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Math 221: LINEAR ALGEBRA Chapter 3. Determinants and Diagonalization - PowerPoint PPT Presentation

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Math 221: LINEAR ALGEBRA Chapter 3. Determinants and Diagonalization 3-3. Determinants and Diagonalization Le Chen 1 Emory University, 2020 Fall (last updated on 10/21/2020) Creative Commons License (CC BY-NC-SA) 1 Slides are adapted from

  1. Example (continued) Therefore, the eigenvalues of A are λ 1 = 3 , λ 2 = 2 , and λ 3 = 1 . x = � Basic eigenvectors corresponding to λ 1 = 3 : solve (3 I − A ) � 0 . − 1    1 0 0  0 4 − 2 0 2  → · · · → − 1 5 − 2 0 − 1 0 1 0    2 − 1 5 − 2 0 0 0 0 0 1 1  2 t    2  = t 1 1 Thus � x = 2 t  , t ∈ R .   2 t 1  1   as a basic eigenvector corresponding to Choosing t = 2 gives us � x 1 = 1  2 λ 1 = 3 .

  2. Example (continued) x = � Basic eigenvectors corresponding to λ 2 = 2 : solve (2 I − A ) � 0 .    1 0 − 2 0  − 1 4 − 2 0  → · · · → − 1 4 − 2 0 0 1 − 1 0    − 1 5 − 3 0 0 0 0 0

  3. Example (continued) x = � Basic eigenvectors corresponding to λ 2 = 2 : solve (2 I − A ) � 0 .    1 0 − 2 0  − 1 4 − 2 0  → · · · → − 1 4 − 2 0 0 1 − 1 0    − 1 5 − 3 0 0 0 0 0 2 s    2   = s Thus � x =  , s ∈ R . s 1   s 1

  4. Example (continued) x = � Basic eigenvectors corresponding to λ 2 = 2 : solve (2 I − A ) � 0 .    1 0 − 2 0  − 1 4 − 2 0  → · · · → − 1 4 − 2 0 0 1 − 1 0    − 1 5 − 3 0 0 0 0 0 2 s    2   = s Thus � x =  , s ∈ R . s 1   s 1  2   as a basic eigenvector corresponding to Choosing s = 1 gives us � x 2 = 1  1 λ 2 = 2 .

  5. Example (continued) x = � Basic eigenvectors corresponding to λ 3 = 1 : solve ( I − A ) � 0 .    1 0 − 1 0  − 2 4 − 2 0  → · · · → − 1 3 − 2 0 0 1 − 1 0    − 1 5 − 4 0 0 0 0 0

  6. Example (continued) x = � Basic eigenvectors corresponding to λ 3 = 1 : solve ( I − A ) � 0 .    1 0 − 1 0  − 2 4 − 2 0  → · · · → − 1 3 − 2 0 0 1 − 1 0    − 1 5 − 4 0 0 0 0 0 r    1   = r Thus � x =  , r ∈ R . r 1   r 1

  7. Example (continued) x = � Basic eigenvectors corresponding to λ 3 = 1 : solve ( I − A ) � 0 .    1 0 − 1 0  − 2 4 − 2 0  → · · · → − 1 3 − 2 0 0 1 − 1 0    − 1 5 − 4 0 0 0 0 0 r    1   = r Thus � x =  , r ∈ R . r 1   r 1  1   as a basic eigenvector corresponding to Choosing r = 1 gives us � x 3 = 1  1 λ 3 = 1 .

  8. Geometric Interpretation of Eigenvalues and Eigenvectors Let A be a 2 × 2 matrix. Then A can be interpreted as a linear transformation from R 2 to R 2 .

  9. Geometric Interpretation of Eigenvalues and Eigenvectors Let A be a 2 × 2 matrix. Then A can be interpreted as a linear transformation from R 2 to R 2 . Problem How does the linear transformation affect the eigenvectors of the matrix?

  10. Geometric Interpretation of Eigenvalues and Eigenvectors Let A be a 2 × 2 matrix. Then A can be interpreted as a linear transformation from R 2 to R 2 . Problem How does the linear transformation affect the eigenvectors of the matrix? Definition � a � be a nonzero vector in R 2 . Then L � Let � v = v is the set of all scalar b multiples of � v, i.e., L � v = R � v = { t � v | t ∈ R } .

