Diagonalization

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  • 1. Announcements Quiz 4 will be on Thurs Feb 18 on sec 3.3, 5.1 and 5.2. Check the grade sheet for any mistakes or omissions.
  • 2. Last Class 1. Characteristic equation and characteristic polynomial of a square matrtix.
  • 3. Last Class 1. Characteristic equation and characteristic polynomial of a square matrtix. 2. Finding eigenvalues of a 2 × 2 matrix and the char.polynomial of a 3 × 3 matrix.
  • 4. Last Class 1. Characteristic equation and characteristic polynomial of a square matrtix. 2. Finding eigenvalues of a 2 × 2 matrix and the char.polynomial of a 3 × 3 matrix. 3. The char.equation of a 2 × 2 matrix is a quadratic equation which can be factorized (or use formula for solving quad. equation) to give the eigenvalues.
  • 5. Section 5.3 Diagonalization 1. To factorize the given matrix A in the form A = PDP −1 where D and P gives information about the eigenvalues and eigenvectors.
  • 6. Section 5.3 Diagonalization 1. To factorize the given matrix A in the form A = PDP −1 where D and P gives information about the eigenvalues and eigenvectors. 2. Useful in computing higher powers of A quickly (without multiplying A many times)
  • 7. Section 5.3 Diagonalization 1. To factorize the given matrix A in the form A = PDP −1 where D and P gives information about the eigenvalues and eigenvectors. 2. Useful in computing higher powers of A quickly (without multiplying A many times) 3. This factorization is very useful in "decoupling" complicated dynamical systems (dierential equations)
  • 8. Section 5.3 Diagonalization 1. To factorize the given matrix A in the form A = PDP −1 where D and P gives information about the eigenvalues and eigenvectors. 2. Useful in computing higher powers of A quickly (without multiplying A many times) 3. This factorization is very useful in "decoupling" complicated dynamical systems (dierential equations) 4. D in the above factorization stands for a diagonal matrix. Properties of diagonal matrices make life a lot easier.
  • 9. Powers of Diagonal Matrices Find A2 and A3 if 4 0 A = . 0 6
  • 10. Powers of Diagonal Matrices Find A2 and A3 if 4 0 A = . 0 6 Solution: 2 4 0 4 0 4.4 + 0.0 4.0 + 0.6 42 0 A = = = . 0 6 0 6 0.4 + 6.0 0.0 + 6.6 0 62
  • 11. Powers of Diagonal Matrices Find A2 and A3 if 4 0 A = . 0 6 Solution: 2 4 0 4 0 4.4 + 0.0 4.0 + 0.6 42 0 A = = = . 0 6 0 6 0.4 + 6.0 0.0 + 6.6 0 62 3 42 0 4 0 42 . 4 + 0. 0 4 . 0 + 0. 6 A = A2 A = = 0 62 0 6 0.4 + 6.0 0.0 + 62 .6 43 0 = . 0 63
  • 12. Powers of Diagonal Matrices What about A23 ?
  • 13. Powers of Diagonal Matrices What about A23 ? Based on the above "pattern": 23 423 0 A = . 0 623
  • 14. Powers of Diagonal Matrices What about A23 ? Based on the above "pattern": 23 423 0 A = . 0 623 For a general exponent k , k= 4k 0 0 6k A .
  • 15. Observations 1. Raising a diagonal matrix to a power is same as raising the diagonal elements to the same power and the result is still a diagonal matrix.
  • 16. Observations 1. Raising a diagonal matrix to a power is same as raising the diagonal elements to the same power and the result is still a diagonal matrix. 2. Please note that this will work for any diagonal matrix (3 × 3 or any size)
  • 17. Observations 1. Raising a diagonal matrix to a power is same as raising the diagonal elements to the same power and the result is still a diagonal matrix. 2. Please note that this will work for any diagonal matrix (3 × 3 or any size) 3. DO NOT do this to a general matrix (even a triangular matrix)
  • 18. Example Find A3 if 1 0 0 0    0 2 0 0  = .   A  0 0 3 0  0 0 0 4
  • 19. Example Find A3 if 1 0 0 0    0 2 0 0  = .   A  0 0 3 0  0 0 0 4 1 0 0 0   3  0 8 0 0  = .   A  0 0 27 0  0 0 0 64
  • 20. Diagonalization A square matrix A is diagonalizable if 1. A is similar to a diagonal matrix D which means
  • 21. Diagonalization A square matrix A is diagonalizable if 1. A is similar to a diagonal matrix D which means 2. We can write A = PDP −1 for some invertible matrix P .
