1. Matrices CHAPTER 8.1 ~ 8.8

2. Ch8.1-8.8_2 Contents  8.1 Matrix Algebra 8.1 Matrix Algebra  8.2 Systems of Linear Algebra Equations 8.2 Systems of Linear Algebra Equations  8.3 Rank of a Matrix 8.3 Rank of a Matrix  8.4 Determinants 8.4 Determinants  8.5 Properties of Determinants 8.5 Properties of Determinants  8.6 Inverse of a Matrix 8.6 Inverse of a Matrix  8.7 Cramer’s Rule 8.7 Cramer’s Rule  8.8 The Eigenvalue Problem 8.8 The Eigenvalue Problem

3. Ch8.1-8.8_3 8.1 Matrix Algebra  Matrix The forms (x 1 x 2 … x n ) or(1) Each array in (1) is called a matrix.

4. Ch8.1-8.8_4  Entry or Element: a mn, the members in the array Size: m  n Square matrix: n  n, n is called the order. Main diagonal entries: a nn A matrix is any rectangular array of numbers or functions (2) DEFINITION 8.1 Matrix

5. Ch8.1-8.8_5 A n  1 matrix is called a column vector. A 1  n matrix is called a row vector. DEFINITION 8.2 Column and Row Vectors Two matrices A, B are equal if aij = bij. DEFINITION 8.4 Equality of Matrices

6. Ch8.1-8.8_6 Example 1 (a) The following matrix are not equal, since they are not of the same size. (b) The following matrix are not equal, since the corresponding entries are not all equal.

7. Ch8.1-8.8_7 The sum of two m  n matrices A, B is A + B = (a ij + b ij ) m  n DEFINITION 8.4 Matrix Addition

8. Ch8.1-8.8_8 Example 2 (a)

9. Ch8.1-8.8_9 Example 2 (2) (b) The sum is not defined.

10. Ch8.1-8.8_10 If k is a real number, then the scalar multiple is DEFINITION 8.5 Scalar Multiple of a Matrix

11. Ch8.1-8.8_11 If A, B, C are m  n matrices, k 1 and k 2 are scalars (i) A + B = B + A (ii) A + (B + C) = (A + B) + C (iii) (k 1 k 2 ) A = k 1 (k 2 A) (iv) 1 A = A (v) k 1 (A + B) = k 1 A + k 1 B (vi) (k 1 + k 2 ) A = k 1 A + k 2 A THEOREM 8.1 Matrix Multiplication

12. Ch8.1-8.8_12 If A is m  p, B is p  n, then the product is DEFINITION 8.6 Matrix Multiplication

13. Ch8.1-8.8_13 Example 3 (a) (b) Note: In general, AB  BA

14. Ch8.1-8.8_14  We can write a system of linear equations as the form of product. eg:  Associate Law: A(BC) = (AB)C  Distributive Law: A(B + C) = AB + BC

15. Ch8.1-8.8_15 The transpose of (2) is DEFINITION 8.7 Transpose of a Matrix

16. Ch8.1-8.8_16  Note: (A + B + C) T = A T + B T + C T (ABC) T = C T B T A T Suppose A and B are matrices and k a scalar, (i) (A T ) T = A (ii) (A + B) T = A T + B T (iii) (AB) T = B T A T (iv) (kA) T = kA T THEOREM 8.2 Properties of Transpose

17. Ch8.1-8.8_17 Zero Matrices  A + 0 = A(4) andA + (–A) = 0(5)

18. Ch8.1-8.8_18 Triangular Matrices  For square n  n matrices

19. Ch8.1-8.8_19 Diagonal Matrices  For square n  n matrices, i≠j, a ij = 0

20. Ch8.1-8.8_20 Scalar Matrices  Diagonal matrices with equal a ii

21. Ch8.1-8.8_21 Identity Matrices  A: m  n, then I m A = A I n = A

22. Ch8.1-8.8_22 Symmetric  An n × n matrix A is said to be symmetric if A T = A.

23. Ch8.1-8.8_23 8.2 Systems of Linear Algebraic Equations  General Form (1)

24. Ch8.1-8.8_24 Solution  A solution of a (1) is said to be consistent if it has at least one solution and inconsistent if it has no solutions.  If a linear system is consistent: (i) a unique solution (ii) infinitely many solutions See Fig 8.2

25. Ch8.1-8.8_25 Fig 8.2

26. Ch8.1-8.8_26 Example 1 Verify that x 1 = 14 + 7t, x 2 = 9 + 6t, x = t, where t is any real number, is a solution of 2x 1 – 3x 2 + x 3 = 1 x 1 – x 2 – x 3 = 5 Solution 2(14 + 7t) – 3(9 + 6t) + 4t = 1 14 + 7t – (9 + 6t) – t = 5 For each number of t, we obtain a different solution.

