ¡ x2; T(x) = ¡2 + 3x + 2x2; and T(x2) = 3x2 so the matrix for T is2 4 ¡4 ¡2 0 3 3 0 ¡1 2 3 3 5: 5.9. MATRIX REPRESENTATIONS FOR LINEAR TRANSFORMATIONS 165 22. [p]B = [a0; a1; a2]T and [T(p)]B = [¡4a0 ¡2a1;3a0 +3a1;¡a0 +2a1 +3a2]T: It is easily verifled that Q[p]B = [T(p)]B: 23. T(1¡3x+7x2) = 2(1¡3x+7x2); T(6¡3x+2x2) =¡3(6¡3x+2x2); and T(x2) = 3x2: Therefore the matrix of T is 2 4 2 0 0 0 ¡3 0 0 0 3 3 5: 24. Let P = [P1 ;P2 ; : : : ;Pk ] be a matrix such that P[u]B = [T(u)]C for every vector u in U: Since [u]B is in Rn and [T(u)]C is in Rm; P is necessarily an (m xn) matrix. Therefore k = n: Suppose Q = [Q1 ;Q2 ; : : : ;Qn ] and assume that B =fu1 ;u2 ; : : : ;ung: Then for 1? j ? n;Pj = Pej = P[uj ]B = [T(uj )]C = Q[uj ]B = Qej = Qj: It follows that P = Q: 25. Let B =fu1 ;u2 ; : : : ;ung and C =fv1 ;v2 ; : : : ;vmg: Suppose T1(u1) = a11v1 + a21v2 +¢¢¢+ am1vm T1(u2) = a12v1 + a22v2 +¢¢¢+ am2vm ... ... T1(un) = a1nv1 + a2nv2 +¢¢¢+ amnvm : Also assume that T2(u1) = b11v1 + b21v2 +¢¢¢+ bm1vm T2(u2) = b12v1 + b22v2 +¢¢¢+ bm2vm ... ... T2(un) = b1nv1 + b2nv2 +¢¢¢+ bmnvm : Then T1 and T2 are represented by the (m xn) matrices Q1 = [aij] and Q2 = [bij]; respectively. To obtain the matrix for T1 + T2 note that (T1 + T2)(u1 ) = T1(u1 )+ T2(u1 ) = (a11 + b11)v1 +(a21 + b21)v2 +¢¢¢+(am1 + bm1)vm (T1 + T2)(u2 ) = T1(u2 )+ T2(u2 ) = (a12 + b12)v1 +(a22 + b22)v2 +¢¢¢+(am2 + bm2)vm ... ... (T1 + T2)(un ) = T1(un )+ T2(un ) = (a1n + b1n)v1 +(a2n + b2n)v2 +¢¢¢+(amn + bmn)vm Therefore the matrix for T1 + T2 is the (m xn) matrix [aij + bij] = Q1 + Q2: 166 CHAPTER 5. VECTOR SPACES AND LINEAR TRANSFORMATIONS 26. Let B =fu1 ; : : : ;ung and C =fv1 ; : : : ;vmg: If T(uj ) =P m i=1 qijvi for 1? j ? n; then the matrix for T is the (m xn) matrix Q = [qij]: Moreover [aT](uj ) = aPmi=1 qijvi = Pmi=1(aqij)vi for 1 ? j ? n: Therefore the matrix for aT is the (m xn) matrix [aqij] = aQ: 27. By assumption Q[u]B = [T(u)]C for every vector u in U: If P is the matrix for aT then P is the unique matrix such that P[u]B= [(aT)(u)]C for every vector u in U. But (aQ)[u]B = a(Q[u]B) = a[T(u)]C = [aT(u)]C = [(aT)(u)]C. It follows that P = aQ. 28. If B = fv1 ;v2 ; : : : ;vn g is a basis for V then IV (vj ) = vj for 1 ? j ? n: Therefore [IV (vj )]B = ej and the matrix representation for IV is [e1 ;e2 ; : : : ;en ] = I: 29. If B = fv1 ;v2 ; : : : ;vng is a basis for V then T0(vj ) = V = 0v1 +0v2 +¢¢¢+0vn for 1 ? j ? n: Thus [T0(vj )]B = (the zero vector in Rn ) and the matrix for T0 is the (n xn) zero matrix. 30. It is an immediate consequence of Theorem 20 and Exercise 28 that PQ = I and QP = I: Therefore P = Q¡1: 31. If Q is the matrix for T then T(1) = 1¡x2; T(x) = x+x2; T(x2) = 2; and T(x3) = x¡x2: Therefore T(a0+a1x+a2x2+a3x3) = a0T(1)+a1(T(x)+a2T(x2)+a3T(x3) = (a0+2a2)+ (a1 + a3)x +(¡a0 + a1 ¡a3)x2: 32. S(a0 + a1x + a2x2) = ? a 0 ¡a2 a1 + a2 2a0 a1 ¡a2 ? : 33. To see that ? is one to one let T : U ! V be a linear transformation and assume that T is in N(?); that is ?(T) is the (m xn) zero matrix. Let B =fu1 ;u2 ; : : : ;ung and C =fv1 ;v2 ; : : : ;vmg be the given bases for U and V; respectively. By assumption, T(uj) = Pmi=1 0vi = V for each vector uj in B;1 ? j ? n: It follows that T(u ) = V for each u in U: Therefore T = T0; where T0 is the zero transformation from U to V (cf. Exercise 29). But T0 is the zero vector in the vector space L(U; V ); and N(?) =fT0g: This proves that ? is one to one. 5.10 Change of Basis and Diagonalization 1. T(u1) = u1 and T(u2) = 3u2: Therefore u1 and u2 areeigenvectorsfor T corresponding to the eigenvalues ?1 = 1 and ?2 = 3; respectively. The matrix of T with respect to C is? 1 0 0 3 ? : 2. The matrix for T with respect to C is 2 4 2 0 0 0 1 0 0 0 1 3 5: 5.10. CHANGE OF BASIS AND DIAGONALIZATION 167 3. T(A1) = 2A1; T(A2) =¡2A2; T(A3) = 3A3; and T(A4) =¡3A4: Therefore A1; A2; A3 and A4 are eigenvectors for T corresponding to the eigenvalues ?1 = 2; ?2 = ¡2; ?3 = 3; and ?4 = ¡3; respectively. The matrix for T with respect to C is given by 2 66 4 2 0 0 0 0 ¡2 0 0 0 0 3 0 0 0 0 ¡3 3 77 5: 4. The transitionmatrixisthematrix P = ? ¡1=2 1=2 1=2 1=2 ? : Notethat Pa =[¡1;3]T = [a]C: Therefore a=¡u1+3u2 : Similarly b= u1¡u2 ;c=¡2u1+7u2 ; and d= (¡a=2+b=2)u1 +(a=2+ b=2)u2 : 5. The transition matrix is the matrix P = 2 4 1 ¡1 ¡1 1 ¡1 0 ¡1 2 1 3 5: Now [p(x)]B = [2;1;0]T and P[p(x)]B = [1;1;0]T = [p(x)]C: Denote the polynomials in C by g1(x); g2(x); g3(x); respectively. It follows that p(x) = g1(x)+ g2(x): Similarly s(x) =¡2g1(x)¡g2(x)+2g3(x); q(x) =¡5g1(x)¡3g2(x)+7g3(x); and r(x) = (a0 ¡a1 ¡a2)g1(x)+(a0 ¡a1)g2(x)+(¡a0 +2a1 + a2)g3(x): 6. The transition matrix is P = 2 66 4 0 0 0 1 0 0 1 0 0 1 ¡1 0 1 0 0 ¡1 3 77 5: A = 4A1 + 3A2 ¡ A3 ¡ 3A4;B = 3A1 + A3 ¡4A4;C = dA1 + cA2 +(b¡c)A3 +(a¡d)A4: 7. Since u1 = (1=3)w1 +(1=3)w2 and u2 = (5=3)w1 ¡(1=3)w2 ; the transition matrix is P = ? 1=3 5=3 1=3 ¡1=3 ? : 8. P = ? 5=3 2=3 ¡4=3 ¡1=3 ? : 9. The transition matrix is P = 2 66 4 ¡1 1 2 3 1 0 0 ¡3 0 0 1 0 0 0 0 1 3 77 5: Since [p(x)]B = [2;¡7;1;0]T; P[p(x)]B = [¡7;2;1;0]T = [p(x)]C: Let the polynomials in C be denoted by g1(x); g2(x); g3(x); and g4(x); respectively. It follows that p(x) = ¡7g1(x)+2g2(x)+g3(x): Similarly q(x) = 13g1(x)¡4g2(x)+g4(x) and r(x) =¡7g1(x)+ 3g2(x)¡2g3(x)+ g4(x): 168 CHAPTER 5. VECTOR SPACES AND LINEAR TRANSFORMATIONS 10. The transition matrix is P = 2 4 1 0 0 0 1 1 0 0 1 3 5: p(x) =¡3+6x+ x(x¡1);q(x) = 8¡4x +2x(x¡1);r(x) =¡5+ x + x(x¡1): 11. Note that T(e1 ) = [2;1]T = 2e1 +e2 and T(e2 ) = [1;2]T = e1 +2e2 : Therefore the matrix of T with respect to B is the matrix Q1 = ? 2 1 1 2 ? : The transition matrix from B to C is the matrix P = ? ¡1 1 1 1 ? and P¡1 = ? ¡1=2 1=2 1=2 1=2 ? : By Theorem 24 the matrix of T with respect to C is the matrix Q2 given by Q2 = P¡1Q1P = ? 1 0 0 3 ? : 12. The matrix for T with respect to B is Q1 = 2 4 2 ¡1 ¡1 1 0 ¡1 ¡1 1 2 3 5: The transition matrix from C to B is P = 2 4 1 1 1 1 0 1 ¡1 1 0 3 5 and P¡1 = 2 4 1 ¡1 ¡1 1 ¡1 0 ¡1 2 1 3 5: By Theorem 24 the matrix of T with respect to C is the matrix Q2 = P¡1Q1P = 2 4 2 0 0 0 1 0 0 0 1 3 5: 13. The matrix of T with respect to B is Q1 = 2 66 4 ¡3 0 0 5 0 3 ¡5 0 0 0 ¡2 0 0 0 0 2 3 77 5: The transition matrix from C to B is P = 2 66 4 1 0 0 1 0 1 1 0 0 1 0 0 1 0 0 0 3 77 5 and P¡1 = 2 66 4 0 0 0 1 0 0 1 0 0 1 ¡1 0 1 0 0 ¡1 3 77 5: By Theorem 24 the matrix of T with respect to C is the matrix Q2 = P¡1Q1P = 2 66 4 2 0 0 0 0 ¡2 0 0 0 0 3 0 0 0 0 ¡3 3 77 5: 14. (a) Q = ? 4 3 ¡2 ¡3 ? : (b) If S = ? ¡3 1 1 ¡2 ? then S¡1QS = R where R = ? 3 0 0 ¡2 ? : (c) C =f¡3+ x;1¡2xg: 5.10. CHANGE OF BASIS AND DIAGONALIZATION 169 (d) P = ? ¡2=5 ¡1=5 ¡1=5 ¡3=5 ? : (e) [w1 ]B = [2;3]T so [w1 ]C = P[w1 ]B = [¡7=5;¡11=5]T: There- fore [T(w1 )]C = R[w1 ]C = [¡21=5;22=5]T: It follows that T(w1 ) = (¡21=5)[¡3+ x] + (22=5)[1 ¡ 2x] = 17 ¡ 13x: Similarly [T(w2 )]C = [3=5;4=5]T so T(w2 ) = (3=5)[¡3+ x]+(4=5)[1¡2x] = ¡1¡ x: Finally, [T(w3 )]C = [¡3=5;6=5]T so T(w3 ) = (¡3=5)[¡3+ x]+(6=5)[1¡2x] = 3¡3x: 15. (a) T(1) = 1; T(x) = 1+2x; and T(x2) = 4x +3x2: Therefore Q = 2 4 1 1 0 0 2 4 0 0 3 3 5: (b) Q has characteristic polynomial p(t) =¡(t¡1)(t¡2)(t¡3): Therefore Q has eigenvalues ?1 = 1; ?2 = 2; ?3 = 3: The corresponding eigenvectors are u1 = [1;0;0]T;u2 = [1;1;0]T and u3 = [2;4;1]T; respectively. If S = [u1 ;u2 ;u3 ] then S¡1QS = R where R = 2 4 1 0 0 0 2 0 0 0 3 3 5: (c) C = fv1 ;v2 ;v3 g where [vi ]B = ui ; the ith column of S: Thus [v1 ]B = [1;0;0]T so v1 = 1;[v2 ]B = [1;1;0]T so v2 = 1+ x;[v3 ]B = [2;4;1]T so v3 = 2+4x + x2: (d) 1 = v1 ; x = ¡v1 +v2 ; and x2 = 2v1 ¡4v2 +v3 so the transition matrix is P = 2 4 1 ¡1 2 0 1 ¡4 0 0 1 3 5: (e) [w1 ]B = [¡8;7;1]T and [w1 ]C = P[w1 ]B = [¡13;3;1]T: There- fore [T(w1 )]C = R[w1 ]C = [¡13;6;3]T: It follows that T(w1) = ¡13v1 +6v2 +3v3 =¡1+18x +3x2: Similarly [T(w2 )]C = [7;¡8;3]T so T(w2 ) = 7v1¡8v2 +3v3 = 5+4x +3x2: Finally, [T(w3)]C = [11;¡22;6]T so T(w3 ) = 11v1¡22v2 +6v3 = 1+2x +6x2: 16. (a) Q = 2 66 4 1 ¡1 0 0 0 2 ¡2 0 0 0 5 ¡3 0 0 0 10 3 77 5: (b) If S = 2 66 4 1 ¡1 1 ¡3 0 1 ¡4 27 0 0 6 ¡108 0 0 0 180 3 77 5 then S¡1QS = R where R = 2 66 4 1 0 0 0 0 2 0 0 0 0 5 0 0 0 0 10 3 77 5: 170 CHAPTER 5. VECTOR SPACES AND LINEAR TRANSFORMATIONS (c) C = ? ? 1 0 0 0 ? ; ? ¡1 1 0 0 ? ; ? 1 ¡4 6 0 ? ; ? ¡3 27 ¡108 180 ? : (d) P = 2 66 4 1 1 1=2 1=6 0 1 2=3 1=4 0 0 1=6 1=10 0 0 0 1=180 3 77 5: (e) T(w1 ) = ? ¡3 6 ¡3 10 ? ; T(w2 ) = ? 5 ¡8 5 0 ? ; T(w3 ) = ? 15 ¡14 ¡6 20 ? : 17. Let Q be the matrix of IV with respect to C and B: Since P is the matrix of IV with respect to B and C it follows from Theorem 20 in Section 4.9 that PQ is the matrix of IV ? IV = IV with respect to C: Thus PQ = I: Similarly, QP is the matix of IV with respect to B; so QP = I: Therefore Q = P¡1 and P is nonsingular. 18. (a) By assumption T(v) = ?v: Therefore Q[v]B = [T(v)]B = [?v]B = ?[v]B: (b) By assumption Qx = ?x and x= [v]B: Thus Qx= Q[v]B = [T(v)]B: It follows that [T(v)]B = ?x= ?[v]B = [?v]B: Therefore T(v) = ?v: 19. Let v be an eigenvector for T corresponding to ?. Then T 2(v) = T(T(v)) = T(?