  • 22. Diagonalization A square matrix A is diagonalizable if 1. A is similar to a diagonal matrix D which means 2. We can write A = PDP −1 for some invertible matrix P . If A = PDP −1 what is A2 ? 2 A = (PDP −1 )(PDP −1 ) = PD (P −1 P ) DP −1 = PD 2 P −1 I
  • 23. Diagonalization A square matrix A is diagonalizable if 1. A is similar to a diagonal matrix D which means 2. We can write A = PDP −1 for some invertible matrix P . If A = PDP −1 what is A2 ? 2 A = (PDP −1 )(PDP −1 ) = PD (P −1 P ) DP −1 = PD 2 P −1 I Similarly, 3 A = (PD 2 P −1 )(PDP −1 ) = PD 2 (P −1 P ) DP −1 = PD 3 P −1 A2 I and so on
  • 24. Taking Advantage of Diagonal Matrix To nd Ak of any square matrix A, 1. "Diagonalize" A, in other words factorize A as PDP −1 for suitable D and invertible P .
  • 25. Taking Advantage of Diagonal Matrix To nd Ak of any square matrix A, 1. "Diagonalize" A, in other words factorize A as PDP −1 for suitable D and invertible P . 2. Raise the diagonal entries of D to k , no change to P and P −1 .
  • 26. Taking Advantage of Diagonal Matrix To nd Ak of any square matrix A, 1. "Diagonalize" A, in other words factorize A as PDP −1 for suitable D and invertible P . 2. Raise the diagonal entries of D to k , no change to P and P −1 . 3. Find the product PD k P −1 . The following theorem says when exactly we can diagonalize a square matrix A. (very important)
  • 27. The Diagonalization Theorem Theorem An n ×n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors.
  • 28. The Diagonalization Theorem Theorem An n ×n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors. A = PDP −1 , where D is a diagonal matrix, if and only if the columns of P are n linearly independent eigenvectors of A.
  • 29. The Diagonalization Theorem Theorem An n ×n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors. A = PDP −1 , where D is a diagonal matrix, if and only if the columns of P are n linearly independent eigenvectors of A. If this is done, the diagonal entries of D are the eigenvalues of A that correspond to the respective eigenvectors in P .
  • 30. Notes 1. You can arrange the eigenvalues of A in any order you like to form D .
  • 31. Notes 1. You can arrange the eigenvalues of A in any order you like to form D . 2. Arrange the linearly independent eigenvectors of A as columns to form P . This should correspond to how you write D .
  • 32. Notes 1. You can arrange the eigenvalues of A in any order you like to form D . 2. Arrange the linearly independent eigenvectors of A as columns to form P . This should correspond to how you write D . 3. This means the rst column in P must be the eigenvector of the rst eigenvalue in D , the second column in P the eigenvector corresponding to the second eigenvalue in D and so on. This is very important.
  • 33. Notes 1. You can arrange the eigenvalues of A in any order you like to form D . 2. Arrange the linearly independent eigenvectors of A as columns to form P . This should correspond to how you write D . 3. This means the rst column in P must be the eigenvector of the rst eigenvalue in D , the second column in P the eigenvector corresponding to the second eigenvalue in D and so on. This is very important. 4. Of course, you could write P rst and arrange the eigenvalues of D accordingly.