  11. Example (revisited) � 4 − 2 � A = has two eigenvalues: λ 1 = 2 and λ 2 = 5 with − 1 3 corresponding eigenvectors � 1 � � − 1 � v 1 = � and � v 2 = 1 1/2

  12. � 4 − 2 � A = − 1 3 y 6 5 4 3 2 1 x − 5 − 4 − 3 − 2 − 1 1 2 3 4 5 6 − 1 − 2 − 3 − 4

  13. � 4 − 2 � A = − 1 3 y 6 5 4 3 2 1 x − 5 − 4 − 3 − 2 − 1 1 2 3 4 5 6 − 1 − 2 − 3 − 4

  14. � 4 − 2 � A = − 1 3 y 6 5 4 3 2 1 x − 5 − 4 − 3 − 2 − 1 1 2 3 4 5 6 − 1 − 2 − 3 − 4

  15. � 4 − 2 � A = − 1 3 y 6 5 4 3 2 1 x − 5 − 4 − 3 − 2 − 1 1 2 3 4 5 6 − 1 − 2 − 3 − 4

  16. � 4 − 2 � A = − 1 3 y 6 5 4 3 2 1 x − 5 − 4 − 3 − 2 − 1 1 2 3 4 5 6 − 1 − 2 − 3 − 4

  17. � 4 − 2 � A = − 1 3 y 6 5 4 3 2 1 x − 5 − 4 − 3 − 2 − 1 1 2 3 4 5 6 − 1 − 2 − 3 − 4

  18. � 4 − 2 � A = − 1 3 y 6 5 4 3 2 1 x − 5 − 4 − 3 − 2 − 1 1 2 3 4 5 6 − 1 − 2 − 3 − 4

  19. � 4 − 2 � A = − 1 3 y 6 5 4 3 2 1 x − 5 − 4 − 3 − 2 − 1 1 2 3 4 5 6 − 1 − 2 − 3 − 4

  20. � 4 − 2 � A = − 1 3 y 6 5 4 3 2 1 x − 5 − 4 − 3 − 2 − 1 1 2 3 4 5 6 − 1 − 2 − 3 − 4

  21. � 4 − 2 � A = − 1 3 y 6 5 4 3 2 1 x − 5 − 4 − 3 − 2 − 1 1 2 3 4 5 6 − 1 − 2 − 3 − 4

  22. � 4 − 2 � A = − 1 3 y 6 5 4 3 2 1 x − 5 − 4 − 3 − 2 − 1 1 2 3 4 5 6 − 1 − 2 − 3 − 4

  23. � 4 − 2 � A = − 1 3 y 6 5 4 3 2 1 x − 5 − 4 − 3 − 2 − 1 1 2 3 4 5 6 − 1 − 2 − 3 − 4

  24. � 4 − 2 � A = − 1 3 y 6 5 4 3 2 1 x − 5 − 4 − 3 − 2 − 1 1 2 3 4 5 6 − 1 − 2 − 3 − 4

  25. � 4 − 2 � A = − 1 3 y 6 5 4 3 2 1 x − 5 − 4 − 3 − 2 − 1 1 2 3 4 5 6 − 1 − 2 − 3 − 4

  26. Definition Let A be a 2 × 2 matrix and L a line in R 2 through the origin. Then L is said to be A-invariant if the vector A � x lies in L whenever � x lies in L,

  27. Definition Let A be a 2 × 2 matrix and L a line in R 2 through the origin. Then L is said to be A-invariant if the vector A � x lies in L whenever � x lies in L, i.e., A � x is a scalar multiple of � x,

  28. Definition Let A be a 2 × 2 matrix and L a line in R 2 through the origin. Then L is said to be A-invariant if the vector A � x lies in L whenever � x lies in L, i.e., A � x is a scalar multiple of � x, i.e., A � x = λ� x for some scalar λ ∈ R ,

  29. Definition Let A be a 2 × 2 matrix and L a line in R 2 through the origin. Then L is said to be A-invariant if the vector A � x lies in L whenever � x lies in L, i.e., A � x is a scalar multiple of � x, i.e., A � x = λ� x for some scalar λ ∈ R , i.e., � x is an eigenvector of A.

  30. Definition Let A be a 2 × 2 matrix and L a line in R 2 through the origin. Then L is said to be A-invariant if the vector A � x lies in L whenever � x lies in L, i.e., A � x is a scalar multiple of � x, i.e., A � x = λ� x for some scalar λ ∈ R , i.e., � x is an eigenvector of A. Theorem (A-Invariance) v � = 0 be a vector in R 2 . Then L � Let A be a 2 × 2 matrix and let � v is A-invariant if and only if � v is an eigenvector of A.