(v) = ?T(v) = ?(?v) = ?2v. 20. Suppose T is one to one and let v be a vector in V such that T(v) = 0v. Then T(v) = so v is in N(T). But N(T) = f g so v = . Therefore 0 is not an eigenvalue for T (since eigenvectors must be nonzero). Next assume that 0 is not an eigenvalue for T and let u be in N(T). Then T(u) = = 0u. Since 0 is not an eigenvalue, it must be the case that u = . Therefore N(T) =f g and T is one to one. 21. Suppose v is an eigenvector for T corresponding to ?. Thus T(v) = ?v and T¡1(?v) = v. But T¡1(?v) = ?T¡1(v) so it follows that T¡1(v) = ?¡1v: 5.11. SUPPLEMENTARY EXERCISES 171 5.11 Supplementary Exercises 1. V is not a vector space. For example, if A = ? 1 1 1 1 ? then 1A 6= A. 3. (a) For arbitrary c, ¡2cA1 ¡3cA2 +cA3 =O. In particular, with c = 1, A3 = 2A1 +3A2. (b) As in (a), p3(x) = 2p1(x)+3p2(x) (c) v3 = 2v1 +3v2. 4. (a) B = ?? 2 1 0 0 ? ; ? ¡3 0 1 0 ? ; ? ¡1 0 0 1 ? is one basis for W. (b) Set A = 2 ? 2 1 0 0 ? + ? ¡3 0 1 0 ? ¡2 ? ¡1 0 1 1 ? = ? 3 2 1 ¡2 ? . 5. (a) Sp(S) =fa + bx + cx2 : 7a¡3b¡5c = 0g: (b) q1(x), q3(x), q4(x) are in Sp(S). (c) A polynomial p(x) = a+bx+cx2 is in Sp(S) if and only if c = (7=5)a¡(3=5)b. Thus, p(x) = a[1+(7=5)x2]+ b[x¡(3=5)x2]. The set B =f1+(7=5)x2; x¡(3=5)x2g is one choice of a basis for Sp(S). Moreover, for this choice of B, [p(x)]B = [a; b]T. (d) [q1(x)]B = [5; 5]T; [q3(x)]B = [0; ¡5]T ; [q4(x)]B = [5; 0]T 6. (a) The dependence relation x1A1 + x2A2 + x3A3 + x4A4 + x5A5 = O has solution x1 = ¡x3 ¡2x5, x2 = x3 ¡3x5, x4 = ¡4x5, x3 and x5 arbitrary. fA1; A2; A4g is a basis for Sp(S). Setting x3 = 1 and x5 = 0 yields ¡A1 + A2 + A3 = O, so A3 = A1 ¡ A2. Setting, x3 = 0 and x5 = 1 gives¡2A1¡3A2¡4A4+A5 =O, so A5 = 2A1+3A2+4A4. (b) Using the same calculations as in (a),fp1(x); p2(x); p4(x)gis a basis for Sp(S), p3(x) = p1(x)¡p2(x), and p5(x) = 2p1(x)+3p2(x)+4p4(x). (c) ff1(x); f2(x); f3(x)g is a basis for Sp(S), f3(x) = f1(x)¡f2(x) and f5(x) = 2f1(x)+ 3f2(x)+4f4(x). 7. ?? 1 0 2 3 0 ¡1 ? ; ? 0 1 ¡1 2 0 3 ? ; ? 0 0 0 0 1 2 ? 8. fp1(x); p2(x); p5(x)g 9. A polynomial p(x) = a+bx+cx2 +dx3 is in Sp(S) if and only if a¡3b¡c+d = 0 and in this case q(x) = (4a¡3b¡2c)p1(x)+(¡3a+3b+c)p2(x)+(¡2a+2b+c)p5(x): Therefore, q(x) is in Sp(S) and q(x) = 2p2(x)+ p5(x). 10. Sp(S) = ?? a b c d ? : a¡3b¡c + d = 0 . ?? 3 1 0 0 ? ; ? 1 0 1 0 ? ; ? ¡1 0 0 1 ? ; , is a basis for Sp(S). 172 CHAPTER 5. VECTOR SPACES AND LINEAR TRANSFORMATIONS 11. (a) The matrix of T is A. (b) rank(T) = 3 and nullity(T) = 3. (c) R(T)= 'p(x) = a + bx + cx2 + dx3 : a¡3b¡c + d = 0?. The set '3+ x;1+ x2;¡1+ x3? is a bais for R(T). (cf Exercise 10). (d) If B = ? 0 2 0 0 1 0 ? then T(B) = q(x) (cf. Exercise 9). (e) T ? a 11 a12 a13 a21 a22 a23 ?¶ = (x) if and only if a11 = ¡2a13 ¡3a21 + a23, a12 = a13 ¡ 2a21 ¡3a23, a22 =¡2a23, a13, a21, a23 arbitrary. Therefore, ?? ¡2 1 1 0 0 0 ? ; ? ¡3 ¡2 0 1 0 0 ? ; ? 1 ¡3 0 0 ¡2 1 ? is a basis for N(T). 12. ? a b ? = (b + a) ? 0 1 ? ¡ a ? ¡1 1 ? , so T ? a b ?¶ = (b + a)T ? 0 1 ?¶ ¡ aT ? ¡1 1 ?¶ = (b + a)(1+2x + x2)¡a(2¡x) = (b¡a)+(3a +2b)x +(a + b)x2. 13. If T(1) = ? u 1 u2 ? then T(a + bx + cx2) = aT(1) + bT(x) + cT(x2) = ? au 1 + b bu2 + c ? . In particular, T(a + bx + cx2) = ? b c ? is one such linear transformation. 14. (a) 2 4 1 ¡1 1 ¡4 0 1 1 3 1 0 2 ¡1 3 5 (b) R(T) = fu + vx + wx2 : u + v ¡ w = 0g. Moreover, if q(x) = u + vx + wx2 is in R(T) then T ? u + v ¡2c + d v ¡c¡3d c d ?¶ = q(x) for arbitrary c and d. If q1(x) = 1+ x2 and q2(x) = x + x2, then S =fq1(x); q2(x)g is a basis for R(T). (c) If A1 = ? 1 0 0 0 ? and A2 = ? 1 1 0 0 ? then T(A1) = q1(x) and T(A2) = q2(x). (d) N(T)= ?? a b c d ? : a =¡2c + d and b =¡c¡3d; c and d arbitrary . Therefore, if A3 = ? ¡2 ¡1 1 0 ? and A4 = ? 1 ¡3 0 1 ? then B2 =fA3; A4g is a basis for N (T). 5.12. CONCEPTUAL EXERCISES 173 5.12 Conceptual Exercises 1. True. u = a¡1(au) = a¡1(av) = v. 2. True. (a¡b)v = and v 6= , so a¡b = 0. 3. False. Each vector v in V has a unique inverse ¡v in V , but as v varies, so does ¡v. 4. False. If n = 1 then p(x) = 1¡x and q(x) = 1+ x are in V but p(x)+ q(x) is not in V . 5. True. Every basis for W is also a basis for V . 6. True. If dim(W) = k then a basis B for W is a linearly independent susbset of V containing k vectors. Therefore, k ? n. 7. True. a = for every nonzero scalar a. 8. True. 9. False. In R2 let S1 =f[1;0]T;[0;1]Tg and S2f[1;0]T;[0;1]T;[1;1]Tg. 10. True. Since V = Sp(S1), dim(V ) ? k. Since S2 is a linearly independent subset of V , l ?dim(V ). 11. u = (1=2)(u+v)+(1=2)(u¡v) and v = (1=2)(u+v)¡(1=2)(u¡v) Chapter 6 Determinants 6.1 Introduction (No exercises) 6.2 Cofactor Expansion of Determinants 1. det(A) = 1(1)¡3(2) =¡5: 2. det(A) =¡31: 3. det(A) = 2(8)¡4(4) = 0;x= a ? ¡2 1 ? ; a 6= 0: 4. det(A) = 2: 5. det(A) = 4(7)¡3(1) = 25: 6. det(A) = 3: 7. det(A) = 4(1)¡1(¡2) = 6: 8. det(A) = 0;x= a ? ¡3 1 ? ; a 6= 0: 9. A11 = (¡1)2 flfl flfl 1 31 1 flfl flfl =¡2;A12 = (¡1)3 flfl flfl 0 32 1 flfl flfl = 6; A13 = (¡1)4 flfl flfl 0 12 1 flfl flfl =¡2;A33 = (¡1)6 flfl flfl 1 20 1 flfl flfl = 1: 10. A11 =¡2;A12 = 4;A13 = 1; A33 =¡4: 11. A11 = flfl flfl 2 22 1 flfl flfl =¡2;A12 =¡ flfl flfl ¡1 23 1 flfl flfl = 7; 176 CHAPTER 6. DETERMINANTS A13 = flfl flfl ¡1 23 2 flfl flfl =¡8;A33 = flfl flfl 2 ¡1¡1 2 flfl flfl = 3: 12. A11 =¡1;A12 = 3;A13 =¡1;A33 = 0: 13. A11 = flfl flfl 1 01 3 flfl flfl = 3;A12 =¡ flfl flfl 2 00 3 flfl flfl =¡6; A13 = flfl flfl 2 10 1 flfl flfl = 2;A33 = flfl flfl ¡1 12 1 flfl flfl =¡3: 14. A11 = 6;A12 =¡8;A13 = 0;A33 = 4: 15. det(A) = A11 +2A12 + A13 =¡2+2(6)+(¡2) = 8: 16. det(A) = 14: 17. det(A) = 2A11 ¡A12 +3A13 = 2(¡2)¡7+3(¡8) =¡35: 18. det(A) = 1: 19. det(A) =¡A11 + A12 ¡A13 =¡3¡6¡2 =¡11: 20. det(A) = 8: 21. det(A) = 2 flfl flfl flfl 0 0 1 1 2 0 1 1 2 flfl flfl flfl+(¡1) flfl flfl flfl 3 0 1 2 2 0 3 1 2 flfl flfl flfl+(¡1) flfl flfl flfl 3 0 1 2 1 0 3 1 2 flfl flfl flfl+ 2(¡1) flfl flfl flfl 3 0 0 2 1 2 3 1 1 flfl flfl flfl = 2 flfl flfl 1 21 1 flfl flfl+(¡1) ? 3 flfl flfl 2 01 2 flfl flfl+ flfl flfl 2 23 1 flfl flfl ? + (¡1) ? 3 flfl flfl 1 01 2 flfl flfl+ flfl flfl 2 13 1 flfl flfl ? +(¡2)(3) flfl flfl 1 21 1 flfl flfl =¡9: 22. det(A) = 2: 23. det(A) = 2 flfl flfl flfl 3 1 2 1 2 1 3 1 4 flfl flfl flfl+2 flfl flfl flfl 1 3 2 0 1 1 0 3 4 flfl flfl flfl = 2 ? 3 flfl flfl 2 11 4 flfl flfl¡ flfl flfl 1 13 4 flfl flfl+2 flfl flfl 1 23 1 flfl flfl ? + 2 ? flfl flfl 1 13 4 flfl flfl¡3 flfl flfl 0 10 4 flfl flfl+2 flfl flfl 0 10 3 flfl flfl ? = 22: 24. det(A) = 4: 6.2. COFACTOR EXPANSION OF DETERMINANTS 177 25. det(A) = a11A11 + a12A12 + a13A13 = (1)(10)+(3)(5)+(2)(¡10) = 5; a21A21 + a22A22 + a23A23 = (¡1)(¡5)+(4)(¡1)+(1)(4) = 5; a31A31+a32A32+a33A33 = (2)(¡5)+(2)(¡3)+ (3)(7) = 5. 26. det(A) = a11A11 + a12A12 + a13A13 = (2)(¡7)+(4)(0)+(1)(7) = ¡7; a21A21 + a22A22 + a23A23 = (3)(¡5)+(1)(2)+(3)(2) = ¡7; a31A31 + a32A32 + a33A33 = (2)(11)+(3)(¡3)+ (2)(¡10) =¡7. 27. a11A21 + a12A22 + a13A23 = (1)(¡5)+(3)(¡1)+(2)(4) = 0; a11A31 + a12A32 + a13A33 = (1)(¡5)+(3)(¡1)+(2)(7) = 0. 28. a11A21 + a12A22 + a13A23 = (2)(¡5) + (4)(2) + (1)(2) = 0; a11A31 + a12A32 + a13A33 = (2)(11)+(4)(¡3)+(1)(¡10) = 0. 29. C = 2 4 10 5 ¡10 ¡5 ¡1 4 ¡5 ¡3 7 3 5, CTA = 2 4 5 0 0 0 5 0 0 0 5 3 5 = [det(A)]I. So A¡1 = (1=5)CT = [1=det(A)]CT. 30. C = 2 4 ¡7 0 7 ¡5 2 2 11 ¡3 ¡10 3 5, CTA = 2 4 ¡7 0 0 0 ¡7 0 0 0 ¡7 3 5 = [det(A)]I. So A¡1 = ¡(1=7)CT = [1=det(A)]CT. 31. det(A) =¡a12 flfl flfl 0 a230 a 33 flfl flfl+ a13 flfl flfl 0 a220 a 32 flfl flfl = 0: 32. det(U) = u11 flfl flfl flfl u22 u23 u24 0 u33 a34 0 0 u44 flfl flfl flfl¡u12 flfl flfl flfl 0 u23 u24 0 u33 u34 0 0 u44 flfl flfl flfl+ u13 flfl flfl flfl 0 u22 u24 0 0 u34 0 0 u44 flfl flfl flfl¡u14 flfl flfl flfl 0 u22 u23 0 0 u33 0 0 0 flfl flfl flfl = u11 flfl flfl flfl u22 u23 u24 0 u33 u34 0 0 u44 flfl flfl flfl = u11u22 flfl flfl u33 u340 u 44 flfl flfl = u11u22u33u44: 33. AT = ? a 11 a21 a12 a22 ? so det(AT) = a11a22 ¡a21a12 = det(A): 34. (a) If A ispositivedeflnitethen0 < e1TAe1= a11: Ifx= [u; v]T andx6= then0 < xTAx = a11u2 +2a12uv+a22v2 (since A is symmetric a12 = a21 ). In particular if u = a12 and v = ¡a11 Then 0 < a11a212 ¡2a212a11 + a22a211 = a11(a11a22 ¡ a212) = a11 det(A): It follows that det(A) > 0: 178 CHAPTER 6. DETERMINANTS (b) Suppose a11 > 0 and det(A) > 0: For x= [u; v]T we have a11(xTAx) = a211u2 +2a11a12uv + a11a22v2 = (a11u + a12v)2+ v2(a11a22 ¡a212) = (a11u + a12v)2 + v2 det(A): For x6= it follows that xTAx > 0: 35. (a) For n = 3 and n = 4; H(n) = n!=2: For some integer k ? 4 suppose we have seen that H(k) = k!=2: If A is a ((k +1) x (k +1)) matrix then det(A) can be obtained by evaluating k +1 (k x k) determinants. Thus the number of (2 x 2) determinants in the expansion of det(A) is (k +1)H(k) = (k +1)!=2: It follows by induction that H(n) = n!