  • 34. Example 2, section 5.3 Let A = PDP −1 . For the given P and D , compute A4 . 2 −3 1 0 P = ,D = −3 5 0 1 /2
  • 35. Example 2, section 5.3 Let A = PDP −1 . For the given P and D , compute A4 . 2 −3 1 0 P = ,D = −3 5 0 1 /2 Solution: Since A = PDP −1 , A4 = PD 4 P −1
  • 36. Example 2, section 5.3 Let A = PDP −1 . For the given P and D , compute A4 . 2 −3 1 0 P = ,D = −3 5 0 1 /2 Solution: Since A = PDP −1 , A4 = PD 4 P −1 4 14 0 1 0 D = = 0 (1/2)4 0 1/16
  • 37. Example 2, section 5.3 Let A = PDP −1 . For the given P and D , compute A4 . 2 −3 1 0 P = ,D = −3 5 0 1 /2 Solution: Since A = PDP −1 , A4 = PD 4 P −1 4 14 0 1 0 D = = 0 (1/2)4 0 1/16 Here det P = 10 − 9 = 1, we can nd P −1 . (interchange the main diagonals, change signs of o diagonals, divide by det P = 1)
  • 38. Example 2, section 5.3 Let A = PDP −1 . For the given P and D , compute A4 . 2 −3 1 0 P = ,D = −3 5 0 1 /2 Solution: Since A = PDP −1 , A4 = PD 4 P −1 4 14 0 1 0 D = = 0 (1/2)4 0 1/16 Here det P = 10 − 9 = 1, we can nd P −1 . (interchange the main diagonals, change signs of o diagonals, divide by det P = 1) −1 5 3 P = 3 2
  • 39. Example 2, section 5.3 A 4 = PD 4 P −1 =⇒ 4 2 −3 1 0 5 3 A = −3 5 0 1/16 3 2 P D4 P −1
  • 40. Example 2, section 5.3 A 4 = PD 4 P −1 =⇒ 4 2 −3 1 0 5 3 A = −3 5 0 1/16 3 2 P D4 P −1 2 −3/16 5 3 = −3 5/16 3 2 PD 4 P −1
  • 41. Example 2, section 5.3 A 4 = PD 4 P −1 =⇒ 4 2 −3 1 0 5 3 A = −3 5 0 1/16 3 2 P D4 P −1 2 −3/16 5 3 = −3 5/16 3 2 PD 4 P −1 10 − 9/16 6 − 6/16 = −15 + 15/16 −9 + 10/16 PD 4 P −1
  • 42. Example 2, section 5.3 A 4 = PD 4 P −1 =⇒ 4 2 −3 1 0 5 3 A = −3 5 0 1/16 3 2 P D4 P −1 2 −3/16 5 3 = −3 5/16 3 2 PD 4 P −1 10 − 9/16 6 − 6/16 = −15 + 15/16 −9 + 10/16 PD 4 P −1 151/16 90/16 1 151 90 = = −225/16 −134/16 16 −225 −134
  • 43. Example 4, section 5.3 (slightly modied) −2 12 Let A = . Use the factorization PDP −1 to compute A6 −1 5 where 3 4 2 0 P = ,D = 1 1 0 1
  • 44. Example 4, section 5.3 (slightly modied) −2 12 Let A = . Use the factorization PDP −1 to compute A6 −1 5 where 3 4 2 0 P = ,D = 1 1 0 1 Solution: Since A = PDP −1 , A6 = PD 6 P −1 6 26 0 64 0 D = = 0 16 0 1
  • 45. Example 4, section 5.3 (slightly modied) −2 12 Let A = . Use the factorization PDP −1 to compute A6 −1 5 where 3 4 2 0 P = ,D = 1 1 0 1 Solution: Since A = PDP −1 , A6 = PD 6 P −1 6 26 0 64 0 D = = 0 16 0 1 Here det P = 3 − 4 = −1, we can nd P −1 . (interchange the main diagonal entries, change signs of o diagonal entries, divide by det P = −1) −1 −1 4 P = 1 −3
  • 46. Example 2, section 5.3 A 6 = PD 6 P −1 =⇒ 6 3 4 64 0 −1 4 A = 1 1 0 1 1 −3 P D6 P −1
  • 47. Example 2, section 5.3 A 6 = PD 6 P −1 =⇒ 6 3 4 64 0 −1 4 A = 1 1 0 1 1 −3 P D6 P −1 192 4 −1 4 = 64 1 1 −3 PD 6 P −1
  • 48. Example 2, section 5.3 A 6 = PD 6 P −1 =⇒ 6 3 4 64 0 −1 4 A = 1 1 0 1 1 −3 P D6 P −1 192 4 −1 4 = 64 1 1 −3 PD 6 P −1 −188 756 = −63 253 PD 6 P −1
  • 49. Example 6, section 5.3 The matrix A is factored in the form PDP −1 . Find the eigenvalues of A and the basis for each eigenspace. 4 0 −2 −2 0 −1 5 0 0 0 0 1        2 5 4 = 0 1 2  0 5 0  2 1 4  0 0 5 1 0 0 0 0 4 −1 0 −2
  • 50. Example 6, section 5.3 The matrix A is factored in the form PDP −1 . Find the eigenvalues of A and the basis for each eigenspace. 4 0 −2 −2 0 −1 5 0 0 0 0 1        2 5 4 = 0 1 2  0 5 0  2 1 4  0 0 5 1 0 0 0 0 4 −1 0 −2 Solution: The eigenvalues of A are the entries of the diagonal matrix D . Here the eigenvalues are λ = 5, 5, 4. Note that 5 has multiplicity 2 (repeated).
