Lesson 15: Diagonalization

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Lesson 15: Diagonalization

  1. 1. Lesson 15 (S&H, Section 14.5) Diagonalization Math 20 October 24, 2007 Announcements Midterm done. Nice job! Problem Set 6 assigned today. Due October 31. OH: Mondays 1–2, Tuesdays 3–4, Wednesdays 1–3 (SC 323) Prob. Sess.: Sundays 6–7 (SC B-10), Tuesdays 1–2 (SC 116)
  2. 2. Outline Concept Review Diagonalization Motivating Example Procedure Computations Uses of Diagonalization
  3. 3. Concept Review Definition Let A be an n × n matrix. The number λ is called an eigenvalue of A if there exists a nonzero vector x ∈ Rn such that Ax = λx. (1) Every nonzero vector satisfying (1) is called an eigenvector of A associated with the eigenvalue λ.
  4. 4. Geometric effect of a non-diagonal linear transformation Example 1/2 3/2 Let A = 3/2 1/2 . Draw the effect of the linear transformation which is multiplication by A. y x
  5. 5. Geometric effect of a non-diagonal linear transformation Example 1/2 3/2 Let A = 3/2 1/2 . Draw the effect of the linear transformation which is multiplication by A. y v1 x
  6. 6. Geometric effect of a non-diagonal linear transformation Example 1/2 3/2 Let A = 3/2 1/2 . Draw the effect of the linear transformation which is multiplication by A. y Av1 v1 x
  7. 7. Geometric effect of a non-diagonal linear transformation Example 1/2 3/2 Let A = 3/2 1/2 . Draw the effect of the linear transformation which is multiplication by A. y Av1 v1 x v2
  8. 8. Geometric effect of a non-diagonal linear transformation Example 1/2 3/2 Let A = 3/2 1/2 . Draw the effect of the linear transformation which is multiplication by A. y Av1 v2 v1 x v2
  9. 9. Geometric effect of a non-diagonal linear transformation Example 1/2 3/2 Let A = 3/2 1/2 . Draw the effect of the linear transformation which is multiplication by A. y Av1 v2 v1 2e2 x v2
  10. 10. Geometric effect of a non-diagonal linear transformation Example 1/2 3/2 Let A = 3/2 1/2 . Draw the effect of the linear transformation which is multiplication by A. y Av1 v2 v1 2e2 x v2
  11. 11. Geometric effect of a non-diagonal linear transformation Example 1/2 3/2 Let A = 3/2 1/2 . Draw the effect of the linear transformation which is multiplication by A. y Av1 v2 v1 2e2 x v2
  12. 12. Geometric effect of a non-diagonal linear transformation Example 1/2 3/2 Let A = 3/2 1/2 . Draw the effect of the linear transformation which is multiplication by A. y Ae2 Av1 v2 v1 2e2 x v2
  13. 13. Geometric effect of a non-diagonal linear transformation Example 1/2 3/2 Let A = 3/2 1/2 . Draw the effect of the linear transformation which is multiplication by A. y Ae2 Av1 v2 v1 2e2 x v2
  14. 14. Geometric effect of a non-diagonal linear transformation Example 1/2 3/2 Let A = 3/2 1/2 . Draw the effect of the linear transformation which is multiplication by A. y Ae2 Av1 v2 v1 2e2 x v2
  15. 15. Geometric effect of a non-diagonal linear transformation Example 1/2 3/2 Let A = 3/2 1/2 . Draw the effect of the linear transformation which is multiplication by A. y Ae2 Av1 v2 v1 2e2 x v2
  16. 16. Methods To find the eigenvalues of a matrix A, find the determinant of A − λI. This will be a polynomial in λ (called the characteristic polynomial of A, and its roots are the eigenvalues. To find the eigenvector(s) of a matrix corresponding to an eigenvalue λ, do Gaussian Elimination on A − λI.
  17. 17. Outline Concept Review Diagonalization Motivating Example Procedure Computations Uses of Diagonalization
  18. 18. The fact that we have eigenvectors corresponding to two different eigenvalues gives us the following: 1 1 1 1 1 1 A = A A = 2 −1 1 −1 1 −1 1 −1 P 1 1 2 0 = 1 −1 0 −1 D
  19. 19. The fact that we have eigenvectors corresponding to two different eigenvalues gives us the following: 1 1 1 1 1 1 A = A A = 2 −1 1 −1 1 −1 1 −1 P 1 1 2 0 = 1 −1 0 −1 D So we have found a matrix P and a diagonal matrix D such that AP = PD
  20. 20. The fact that we have eigenvectors corresponding to two different eigenvalues gives us the following: 1 1 1 1 1 1 A = A A = 2 −1 1 −1 1 −1 1 −1 P 1 1 2 0 = 1 −1 0 −1 D So we have found a matrix P and a diagonal matrix D such that AP = PD Since P is invertible (det P = −2), we have A = PDP−1
  21. 21. Diagonalization Procedure Find the eigenvalues and eigenvectors. Arrange the eigenvectors in a matrix P and the corresponding eigenvalues in a diagonal matrix D. If you have “enough” eigenvectors so that the matrix P is square and invertible, the original matrix is diagonalizable and equal to PDP−1 .