=2 for every positive integer n; n ?2: (b) Note that evaluating a single (2 x 2) determinant requires 3 operations, two multi- plications and one subtraction. n H(n) Timerequired ¡¡ ¡¡¡¡¡¡ ¡¡¡¡¡¡¡ 2 1 3seconds 5 60 3minutes 10 1;814;400 1512hours: 6.3 Elementary Operations and Determinants 1. flfl flfl flfl 1 2 1 2 0 1 1 ¡1 1 flfl flfl flfl C2 ¡2C1 C3 ¡C1 = flfl flfl flfl 1 0 0 2 ¡4 ¡1 1 ¡3 0 flfl flfl flfl = flfl flfl ¡4 ¡1¡3 0 flfl flfl =¡3: 2. flfl flfl flfl 2 4 ¡2 0 2 3 1 1 2 flfl flfl flfl C2 ¡2C1 C3 + C1 = flfl flfl flfl 2 0 0 0 2 3 1 ¡1 3 flfl flfl flfl = 2 flfl flfl 2 3¡1 3 flfl flfl = 18: 3. flfl flfl flfl 0 1 2 3 1 2 2 0 3 flfl flfl flfl C1 $ C2 = ¡ flfl flfl flfl 1 0 2 1 3 2 0 2 3 flfl flfl flfl C3 ¡2C1 = ¡ flfl flfl flfl 1 0 0 1 3 0 0 2 3 flfl flfl flfl = ¡ flfl flfl 3 02 3 flfl flfl =¡9: 4. flfl flfl flfl 2 2 4 1 0 1 2 1 2 flfl flfl flfl = flfl flfl flfl 2 0 0 1 ¡1 ¡1 2 ¡1 ¡2 flfl flfl flfl = 2: 5. flfl flfl flfl 0 1 3 2 1 2 1 1 2 flfl flfl flfl C1 $ C2 = ¡ flfl flfl flfl 1 0 3 1 2 2 1 1 2 flfl flfl flfl C3 ¡3C1 = 6.3. ELEMENTARY OPERATIONS AND DETERMINANTS 179 ¡ flfl flfl flfl 1 0 0 1 2 ¡1 1 1 ¡1 flfl flfl flfl =¡ flfl flfl 2 ¡11 ¡1 flfl flfl = 1: 6. flfl flfl flfl 1 1 1 2 1 2 3 0 2 flfl flfl flfl C2 ¡C1 C3 ¡C1 = flfl flfl flfl 1 0 0 2 ¡1 0 3 ¡3 ¡1 flfl flfl flfl = 1: 7. det(B) =¡2det(A) =¡6: 8. det(B) =¡6det(A) =¡18: 9. det(B) = det(A) = 3: 10. det(B) = 2det(A) = 6: 11. det(B) = det(A) = 3: 12. det(B) = 4det(A) = 12: 13. flfl flfl flfl flfl 1 0 0 0 2 0 0 3 1 1 0 1 1 4 2 2 flfl flfl flfl flfl C2 $ C4 = ¡ flfl flfl flfl flfl 1 0 0 0 2 3 0 0 1 1 0 1 1 2 2 4 flfl flfl flfl flfl C3 $ C4 = flfl flfl flfl flfl 1 0 0 0 2 3 0 0 1 1 1 0 1 2 4 2 flfl flfl flfl flfl = 6: 14. flfl flfl flfl flfl 0 0 2 0 0 0 1 3 0 4 1 3 2 1 5 6 flfl flfl flfl flfl =¡ flfl flfl flfl flfl 2 0 0 0 1 3 0 0 1 3 4 0 5 6 1 2 flfl flfl flfl flfl =¡48: 15. flfl flfl flfl flfl 0 1 0 0 0 2 0 3 2 1 0 6 3 2 2 4 flfl flfl flfl flfl C1 $ C2 C3 $ C4 = flfl flfl flfl flfl 1 0 0 0 2 0 3 0 1 2 6 0 2 3 4 2 flfl flfl flfl flfl C2 $ C3 = ¡ flfl flfl flfl flfl 1 0 0 0 2 3 0 0 1 6 2 0 2 4 3 2 flfl flfl flfl flfl =¡12: 180 CHAPTER 6. DETERMINANTS 16. flfl flfl flfl flfl 1 2 0 3 2 5 1 1 2 0 4 3 0 1 6 2 flfl flfl flfl flfl = flfl flfl flfl flfl 1 0 0 0 2 1 0 0 2 ¡4 8 ¡23 0 1 5 7 flfl flfl flfl flfl = flfl flfl 8 ¡235 7 flfl flfl = 171: 17. flfl flfl flfl flfl 2 4 ¡2 ¡2 1 3 1 2 1 3 1 3 ¡1 2 1 2 flfl flfl flfl flfl C2 ¡2C1 C3 + C2 C4 + C1 = flfl flfl flfl flfl 2 0 0 0 1 1 2 3 1 1 2 4 ¡1 4 0 1 flfl flfl flfl flfl C3 ¡2C2 C4 ¡3C2 = flfl flfl flfl flfl 2 0 0 0 1 1 0 0 1 1 0 1 ¡1 4 ¡8 ¡11 flfl flfl flfl flfl = 2 flfl flfl 0 1¡8 ¡11 flfl flfl = 16: 18. flfl flfl flfl flfl 1 1 2 1 0 1 4 1 2 1 3 0 2 2 1 2 flfl flfl flfl flfl = flfl flfl flfl flfl 1 0 0 0 0 1 0 0 2 ¡1 3 ¡1 2 0 ¡3 0 flfl flfl flfl flfl =¡3: 19. flfl flfl flfl flfl 1 2 0 3 2 5 1 1 2 0 4 3 0 1 6 2 flfl flfl flfl flfl R2 ¡2R1 R3 ¡2R1 = flfl flfl flfl flfl 1 2 0 3 0 1 1 ¡5 0 ¡4 4 ¡3 0 1 6 2 flfl flfl flfl flfl R3 +4R4 R4 ¡R2 = flfl flfl flfl flfl 1 2 0 3 0 1 1 ¡5 0 0 8 ¡23 0 0 5 7 flfl flfl flfl flfl = flfl flfl 8 ¡235 7 flfl flfl = 171: 20. flfl flfl flfl flfl 2 4 ¡2 ¡2 1 3 1 2 1 3 1 3 ¡1 2 1 2 flfl flfl flfl flfl = 2 flfl flfl flfl flfl 1 2 ¡1 ¡1 1 3 1 2 1 3 1 3 ¡1 2 1 2 flfl flfl flfl flfl = 2 flfl flfl flfl flfl 1 2 ¡1 ¡1 0 1 2 3 0 0 0 1 0 0 ¡8 ¡11 flfl flfl flfl flfl = 16: 21. flfl flfl flfl flfl 1 1 2 1 0 1 4 1 2 1 3 0 2 2 1 2 flfl flfl flfl flfl R3 ¡2R1 R4 ¡2R1 = flfl flfl flfl flfl 1 1 2 1 0 1 4 1 0 ¡1 ¡1 ¡2 0 0 ¡3 0 flfl flfl flfl flfl R3 + R2 = 6.3. ELEMENTARY OPERATIONS AND DETERMINANTS 181 flfl flfl flfl flfl 1 1 2 1 0 1 4 1 0 0 3 ¡1 0 0 ¡3 0 flfl flfl flfl flfl = flfl flfl 3 ¡1¡3 0 flfl flfl =¡3: 22. If A = ? 1 0 0 0 flfl flfl and B = ? 0 0 0 1 ? then det(A) = det(B) = 0 whereas det(A+B) = 1: If A = ? 1 0 0 0 ? and B = ? 0 1 0 0 ? then det(A) = det(B) = 0 and det(A + B) = 0 = det(A)+det(B): 23. flfl flfl flfl a +1 a +4 a +7 a +2 a +5 a +8 a +3 a +6 a +9 flfl flfl flfl 8 < : C3 ¡C2 = C2 ¡C1 9 = ; flfl flfl flfl a +1 3 3 a +2 3 3 a +3 3 3 flfl flfl flfl = 0; flfl flfl flfl a 4a 7a 2a 5a 8a 3a 6a 9a flfl flfl flfl 8 < : C3 ¡C2 = C2 ¡C1 9 = ; flfl flfl flfl a 3a 3a 2a 3a 3a 3a 3a 3a flfl flfl flfl = 0; flfl flfl flfl a a4 a7 a2 a5 a8 a3 a6 a9 flfl flfl flfl = (a)(a4)(a7) flfl flfl flfl 1 1 1 a a a a2 a2 a2 flfl flfl flfl = 0. 24. (a) Set B = [B1;B2;B3]. Then AB = [AB1; AB2; AB3] where AB1 = A 2 4 2 3 1 3 5 = 2A1 + 3A2 +A3, AB2 = A 2 4 0 ¡1 3 3 5 =¡A2 +3A3, AB3 = A 2 4 0 0 4 3 5 = 4A3. (b) det(AB) = det[2A1 +3A2 +A3;¡A2 +3A3;4A3] (C3=4) : = (4)det[2A1 +3A2 +A3;¡A2 +3A3;A3] (C2 ¡3C3) : = (4)det[2A1 +3A2 +A3;¡A2;A3] (¡C2) : = (¡4)det[2A1 +3A2 +A3;A2;A3] (C1 ¡C3) : = (¡4)det[2A1 +3A2;A2;A3] (C1 ¡3C2) : = (¡4)det[2A1;A2;A3] (C1=2) : = (¡8)det[A1;A2;A3] = (¡8)det(A): (c) det(B) =¡8, so det(AB) = det(A)det(B). 25. It follows from Theorem 3 (with c = 0 ) that if A has a zero column then det(A) = 0: Since the flrst column of U1j contains only zeros for 2? j ? n;det(U1j) = 0: 182 CHAPTER 6. DETERMINANTS 26. If U is a (2 x 2) upper triangular matrix then U = ? u 11 u12 0 u22 ? and det(U) = u11u22: Suppose we have seen that det(U) = u11u22¢¢¢ukk for a (kxk) upper triangular matrix U: If U is a ((k +1)x(k+1)) upper triangular matrix then det(U) = u11U11 + u12U12 +¢¢¢+ u1;k+1U1;k+1 = u11U11: But U11 = flfl flfl flfl flfl fl u22 u23 ¢¢¢ u2;k+1 0 u33 ¢¢¢ u3;k+1 ... 0 0 ¢¢¢ uk+1;k+1 flfl flfl flfl flfl fl = u22u33¢¢¢uk+1;k+1 by assumption. Thus det(U) = u11u22¢¢¢uk+1;k+1: It follows by induction that if U = (uij) is an (nxn) upper triangular matrix, n ?2; then det(U) = u11u22¢¢¢unn: 27. First note that flfl flfl flfl x y 1 x1 y1 1 x2 y2 1 flfl flfl flfl = 0 is a linear equation in x and y: It follows from Theorem 5 that x = x1; y = y1 and x = x2; y = y2 are solutions. Consequently the equation describes the line through the points (x1; y1) and (x2; y2): 28. Consider the case represented by the flgure below. Clearly area(ABC) = area(ADEC)+area(CEFB)¡ area(ADFB): Therefore area(ABC) = (1=2)(x3 ¡x1)(y1 +y3)+ (1=2)(x2 ¡x3)(y2 +y3)¡ (1=2)(x2 ¡x1)(y1 + y2) = (1=2)[x1y2 ¡x1y3 ¡x2y1 + x2y3 + x3y1 ¡x3y2] = (1=2) flfl flfl flfl x1 y1 1 x2 y2 1 x3 y3 1 flfl flfl flfl: 29. Let x= [x1; x2; x3]T and y= [y1; y2; y3]T and let B = xyT = [B1;B2;B3] where Bj = [xjy1; xjy2; xjy3]T: Then A = [B1 + e1;B2 + e2;B3 + e3]T: Repeated applications of Theorem 4 yield; det(A) = det[B1;B2;B3]+det[B1;B2;e3]+ 6.4. CRAMER?S RULE 183 det[B1;e2;B3 ]+det[e1;B2;B3 ]+det[B1;e2;e3 ]+ det[e1;B2;e3 ]+det[e1;e2;B3 ]+det[e1;e2;e3 ]: Since each Bj; 1? j ?3; is a scalar multiple of y; Theorem 5 implies that det(A) = det[B1;e2 ;e3]+det[e1;B2;e3 ]+det[e1;e2;B3 ]+ det[e1;e2;e3 ] = x1y1 + x2y2 + x3y3 +1 = 1+yTx: 30. flfl flfl flfl 1 a a2 1 b b2 1 c c2 flfl flfl flfl C3 ¡aC2 C2 ¡aC1 = flfl flfl flfl 1 0 0 1 b¡a b(b¡a) 1 c¡a c(c¡a) flfl flfl flfl = (b¡a)(c¡a) flfl flfl 1 b1 c flfl flfl = (b¡a)(c¡a)(c¡b): 31. flfl flfl flfl flfl 1 a a2 a3 1 b b2 b3 1 c c2 c3 1 d d2 d3 flfl flfl flfl flfl C4 ¡aC3 C3 ¡aC2 C2 ¡aC1 = flfl flfl flfl flfl 1 0 0 0 1 b¡a b(b¡a) b2(b¡a) 1 c¡a c(c¡a) c2(c¡a) 1 d¡a d(d¡a) d2(d¡a) flfl flfl flfl flfl = (b¡a)(c¡a)(d¡a) flfl flfl flfl 1 b b2 1 c c2 1 d d2 flfl flfl flfl = (b¡a)(c¡a)(d¡a)(c¡b)(d¡b)(d¡c): 32. Write A = [A1;A2; : : : ;An]: Then cA = [cA1;cA2; : : : ;cAn]: By Theorem 3, det(cA) = c det[A1;cA2; : : : ;cAn] = c2 det[A1;A2; : : : ;cAn] =¢¢¢= cn det[A1;A2; : : : ;An] = cn det(A): 6.4 Cramer?s Rule 1. flfl flfl flfl 0 1 3 1 2 1 3 4 1 flfl flfl flfl fC1 $ C2g = ¡ flfl flfl flfl 1 0 3 2 1 1 4 3 1 flfl flfl flfl fC3 ¡3C1g = ¡ flfl flfl flfl 1 0 0 2 1 ¡5 4 3 ¡11 flfl flfl flfl fC3 +5C2g = ¡ flfl flfl flfl 1 0 0 2 1 0 4 3 4 flfl flfl flfl =¡4: 2. flfl flfl flfl 1 2 1 2 4 3 2 1 3 flfl flfl flfl =¡ flfl flfl flfl 1 0 0 2 1 0 2 1 ¡3 flfl flfl flfl = 3: 184 CHAPTER 6. DETERMINANTS 3. flfl flfl flfl 2 2 4 1 3 4 ¡1 2 1 flfl flfl flfl ? C 2 ¡C1 C3 ¡2C1 = flfl flfl flfl 2 0 0 1 2 2 ¡1 3 3 flfl flfl flfl fC3 ¡C2g = flfl flfl flfl 2 0 0 1 2 0 ¡1 3 0 flfl flfl flfl = 0: 4. flfl flfl flfl 1 0 1 2 1 1 1 2 1 flfl flfl flfl = 2 flfl flfl flfl 1 0 0 0 1 0 0 0 1 flfl flfl flfl = 2: 5. flfl flfl flfl 1 0 ¡2 3 1 3 0 1 2 flfl flfl flfl fC3 +2C1g = flfl flfl flfl 1 0 0 3 1 9 0 1 2 flfl flfl flfl fC3 ¡9C2g = flfl flfl flfl 1 0 0 3 1 0 0 1 ¡7 flfl flfl flfl f(¡1=7)C3g = ¡7 flfl flfl flfl 1 0 0 3 1 0 0 1 1 flfl flfl flfl fC2 ¡C3g = ¡7 flfl flfl flfl 1 0 0 3 1 0 0 0 1 flfl flfl flfl fC1 ¡3C2g = ¡7 flfl flfl flfl 1 0 0 0 1 0 0 0 1 flfl flfl flfl =¡7: 6. A is singular. 7. (a) det(AB) = det(A)det(B) = (2)(3) = 6 (b) det(AB2) = det(A)[det(B)]2 = (2)(9) = 18 (c) det(A¡1B) = det(B)=det(A) = 3=2 (d) det(2A¡1) = 8det(A¡1) = 8=det(A) = 4 (e) det(2A)¡1 = det((1=2)A¡1) = (1=8)det(A¡1) = 1=[8det(A)] = 1=16. 8. Both matrices have determinant sin2 +cos2 = 1, so the matrices are nonsingular for all values of . 9. det(B(?)) = 2?¡?2 = ?(2¡?);B(?) is singular provided ? = 0 or ? = 2: 10. det(B(?)) = ?2 ¡1;B(?) is singular for ? =§1: 11. det(B(?)) = 4¡?2;B(?) is singular for ? =§2: 12. det(B(?)) = 2(1¡?)(3¡?);B(?) is singular provided ? = 1 or ? = 3: 13. det(B(?)) = (?¡1)2(? +2);B(?) is singular provided ? = 1 or ? =¡2: 14. det(B(?)) = ?(?