27. Ch8.1-8.8_27 Solving Systems: Elementary Operations  (i) Multiply an equation by a nonzero constant. (ii) Interchange the position of equations (iii) Add a nonzero multiple of one equation to another one.

28. Ch8.1-8.8_28 Example 2 Solve 2x 1 + 6x 2 + x 3 = 7 x 1 + 2x 2 – x 3 = –1 5x 1 + 7x 2 – 4x 3 = 9 Solution (i) Interchange positions x 1 + 2x 2 – x 3 = –1 …(a) 2x 1 + 6x 2 + x 3 = 7…(b) 5x 1 + 7x 2 – 4x 3 = 9…(c) (ii) −2  (a) + (b) x 1 + 2x 2 – x 3 = –1 …(a) 2x 2 + 3x 3 = 9…(b)’ 5x 1 + 7x 2 – 4x 3 = 9…(c)

29. Ch8.1-8.8_29 Example 2 (2) (iii) −5  (a) + (c) x 1 + 2x 2 – x 3 = –1 …(a) 2x 2 + 3x 3 = 9…(b)’ –3x 2 + x 3 = 14…(c)’ (iv) From (b)’ and (c)’, we have x 2 = −3, x 3 = 5, then x 1 = 10

30. Ch8.1-8.8_30 Augmented Matrix  To solve (1) we can use the augmented matrix (2)

31. Ch8.1-8.8_31 Example 3  (a) The augmented matrix represents x 1 – 3x 2 + 5x 3 = 2 4x 1 + 7x 2 – x 3 = 8  (b)x 1 – 5x 3 = – 1 x 1 + 0x 2 – 5x 3 = – 1 2x 1 + 8x 2 = 7 and2x 1 + 8x 2 + 0x 3 = 7 x 2 + 9x 3 = 10x 1 + x 2 + 9x 3 = 1 are the same. Thus the matrix of the system is

32. Ch8.1-8.8_32 Elementary Row Operations  (i) Multiply a row by a nonzero constant. (ii) Interchange the position of rows (iii) Add a nonzero multiple of one row to another one.  Matrices after elementary row operations are called row equivalent. The procedure is called row reduction.

33. Ch8.1-8.8_33 Example 4  (a) and are in row-echelon form.  (b) and are in reduced row-echelon form.

34. Ch8.1-8.8_34 Example 5 Solve the equations in example 2. Solution (a)

35. Ch8.1-8.8_35 Example 5 (2) and x 3 = 5, x 2 = –3, x 1 = 10

36. Ch8.1-8.8_36 Example 5 (3) (b) we have the same solution.

37. Ch8.1-8.8_37 Example 6 Use Gauss-Jordan method to solve x 1 + 3x 2 – 2x 3 = – 7 4x 1 + x 2 + 3x 3 = 5 2x 1 – 5x 2 + 7x 3 = 19 Solution

38. Ch8.1-8.8_38 Example 6 (2) We havex 2 – x 3 = –3 x 1 + x 3 = 2 Let x 3 = t, then x 2 = –3 + t, x 1 = 2 – t.

39. Ch8.1-8.8_39 Example 7 Solve x 1 + x 2 = 1 4x 1 − x 2 = −6 2x 1 – 3x 2 = 8 Solution We have 0 + 0 = 16, no solutions.

40. Ch8.1-8.8_40 Networks  From Fig 8.3, we have or(3)

41. Ch8.1-8.8_41 Fig 8.3

42. Ch8.1-8.8_42 Example 8 Solve (3) where R 1 = 10 ohms, R 2 = 20 ohms, R 2 = 10 ohms, E = 12 volts Solution From the data, we have i 1 – i 2 – i 3 = 0 10i 1 + 20i 2 = 12 20i 2 – 10i 3 = 0

43. Ch8.1-8.8_43 Example 8 (2) Use Gauss-Jordan method We have i 1 = 18/25, i 2 = 6/25, i 3 = 12/25.

44. Ch8.1-8.8_44 Homogeneous Systems  The system of equations (4) is always consistent, since x 1 = x 2 = … = x n = 0 will satisfy the system.