  31. � 4 − 2 � A = − 1 3 y 6 5 4 3 2 1 x − 5 − 4 − 3 − 2 − 1 1 2 3 4 5 6 − 1 − 2 − 3 − 4

  32. � 4 − 2 � A = − 1 3 y 6 5 4 3 2 1 x − 5 − 4 − 3 − 2 − 1 1 2 3 4 5 6 − 1 − 2 − 3 − 4

  33. � 4 − 2 � A = − 1 3 y 6 5 4 3 2 1 x − 5 − 4 − 3 − 2 − 1 1 2 3 4 5 6 − 1 − 2 − 3 − 4

  34. Problem Let m ∈ R and consider the linear transformation Q m : R 2 → R 2 , i.e., reflection in the line y = mx. y y = mx x Recall that this is a matrix transformation induced by � 1 − m 2 1 � 2 m A = . m 2 − 1 1 + m 2 2 m Find the lines that pass through origin and are A-invariant. Determine corresponding eigenvalues.

  35. Solution y y = mx x

  36. Solution y y = mx x

  37. Solution y y = mx x

  38. Solution y y = mx x

  39. Solution y y = mx x � 1 � Let � x 1 = . Then L � x 1 is A-invariant, that is, � x 1 is an eigenvector. m Since the vector won’t change, its eigenvalue should be 1 . Indeed, one can verify that � 1 − m 2 � � 1 � 1 1 � � 2 m A � x 1 = = ... = = � x 1 . m 2 − 1 1 + m 2 2 m m m

  40. Solution (continued) y y = mx x

  41. Solution (continued) y y = mx x

  42. Solution (continued) y y = mx x

  43. Solution (continued) y y = mx x

  44. Solution (continued) y y = mx x � − m � Let � x 2 = . Then L � x 2 is A-invariant, that is, � x 2 is an eigenvector. 1 Since the vector won’t change the size, only flip the direction, its eigenvalue should be − 1 . Indeed, one can verify that � 1 − m 2 � m 1 � � − m � � 2 m A � x 2 = = · · · = = − � x 2 . m 2 − 1 1 − 1 1 + m 2 2 m

  45. Example Let θ be a real number, and R θ : R 2 → R 2 rotation through an angle of θ , induced by the matrix � cos θ � − sin θ A = . sin θ cos θ

  46. Example Let θ be a real number, and R θ : R 2 → R 2 rotation through an angle of θ , induced by the matrix � cos θ � − sin θ A = . sin θ cos θ Claim: A has no real eigenvalues unless θ is an integer multiple of π , i.e., ± π, ± 2 π, ± 3 π , etc.

  47. Example Let θ be a real number, and R θ : R 2 → R 2 rotation through an angle of θ , induced by the matrix � cos θ � − sin θ A = . sin θ cos θ Claim: A has no real eigenvalues unless θ is an integer multiple of π , i.e., ± π, ± 2 π, ± 3 π , etc. Consequence: a line L in R 2 is A invariant if and only if θ is an integer multiple of π .

  48. Diagonalization Denote an n × n diagonal matrix by  a 1 0 0 · · · 0 0  0 a 2 0 · · · 0 0    0 0 a 3 · · · 0 0    diag ( a 1 , a 2 , . . . , a n ) =  . . . . . .  . . . . . .   . . . . . .     0 0 0 · · · a n − 1 0   0 0 0 · · · 0 a n

  49. Diagonalization Denote an n × n diagonal matrix by  a 1 0 0 · · · 0 0  0 a 2 0 · · · 0 0    0 0 a 3 · · · 0 0    diag ( a 1 , a 2 , . . . , a n ) =  . . . . . .  . . . . . .   . . . . . .     0 0 0 · · · a n − 1 0   0 0 0 · · · 0 a n Recall that if A is an n × n matrix and P is an invertible n × n matrix so that P − 1 AP is diagonal, then P is called a diagonalizing matrix of A, and A is diagonalizable.

  50. columns vectors into a matrix: Pack the above ... ◮ Suppose we have n eigenvalue-eigenvector pairs: A � x j = λ j � x j , j = 1 , 2 , . . . , n

  51. ... ◮ Suppose we have n eigenvalue-eigenvector pairs: A � x j = λ j � x j , j = 1 , 2 , . . . , n ◮ Pack the above n columns vectors into a matrix: � � � � A � x 1 A � x 2 · · · A � x n = λ 1 � x 1 λ 2 � x 2 · · · λ n � x n || � � A � x 1 � x 2 · · · � x n ||  λ 1  λ 2   � x 1 x 2 · · · x n � � � �       λ n

  52. ◮ By denoting: � � P = � x 1 � x 2 · · · � x n and D = diag ( λ 1 , · · · , λ n ) we see that AP = PD

  53. ◮ By denoting: � � P = � x 1 � x 2 · · · � x n and D = diag ( λ 1 , · · · , λ n ) we see that AP = PD ◮ Hence, provided P is invertible, we have A = PDP − 1 D = P − 1 AP or equivalently

  54. ◮ By denoting: � � P = � x 1 � x 2 · · · � x n and D = diag ( λ 1 , · · · , λ n ) we see that AP = PD ◮ Hence, provided P is invertible, we have A = PDP − 1 D = P − 1 AP or equivalently that is, A is diagonalizable.