¡3)(? +1);B(?) is singular provided ? = 0; ? = 3; or ? =¡1: 15. det(A) = flfl flfl 1 11 ¡1 flfl flfl =¡2;det(B1) = flfl flfl 3 1¡1 ¡1 flfl flfl =¡2; 6.4. CRAMER?S RULE 185 det(B2) = flfl flfl 1 31 ¡1 flfl flfl =¡4: x1 = det(B1)=det(A) = 1;x2 = det(B2)=det(A) = 2: 16. x1 = x2 = 1: 17. det(A) = flfl flfl flfl 1 ¡2 1 1 0 1 1 ¡2 0 flfl flfl flfl =¡2;det(B1) = flfl flfl flfl ¡1 ¡2 1 3 0 1 0 ¡2 0 flfl flfl flfl =¡8; det(B2) = flfl flfl flfl 1 ¡1 1 1 3 1 1 0 0 flfl flfl flfl =¡4;det(B3) = flfl flfl flfl 1 ¡2 ¡1 1 0 3 1 ¡2 0 flfl flfl flfl = 2: x1 = det(B1)=det(A) = 4;x2 = det(B2)=det(A) = 2; x3 = det(B3)=det(A) =¡1: 18. x1 =¡1; x2 = 0; x3 = 3: 19. det(A) = flfl flfl flfl flfl 1 1 1 ¡1 0 1 ¡1 1 0 0 1 ¡1 0 0 1 2 flfl flfl flfl flfl = 3;det(B1) = flfl flfl flfl flfl 2 1 1 ¡1 1 1 ¡1 1 0 0 1 ¡1 3 0 1 2 flfl flfl flfl flfl = 3; det(B2) flfl flfl flfl flfl 1 2 1 ¡1 0 1 ¡1 1 0 0 1 ¡1 0 3 1 2 flfl flfl flfl flfl = 3;det(B3) = flfl flfl flfl flfl 1 1 2 ¡1 0 1 1 1 0 0 0 ¡1 0 0 3 2 flfl flfl flfl flfl = 3; det(B4) = flfl flfl flfl flfl 1 1 1 2 0 1 ¡1 1 0 0 1 0 0 0 1 3 flfl flfl flfl flfl = 3: x1 = det(B1)=det(A) = 1;x2 = det(B2)=det(A) = 1; x3 = det(B3)=det(A) = 1;x4 = det(B4)=det(A) = 1: 20. x1 = 2; x2 = 1; x3 = 0: 21. det(A) = flfl flfl flfl 1 1 1 0 1 1 0 0 1 flfl flfl flfl = 1;det(B1) = flfl flfl flfl a 1 1 b 1 1 c 0 1 flfl flfl flfl = a¡b; det(B2) = flfl flfl flfl 1 a 1 0 b 1 0 c 1 flfl flfl flfl = b¡c;det(B3) = flfl flfl flfl 1 1 a 0 1 b 0 0 c flfl flfl flfl = c: x1 = a¡b; x2 = b¡c; x3 = c: 186 CHAPTER 6. DETERMINANTS 22. det(A)2 = det(A2) = det(I) = 1 so det(A) =§1: 23. Suppose ^B is produced by interchanging the ith and the jth columns of B: Thus A ^B = A[B1; : : : ;Bj; : : : ;Bi; : : : ;Bn] = [AB1 ; : : : ; ABj ; : : : ; ABi : : : : ; ABn ] = ^C where ^C is obtained by interchanging the ith and jth columns of C = AB: Suppose ^B is produced by replacing Bi with Bi +aBj : Then A ^B = [AB1 ; : : : ; ABi+aABj ; : : : ; ABn] = ^C where C = AB and ^C is obtained by adding a times the jth column of C to the ith column of C: Finally suppose ^B is produced by replacing Bi with aBi : Then A ^B = [AB1 ; : : : ; aABi ; : : : ; ABn ] = ^C where C = AB and ^C is obtained by multiplying the ith column of C by a:. 24. det(AB) = det(A)det(B) = det(B)det(A) = det(BA): 25. det(B) = det(SAS¡1) = det(S)det(A)det(S¡1) = det(S)det(S)¡1 det(A) = det(A): 26. Either det(A) = 0 or det(A) = 1: 27. det(A5) = det(A)5 = 35 = 243: 28. Set x = det(A). Then x¡1 = det(A¡1). Moreover, since A and A¡1 both have only integer entries, both det(A) and det(A¡1) are integers. It follows that x =§1. 29. det(Q) =¡33 = (¡3)(11) = flfl flfl 1 22 1 flfl flfl flfl flfl flfl 1 2 2 3 5 1 1 4 1 flfl flfl flfl: 30. If A = ? 1 2 2 1 ? and B = 2 4 1 2 2 3 5 1 1 4 1 3 5 then det(A)det(B) = (¡3)(11) =¡33 = det(Q). 6.5 Applications of Determinants 1. flfl flfl flfl 1 2 1 2 3 2 ¡1 4 1 flfl flfl flfl ? R 2 ¡2R1 R3 + R1 = flfl flfl flfl 1 2 1 0 ¡1 0 0 6 2 flfl flfl flfl fR3 +6R2g = flfl flfl flfl 1 2 1 0 ¡1 0 0 0 2 flfl flfl flfl =¡2: 6.5. APPLICATIONS OF DETERMINANTS 187 2. flfl flfl flfl 0 3 1 1 2 1 2 ¡2 2 flfl flfl flfl =¡ flfl flfl flfl 1 2 1 0 3 1 0 0 2 flfl flfl flfl =¡6: 3. flfl flfl flfl 0 1 3 1 2 2 3 1 0 flfl flfl flfl fR1 $ R2g = ¡ flfl flfl flfl 1 2 2 0 1 3 3 1 0 flfl flfl flfl fR3 ¡3R1g = ¡ flfl flfl flfl 1 2 2 0 1 3 0 ¡5 ¡6 flfl flfl flfl fR3 +5R2g = flfl flfl flfl 1 2 2 0 1 3 0 0 9 flfl flfl flfl =¡9: 4. flfl flfl flfl 1 0 1 0 2 4 3 2 1 flfl flfl flfl = flfl flfl flfl 1 0 1 0 2 4 0 0 ¡6 flfl flfl flfl =¡12: 5. adj(A) = ? 4 ¡2 ¡3 1 ? ;det(A) =¡2;A¡1 = (¡1=2) ? 4 ¡2 ¡3 1 ? : 6. adj(A) = ? d ¡b ¡c a ? ;det(A) = ad¡bc; A¡1 = [1=(ad¡bc)] ? d ¡b ¡c a ? : 7. adj(A) = 2 4 0 1 ¡1 ¡2 1 0 1 ¡1 1 3 5;det(A) = 1;A¡1 = adj(A): 8. adj(A) = 2 4 ¡1 ¡1 1 ¡3 2 ¡2 3 ¡2 ¡3 3 5;det(A) =¡5; A¡1 = (¡1=5) 2 4 ¡1 ¡1 1 ¡3 2 ¡2 3 ¡2 ¡3 3 5: 9. adj(A) = 2 4 ¡4 2 0 1 0 ¡1 1 ¡2 1 3 5;det(A) =¡2; A¡1 = (¡1=2) 2 4 ¡4 2 0 1 0 ¡1 1 ¡2 1 3 5: 188 CHAPTER 6. DETERMINANTS 10. adj(A) = 2 4 1 ¡2 1 0 1 ¡2 0 0 1 3 5;det(A) = 1;A¡1 = adj(A): 11. W(x) = flfl flfl flfl 1 x x2 0 1 2x 0 0 2 flfl flfl flfl = 2: Since W(0) = 2 the given set of functions is linearly independent. 12. W(x) = flfl flfl flfl ex e2x e3x ex 2e2x 3e3x ex 4e2x 9e3x flfl flfl flfl = e6x flfl flfl flfl 1 1 1 1 2 3 1 4 9 flfl flfl flfl = 2e6x: Since W(0) = 6 6= 0; the set of functions is linearly independent. 13. W(x) = flfl flfl flfl 1 cos2 x sin2 x 0 ¡2cosxsinx 2cosxsinx 0 2sin2 x¡2cos2 x 2cos2 x¡2sin2 x flfl flfl flfl = 4cosxsinxcos2x flfl flfl flfl 1 cos2 x sin2 x 0 ¡1 1 0 ¡1 1 flfl flfl flfl = 0. The Wronskian gives no information, but 1¡ cos2 x¡sin2 x = 0 so the set is linearly dependent. 14. W(x) = 4sinxcos2x ¡2sin2xcosx:W(?=4) 6= 0 so the set of functions is linearly inde- pendent. 15. Note that x j x j= x2 for x ? 0 and x j x j= ¡x2 for x < 0: Therefore W(x) =fl flfl fl x2 x2 2x 2x flfl flfl = 0 if x ? 0 and W(x) = flfl flfl x2 ¡x22x ¡2x flfl flfl when x < 0: Since W(x) = 0 for all x;¡1 ? x ? 1; the Wronskian test is inconclusive. Thus suppose that c1x2 + c2x jxj= 0 for all x;¡1 ? x ? 1: Then for x = 1 we have c1 + c2 = 0 and for x = ¡1 we obtain c1 ¡c2 = 0: It follows that c1 = c2 = 0 and the set fx2; x jxjg is linearly independent. 16. W(x) = 0 for all x;¡1? x ?1: The set is linearly dependent since 3x2 ¡2(1+x2)+(2¡ x2) = 0: 17. The column operations C1 $ C2; C3 ¡ 3C1; C3 + 2C2 reduce A to the matrix L =2 4 1 0 0 2 1 0 2 2 ¡1 3 5: Therefore Q = E1E2E3 where E1 = 2 4 0 1 0 1 0 0 0 0 1 3 5; E2 = 2 4 1 0 ¡3 0 1 0 0 0 1 3 5; and E3 = 2 4 1 0 0 0 1 2 0 0 1 3 5: 6.5. APPLICATIONS OF DETERMINANTS 189 Multiplication yields Q = 2 4 0 1 2 1 0 ¡3 0 0 1 3 5: It is easily seen that det(Q) = det(QT) =¡1: 18. E1 = 2 4 0 1 0 1 0 0 0 0 1 3 5;E2 = 2 4 1 0 2 0 1 0 0 0 1 3 5;E3 = 2 4 1 0 0 0 1 ¡5 0 0 1 3 5; Q = E1E2E3 = 2 4 0 1 ¡5 1 0 2 0 0 1 3 5;AQ = L = 2 4 ¡1 0 0 3 1 0 2 1 0 3 5; det(Q) = det(QT) =¡1: 19. Thecolumnoperations C2¡2C1; C3+C1; C3+4C4 transform A to L = 2 4 1 0 0 3 ¡1 0 4 ¡8 ¡26 3 5: If E1 = 2 4 1 ¡2 0 0 1 0 0 0 1 3 5; E2 = 2 4 1 0 1 0 1 0 0 0 1 3 5; E3 = 2 4 1 0 0 0 1 4 0 0 1 3 5; and Q = E1E2E3 = 2 4 1 ¡2 ¡7 0 1 4 0 0 1 3 5 then AQ = L and det(Q) = det(QT) = 1: 20. E1 = 2 4 1 ¡2 0 0 1 0 0 0 1 3 5;E2 = 2 4 1 0 3 0 1 0 0 0 1 3 5;E3 = 2 4 1 0 0 0 1 4 0 0 1 3 5; Q = E1E2E3 = 2 4 1 ¡2 ¡5 0 1 4 0 0 1 3 5;AQ = L = 2 4 2 0 0 1 ¡1 0 3 ¡4 ¡6 3 5; det(Q) = det(QT) = 1: 21. det(A(x)) = x2 + 1 > 0 for all real x: adj(A(x)) = ? x ¡1 1 x ? so A¡1 = [1=(x2 + 1)] ? x ¡1 1 x ? : 22. det(A(x)) = x2 +2 > 0 for all real x: A¡1 = [1=(x2 +2)] ? 2 ¡x x 1 ? : 190 CHAPTER 6. DETERMINANTS 23. det(A(x)) = 4x2 +8 > 0 for all real x: adj(A) =2 4 x2 +4 ¡2x x2 2x 4 ¡2x x2 2x x2 +4 3 5 so A¡1 = [1=(4x2 +8)]adj(A): 24. det(A(x)) = 1 for all x:A¡1 = adj(A) = 2 4 sinx 0 ¡cosx 0 1 0 cosx 0 sinx 3 5: 25. det(L) = 1 so L¡1 = adj(L) = 2 4 1 0 0 ¡a 1 0 ac¡b ¡c 1 3 5:det(U) = 1 so U¡1 = adj(U) = 2 4 1 ¡a ac¡b 0 1 ¡c 0 0 1 3 5: 26. Let L = [lij] be a (4 x 4) nonsingular, lower-triangular matrix. Direct calculations show that adj(L) is also a lower-triangular matrix. By Theorem 14, L¡1 is lower-triangular. 27. Clearly each cofactor, Aij of A is an integer. Therefore A¡1 = adj(A) contains only integer entries. 28. (a) Let A = [A1; : : : ;An ]: By assumption E = [e1; : : : ;ej; : : : ;ei; : : : ;en]: Therefore AE = [Ae1 : : : ;Aej; : : : ;Aei; : : : ;Aen] = [A1; : : : ;Aj; : : : ;Ai; : : : ;An]: (b) Let E = [ers]: If su?ces to note that ers = esr when s 6= r: If (r; s) 6= (i; j) and (r; s)6= (j; i) then ers = esr = 0: But eij = eji = 1 so E is symmetric. 29. If A is an (nxn) skew symmetric matrix then det(AT) = det(¡A) = (¡1)n det(A): But det(A) = det(AT) so det(A) ¡ det(AT) = det(A) ¡ (¡1)n det(A) = : For n odd this implies that 2det(A) = 0: Therefore det(A) = 0 and A is singular. 30. Let x = det(A). Then x = det(AT) = det(A¡1) = 1=x. Thus, x2 = 1 and it follows that x =§1. 31. Set c = det(A). Since A is nonsingular, c 6= 0. Moreover, Adj(A) = det(A)A¡1 = cA¡1, so det[Adj(A)] = det(cA¡1) = cndet(A¡1) = cn=det(A) = cn=c = cn¡1. 32. (a) I = AA¡1 = A[ 1det(A) Adj(A)] = [ 1det(A)A]Adj(A). Therefore, [Adj(A)]¡1 = [1=det(A)]A. (b) A = (A¡1)¡1 = £1=det(A¡1)?Adj(A¡1) = det(A)Adj(A¡1). Therefore, Adj(A¡1) = [1=det(A)]A. 6.6. SUPPLEMENTARY EXERCISES 191 6.6 Supplementary Exercises 1. flfl flfl a11 a12a 21 a22 flfl flfl+ flfl flfl a11 a12b 21 b22 flfl flfl+ flfl flfl b11 b12a 21 a22 flfl flfl+ flfl flfl b11 b12b 21 b22 flfl flfl. 2. Note that if [An¡2; : : : ;A1] can be obtained from [A1; : : : ;An¡2] with m column inter- changes, then [A1;An¡1; : : : ;A2;An] can be obtained from [A1;A2; : : : ;An] with m col- umn interchanges. Thus, m+1 column interchanges yields [An;An¡1; : : : ;A1]. Therefore, if n = 4k or n = 4k + 1 then an even number of column interchanges is required and det(B) = det(A). If n = 2k or n = 2k +1, where k is odd, then an odd number of column interchanges is required and det(B) =¡det(A). 3. If x = det(A) then x3 = x so x = 0, x = 1, or x =¡1. 4. Let x = det(A). Then x 6= 0, x = det(AT), and if A is a (2£2) matrix, det(cA) = c2x. From x = c2x it follows that c =§1. Similarly, if A is a (3£3) matrix, c3 = 1 so c = 1. 