45. Ch8.1-8.8_45 The system (4) possesses nontrivial solutions if the number m of equations is less than the number n of unknowns. THEOREM 8.3 Existence of Nontrivial Solutions

46. Ch8.1-8.8_46 Example 9 Solve 2x 1 − 4x 2 + 3x 3 = 0 x 1 + x 2 − 2x 3 = 0 Solution Use Gauss-Jordan method

47. Ch8.1-8.8_47 Example 9 (2) Let x 3 = t, then x 1 = (5/6)t, x 2 = (7/6)t, and the system has nontrivial solutions.

48. Ch8.1-8.8_48 Example 10: Chemical Equations Balance C 2 H 6 + O 2  CO 2 + H 2 O Solution Assuming x 1 C 2 H 6 + x 2 O 2  x 3 CO 2 + x 4 H 2 O, we have2x 1 = x 3 (for C) 6x 1 = 2x 4 (for H) 2x 2 = 2x 3 + 2x 4 (for O) Then

49. Ch8.1-8.8_49 Thus x 4 = t, x 1 = t/3, x 2 = 7t/6, x 3 = 2t/3. We choose t = 6, then x 1 = 2, x 2 = 7, x 3 = 4, x 4 = 6 2C 2 H 6 + 7O 2  4CO 2 + 6H 2 O Example 10 (2)

50. Ch8.1-8.8_50 8.3 Rank of a Matrix  Introduction Row vectors: u 1 = (a 11 a 12 … a 1n ), u 2 = (a 21 a 22, … a 2n ),…, u m = (a m1 a m2 … a mn )

51. Ch8.1-8.8_51 Column Vectors:

52. Ch8.1-8.8_52 The rank of a m  n matrix A, denoted by rank(A), is the maximum linearly independent row vectors. DEFINITION 8.8 Rank of a Matrix

53. Ch8.1-8.8_53 Example 1 Consider (1) The row vectors are in turn denoted as u 1, u 2, u 3. Since 4u 1 – ½u 2 + u 3 = 0, they are linearly dependent. In addition, u 1 and u 2 are not constant multiple couples, so they are linearly independent. rank(A) = 2

54. Ch8.1-8.8_54 Row Space  As in Example 1, Span(u 1, u 2, u 3 ) is called the Row Space of A. If a matrix A is row equivalent to a row-echelon form B, then (i) Row space of A = Row space of B (ii) The nonzero rows form a basis for the row space of A (iii)rank(A) = the number of nonzero rows in B THEOREM 8.4 Rank of a Matrix by Row Reduction

55. Ch8.1-8.8_55 Example 2 Consider the matrix in (1) rank(A) = 2

56. Ch8.1-8.8_56 Example 3 Determine whether the set of u 1 =, u 2 =, u 3 = is linearly independent. Solution Form a matrix A by using u 1, u 2, u 3, and reduce it. We have rank(A) = 3, so the set of vectors is linearly independent.

57. Ch8.1-8.8_57 Rank and Linear Systems  Consider Make an augmented matrix and reduce it:

58. Ch8.1-8.8_58 We have rank(A|B) = 3. Since then rank(A) = 2

59. Ch8.1-8.8_59 AX = B is consistent if and only if rank(A|B) = rank(A) THEOREM 8.5 Consistence of AX = B Suppose AX = B with m equations and n unknowns is consistent. If rank(A) = r, then the solution contains n – r parameters. THEOREM 8.6 Number of Parameters in a Solution

60. Ch8.1-8.8_60 Example x 1 + 3x 2 – 2x 3 = –7 4x 1 + x 2 + 3x 3 = 5(3) 2x 1 – 5x 2 + 7x 3 = 19 We already know that (3) is consistent and has infinitely many solutions. Since We have rank(A|B) = rank(A) = 2, then the number of parameter of the solution is 3 – 2 = 1.