  • 51. Example 6, section 5.3 The matrix A is factored in the form PDP −1 . Find the eigenvalues of A and the basis for each eigenspace. 4 0 −2 −2 0 −1 5 0 0 0 0 1        2 5 4 = 0 1 2  0 5 0  2 1 4  0 0 5 1 0 0 0 0 4 −1 0 −2 Solution: The eigenvalues of A are the entries of the diagonal matrix D . Here the eigenvalues are λ = 5, 5, 4. Note that 5 has multiplicity 2 (repeated). The eigenvectors of λ = 5 are (the rst 2 columns of P ) −2 0      0 , 1  1 0
  • 52. Example 6, section 5.3 The matrix A is factored in the form PDP −1 . Find the eigenvalues of A and the basis for each eigenspace. 4 0 −2 −2 0 −1 5 0 0 0 0 1        2 5 4 = 0 1 2  0 5 0  2 1 4  0 0 5 1 0 0 0 0 4 −1 0 −2 Solution: The eigenvalues of A are the entries of the diagonal matrix D . Here the eigenvalues are λ = 5, 5, 4. Note that 5 has multiplicity 2 (repeated). The eigenvectors of λ = 5 are (the rst 2 columns of P ) −2 0      0 , 1  1 0 −1   An eigenvector of λ = 4 is (the last column of P )  2  0
  • 53. Steps to Diagonalize an n × n Matrix 1. First nd the eigenvalues of A using the char. equation (from sec 5.2). Eigenvalues will be provided in the problem for dicult 3 × 3 matrices and larger matrices that are not triangular.
  • 54. Steps to Diagonalize an n × n Matrix 1. First nd the eigenvalues of A using the char. equation (from sec 5.2). Eigenvalues will be provided in the problem for dicult 3 × 3 matrices and larger matrices that are not triangular. 2. Find the eigenvectors for each eigenvalue (based on sec 5.1).
  • 55. Steps to Diagonalize an n × n Matrix 1. First nd the eigenvalues of A using the char. equation (from sec 5.2). Eigenvalues will be provided in the problem for dicult 3 × 3 matrices and larger matrices that are not triangular. 2. Find the eigenvectors for each eigenvalue (based on sec 5.1). 3. Make sure you have n linearly independent eigenvectors. Otherwise you cannot diagonalize.
  • 56. Steps to Diagonalize an n × n Matrix 1. First nd the eigenvalues of A using the char. equation (from sec 5.2). Eigenvalues will be provided in the problem for dicult 3 × 3 matrices and larger matrices that are not triangular. 2. Find the eigenvectors for each eigenvalue (based on sec 5.1). 3. Make sure you have n linearly independent eigenvectors. Otherwise you cannot diagonalize. 4. If you are successful with step 3, write P and D carefully. (Make sure that the columns of P and D correspond to eachother)
  • 57. Steps to Diagonalize an n × n Matrix 1. First nd the eigenvalues of A using the char. equation (from sec 5.2). Eigenvalues will be provided in the problem for dicult 3 × 3 matrices and larger matrices that are not triangular. 2. Find the eigenvectors for each eigenvalue (based on sec 5.1). 3. Make sure you have n linearly independent eigenvectors. Otherwise you cannot diagonalize. 4. If you are successful with step 3, write P and D carefully. (Make sure that the columns of P and D correspond to eachother) 5. For a 2 × 2 matrix, compute P −1 . For 3 × 3 and larger matrices, compute the products AP and PD and make sure they are exactly the same.
  • 58. Important Theorem An n ×n matrix with n distinct eigenvalues is diagonalizable.
  • 59. Important Theorem An n ×n matrix with n distinct eigenvalues is diagonalizable. This is because (from section 5.1) Theorem The eigenvectors corresponding to distinct eigenvalues are linearly independent
  • 60. Important 1. If there are no repeated eigenvalues, diagonalization is guaranteed.
  • 61. Important 1. If there are no repeated eigenvalues, diagonalization is guaranteed. 2. Presence of repeated eigenvalues immediately does not mean that diagonalization fails.
  • 62. Important 1. If there are no repeated eigenvalues, diagonalization is guaranteed. 2. Presence of repeated eigenvalues immediately does not mean that diagonalization fails. 3. If you can get enough linearly independent eigenvectors from the repeated eigenvalue, we can still diagonalize.