5. AB = 2 4 0 0 2 0 2 0 2 0 0 3 5 6. Det(A) = a11A11 + a12A12 + a13A13 = ¡1 and A¡1 = (1=det(A))Adj(A) = ¡CT. There- fore, A = (¡CT)¡1 = 2 4 1 2 ¡1 3 1 4 2 2 1 3 5. 7. Det(B + I) = det[b+e1;b+e2; : : : ;b+en] = Pni=1 det(Ai)+det(I) = b1 +¢¢¢+ bn +1. 8. (x5 + x3)ex. 6.7 Conceptual Exercises 1. True. A is nonsingular so B = A¡1(AB) = A¡1(AC) = C. 2. True. Det(AB) = det(A)det(B) = det(B)det(A) = det(AB). 3. False. If A = In then det(cIn ¡A) = (c¡1)n. 4. False. Det(cA) = cndet(A). 5. True. 0 = det(Ak) = (det(A))k, so det(A) = 0. 6. True. 06= det(B) = det(A1)¢¢¢det(Am), so det(Ai)6= 0 for 1? i ? m. 7. True. If C = [Aij] is the cofactor matrix for A then C is symmetric. 192 CHAPTER 6. DETERMINANTS 8. True. A¡1 = Adj(A). 9. If x = det(A) and A2 =¡I then x2 =¡1, which is not possible. 10. A is nonsingular (cf. Exercise 6) so A¡1 exists. Thus I = A¡1IA = A¡1(AB)A = BA. 11. AT ¡cI = (A¡cI)T. 12. (a) Det(B¡1AB¡cI) = det[B¡1(A¡cI)B] = det(B¡1)det(A¡cI)det(B) = det(A¡cI). (b) Det(AB ¡cI) = det(B¡1(BA)B ¡cI) = det(BA¡cI). 13. A[Adj(A)] = (det(A))I so, by Exercise 6, Adj(A) is nonsingular. 14. (a) If A is nonsingular then B = IB = A¡1(AB) = A¡1O = O. Similarly, if B is nonsingular, then A =O. If follows that A and B are both singular. (b) If A is singular then det(A) = 0 so A[Adj(A)] =O. By (a), Adj(A) is singular. 15. AT = A¡1 = (1=det(A))Adj(A) so Adj (A)T = (det(A))A. Chapter 7 Eigenvalues and Applications 7.1 Quadratic Forms 1. A = ? 2 2 2 ¡3 ? : 2. A = ? ¡1 3 3 1 ? : 3. A = 2 4 1 1 ¡3 1 ¡4 4 ¡3 4 3 3 5: 4. A = 2 66 4 1 1 5 ¡2 1 0 2 ¡1 5 2 4 3 ¡2 ¡1 3 ¡1 3 77 5: 5. A = ? 2 2 2 1 ? : 6. A = 2 4 1 4 2 4 2 3 2 3 1 3 5: 7. q(x) = xTAx where A = ? 2 3 3 2 ? : A has eigenvalues ?1 = 5; ?2 =¡1 with corresponding eigenvectors a ? 1 1 ? ; b ? 1 ¡1 ? ; respectively, where a 6= 0 and b 6= 0: In particular Q = (1=p2) ? 1 1 1 ¡1 ? : The form is indeflnite. 194 CHAPTER 7. EIGENVALUES AND APPLICATIONS 8. q(x) = xTAx for A = ? 5 ¡2 ¡2 5 ? : A has eigenvalues ?1 = 7 and ?2 = 3: We may take Q = (1=p2) ? ¡1 1 1 1 ? : The form is positive deflnite. 9. q(x) = xTAx for A = 2 4 1 2 2 2 1 2 2 2 1 3 5: The eigenvalues for A are ?1 = 5 and ?2 = ¡1 (algebraic multiplicity 2). An eigenvector for ?1 = 5 is u1 = [1;1;1]T: The vectors w2 = [¡1;1;0]T and w3 = [¡1;0;1]T are eigenvectors for ?2 = ¡1: The Gram-Schmidt process yields orthogonal eigenvectors u2 = w2 = [¡1;1;0]T and u3 = [¡1;¡1;2]T: We form Q by normalizing the set fu1 ;u2 ;u3g of eigenvectors; Q = 2 4 1=p3 ¡1=p2 ¡1=p6 1=p3 1=p2 ¡1=p6 1=p3 0 2=p6 3 5: The form is indeflnite. 10. q(x) = xTAx for A = 2 4 1 1 1 1 1 1 1 1 1 3 5: A has eigenvalues ?1 = 3 and ?2 = 0 (algebraic multiplicity 2). One choice for Q is Q = 2 4 1=p3 ¡1=p2 ¡1=p6 1=p3 1=p2 ¡1=p6 1=p3 0 2=p6 3 5: The form is positive semideflnite. 11. q(x) = xTAx where A = ? 3 ¡1 ¡1 3 ? : A has eigenvalues ?1 = 2 and ?2 = 4 with corresponding eigenvectors u1 = [1;1]T and u2 = [¡1;1]T; respectively. The set fu1 ;u2 g is orthogonal. We normalize u1 and u2 to obtain Q;Q = (1=p2) ? 1 ¡1 1 1 ? : The form is positive deflnite. 12. q(x) = xTAx where A = 2 66 4 1 ¡1 ¡1 ¡1 ¡1 1 ¡1 ¡1 ¡1 ¡1 1 ¡1 ¡1 ¡1 ¡1 1 3 77 5: A has eigenvalues ?1 = 2 (with algebraic 7.1. QUADRATIC FORMS 195 multiplicity3)and ?2 =¡2: Wemaytake Q = 2 66 4 ¡1=p2 ¡1=p6 ¡1=2p3 1=2 1=p2 ¡1=p6 ¡1=2p3 1=2 0 2=p6 ¡1=2p3 1=2 0 0 3=2p3 1=2 3 77 5: The form is indeflnite. 13. Set q(x) = 2x2 +p3xy + y2: Then q(x) = xTAx for A = ? 2 p3=2 p3=2 1 ? : A has eigenvalues ?1 = 1=2; ?2 = 5=2 with corresponding eigenvectors u1 = [¡1;p3]T and u2 = [p3;1]T; respect- ively. Since fu1 ;u2 g is an orthogonal set we may normalize and obtainQ = ? ¡1=2 p3=2 p3=2 1=2 ? : The substitution x= Qy yields q(x) = (1=2)u2 +(5=2)v2 = 10: The graph corresponds to the ellipse u2=20+ v2=4 = 1: 14. Q = (1=p2) ? 1 ¡1 1 1 ? and the graph corresponds to the ellipse u2=2+ v2=4 = 1: 15. Set q(x) = x2 +6xy ¡7y2: Then q(x) = xTAx where A = ? 1 3 3 ¡7 ? : A has eigenvalues ?1 = ¡8 and ?2 = 2 with corresponding eigenvectors u1 = [¡1;3]T and u2 = [3;1]T; respectively. Since fu1 ;u2 g is an orthogonal set we may normalize to obtain Q = (1=p10) ? ¡1 3 3 1 ? : The substitution x= Qy yields q(x) = ¡8u2 +2v2 = 8: The graph corresponds to the hyperbola v2=4¡u2 = 1: 16. Q = (1=p2) ? 1 ¡1 1 1 ? and the graph corresponds to the ellipse u2 + v2=5 = 1: 17. If q(x) = xy then q(x) = xTAy where A = ? 0 1 1 0 ? : The mat- rix A is diagonalized by Q = (1=p2) ? 1 ¡1 1 1 ? and QTAQ = D = ? 1 0 0 ¡1 ? : The substitution x= Qy yields q(x) = u2 ¡v2 = 4: The graph corresponds to the hyperbola u2=4¡v2=4 = 1: 18. Q = ? p3=2 1=2 1=2 ¡p3=2 ? : The transformed equation is the parabola v =¡2u2 ¡p3u +2: 196 CHAPTER 7. EIGENVALUES AND APPLICATIONS 19. If q(x) = 3x2 ¡2xy +3y2 then q(x) = xTAx where A = ? 3 ¡1 ¡1 3 ? : A is diagonalized by Q = (1=p2) ? ¡1 1 1 1 ? and QTAQ = D where D = ? 4 0 0 2 ? : The transformed equation is the ellipse 4u2 +2v2 = 16; or u2=4+ v2=8 = 1: 20. Q = (1=p2) ? 1 ¡1 1 1 ? and the transformed equation is 2u2 =¡1: 21. Note that aii = eiTAei = eiTCei = cii for 1 ? i ? n: For r 6= s set x= er + es: Then xTAx= arr + ars + asr + ass = arr +2ars + ass; since A is symmetric. Similarly xTCx = crr +2crs + css it follows that ars = crs: 22. (a) Suppose that Ax= ?ix where x6= . Then 0 < xTAx= ?ixTx= ?i kxk2: It follows that ?i > 0: (b) By Equation (3) q(x) = ?1y21 + ?2y22 +¢¢¢+ ?ny2n where y = [y1; : : : ; yn]T and x = Qy: If x6= then y6= so q(x) > 0: 23. See Exercise 22. 24. See Exercise 22. 25. If q(x) = xTAx is indeflnite then it follows from properties (a) - (d) of Theorem 2 that A has positive and negative eigenvalues. Conversely assume that A has eigenvalues ?1 > 0 and ?2 < 0 and let x1 and x2 be corresponding eigenvectors, respectively. Then x1TAx1 = ?1 kxk21> 0 and x2TAx2 = ?2 kxk22< 0. This shows that q(x) is indeflnite. 26. Following the hint we obtain R(x) = (Pni=1 a21?i)=(Pni=1 a2i): There- fore ?1 = (?1 Pni=1 a2i)=(Pni=1 a2i)?(Pni=1 a2i?i)=(Pni=1 a2i) = R(x)?(?nPni=1 a2i)=(Pni=1 a2i) = ?n: 27. If kxk= 1 then R(x) = xTAx= q(x) [cf. Exercise 26]. By Exercise 26, ?1 ? R(x)? ?n: Let u1 and un be eigenvectors corresponding to ?1 and ?n; respectively, where ku1k =kunk= 1: Then R(u1) = u1TAu1 = ?1 and, similarly, R(un) = ?n: 28. (a) BT = (STAS)T = STATSTT = STAS = B: (b) Suppose that q1 is positive deflnite and assume x6= . Since S is nonsingular y = Sx6= . Therefore q2(x) = xTBx= xTSTASx= (Sx)TA(Sx) = yTAy= q1(y) > 0: This shows that q2 is positive deflnite. The reverse argument is similar. 7.2. SYSTEMS OF DIFFERENTIAL EQUATIONS 197 7.2 Systems of Difierential Equations 1. The given system has matrix equation x 0(t) = Ax(t) where x(t) = [u(t); v(t)]T and A = ? 5 ¡2 6 ¡2 ? : The eigenvalues for A are ?1 = 1 and ?2 = 2 and the corresponding eigenvectors are u1 = [1;2]T;u2 = [2;3]T: Thus x1 (t) = etu1 and x2 (t) = e2tu2 are solutions. The general solution is given by x (t) = b1x1 (t)+ b2x2 (t); that is x(t) = b1et ? 1 2 ? + b2et ? 2 3 ? : It is easily seen that x0 = u1 +2u2 so the solution x(t) = et ? 1 2 ? +2e2t ? 2 3 ? = ? et +4e2t 2et +6e2t ? satisfles the initial condition. 2. A = ? 2 ¡1 ¡1 2 ? : The general solution is x(t) = b1et ? 1 1 ? + b2e3t ? ¡1 1 ? : The solution that satisfles the initial condition is x(t) = (1=2)et ? 1 1 ? ¡(3=2)e3t ? ¡1 1 ? : 3. The system is x 0(t) = Ax(t) where A = ? 1 1 2 2 ? : A has eigen- values ?1 = 0 and ?2 = 3 with corresponding eigenvectors u1 = [¡1;1]T and u2 = [1;2]T: The general solution is given by x(t) = b1 ? ¡1 1 ? + b2e3t ? 1 2 ? : The particular solution that satisfles the initial condition is x(t) =¡3 ? ¡1 1 ? +2e3t ? 1 2 ? = ? 3+2e3t ¡3+4e3t ? : 4. A = ? 5 ¡6 3 ¡4 ? : The general solution is x(t) = b1e2t ? 2 1 ? + b2e¡t ? 1 1 ? : The solution that satisfles the initial condition is x(t) = e2t ? 2 1 ? + e¡t ? 1 1 ? = ? 2e2t + e¡t e2t + e¡t ? : 5. The system is x0(t) = Ax(t) where A = ? 0:5 0:5 ¡0:5 0:5 ? : A has eigenvalues ?1 = 0:5+0:5i and ?2 = 0:5¡0:5i with corresponding eigenvectors u1 = [¡i;1]T and u2 = [i;1]T: The general solution is x(t) = b1e?1tu1 +b2e?2tu2 where e?1t = e(0:5+0:5i)t = et=2[cos(t=2)+isin(t=2)] and e?2t = e(0:5¡0:5i)t = et=2[cos(t=2)¡isin(t=2)]: The equation 198 CHAPTER 7. EIGENVALUES AND APPLICATIONS x0 = b1u1 +b2u2 has solution b1 = 2+2i and b2 = 2¡2i so the particular solution that satisfles the initial condition is x(t) = 4e(t=2) ? cos(t=2)+sin(t=2) cos(t=2)¡sin(t=2) ? : 6. A = ? 6 8 ¡1 2 ? : The general solution is x(t) = b1e(4+2i)t ? ¡2¡2i 1 ? +b2e(4¡2i)t ? ¡2+2i 1 ? : The solution that satisfles the initial condition is x(t) = 2ie(4+2i)t ? ¡2¡2i 1 ? ¡ 2ie(4¡2i)t ? ¡2+2i 1 ? = 4e4t ? 2cos2t +4sin2t ¡sin2t ? : 7. The system is x0(t) = Ax(t) where A = 2 4 4 0 1 ¡2 1 0 ¡2 0 1 3 5: A has eigenvalues ?1 = 1; ?2 = 2; and ?3 = 3 with corresponding eigenvectors u1 = [0;1;0]T;u2 = [1;¡2;¡2]T; and u3 = [¡1;1;1]T; respectively. Therefore the general solution is x(t) = b1et 2 4 0 1 0 3 5 + b2e2t 2 4 1 ¡2 ¡2 3 5 + b3e3t 2 4 ¡1 1 1 3 5: Since x0 = u1 +u2 +2u3 the solution x (t) = et 2 4 0 1 0 3 5+ e2t 2 4 1 ¡2 ¡2 3 5+2e3t 2 4 ¡1 1 1 3 5 satisfles the initial condition. 