61. Ch8.1-8.8_61 Flow chart AX = 0 Always consistent Unique solution X = 0 rank(A) = n Infinity of solutions Rank(A) < n n – r parameters

62. Ch8.1-8.8_62 AX = B, B≠0 Inconsistent rank(A) < rank(A│B) consistent rank(A) = rank(A│B) Unique solution rank(A) = n Infinity of solutions rank(A) < n n – r parameters

63. Ch8.1-8.8_63 8.4 Determinants  Notation

64. Ch8.1-8.8_64  eg: The determinant is (1) DEFINITION 8.9 Determinant of 2 × 2 Matrix

65. Ch8.1-8.8_65 The determinant is (1) DEFINITION 8.10 Determinant of 3 × 3 Matrix

66. Ch8.1-8.8_66 In view of (1), we have (4) where A has been expanded by cofactors along the first row, with the cofactors of a 11, a 12, a 13 : Thus det A = a 11 C 11 + a 12 C 12 + a 13 C 13 (5)

67. Ch8.1-8.8_67 In general, the cofactors of a ij is C ij = (–1) i+ j M ij (6) where M ij is called a minor determinant. From (3), we have (8) Similarly, we can expand by cofactors along the third rows: det A = a 31 C 31 + a 32 C 32 + a 33 C 33 (9) Note: We can expand by cofactors along any rows or any columns.

68. Ch8.1-8.8_68 Example 1 Find the determinant of Solution Along the first row:

69. Ch8.1-8.8_69 Example 1 (2)

70. Ch8.1-8.8_70  We also can expanded along the second row, since it has zero entry.

71. Ch8.1-8.8_71 Example 2 Find the determinant of Solution Along the third column:

72. Ch8.1-8.8_72 Let A = (a ij ) n × n ne an n × n matrix. For each the cofactor expansion of det A along the ith row is For each the cofactor expansion of det A along the ith column is THEOREM 8.7 Consistence of AX = B

73. Ch8.1-8.8_73 Example 3 Find the determinant of Solution Along the fourth row where

74. Ch8.1-8.8_74 Example 3 (2)

75. Ch8.1-8.8_75 Example 3 (3)

76. Ch8.1-8.8_76 8.5 Properties of Determinants If A T is the transpose of the n × n matrix A, then det A T = det A THEOREM 8.8 Determinant of a Transpose If any two rows (columns) of an n × n matrix A are the same, then det A = 0. THEOREM 8.9 Two Identical Rows

77. Ch8.1-8.8_77 Example 1

78. Ch8.1-8.8_78 If all the entries in a row (column) of an n × n matrix A are all zeros, then det A = 0. THEOREM 8.10 Zero Row or Column If B is the matrix obtained by interchanging any two rows (columns) of an n × n matrix A, then det B = −det A THEOREM 8.11 Interchanging Rows

79. Ch8.1-8.8_79  If B is obtained by interchanging the first and third row of

80. Ch8.1-8.8_80 If B is obtained from an n × n matrix A by multiplying a row (column) by a nonzero real number k, then det B = k det A THEOREM 8.12 Constant Multiple of a Row

81. Ch8.1-8.8_81 Example 2  (a)  (b)

82. Ch8.1-8.8_82 If A and B are both n × n matrices, then det AB = det A  det B. THEOREM 8.13 Determinant of a Matrix Product

83. Ch8.1-8.8_83 Example 3 Suppose then Now det AB = −24, det A = −8, det B = 3, Thus det AB = det A  det B.

84. Ch8.1-8.8_84 Suppose B is the matrix obtained from an n × n matrix A by multiplying the entries in a row (column) by Nonzero real number k and adding the result to the corresponding entries in another row (column). Then det B = det A. THEOREM 8.14 Determinant is Unchanged

85. Ch8.1-8.8_85 Example 4 We have det A = 45 = det B = 45.

86. Ch8.1-8.8_86 Proof Suppose A is an n × n matrix (upper or lower). Then det A = a 11 a 22 … a nn where a 11, a 22, …, a nn are the entries on the main diagonal of A. THEOREM 8.15 Determinant of a Triangular Product

87. Ch8.1-8.8_87 Example 5  (a)

88. Ch8.1-8.8_88  (b)

89. Ch8.1-8.8_89 Example 6 Find the determinant of Solution

90. Ch8.1-8.8_90 Example 6 (2)

91. Ch8.1-8.8_91 Suppose A is a n  n matrix. If a i1, a i2, …, a in are the entries in the ith row and C k1, C k2, …, C kn are the cofactors in the kth row, then a i1 C k1 + a i2 C k2 + …+ a in C kn = 0, for i  k If a 1j, a 2j, …, a nj are the entries in the jth column and C 1k, C 2k, …, C nk are the cofactors in the kth column, then a 1j C 1k + a 2j C 2k + …+ a nj C nk = 0, for j  k THEOREM 8.16 A Property of Cofactors