  • 63. Important 1. If there are no repeated eigenvalues, diagonalization is guaranteed. 2. Presence of repeated eigenvalues immediately does not mean that diagonalization fails. 3. If you can get enough linearly independent eigenvectors from the repeated eigenvalue, we can still diagonalize. 4. For example, suppose a 3 × 3 matrix has eigenvalues 2, 2, and 4. If we can get 2 linearly independent eigenvectors for eigenvalue 2, we are good. If the eigenvalue 2 gives only one eigenvector, diagonalization fails.
  • 64. Example 8, section 5.3 5 1 Diagonalize A = if possible. 0 5
  • 65. Example 8, section 5.3 5 1 Diagonalize A = if possible. 0 5 Solution: What are the eigenvalues of A?
  • 66. Example 8, section 5.3 5 1 Diagonalize A = if possible. 0 5 Solution: What are the eigenvalues of A? We can write the char.equation and solve if necessary. Look carefully at A. It is triangular. The eigenvalues are thus λ = 5, 5.
  • 67. Example 8, section 5.3 5 1 Diagonalize A = if possible. 0 5 Solution: What are the eigenvalues of A? We can write the char.equation and solve if necessary. Look carefully at A. It is triangular. The eigenvalues are thus λ = 5, 5. Since 5 is a repeated eigenvalue there is a possibility that diagonalization may fail. But we have to nd the eigenvectors to conrm this. Start with the matrix A − 5I . 5 1 5 0 0 1 A − 5I = − = 0 5 0 5 0 0
  • 68. Example 8, section 5.3 From the rst row, x2 = 0 and x1 is free.
  • 69. Example 8, section 5.3 From the rst row, x2 = 0 and x1 is free. Thus an eigenvector is x1 x1 1 = = x1 . x2 0 0
  • 70. Example 8, section 5.3 From the rst row, x2 = 0 and x1 is free. Thus an eigenvector is x1 x1 1 = = x1 . x2 0 0 1 Fix x1 = 1 and an eigenvector is . 0
  • 71. Example 8, section 5.3 From the rst row, x2 = 0 and x1 is free. Thus an eigenvector is x1 x1 1 = = x1 . x2 0 0 1 Fix x1 = 1 and an eigenvector is . 0 We are unable to nd another eigenvector for λ = 5 so that we have 2 linearly independent eigenvectors. So A is NOT diagonalizable.
  • 72. Example 10, section 5.3 2 3 Diagonalize A = if possible. 4 1 Solution: We have to write the char.equation and solve to nd the eigenvalues.
  • 73. Example 10, section 5.3 2 3 Diagonalize A = if possible. 4 1 Solution: We have to write the char.equation and solve to nd the eigenvalues. So, 2−λ 3 =0 4 1−λ
  • 74. Example 10, section 5.3 2 3 Diagonalize A = if possible. 4 1 Solution: We have to write the char.equation and solve to nd the eigenvalues. So, 2−λ 3 =0 4 1−λ =⇒ (2 − λ)(1 − λ) − 12 = 0 =⇒ 2 − 3λ + λ2 − 12 = 0
  • 75. Example 10, section 5.3 2 3 Diagonalize A = if possible. 4 1 Solution: We have to write the char.equation and solve to nd the eigenvalues. So, 2−λ 3 =0 4 1−λ =⇒ (2 − λ)(1 − λ) − 12 = 0 =⇒ 2 − 3λ + λ2 − 12 = 0 =⇒ λ2 − 3λ − 10 = 0
  • 76. Example 10, section 5.3 2 3 Diagonalize A = if possible. 4 1 Solution: We have to write the char.equation and solve to nd the eigenvalues. So, 2−λ 3 =0 4 1−λ =⇒ (2 − λ)(1 − λ) − 12 = 0 =⇒ 2 − 3λ + λ2 − 12 = 0 =⇒ λ2 − 3λ − 10 = 0 =⇒ (λ − 5)(λ + 2) = 0 =⇒ λ = 5, λ = −2
  • 77. Example 10, section 5.3 2 3 Diagonalize A = if possible. 4 1 Solution: We have to write the char.equation and solve to nd the eigenvalues. So, 2−λ 3 =0 4 1−λ =⇒ (2 − λ)(1 − λ) − 12 = 0 =⇒ 2 − 3λ + λ2 − 12 = 0 =⇒ λ2 − 3λ − 10 = 0 =⇒ (λ − 5)(λ + 2) = 0 =⇒ λ = 5, λ = −2 Since we have distinct eigenvalues, we can surely diagonalize A. First nd an eigenvector for each eigenvalue.