  55. Theorem (Matrix Diagonalization) Let A be an n × n matrix. 1. A is diagonalizable if and only if it has eigenvectors � x 1 ,� x 2 , . . . ,� x n so that � � P = � x 1 � x 2 · · · � x n is invertible.

  56. Theorem (Matrix Diagonalization) Let A be an n × n matrix. 1. A is diagonalizable if and only if it has eigenvectors � x 1 ,� x 2 , . . . ,� x n so that � � P = � x 1 x 2 � · · · � x n is invertible. 2. If P is invertible, then P − 1 AP = diag ( λ 1 , λ 2 , . . . , λ n ) where λ i is the eigenvalue of A corresponding to the eigenvector � x i , i.e., A � x i = λ i � x i .

  57. Example   3 − 4 2  has eigenvalues and corresponding basic eigenvectors A = 1 − 2 2  1 − 5 5   1  ; λ 1 = 3 and � x 1 = 1  2  2   ; λ 2 = 2 and x 2 = 1 �  1  1   . λ 3 = 1 and � x 3 = 1  1

  58. Example (continued)   1 2 1 � x 1 x 2 x 3 � 1 1 1 Let P = � � � =  .  2 1 1

  59. Example (continued)   1 2 1 � x 1 x 2 x 3 � 1 1 1 Let P = � � � =  .  2 1 1

  60. Example (continued)   1 2 1 � x 1 x 2 x 3 � 1 1 1 Let P = � � � =  . Then P is invertible (check  2 1 1 this!), so by the above Theorem,  3 0 0  P − 1 AP = diag (3 , 2 , 1) =  . 0 2 0  0 0 1

  61. Remark It is not always possible to find n eigenvectors so that P is invertible.

  62. Remark It is not always possible to find n eigenvectors so that P is invertible. Example   1 − 2 3 Let A = 2 6 − 6  .  1 2 − 1

  63. Remark It is not always possible to find n eigenvectors so that P is invertible. Example   1 − 2 3 Let A = 2 6 − 6  .  1 2 − 1 Then � � x − 1 2 − 3 � � � � = · · · = ( x − 2) 3 . − 2 x − 6 6 c A ( x ) = � � � � − 1 − 2 x + 1 � �

  64. Remark It is not always possible to find n eigenvectors so that P is invertible. Example   1 − 2 3 Let A = 2 6 − 6  .  1 2 − 1 Then � � x − 1 2 − 3 � � � � = · · · = ( x − 2) 3 . − 2 x − 6 6 c A ( x ) = � � � � − 1 − 2 x + 1 � � A has only one eigenvalue, λ 1 = 2 , with multiplicity three. Sometimes, one writes λ 1 = λ 2 = λ 3 = 2 .

  65. Example (continued) x = � To find the 2 -eigenvectors of A, solve the system (2 I − A ) � 0 .     1 2 − 3 0 1 2 − 3 0  → · · · → − 2 − 4 6 0 0 0 0 0    − 1 − 2 3 0 0 0 0 0

  66. Example (continued) x = � To find the 2 -eigenvectors of A, solve the system (2 I − A ) � 0 .     1 2 − 3 0 1 2 − 3 0  → · · · → − 2 − 4 6 0 0 0 0 0    − 1 − 2 3 0 0 0 0 0 The general solution in parametric form is       − 2 s + 3 t − 2 3  = s  + t  , s , t ∈ R . � x = s 1 0    t 0 1

  67. Example (continued) x = � To find the 2 -eigenvectors of A, solve the system (2 I − A ) � 0 .     1 2 − 3 0 1 2 − 3 0  → · · · → − 2 − 4 6 0 0 0 0 0    − 1 − 2 3 0 0 0 0 0 The general solution in parametric form is       − 2 s + 3 t − 2 3  = s  + t  , s , t ∈ R . � x = s 1 0    t 0 1 Since the system has only two basic solutions, there are only two basic eigenvectors, implying that the matrix A is not diagonalizable.

  68. det Example   1 0 1 Diagonalize, if possible, the matrix A = 0 1 0  .  0 0 − 3

  69. Example   1 0 1 Diagonalize, if possible, the matrix A = 0 1 0  .  0 0 − 3 � � x − 1 0 − 1 � � � � = ( x − 1) 2 ( x + 3) . c A ( x ) = det ( xI − A ) = 0 x − 1 0 � � � � 0 0 x + 3 � �

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