92. Ch8.1-8.8_92 THEOREM 8.16 Proof Let B be the matrix obtained from A by letting the entries in the ith row of A be the same as the ones in the kth row, that is, a i1 = a k1, a i2 = a k2, …, a in = a kn then there are two identical rows in B, det B = 0, then

93. Ch8.1-8.8_93 Example 7 Consider the matrix we have

94. Ch8.1-8.8_94 8.6 Inverse of a Matrix  Note: If A has no inverse, then A is called singular. Let A be an n  n matrix. If there exists an n  n matrix B such that AB = BA = I(1) where I is the n  n identity, then A is said to be nonsingular or invertible. Then B is said to be the inverse of A. DEFINITION 8.11 Inverse of a Matrix

95. Ch8.1-8.8_95 Proof (i)A -1 A = AA -1 = I, A = (A -1 ) -1 (ii)(AB)(AB) -1 = I, B -1 A -1 (AB)(AB) -1 = B -1 A -1 (AB) -1 = B -1 A -1 Let A, B be nonsingular matrices. (i)(A -1 ) -1 = A (ii)(AB) -1 = B -1 A -1 (iii)(A T ) -1 = (A -1 ) T THEOREM 8.17 Properties of the Inverse

96. Ch8.1-8.8_96 Let A be an n × n matrix. The matrix that is the Transpose of the matrix of cofactors corresponding to the entries of A: is called the adjoint of A and is denoted by adj A. DEFINITION 8.12 Adjoint Matrix

97. Ch8.1-8.8_97 Proof For n = 3, (3) Let A be an n × n matrix. If det A  0, then (2) THEOREM 8.18 Finding the Inverse

98. Ch8.1-8.8_98 From Theorem 8.16, we have Thus A -1 = adj A/det A

99. Ch8.1-8.8_99  For 2  2 matrix (4)

100. Ch8.1-8.8_100  For 3  3 matrix (5)

101. Ch8.1-8.8_101 Example 1 Find the inverse of Solution Check

102. Ch8.1-8.8_102 Example 2 Find the inverse of Solution Since

103. Ch8.1-8.8_103 Example 2 (2) We have

104. Ch8.1-8.8_104 An n  n matrix A is nonsingular if and only if det A  0 THEOREM 8.19 Nonsingular Matrices and det A

105. Ch8.1-8.8_105 Example 3 has no inverse. A is singular, since det A = 0

106. Ch8.1-8.8_106 An n  n matrix A can be transformed into the n  n identity matrix I by a sequence of elementary row operations, then A is nonsingular. The same operations that transforms A into the identity I will also transform I into A -1. THEOREM 8.20 Finding the Inverse

107. Ch8.1-8.8_107  First we construct the augmented matrix (A|I), and the process for finding A -1 is outlined.

108. Ch8.1-8.8_108 Row operations on A until I is obtained )IA(| )A|I( 1  Simultaneously applying the same operations

109. Ch8.1-8.8_109 Example 4 Find the inverse of Solution

110. Ch8.1-8.8_110 Example 4 (2)

111. Ch8.1-8.8_111 Example 4 (3) Thus

112. Ch8.1-8.8_112 Example 5 Find the inverse of Solution

113. Ch8.1-8.8_113 Example 5 (2) There is a row of zeros, it is singular

114. Ch8.1-8.8_114 Using the Inverse  A system of m linear equations in n unknowns (6) can be written as AX = B, where

115. Ch8.1-8.8_115 Special Case  When m = n, if A is nonsingular, then X = A -1 B(7)

116. Ch8.1-8.8_116 Example 6 Use the inverse to solve Solution The system can be written as Since, it is nonsingular. From (4)

117. Ch8.1-8.8_117 Example 6 (2) Using (7) Thus

118. Ch8.1-8.8_118 Example 7 Use the inverse to solve Solution From the equations we have

119. Ch8.1-8.8_119 Example 7 (2) Thus (7) gives Thus

120. Ch8.1-8.8_120 A homogeneous system of n linear equations in n unknowns AX = 0 has only the trivial solution if and only if A is nonsingular. THEOREM 8.21 Trivial Solution Only A homogeneous system of n linear equations in n unknowns AX = 0 has a nontrivial solution if and only if A is singular. THEOREM 8.22 Existence of Nontrivial Solutions