  • 78. Example 10, section 5.3 For λ = 5, 2 3 5 0 −3 3 A − 5I = − = 4 1 0 5 4 −4
  • 79. Example 10, section 5.3 For λ = 5, 2 3 5 0 −3 3 A − 5I = − = 4 1 0 5 4 −4 Divide the rst row by -3, second row by 4 1 −1 R 2−R 1 1 −1 A − 5I = −− − − −→ 1 −1 0 0
  • 80. Example 10, section 5.3 For λ = 5, 2 3 5 0 −3 3 A − 5I = − = 4 1 0 5 4 −4 Divide the rst row by -3, second row by 4 1 −1 R 2−R 1 1 −1 A − 5I = −− − − −→ 1 −1 0 0 x2 is a free variable and from rst row, x1 = x2 . x1 x2 1 = = x2 . x2 x2 1
  • 81. Example 10, section 5.3 For λ = 5, 2 3 5 0 −3 3 A − 5I = − = 4 1 0 5 4 −4 Divide the rst row by -3, second row by 4 1 −1 R 2−R 1 1 −1 A − 5I = −− − − −→ 1 −1 0 0 x2 is a free variable and from rst row, x1 = x2 . x1 x2 1 = = x2 . x2 x2 1 1 An eigenvector for λ = 5 is . 1
  • 82. Example 10, section 5.3 For λ = −2, 2 3 2 0 4 3 A + 2I = + = 4 1 0 2 4 3
  • 83. Example 10, section 5.3 For λ = −2, 2 3 2 0 4 3 A + 2I = + = 4 1 0 2 4 3 4 3 R 2−R 1 4 3 A + 2I = −− − − −→ 4 3 0 0
  • 84. Example 10, section 5.3 For λ = −2, 2 3 2 0 4 3 A + 2I = + = 4 1 0 2 4 3 4 3 R 2−R 1 4 3 A + 2I = −− − − −→ 4 3 0 0 x2 is a free variable and from rst row, x1 = − 3 x2 . 4 x1 − 3 x2 4 −3 4 = = x2 . x2 x2 1
  • 85. Example 10, section 5.3 For λ = −2, 2 3 2 0 4 3 A + 2I = + = 4 1 0 2 4 3 4 3 R 2−R 1 4 3 A + 2I = −− − − −→ 4 3 0 0 x2 is a free variable and from rst row, x1 = − 3 x2 . 4 x1 − 3 x2 4 −3 4 = = x2 . x2 x2 1 −3 Pick x2 = 4 and an eigenvector for λ = −2 is . 4
  • 86. Example 10, section 5.3 We can now write P using these 2 eigenvectors as columns. 1 −3 P = . 1 4 D would be the eigenvalues written as diagonal entries, in the same order 5 0 D = . 0 −2
  • 87. Example 10, section 5.3 We can now write P using these 2 eigenvectors as columns. 1 −3 P = . 1 4 D would be the eigenvalues written as diagonal entries, in the same order 5 0 D = . 0 −2 Also since det P = 7, −1 4 /7 3 /7 P = −1/7 1/7
  • 88. Example 10, section 5.3 We can now write P using these 2 eigenvectors as columns. 1 −3 P = . 1 4 D would be the eigenvalues written as diagonal entries, in the same order 5 0 D = . 0 −2 Also since det P = 7, −1 4 /7 3 /7 P = −1/7 1/7 −1 1 −3 5 0 4/7 3/7 PDP = 1 4 0 −2 −1/7 1/7
  • 89. Example 10, section 5.3 We can now write P using these 2 eigenvectors as columns. 1 −3 P = . 1 4 D would be the eigenvalues written as diagonal entries, in the same order 5 0 D = . 0 −2 Also since det P = 7, −1 4 /7 3 /7 P = −1/7 1/7 −1 1 −3 5 0 4/7 3/7 PDP = 1 4 0 −2 −1/7 1/7 5 6 4/7 3/7 2 3 = = 5 −8 −1/7 1/7 4 1
  • 90. Example 12, section 5.3 4 2 2   Diagonalize A =  2 4 2  if possible if λ = 2, 8 are the 2 2 4 eigenvalues.
  • 91. Example 12, section 5.3 4 2 2   Diagonalize A =  2 4 2  if possible if λ = 2, 8 are the 2 2 4 eigenvalues. Solution: Only 2 eigenvalues λ = 2, 8 are given. This means one of these could be repeated. One way to check is to nd the trace of the matrix which is 4+4+4=12 and the sum of the eigenvalues which is 2+8+?. Since they must be same ? must be 2.