8. A = 2 4 3 1 ¡2 ¡1 2 1 4 1 ¡3 3 5: The general solution is x(t) = b1et 2 4 1 0 1 3 5+ b2e¡t 2 4 7 ¡2 13 3 5+ b3e2t 2 4 1 1 1 3 5: The solution x(t) = 3et 2 4 1 0 1 3 5¡ e¡t 2 4 7 ¡2 13 3 5+2e2t 2 4 1 1 1 3 5 satisfles the initial condition. 9. (a) The system is x0(t) = Ax(t) for A = ? 1 ¡1 1 3 ? : A has only one eigenvalue, ? = 2; with corresponding eigenvector u = [1;¡1]T: Therefore x1 (t) = e2t ? 1 ¡1 ? is a solution for the system. 7.3. TRANSFORMATION TO HESSENBERG FORM 199 (b) Set x2 (t) = te?tu+e?ty0 : Then x2 0(t) = e?t(t?u+u+ ?y0) whereas Ax2 (t) = e?t(t?u+Ay0 ): Therefore we require that Ay0 = u+?y0 ; that is (A¡?I)y0 = u: One choice is y0 = [¡2;1]T: Thus x2 (t) = te2t ? 1 ¡1 ? + e2t ? ¡2 1 ? is a solution. (c) If y(t) = c1x1(t)+c2x2(t) note that y(0) = c1u+c2y0 : Since fu;y0g is a linearly independent set for every x0 in R2 we may flnd c1 and c2 such that x0= c1u+c2y0 : 10. (a) A = ? 2 ¡1 4 6 ? and x1 (t) = e4t ? 1 ¡2 ? : (b) One choice for y0 is y0 = [0;¡1]T: In this case x2 (t) = te4t ? 1 ¡2 ? + e4t ? 0 ¡1 ? : (c) The solution x(t) = x1 (t)¡3x2 (t) = ? e4t(1¡3t) e4t(1+6t) ? satisfles the initial condition. 7.3 Transformation to Hessenberg Form 1. The desired elementary row operation is R3 ¡4R2: Performing this operation on the (3 x 3) identity matrix yields Q1 = 2 4 1 0 0 0 1 0 0 ¡4 1 3 5: Q¡11 = 2 4 1 0 0 0 1 0 0 4 1 3 5 and Q1AQ¡11 = H = 2 4 ¡7 16 3 8 9 3 0 1 1 3 5: 2. Q1 = 2 4 1 0 0 0 1 0 0 2 1 3 5;H = Q1AQ¡11 2 4 ¡6 31 ¡14 ¡1 6 ¡2 0 2 1 3 5: 3. Let Q1 denote the permutation matrix Q1 = 2 4 1 0 0 0 0 1 0 1 0 3 5: Then Q1 = Q¡11 and Q1A interchanges the second and third rows of A: Further (Q1A)Q1 interchanges the second and third columns of Q1A: Therefore H = Q1AQ¡11 = 2 4 1 1 3 1 3 1 0 4 2 3 5: 4. Q1 = 2 4 1 0 0 0 1 0 0 2 1 3 5 and H = Q1AQ¡11 = 2 4 1 4 ¡1 3 0 1 0 ¡5 5 3 5: 200 CHAPTER 7. EIGENVALUES AND APPLICATIONS 5. The desired elementary row operation is R3 +3R2: Performing this operation on the (3 x 3) identity matrix yields Q1 = 2 4 1 0 0 0 1 0 0 3 1 3 5: Q¡11 = 2 4 1 0 0 0 1 0 0 ¡3 1 3 5 and H = Q1AQ¡11 = 2 4 3 2 ¡1 4 5 ¡2 0 20 ¡6 3 5: 6. Q1 = 2 4 1 0 0 0 0 1 0 1 0 3 5 and H = Q1AQ¡11 = 2 4 4 3 0 3 1 2 0 2 1 3 5: 7. Performing the elementary row operations R3 ¡R2 and R4 ¡R2 on the (4 x 4) identity matrix yields Q1 = 2 66 4 1 0 0 0 0 1 0 0 0 ¡1 1 0 0 ¡1 0 1 3 77 5: Q¡11 = 2 66 4 1 0 0 0 0 1 0 0 0 1 1 0 0 1 0 1 3 77 5 and H = Q1AQ¡11 = 2 66 4 1 ¡3 ¡1 ¡1 ¡1 ¡1 ¡1 ¡1 0 0 2 0 0 0 0 2 3 77 5: 8. Let Q1 = 2 66 4 1 0 0 0 0 1 0 0 0 ¡4 1 0 0 ¡4 0 1 3 77 5 and Q2 = 2 66 4 1 0 0 0 0 1 0 0 0 0 1 0 0 0 ¡1 1 3 77 5: Then H = Q2Q1AQ¡11 Q¡12 = 2 66 4 6 33 8 4 1 38 8 4 0 ¡120 ¡25 ¡15 0 0 0 5 3 77 5: 9. If Q1 = 2 66 4 1 0 0 0 0 0 0 1 0 0 1 0 0 1 0 0 3 77 5 then Q1 = Q¡11 and Q1A interchanges the second and fourth rows of A whereas (Q1A)Q1 interchanges the second and fourth columns 0f Q1A: Therefore Q1AQ1 = 2 66 4 1 3 1 2 1 2 0 2 0 1 1 3 0 2 1 1 3 77 5: 7.3. TRANSFORMATION TO HESSENBERG FORM 201 Now the desired elementary row operation is R4 ¡2R3 so set Q2 = 2 66 4 1 0 0 0 0 1 0 0 0 0 1 0 0 0 ¡2 1 3 77 5: Then Q¡12 = 2 66 4 1 0 0 0 0 1 0 0 0 0 1 0 0 0 2 1 3 77 5 and H = Q2Q1AQ¡11 Q¡12 = 2 66 4 1 3 5 2 1 2 4 2 0 1 7 3 0 0 ¡11 ¡5 3 77 5: 10. If Q1 = 2 66 4 1 0 0 0 0 1 0 0 0 2 1 0 0 1 0 1 3 77 5 and Q2 = 2 66 4 1 0 0 0 0 1 0 0 0 0 1 0 0 0 ¡5=3 1 3 77 5 Then H = Q2Q1AQ¡11 Q¡12 = 2 66 4 2 ¡1 ¡5=3 ¡1 ¡1 2 ¡1=3 1 0 0 7 6 0 0 0 0 3 77 5: 11. Since ? is an eigenvalue for H; the matrix H¡?I is singular; that is nullity(H¡?I)?1: It follows that rank(H¡?I)?3: But H¡?I = 2 66 4 a1 ¡? b1 c1 d1 a2 b2 ¡? c2 d2 0 b3 c3 ¡? d3 0 0 c4 d4 ¡? 3 77 5 and clearly the flrst three columns of H ¡?I are linearly independent. Therefore rank(H ¡?I) ? 3: Thus rank(H ¡?I) = 3 and so, nullity(H ¡?I) = 1: Hence, ? has geometric multiplicity equal to 1: 12. Since H is similar to a symmetric matrix, H is diagonizable. Therefore the algebraic multiplicity for ? equals the geometric multiplicity. Now apply Exercise 11. 13. p(t) = (t¡2)3(t +2) is the characteristic polynomial for H: Since H and A are similar, p(t) is also the characteristic polynomial for A: Therefore A has eigenvalues ?1 = 2 and ?2 =¡2 and ?1 = 2 has algebraic (and hence geometric) multiplicity 3 (cf. Exercise 12). 14. p(t) = (t¡5)2(t¡15)(t +1) so A has eigenvalues ?1 = 5; ?2 = 15; and ?3 =¡1: 15. [e1;e2;e3];[e1;e3;e2];[e2;e1;e3];[e2;e3;e1];[e3;e1;e2]; [e3;e2;e1]: 202 CHAPTER 7. EIGENVALUES AND APPLICATIONS 16. [e1;e2;e3;e4];[e1;e2;e4;e3];[e1;e3;e2;e4];[e1;e3;e4;e2]; [e1;e4;e2;e3];[e1;e4;e3;e2];[e2;e1;e3;e4];[e2;e1;e4;e3]; [e2;e3;e1;e4];[e2;e3;e4;e1];[e2;e4;e1;e3];[e2;e4;e3;e1]; [e3;e1;e2;e4];[e3;e1;e4;e2];[e3;e2;e1;e4];[e3;e2;e4;e1]; [e3;e4;e1;e2];[e3;e4;e2;e1];[e4;e1;e2;e3];[e4;e1;e3;e2]; [e4;e2;e1;e3];[e4;e2;e3;e1];[e4;e3;e1;e2];[e4;e3;e2;e1]: 17. There are n! (n xn) permutation matrices. 18. Since the columns of P are some ordering of e1;e2; : : : ;en; they form an orthonormal set. 19. AP = A[ei;ej;ek; : : : ;er] = [Aei ; Aej ; Aek ; : : : ; Aer ] = [Ai;Aj;Ak; : : : ;Ar]: 20. Let A = 2 66 64 a1 a2 ... an 3 77 75 where aj is the jth row of A: By Exercise 19, ATP = [aiT;ajT;akT; : : : ;arT]: Therefore P TA = (ATP)T = 2 66 66 64 ai aj ak ... ar 3 77 77 75: 21. Apply Exercise 19. 22. By Exercise 21 each of the matrices P 1; P2; P3; : : : is a permutation matrix. By Exercise 17 there are n! distinct (nxn) permutation matrices. Therefore there exists integers r and s such that r > s and Pr = Ps: Since P is nonsingular this implies that Pr¡s = I: 7.4 Eigenvalues of Hessenberg Matrices 1. Note that the given matrix H is in unreduced Hessenberg form. We have w0 = e1 = [1;0]T;w1 = Hw0 = [2;1]T; and w2 = Hw1 = [4;3]T: The vector equation a0w0 +a1w1 =¡w2 is equivalent to the system a0 + 2a1 = ¡4 a1 = ¡3 : The system has solution a0 = 2; a1 =¡3 so p(t) = 2¡3t + t2: 7.4. EIGENVALUES OF HESSENBERG MATRICES 203 2. w0 = e1 = [1;0]T;w1 = [0;3]T;w2 = [0;0]T; The vector equation a0w0 +a1w1 = ¡w2 has solution a0 = a1 = 0; so p(t) = t2: 3. Note that the given matrix H is in unreduced Hessenberg form. We have w0 = e1 = [1;0;0]T;w1 = Hw0 = [1;2;0]T;w2 = Hw1 = [1;4;2]T; and w3 = Hw2 = [3;6;8]T: The vector equation a0w0 + a1w1 + a2w2 =¡w3 is equivalent to the system of equations a0 + a1 + a2 = ¡3 2a1 + 4a2 = ¡6 2a2 = ¡8 : The system has unique solution a0 =¡4; a1 = 5; a2 =¡4 so p(t) =¡4+5t¡4t2 + t3: 4. w0 = e1 = [1;0;0]T;w1 = [1;1;0]T;w2 = [3;4;1]T;w3 = [12;14;6]T: The vector equation a0w0 +a1w1 +a2w2 =¡w3 has solution a0 =¡4; a1 = 10; a2 =¡6; so p(t) =¡4+10t¡6t2 + t3: 5. Note that the given matrix H is in unreduced Hessenberg form. We have w0 = e1 = [1;0;0]T;w1 = Hw0 = [2;1;0]T;w2 = Hw1 = [8;3;1]T; and w3 = Hw2 = [29;14;8]T: The vector equation a0w0 +a1w1 +a2w2 =¡w3 is equivalent to the system of equations a0 + 2a1 + 8a2 = ¡29 a1 + 3a2 = ¡14 a2 = ¡8 : The system has unique solution a0 = 15; a1 = 10; a2 =¡8 so p(t) = 15+10t¡8t2 + t3: 6. w0 = e1;w1 = e2;w2 = e3;w3 = e1: The vector equation a0w0 ++a1w1 +a2w2 =¡w3 has solution a0 =¡1; a1 = a2 = 0: Therefore p(t) =¡1+ t3: 7. Note that the given matrix H is in unreduced Hessenberg form. We have w0 = e1 = [1;0;0;0]T;w1 = Hw0 = [0;1;0;0]T; w2 = Hw1 = [1;2;1;0]T;w3 = Hw2 = [2;6;2;2]T; and w4 = Hw3= [8;18;8;6]T: Thevectorequation a0w0+a1w1+a2w2+a3w3 =¡w4 isequivalent to the system of equations a1 + a2 + 2a3 = ¡8 a1 + 2a2 + 6a3 = ¡18 a2 + 2a3 = ¡8 2a3 = ¡6 : Thesystemhasuniquesolution a0 = 0; a1 = 4; a2 =¡2; a3 =¡3; so p(t) = 4t¡2t2¡3t3+t4: 204 CHAPTER 7. EIGENVALUES AND APPLICATIONS 8. w0 = e1 = [1;0;0;0]T;w1 = [0;1;0;0]T;w2 = [2;0;2;0]T;w3 = [2;4;0;2]T; and w4 = [12;0;12;2]T: The vector equation a0w0 + a1w1 +a2w2 +a3w3 =¡w4 has solution a0 = 2; a1 = 4; a2 =¡6; a3 =¡1: Therefore p(t) = 2+4t¡6t2 ¡t3 + t4: 9. H = ? B 11 B12 O B22 ? where B11 = ? 1 ¡1 1 3 ? ; B12 = ? 1 4 ¡2 1 ? and B22 = ? 2 ¡1 ¡1 2 ? : B11 has eigenvalue ?1 = 2 (with algebraic multiplicity 2) with corresponding eigenvector u1 = [¡1;1]T:B22 has eigenvalues ?2 = 1; ?3 = 3 with corresponding eigenvectors v2= [1;1]T and v3= [¡1;1]T; respectively. Thus H haseigenvalues ?1 = 2; ?2 = 1; ?3 = 3: The vector x1 = ? u 1 ? = [¡1;1;0;0]T is an eigenvector for H corresponding to ?1 = 2: The system of equations (B11 ¡ I)u = ¡B12v2 has solution u2 = [¡9;5]T; so x2 = ? u 2 v2 ? = [¡9;5;1;1]T is an eigenvector of H corresponding to ?2 = 1: Similarly (B11 ¡3I)u= ¡B12v3 has solution u3 = [¡3;9]T so x3 = ? u 3 v3 ? =[¡3;9;¡1;1]T is an eigenvector of H corresponding to ?3 = 3: 10. B11 = ? 1 1 1 1 ? and B22 = ? 3 0 1 4 ? : B11 has eigenvalues ?1 = 0 and ?2 = 2 and B22 has eigenvalues ?3 = 3; ?4 = 4: The corresponding eigenvectors are x1 = [¡1;1;0;0]T;x2 = [1;1;0;0]T;x3 = [0;1;¡1;1]T and x4 = [3=4;5=4;0;1]T: 11. H = ? B 11 B12 O B22 ? where B11 = 2 4 ¡2 0 ¡2 ¡1 1 ¡2 0 1 ¡1 3 5; B12 = 2 4 1 3 ¡2 3 5; and B22 = [2]: B11 has eigenvalues ?1 = 0 and ?2 = ¡1 (algebraic multiplicity 2) with corresponding eigenvectors u1= [¡1;1;1]T and u2= [¡2;0;1]T:B22 has eigenvalue ?3 = 2 with corresponding eigenvector v3 = [1]: Thus H has eigenvalues ?1 = 0; ?2 = ¡1; and ?3 = 2: The vectors x1 = ? u 1 ? = [¡1;1;1;0]T and x2 = ? u 2 ? = [¡2;0;1;0]T are eigenvectors for H corresponding to ?1 = 0 and ?2 = ¡1; respectively. The system of equations (B11 ¡2I)u1 =¡B12v3 has solution u3 = [1=6;15=6;1=6]T so x3 = ? u 3 v3 ? = [1=6;15=6;1=6;1]T is an eigenvector for H corresponding to ?3 = 2: 12. B11 = ? 