121. Ch8.1-8.8_121 8.7 Cramer’s Rule  Introduction For example, the equations (1) possesses the solution (2) where a 11 a 22 – a 12 a 21  0

122. Ch8.1-8.8_122 Rewrite (2) as determinant forms, we have (3)

123. Ch8.1-8.8_123 Special Matrix  For the following system (4)

124. Ch8.1-8.8_124  We define a special matrix

125. Ch8.1-8.8_125 Let A be the coefficient matrix of the system (1). If det A  0, then the solution of (1) is given by where A k, k = 1, 2, …, n, is defined in (5) THEOREM 8.23 Cramer’s Rule

126. Ch8.1-8.8_126 Proof

127. Ch8.1-8.8_127 Now the entry in the kth row is (7)

128. Ch8.1-8.8_128 Example 1 Solve Solution

129. Ch8.1-8.8_129 Example 1 (2) From (3), we have

130. Ch8.1-8.8_130 8.8 The Eigenvalue Problems Let A be n  n matrix. A number is said to be an eigenvalue of A if there exists a nonzero solution vector K of AK = K(1) The solution vector K is said to be an eigenvector corresponding to the eigenvalue. DEFINITION 8.13 Eigenvalues and Eigenvectors

131. Ch8.1-8.8_131 Example 1 Verify that is an eigenvector of the matrix Solution Since we conclude that K is an eigenvector of A.

132. Ch8.1-8.8_132  From (1), we have (A – I)K = 0(2) However, (2) is the same as a homogeneous system of linear equations. Since we want K to be nontrivial, we must have det (A – I) = 0(4) Inspection of (4) shows det (A – I) results in an nth- degree polynomial in, and is called the characteristic equation.

133. Ch8.1-8.8_133 Example 2 Find the eigenvalues and eigenvectors of Solution

134. Ch8.1-8.8_134 Example 2 (2) We have – 3 – 2 + 12 = 0 or ( + 4)( – 3) = 0 then = 0, −4, 3. To find the eigenvectors, (i) 1 = 0,

135. Ch8.1-8.8_135 Example 2 (3) Choose k 3 = −13, then

136. Ch8.1-8.8_136 Example 2 (4) (ii) For 2 = −4,

137. Ch8.1-8.8_137 Example 2 (5) implies k 1 = −k 3, k 2 = 2k 3. Choose k 3 = 1, then

138. Ch8.1-8.8_138 Example 2 (6) (iii) 2 = 3, implies k 1 = – k 3, k 2 = –(3/2)k 3. Choose k 3 = –2,

139. Ch8.1-8.8_139 Example 3 Find the eigenvalues and eigenvectors of Solution We see 1 = 2 = 5 is an eigenvalue of multiplicity 2. From (A – 5I|0), we get

140. Ch8.1-8.8_140 Example 3 (2) Choose k 2 = 1, we have k 1 = 2, then We can have only one eigenvector though A is a 2  2 matrix.

141. Ch8.1-8.8_141 Example 4 Find the eigenvalues and eigenvectors of Solution We see 1 = 11, 2 = 3 = 8 is of multiplicity 2.

142. Ch8.1-8.8_142 Example 4 (2) (i) For 1 = 11, Gauss-Jordan method gives Hence k 1 = k 3, k 2 = k 3. If k 3 = 1, then

143. Ch8.1-8.8_143 Example 4 (3) (ii) Now for 2 = 8, we have For k 1 + k 2 + k 3 = 0, we can select two of them arbitrarily. We choose: k 2 = 1, k 3 = 0, and k 2 = 0, k 3 = 1, then

144. Ch8.1-8.8_144 Proof Since AK = K, The proof is completed. Let A be a square matrix with real entries. If =  + i ,   0, is a complex eigenvalue of A, then its conjugate is also an eigenvalue of A. If K is an eigenvector corresponding to, then is an eigenvector corresponding to. THEOREM 8.24 Complex Eignvalues and Eigenvectors

145. Ch8.1-8.8_145 Example 5 Find the eigenvalues and eigenvectors of Solution For 1 = 5 + 2i,

146. Ch8.1-8.8_146 Example 5 (2) Since k 2 = (1 – 2i) k 1, after choosing k 1 = 1, then From Theorem 8.24, then

147. Ch8.1-8.8_147 The eigenvalues of an upper triangular, lower triangular, or diagonal matrix are the main diagonal entries. THEOREM 8.25 Triangular and Diagonal Matrices