  • 92. Example 12, section 5.3 4 2 2   Diagonalize A =  2 4 2  if possible if λ = 2, 8 are the 2 2 4 eigenvalues. Solution: Only 2 eigenvalues λ = 2, 8 are given. This means one of these could be repeated. One way to check is to nd the trace of the matrix which is 4+4+4=12 and the sum of the eigenvalues which is 2+8+?. Since they must be same ? must be 2. Since we have repeated eigenvalue 2, it is possible (not already sure) that A may not be diagonalizable. Finding the eigenvectors for λ = 2 is the only way to nd out.
  • 93. Example 12, section 5.3 For λ = 5, 4 2 2 2 0 0 2 2 2       A − 2I =  2 4 2 − 0 2 0 = 2 2 2  2 2 4 0 0 2 2 2 2
  • 94. Example 12, section 5.3 For λ = 5, 4 2 2 2 0 0 2 2 2       A − 2I =  2 4 2 − 0 2 0 = 2 2 2  2 2 4 0 0 2 2 2 2 Divide all rows by 2 1 1 1   A − 2I =  1 1 1 1 1 1
  • 95. Example 12, section 5.3 For λ = 5, 4 2 2 2 0 0 2 2 2       A − 2I =  2 4 2 − 0 2 0  =  2 2 2  2 2 4 0 0 2 2 2 2 Divide all rows by 2 1 1 1 1 1 1     R 2−R 1,R 3−R 1  A − 2I =  1 1 1  − −−−−−−−− − → 0 0 0  1 1 1 0 0 0
  • 96. Example 12, section 5.3 For λ = 5, 4 2 2 2 0 0 2 2 2       A − 2I =  2 4 2 − 0 2 0  =  2 2 2  2 2 4 0 0 2 2 2 2 Divide all rows by 2 1 1 1 1 1 1     R 2−R 1,R 3−R 1  A − 2I =  1 1 1  − −−−−−−−− − → 0 0 0  1 1 1 0 0 0 x2 and x3 are free variable and from rst row, x1 = −x2 − x3 .
  • 97. Example 12, section 5.3 For λ = 5, 4 2 2 2 0 0 2 2 2       A − 2I =  2 4 2 − 0 2 0  =  2 2 2  2 2 4 0 0 2 2 2 2 Divide all rows by 2 1 1 1 1 1 1     R 2−R 1,R 3−R 1  A − 2I =  1 1 1  − −−−−−−−− − → 0 0 0  1 1 1 0 0 0 x2 and x3 are free variable and from rst row, x1 = −x2 − x3 . −1 −1         x1 −x2 − x3  x2  =  x2  = x2  1  + x3  0  . x3 x3 0 1
  • 98. Example 12, section 5.3 For λ = 5, 4 2 2 2 0 0 2 2 2       A − 2I =  2 4 2 − 0 2 0  =  2 2 2  2 2 4 0 0 2 2 2 2 Divide all rows by 2 1 1 1 1 1 1     R 2−R 1,R 3−R 1  A − 2I =  1 1 1  − −−−−−−−− − → 0 0 0  1 1 1 0 0 0 x2 and x3 are free variable and from rst row, x1 = −x2 − x3 . −1 −1         x1 −x2 − x3  x2  =  x2  = x2  1  + x3  0  . x3 x3 0 1 A linearlyindependent set of eigenvectors for λ = 2 is −1 −1     1  ,  0 . 0 1
  • 99. This means A is diagonalizable. We have to nd an eigenvector for λ = 8. For λ = 8, 4 2 2 8 0 0 −4 2 2       A − 8I =  2 4 2  −  0 8 0  =  2 −4 2  2 2 4 0 0 8 2 2 −4
  • 100. This means A is diagonalizable. We have to nd an eigenvector for λ = 8. For λ = 8, 4 2 2 8 0 0 −4 2 2       A − 8I =  2 4 2  −  0 8 0  =  2 −4 2  2 2 4 0 0 8 2 2 −4 Divide all rows by 2 and interchange the rst 2 rows 1 −2 1   A−8I =  −2 1 1  1 1 −2
  • 101. This means A is diagonalizable. We have to nd an eigenvector for λ = 8. For λ = 8, 4 2 2 8 0 0 −4 2 2       A − 8I =  2 4 2  −  0 8 0  =  2 −4 2  2 2 4 0 0 8 2 2 −4 Divide all rows by 2 and interchange the rst 2 rows 1 −2 1 1 −2 1     R 2+2R 1 A−8I =  −2 1 1  =⇒  0 −3 3  1 1 −2 R 3−R 1 0 3 −3