2 3 3 2 ? and B22 = ? 3 0 1 3 ? : B11 has eigenvalues ?1 = 5 and ?2 = ¡1 and B22 has eigenvalue ?3 = 3: The corresponding eigenvectors for H are x1 = [1;1;0;0]T;x2 = [¡1;1;0;0]T; and x3 = [¡7=8;¡13=8;0;1]T: 7.4. EIGENVALUES OF HESSENBERG MATRICES 205 13. det(B) = afwz ¡afyx¡ebwz + ebyx = (af ¡eb)(wz ¡yx) = det(B11)det(B22): 14. det(H) = flfl flfl 1 ¡11 3 flfl flfl x flfl flfl 2 ¡1¡1 2 flfl flfl = (4)(3) = 12: 15. P = [e2 ;e3 ;e1 ]: 16. P = [e2 ;e3 ;e4 ;e1 ]: 17. P = [e2 ;e3 ; : : : ;en ;e1 ]: 18. Write P = [P1;P2; : : : ;Pn] where, as shown in Exercise 17, P1 = e2;P2 = e3; : : : ;Pn¡1 = en; and Pn = e1: Thus w0 = e1;w1 = Pw0 = Pe1 = P1 = e2;w2 = Pw1 = Pe2 = P2 = e3; : : : ; wn¡1 = Pwn¡2 = Pen¡1 = Pn¡1 = en; and wn = Pwn¡1 = Pen = Pn = e1: Obviously the vector equation a0w0 + a1w1+ ¢¢¢+an¡1wn¡1 =¡wn hassolution a0 =¡1; a1 =¢¢¢= an¡1 = 0: Therefore p(t) = tn¡1: 19. Let H = [hij] and let ? be an eigenvalue for H; Then H ¡?I = 2 66 66 66 64 h11 ¡? h12 ¢¢¢ h1;n¡1 h1n h21 h22 ¡? h2;n¡1 h2n 0 h32 h3;n¡1 h3n ... ... ... ... 0 0 hn¡1;n¡1 ¡? hn¡1;n 0 0 hn;n¡1 hnn ¡? 3 77 77 77 75 : Since h21; h32; : : : ; hn;n¡1 are nonzero, the flrst n¡1 columns of H ¡?I form a linearly independent set. Therefore rank(H ¡?I)? n¡1: It follows that nullity(H ¡?I)?1: Since ? is an eigenvalue for H it follows that nullity(H ¡?I) = 1; that is, ? has geometric multiplicity 1. 20. Since H is symmetric, it is diagonalizable. Therefore the algebraic multiplicity of ? equals the geometric multiplicity. It follows from Exercise 19 that ? has algebraic multiplicity 1. Therefore H neces- sarily has n distinct eigenvalues. 21. If H is unreduced then b 6= 0: Thus p(t) = t2 ¡(a+c)t¡b2: The eigenvalues for H are ? = [(a + c)§p(a + c)2 +4b2]=2: Since (a + c)2 +4b2 > 0; H has two distinct eigenvalues. 22. Set u= [u1; u2; : : : ; un]T and assume that un = 0: Set H = [h1;h2; : : : ;hn]: Thus ?u = Hu = u1h1 + u2h2 +¢¢¢+ un¡1hn¡1: There- 206 CHAPTER 7. EIGENVALUES AND APPLICATIONS fore the nth component of Hu is un¡1hn;n¡1: Since hn;n¡1 6= 0 it follows that un¡1 = 0: Repetition of this argument yields u1 = u2 =¢¢¢= un = 0; so u = . Therefore if u 6= then un 6= 0: 23. Let k be an integer, 1 ? k ? n; and suppose we have shown that wk¡1 has the form wk¡1 = [a1 : : : ; ak;0; : : : ;0]T; where ak 6= 0: If H = [h1 ; : : : ;hn ] then wk = Hwk¡1 = a0h1 +¢¢¢+akhk : But H is in Hessenberg form so hij = 0 when i > j +1: Therefore the k +1 component of wk is akhk+1;k and is nonzero since H is unreduced. Thus wk has the form wk = [b1; : : : ; bk+1;0; : : : ;0]T; where bk+1 6= 0: 7.5 Householder Transformations 1. Qx= x¡ u where = 2uTx=uTu= (¡2)(2)=4 =¡1: Thus Qx= [4;1;6;7]T: 2. Qx= [4;¡3;5;4[T: 3. Set 1 = 2uTA1 =uTu = ¡1 and 2 = 2uTA2 =uTu = ¡2: Then QA1 = A1 ¡ 1u = [3;5;5;1]T and QA2 = A2¡ 2u= [3;1;4;2]T: Therefore QA = 2 66 4 3 3 5 1 5 4 1 2 3 77 5: 4. QA = 2 66 4 2 3 1 0 0 2 3 6 2 1 5 3 3 77 5: 5. Set = 2uTx=uTu=¡1: Then Qx= x¡ u= x+u= [4;1;3;4]T: Thus xTQ = (Qx)T = [4;1;3;4]: 6. xTQ = [2;2;3;1]: 7. Set x= [2;1;2;1]T and y= [1;0;1;4]T: Then QAT = Q[x;y] = [Qx; Qy] = 2 66 4 1 2 2 ¡1 1 2 2 3 3 77 5: Therefore AQ = (QAT)T = ? 1 2 1 2 2 ¡1 2 3 ? : 7.5. HOUSEHOLDER TRANSFORMATIONS 207 8. BQ = 2 66 4 1=2 5=2 7=2 5=2 6 0 ¡1 ¡1 ¡7=2 13=2 11=2 5=2 3 ¡7 1 5 3 77 5: 9. Set u1 = 0: If a = ¡p4+4+1 = ¡3 then u2 = v2 ¡ a = 2 + 3 = 5: Finally take u3 = v3 = 2 and u4 = v4 = 1: Thus u= [0;5;2;1]T: 10. a =¡2;u= [3;1;1;1]T: 11. a = ¡p42 +32 = ¡5;u1 = u2 = 0;u3 = v3 ¡ a = 4 + 5 = 9;u4 = v4 = 3: Therefore u = [0;0;9;3]T: 12. a = 3;u= [0;0;¡5;2;1]T: 13. a = p(¡3)2 +42 = 5;u1 = u2 = u3 = 0;u4 = v4 ¡ a = ¡8;u5 = v5 = 4: Therefore u = [0;0;0;¡8;4]T: 14. a =¡4;u= [0;0;8;0;0]T: 15. We want QA1 = [1; a;0]T: Therefore a =¡p32 +42 =¡5 and u= [u1; u2; u3]T where u1 = 0; u2 = 3¡(¡5) = 8; and u3 = 4: Then u= [0;8;4]T: 16. u= [0;¡5;5]T: 17. We want QA1 = 2 4 0 a 0 3 5: Therefore a = p(¡4)2 +32 = 5 and u= [u1; u2; u3]T where u1 = 0; u2 =¡4¡5 =¡9; and u3 = 3: Thus u= [0;¡9;3]T: 18. u= [0;0;8;4]T: 19. We want QA2 = 2 66 4 1 4 a 0 3 77 5 so a = p(¡3)2 +42 = 5:u= [u1; u2; u3; u4]T where u1 = u2 = 0; u3 =¡3¡5 =¡8; and u4 = 4: Thus u= [0;0;¡8;4]T: 20. u= [0;0;¡1;1]T: 21. QT = (I ¡buuT)T = IT ¡(buuT)T = I ¡buTTu T = I ¡buuT = Q: 208 CHAPTER 7. EIGENVALUES AND APPLICATIONS 22. Set b = 2=uTu: Then Qu= (I ¡buuT)u= Iu¡(buuT )u= u¡bu(uTu) = u¡2u=¡u: If uTv= 0 then Qv= (I ¡buuT)v= Iv¡bu(uTv) = v: 23. Let fu;w2; : : : ;wng be as given in the hint. By Exercise 22, Qu= ¡u and Qwi = wi for 2 ? i ? n: Thus Rn has a basis consisting of eigenvectors for Q; that is Q is diagonalizable. Moreover Q is similar to the (nxn) diagonal matrix D with diagonal entries d11 = 1; d22 = ¢¢¢ = dnn = ¡1: Since Q and D have the same eigenvalues, Q has eigenvalues 1 and ¡1: 24. To prove (a) note that Q¡1 = (Qn¡2¢¢¢Q2Q1)¡1 = Q¡11 Q¡12 ¢¢¢Q¡1n¡2 = QT1 QT2 ¢¢¢QTn¡2 = (Qn¡2¢¢¢Q2Q1)T = QT: To prove (b) assume that A is symmetric. Thus HT = (QAQT)T = QTTATQT = QAQT = H and H is symmetric. 25. (a) Set BT = [v1;v2;v3;v4]: Then QBT = [Qv1;Qv2;Qv3;Qv4]; and for 1? j ?4; Qvj = vj ¡ ju; where j is a constant. Since u= [0; a; b; c]T it follows that Qvj and vj have the same flrst coordinate. Thus BT and QBT have the same flrst row. It follows that B and BQ = (QBT)T have the same flrst column. (b) It follows from (a) that x= [b11; b12;0;0]T is the flrst column of BQ: Thus Qx is the flrst column of QBQ: But Qx= x¡ u where = 2uTx =uTu= 0; that is Qx = x: 7.6 QR Factorization & Least-Squares 1. x? is the unique solution to Rx= c; where R = ? 1 2 0 1 ? and c= ? 3 1 ? : Thus x?= [1;1]T: 2. x?= [2;¡1]T: 3. x? is the unique solution to Rx= c where R = 2 4 1 2 1 0 1 3 0 0 2 3 5 and c= 2 4 6 7 4 3 5: Thus x?= [2;1;2]T: 4. x?= [1;2;1]T: 7.6. QR FACTORIZATION & LEAST-SQUARES 209 5. We require that SA1 = [a;0]T: Therefore a =§pa211 + a221 = ¡5; u1 = a11 ¡ a = 8; and u2 = a21 = 4: Consequently u= [8;4]T and SA1 = A1 ¡u = [¡5;0]T: SA2 = A2¡2u= [¡11;2]T; so SA = R = ? ¡5 ¡11 0 2 ? : 6. u= [1;1]T;R = ? ¡1 ¡5 0 ¡3 ? : 7. We require that SA1 = ? a 0 ? so take a =¡pa211 + a221 =¡4; u1 = a11 ¡a = 4; and u2 = a21 = 4: Thus u= [4;4]T and SA1 = A1 ¡u= [¡4;0]T: Also SA2 = A2¡2u= [¡6;¡2]T: Therefore SA = R = ? ¡4 ¡6 0 ¡2 ? : 8. u= [¡22;4]T;R = ? 5 ¡22 0 4 ? : 9. We require that SA2 = [2; a;0]T so set a =¡pa222 + a232 =¡1;u1 = 0; u2 = a22 ¡a = 1; and u3 = a32 = 1: Therefore u= [0;1;1]T; SA1 = A1 ; SA2 = A2 ¡u= [2;¡1;0]T; and SA3 = A3¡14u= [1;¡8;¡6]T: Consequently SA = R = 2 4 1 2 1 0 ¡1 ¡8 0 0 ¡6 3 5: 10. u= [0;8;4]T;R = 2 4 3 1 2 0 ¡5 ¡11 0 0 2 3 5: 11. We flrst require Q1 such that Q1A1 = [a;0;0;0]T so take u1 = [6;2;2;4]T: Then Q1A1 = A1¡u1 = [¡5;0;0;0]T and Q1A2 = A2¡3u= [¡59=3;6;2;3]T: We now require Q2 such that Q2(Q1A2 ) = [¡59=3; a;0;0]T: Thus set u2 = [0;13;2;3]T: Then Q2(Q1A1 ) = Q1A1 and Q2(Q1A2 ) = Q1A2¡u2 = [¡59=3;¡7;0;0]T: Therefore Q2Q1A = 2 66 4 ¡5 ¡59=3 0 ¡7 0 0 0 0 3 77 5: 12. u1 = [3;1;1;1]T and Q1A = 2 66 4 ¡2 ¡7 0 0 0 0 0 3 3 77 5:u2 = [0;3;0;3]T and 210 CHAPTER 7. EIGENVALUES AND APPLICATIONS Q2Q1A = 2 66 4 ¡2 ¡7 0 ¡3 0 0 0 0 3 77 5: 13. We require a matrix Q such that QA2 = [4; a;0;0]T: Thus u= [0;8;0;4]T and QA =2 66 4 2 4 0 ¡5 0 0 0 0 3 77 5: 14. u= [0;5;1;2]T and QA = 2 66 4 3 5 0 ¡3 0 0 0 0 3 77 5: 15. Let Q1;u1 ; Q2;u2 be as in Exercise 11. Then Q1b = b¡u1 = [¡5;8;¡2;¡3]T and Q2(Q1b) = Q1b¡u2 = [¡5;¡5;¡4;¡6]T: Theleast-squaressolutionistheuniquesolution x? to Rx= c where R = ? ¡5 ¡59=3 0 ¡7 ? and c= [¡5;¡5]T: Thus x?= [¡38=21;15=21]T: 16. Q2Q1b= [¡4;2;¡1;3]: x? is the unique solution to Rx= c where R = ? ¡2 ¡7 0 ¡3 ? and c= [¡4;2]T: Thus x?= [13=3;¡2=3]T: 17. With Q and u as in Exercise 13, Qb= b¡(12=5)u= [2;¡56=5;16;¡8=5]T: Therefore x? is the unique solution to Rx= c where R = ? 2 4 0 ¡5 ? and c= [2;¡56=5]T: Solving yields x?= [¡87=25;56=25]T: 18. Qb = [5;¡10=3;¡19=3;¡8=15]T: x? is the unique solution to Rx = c where R =? 3 5 0 ¡3 ? and c= [5;¡10=3]T: Solving yields x?= [¡5=27;10=9]T: 19. Write [A1;A2; : : : ;An]; where fA1;A2; : : : ;Ang is a linearly independent subset of Rn: Now SA = [SA1;SA2; : : : ;SAn]: Suppose c1; c2; : : : ; cn are scalars such that =c1SA1 + c2SA2 +¢¢¢+ cnSAn: 7.7. MATRIX POLYNOMIALS & THE CAYLEY-HAMILTON THEOREM 211 Then = S(c1A1 + c2A2 +¢¢¢+ cnAn) and S is nonsingular. There- fore = c1A1 +c2A2 +¢¢¢+cnAn: It follows that c1 = c2 =¢¢¢= cn = 0 and hence, the set fSA1;SA2; : : : ;SAng is linearly independent. For each j;1 ? j ? n; SAj = ? R j ? where Rj is the jth column of R and is in Rm¡n: Therefore the set fR1;R2; : : : ; Rng is linearly independent in Rn and the matrix R is nonsingular. 7.7 Matrix Polynomials & The Cayley-Hamilton Theorem 1. q(A) = A2 ¡4A+3I = ? ¡1 0 0 ¡1 ? ;q(B) = B2 ¡4B +3I = ? 