  • 102. This means A is diagonalizable. We have to nd an eigenvector for λ = 8. For λ = 8, 4 2 2 8 0 0 −4 2 2       A − 8I =  2 4 2  −  0 8 0  =  2 −4 2  2 2 4 0 0 8 2 2 −4 Divide all rows by 2 and interchange the rst 2 rows 1 −2 1 1 −2 1 1 −2 1       A−8I =  −2 1 1  R 2+2R 1  0 −3 3  R=⇒ 2  0 =⇒ 3+R −3 3 1 1 −2 R 3−R 1 0 3 −3  0 0 0
  • 103. This means A is diagonalizable. We have to nd an eigenvector for λ = 8. For λ = 8, 4 2 2 8 0 0 −4 2 2       A − 8I =  2 4 2  −  0 8 0  =  2 −4 2  2 2 4 0 0 8 2 2 −4 Divide all rows by 2 and interchange the rst 2 rows 1 −2 1 1 −2 1 1 −2 1       A−8I =  −2 1 1  R 2+2R 1  0 −3 3  R=⇒ 2  0 =⇒ 3+R −3 3 1 1 −2 R 3−R 1 0 3 −3  0 0 0 x3 is a free variable. From second row, x2 = x3 . From rst row, x1 = 2x2 − x3 = x3
  • 104. This means A is diagonalizable. We have to nd an eigenvector for λ = 8. For λ = 8, 4 2 2 8 0 0 −4 2 2       A − 8I =  2 4 2  −  0 8 0  =  2 −4 2  2 2 4 0 0 8 2 2 −4 Divide all rows by 2 and interchange the rst 2 rows 1 −2 1 1 −2 1 1 −2 1       A−8I =  −2 1 1  R 2+2R 1  0 −3 3  R=⇒ 2  0 =⇒ 3+R −3 3 1 1 −2 R 3−R 1 0 3 −3  0 0 0 x3 is a free variable. From second row, x2 = x3 . From rst row, x1 = 2x2 − x3 = x3 1       x1 x3  x2  =  x3  = x3  1  x3 x3 1
  • 105. This means A is diagonalizable. We have to nd an eigenvector for λ = 8. For λ = 8, 4 2 2 8 0 0 −4 2 2       A − 8I =  2 4 2  −  0 8 0  =  2 −4 2  2 2 4 0 0 8 2 2 −4 Divide all rows by 2 and interchange the rst 2 rows 1 −2 1 1 −2 1 1 −2 1       A−8I =  −2 1 1  R 2+2R 1  0 −3 3  R=⇒ 2  0 =⇒ 3+R −3 3 1 1 −2 R 3−R 1 0 3 −3  0 0 0 x3 is a free variable. From second row, x2 = x3 . From rst row, x1 = 2x2 − x3 = x3 1       x1 x3  x2  =  x3  = x3  1  x3 x3 1 1   An eigenvector for λ = 8 is  1 . 1
  • 106. Example 10, section 5.3 We can now write P using these 3 eigenvectors as columns. −1 −1 1   P = 1 0 1 . 0 1 1
  • 107. Example 10, section 5.3 We can now write P using these 3 eigenvectors as columns. −1 −1 1   P = 1 0 1 . 0 1 1 D would be the eigenvalues written as diagonal entries, in the same order 2 0 0   D = 0 2 0 . 0 0 8
  • 108. Example 10, section 5.3 We can now write P using these 3 eigenvectors as columns. −1 −1 1   P = 1 0 1 . 0 1 1 D would be the eigenvalues written as diagonal entries, in the same order 2 0 0   D = 0 2 0 . 0 0 8 Find the products AP and PD (you must nd these clearly). 4 2 2 −1 −1 1 −2 −2 8      AP =  2 4 2  1 0 1  =  2 0 8  2 2 4 0 1 1 0 2 8
  • 109. Example 10, section 5.3 We can now write P using these 3 eigenvectors as columns. −1 −1 1   P = 1 0 1 . 0 1 1 D would be the eigenvalues written as diagonal entries, in the same order 2 0 0   D = 0 2 0 . 0 0 8 Find the products AP and PD (you must nd these clearly). 4 2 2 −1 −1 1 −2 −2 8      AP =  2 4 2  1 0 1  =  2 0 8  2 2 4 0 1 1 0 2 8 −1 −1 1 2 0 0 −2 −2 8      PD = 1 0 1  0 2 0 = 2 0 8  0 1 1 0 0 8 0 2 8