0 0 0 0 ? ;q(C) = C2 ¡4C + 3I = 2 4 15 ¡2 14 5 ¡2 10 ¡1 ¡4 6 3 5: 2. (a) p(A) = (A¡I)3 =O; p(B) = (B ¡I)3 =O; p(C) = (C ¡I)3 =O; p(I) = (I ¡I)3 =O3 =O (b) Set q(t) = (t¡1)2 = t2 ¡2t +1: 3. (a) q(t) = s(t)p(t)+ r(t) where s(t) = t3 + t¡1 and r(t) = t +2: (b) q(B) = s(B)p(B)+r(B) = r(B) since p(B) =O: Thus q(B) = B+2I = ? 4 ¡1 ¡1 4 ? : 4. H11 = ? 2 3 5 7 ? ; H22 = 2 4 4 1 3 6 1 2 0 4 1 3 5 and H33 = ? 6 5 7 3 ? : Using Algorthim 1 we obtain p1(t) = t2 ¡9t¡1; p2(t) = t3 ¡6t2 ¡5t¡38; and p3(t) = t2 ¡9t¡17: 5. Note that H2 = (SAS¡1)(SAS¡1) = (SA2S¡1): For some positive integer k ? 2 sup- pose we have shown that Hk = SAkS¡1: Then Hk+1 = HkH = (SAkS¡1)(SAS¡1) = SAk+1S¡1: It follows by induction that Hn = SAnS¡1 for each positive integer n: Now let q(t) = antn+an¡1tn¡1+¢¢¢+a1t+a0: Then q(H) = anHn+an¡1Hn¡1+¢¢¢+a1H+a0I = anSAnS¡1 + an¡1SAn¡1S¡1 +¢¢¢+ a1SAS¡1 + a0SIS¡1 = S(anAn + an¡1An¡1 +¢¢¢+ a1A + a0I)S¡1 = Sq(A)S¡1: 6. Since u1TAu2 is a (1 x 1) matrix, u1TAu2 = (u1TAu2)T = u2TATu1TT = u2TAu1: Now u1TAu2 = ?2u1Tu2 whereas u2TAu1 = ?1u2Tu1 = ?1u1Tu2: Therefore ?1u1Tu2 = ?2u1Tu2: Since ?1 6= ?2 it follows that u1Tu2 = 0: 212 CHAPTER 7. EIGENVALUES AND APPLICATIONS 7. (a) By assumption, Ax0 is in W: For some positive integer k; suppose we have shown that Akx0 is in W: Then Ak+1x0 = A(Akx0) is in W by the assumed property of A: By induction, Anx0 is in W for each positive integer n: (b) = m(A)x0 = (A¡rI)s(A)x0: Since s(t) has degree k ¡1; S(A)x0 6= . Thus if u = S(A)x0 then (A ¡ rI)u = . It follows that u is an eigenvectorof A correspondingtotheeigenvalue r: Nowif s(t) = bk¡1tk¡1+¢¢¢+b1t+b0 then S(A)x0 = bk¡1Ak¡1x0 +¢¢¢+ b1Ax0 + b0x0: Since r is an eigenvalue for A; r is real. It follows that b0; : : : ; bk¡1 are real. Thus S(A)x0 is a linear combination of the vectors Ak¡1x0,: : :, Ax0, x0 in W. Hence S(A)x0 is in W: 8. (a) Let x and y be in W; that is, xuiT = 0 and yuiT = 0 for 1 ? i ? k: Therefore (x+y)uiT = xuiT + yuiT = 0+0 = 0 for 1 ? i ? k: Consequently, x+y is in W: Likewise if c is a scalar then (cx)uiT = c(xuiT) = c0 = 0; for 1? i ? k; so cx is in W: Certainly is in W; so W is a subspace of Rn: (b) Suppose that Aui = ?iui;1? i ? k; and let x be in W: Thus xTui= 0 for 1? i ? k: Now (Ax)Tui = xTATui = xTAui = xT(?iui) = ?i(xTui) = 0 for 1? i ? k: Therefore Ax is in W: It now follows from Exercise 7 that A has an eigenvector uk+1 in W: By deflnition of W;fu1;u2; : : : ;uk;uk+1g is an orthogonal set of eigenvectors for A: 7.8 Generalized Eigenvectors & Difierential Equations 1. (a) The given matrix H has characteristic polynomial p(t) = (t ¡2)2; so ? = 2 is the only eigenvalue and it has algebraic multiplicity 2. The vector v1 = [1;¡1]T is an eigenvector corresponding to ? = 2: If we solve the system of equations (H ¡2I)x = v1 we see that x = [¡1¡ a; a]T; where a is arbitrary. Taking a = 0 we obtain a generalized eigenvector v2 = [¡1;0]T: (b) The given matrix H has characteristic polynomial p(t) = t(t + 1)2: The eigenvalue ? = ¡1 has corresponding eigenvector v1 = [¡2;0;1]T: Solving the system (H¡(¡1)I)x = v1 yields x= [2¡2a;1; a]T where a is arbitrary. Thus v2 = [0;1;1]T is a generalized eigenvector for ? = ¡1: The eigenvalue ? = 0 has corresponding eigenvector w1 = [¡1;1;1]T: (c) The given matrix H has characteristic polynomial p(t) = (t¡1)2(t +1): The eigenvalue ? = 1 has corresponding eigenvector v1 = [¡2;0;1]T: Solving (H ¡ I)x = v1 yields x= [(5=2)¡2a;1=2; a]T; where a is arbitrary. Thus v2 = [5=2;1=2;0]T is a generalized eigenvector of ? = 1: The eigenvalue ? =¡1 has corresponding eigenvector w1 = [¡9;¡1;1]T: 7.8. GENERALIZED EIGENVECTORS & DIFF. EQNS. 213 2. For A; ? = 1 is the only eigenvalue. Corresponding generalized eigenvectors are v1 = [0;0;0;1]T;v2 = [0;0;1;0]T;v3 = [0;1;0;0]T; and v4 = [1;0;0;0]T: For B generalized eigenvectors are v1 = [¡3;¡5;¡1;2]T;v2 = [0;0;0;1]T;v3 = [0;1=2;0;1=2]T; and v4 = [1=4;1=4;0;1=4]T: 3. (a) If Q = 2 4 1 0 0 0 1 0 0 3 1 3 5 then Q¡1 = 2 4 1 0 0 0 1 0 0 ¡3 1 3 5 and H = QAQ¡1 = 2 4 8 ¡69 21 1 ¡10 3 0 ¡4 1 3 5 is in unreduced Hessenberg form. H has characteristic polynomial p(t) = (t+1)2(t¡1): Theeigenvalue ? =¡1 hascorrespondingeigenvector v1 = [3;1;2]T: Solving the system (H ¡(¡1)I)u = v1 yields u= [¡(7=2)+(3=2)a;(¡1=2)+(1=2)a; a]T; where a is arbitrary. Therefore v2 = [¡2;0;1]T is a generalized eigenvector for ? = ¡1: The eigenvalue ? = 1 has corresponding eigenvector w1 = [¡3;0;1]T: Set y(t) = Qx(t) and y0 = Qx0 = [¡1;¡1;¡2]T: The system y0 =Hy has general solution y(t) = c1e¡tv1+ c2e¡t(v2 + tv1)+ c3etw1 and y0 = y(0) = c1v1 +c2v2 +c3w1 : Solving we obtain c1 =¡1; c2 = 2; c3 =¡2; so y(t) =2 4 e¡t(6t¡7) + 6et e¡t(2t¡1) e¡t(4t) ¡ 2et 3 5: Therefore x(t) = Q¡1y(t) = 2 4 e¡t(6t¡7)+6et e¡t(2t¡1) e¡t(¡2t +3)¡2et 3 5: (b) If Q = 2 4 1 0 0 0 1 0 0 3 1 3 5 then Q¡1 = 2 4 1 0 0 0 1 0 0 ¡3 1 3 5 and H = QAQ¡1 = 2 4 2 4 ¡1 ¡3 ¡4 1 0 3 ¡1 3 5 isinunreducedHessenbergform. H hascharacter- istic polynomial p(t) = (t+1)3: The eigenvalue ? =¡1 has corresponding eigenvector v1 = [1;0;3]T: The system (H ¡(¡1)I)u = v1 has solution u= [¡1+(1=3)a;1; a]T; where a is arbitrary. Therefore v2= [0;1;3]T is a generalized eigenvector of order 2 for ? =¡1: The system (H ¡(¡1)I)u = v2 has solution u= [(¡4=3)+(1=3)a;1; a]T so v3 = [0;1;4]T is a generalized eigenvector of order 3 for ? =¡1: 214 CHAPTER 7. EIGENVALUES AND APPLICATIONS Set y (t) = Qx (t) and y0 = Qx0 = [¡1;¡1;¡2]T: The system y0 = Hy has general solution y(t) = c1e¡tv1 +c2e¡t(v1 +tv1)+c3e¡t(v3 +tv2 +(t2 =2)v1) and y0 = y(0) = c1v1 + c2v2 + c3v3: Solving we obtain c1 = ¡1; c2 = ¡5; c3 = 4; so y (t) = 2 4 e¡t(2t2 ¡5t¡1) e¡t(4t¡1) e¡t(6t2 ¡3t¡2) 3 5: Therefore x(t) = Q¡1y(t) = 2 4 e¡t(2t2 ¡5t¡1) e¡t(4t¡1) e¡t(6t2 ¡15t +1) 3 5: (c) If Q = 2 4 1 0 0 0 1 0 0 3 1 3 5 then Q¡1 = 2 4 1 0 0 0 1 0 0 ¡3 1 3 5 and H = QAQ¡1 = 2 4 1 4 ¡1 ¡3 ¡5 1 0 3 ¡2 3 5 is in unreduced Hessenberg form. H has characteristic polynomial p(t) = (t + 2)3: The eigenvalue ? = ¡2 has eigenvector v1 = [1;0;3]T; generalized eigenvector v2 = [0;1;3]T of order two, and generalized eigenvector v3 = [0;1;4]T of order 3. Set y(t) = Qx(t) and y0 = Qx0 = [¡1;¡1;¡2]T: The system y0 = Hy has general solution y(t) = e¡2t[c1v1 + c2(v2 + tv1) + c3(v3 + tv2 + (t2 =2)v1)] and y0 = y(0) = c1v1 + c2v2 + c3v3: Solving yields c1 =¡1; c2 =¡5; and c3 = 4; so y(t) = 2 4 e¡2t(2t2 ¡5t¡1) e¡2t(4t¡1) e¡2t(6t2 ¡3t¡2) 3 5: Therefore x(t) = Q¡1y(t) = 2 4 e¡2t(2t2 ¡5t¡1) e¡2t(4t¡1) e¡2t(6t2 ¡15t +1) 3 5: 4. x(t) = 2 66 4 etc4 et(c4t + c3) et(c4t2=2+ c3t + c2) et(c4t3=6+ c3t2=2+ c2t + c1) 3 77 5: 5. We see that from part(c) of Exercise 1 that x(t) = c1etv1 +c2et(v2 +tv1 )+ c3e¡tw1 = c1et 2 4 ¡2 0 1 3 5+ c2et 0 @ 2 4 5=2 1=2 0 3 5+ t 2 4 ¡2 0 1 3 5 1 A+ c3e¡t 2 4 ¡9 ¡1 1 3 5: 6. Note that (H ¡?I)v2 6= since v1 6= . But (H ¡?I)2v2 = (H ¡ ?I)v1 = , so v2 is a generalized eigenvector of order 2. Suppose we have seen that vj is a generalized eigenvector of order j for 1 ? j ? k where 1 ? k < m: Then (H ¡?I)kvk+1 = (H ¡?I)k¡1vk 6= whereas (H ¡?I)k+1vk+1 = (H ¡?I)kvk = . 7.8. GENERALIZED EIGENVECTORS & DIFF. EQNS. 215 Therefore vk+1 is a generalized eigenvector of order k +1: It follows by induction that vr is a generalized eigenvector of order r for 1? r ? m: Clearly the set fv1g is linearly independent since v1 6= . Suppose we have seen that the set fv1; : : : ;vkg is linearly independent for some k;1 ? k < m: Now assume that c1v1 +¢¢¢+ ckvk+ ck+1vk+1 = . Note that (H ¡?I)kvj = for 1? j ? k whereas (H ¡?I)kvk+1 = v1: It follows that = (H ¡ ?I)k =(H ¡ ?I)k(c1v1 +¢¢¢+ ckvk + ck+1vk+1) = ck+1v1: Therefore ck+1 = 0: Since the set fv1; : : : ;vkg is linearly independent, c1 = ¢¢¢ = ck = 0: This proves that fv1; : : : ;vk;vk+1g is a linearly independent set. It follows by induction that fv1; : : : ;vmg is linearly independent. 7. Note that Hvj = ?vj +vj¡1 for 2? j ? r whereas Hv1 = ?v1: It is straightforward to see that xr 0(t) = Hxr(t): 8. First note that q(H)(H ¡?1I)m1¡1 = (H ¡?1I)m1¡1q(H): It follows from the equations in (5) that if Eq.(9) is multiplied by (H ¡?1I)m1¡1 then we obtain am1q(H)v1 = . Since v1 is an eigenvector corresponding to ?1; q(H)v1 = q(?1)v1 6= since v1 6= and q(?1) 6= 0: Therefore am1 = 0: By a similar argument, multiplication of (9) by (H¡?1I)m1¡2 shows that am1¡1 = 0: We may continue the process to show that aj = 0 for each j;1? j ? m1: 216 CHAPTER 7. EIGENVALUES AND APPLICATIONS 7.9 Supplementary Exercises 1. A = ? 1 a 3¡a 1 ? , a arbitrary. 2. a = 0 or a =¡6. 3. (a) If x = [1; ¡1]T then q(x) =¡2. (b) The matrix B is not symmetric. 5. (a) L = ? 2 0 3 1 ? (b) L = 2 4 1 0 0 3 2 0 1 2 1 3 5 7.10 Conceptual Exercises 1. Let A have characterstic polynomial p(t) = t3 +ut2 +vt+w: Since A is nonsingular, ? = 0 is not an eigenvalue for A. Therefore, w 6= 0. Since A3 + uA2 + vA + wI = O, it follows that [(¡1=w)A2 ¡(u=w)A¡(v=w)I]A = I. 2. If B = P¡1AP then Bn = (P¡1AP)n = P¡1AnP. It follows that p(B) = p(P¡1AP) = P¡1p(A)P. 3. For 1? i ? n, aii = eTi Aei > 0. solMain.dvi