  22. 22. Outline Concept Review Diagonalization Motivating Example Procedure Computations Uses of Diagonalization
  23. 23. Example Example (Worksheet Problem 1) Let 0 −2 A= . −3 1 Find an invertible matrix P and a diagonal matrix D such that A = PDP−1 .
  24. 24. Example Example (Worksheet Problem 1) Let 0 −2 A= . −3 1 Find an invertible matrix P and a diagonal matrix D such that A = PDP−1 . Solution We found that −2 and 3 are the eigenvalues for A. The eigenvalue 1 −2 has an associated eigenvector , and the eigenvalue 3 has 1 −2 eigenvector . Thus 3 1 −2 −2 0 P= D= . 1 3 0 3
  25. 25. Checking the Solution 1 −2 −2 0 P= D= . 1 3 0 3 Check this: We have 1 3 2 P−1 = . 5 −1 1 1 1 −2 −2 0 3 2 PDP−1 = 5 1 3 0 3 −1 1 1 1 −2 −6 −4 = 5 1 3 −3 3 1 0 −10 0 −2 = = . 5 −15 5 −3 1
  26. 26. Example (Worksheet Problem 2) −7 4 Let B = . Find an invertible matrix P and a diagonal −9 5 matrix D such that B = PDP−1 .
  27. 27. Example (Worksheet Problem 2) −7 4 Let B = . Find an invertible matrix P and a diagonal −9 5 matrix D such that B = PDP−1 . Solution The characteristic polynomial of B is (λ + 1)2 , which has the 2 double root −1. There is one eigenvector, , but nothing more. 3 So there is no diagonal D which works.
  28. 28. Example (Worksheet Problem 3) 0 1 Let B = . Find an invertible matrix P and a diagonal −1 0 matrix D such that B = PDP−1 .
  29. 29. Example (Worksheet Problem 3) 0 1 Let B = . Find an invertible matrix P and a diagonal −1 0 matrix D such that B = PDP−1 . Solution The characteristic polynomial of B is λ2 + 1, which has no real roots. The eigenvalues are i and −i. We could consider the complex eigenvectors i −i z1 = and z2 = 1 1 but scaling by a complex number is more complicated than it looks.
  30. 30. Outline Concept Review Diagonalization Motivating Example Procedure Computations Uses of Diagonalization
  31. 31. Geometric effects of powers of matrices Example 2 0 Let D = . Draw the effect of the linear transformation 0 −1 which is multiplication by Dn . y S x
  32. 32. Geometric effects of powers of matrices Example 2 0 Let D = . Draw the effect of the linear transformation 0 −1 which is multiplication by Dn . y S x D(S)
  33. 33. Geometric effects of powers of matrices Example 2 0 Let D = . Draw the effect of the linear transformation 0 −1 which is multiplication by Dn . y S D2 (S) x D(S)
  34. 34. Geometric effects of powers of matrices Example 2 0 Let D = . Draw the effect of the linear transformation 0 −1 which is multiplication by Dn . y S D2 (S) x D(S) D3 (S)
  35. 35. Geometric effect of a non-diagonal linear transformation Example 1/2 3/2 Let A = 3/2 1/2 . Draw the effect of the linear transformation which is multiplication by An . y S x
  36. 36. Geometric effect of a non-diagonal linear transformation Example 1/2 3/2 Let A = 3/2 1/2 . Draw the effect of the linear transformation which is multiplication by An . y S x
  37. 37. Geometric effect of a non-diagonal linear transformation Example 1/2 3/2 Let A = 3/2 1/2 . Draw the effect of the linear transformation which is multiplication by An . y A(S) S x
  38. 38. Geometric effect of a non-diagonal linear transformation Example 1/2 3/2 Let A = 3/2 1/2 . Draw the effect of the linear transformation which is multiplication by An . y A2 (S) A(S) S x
  39. 39. Computing An with diagonalization Example (Worksheet Problem 4) 1/2 3/2 Let A = 3/2 1/2 . Find A100 .
  40. 40. Computing An with diagonalization Example (Worksheet Problem 4) 1/2 3/2 Let A = 3/2 1/2 . Find A100 . Solution 1 1 2 0 We know A = PDP−1 , where P = and D = . 1 −1 0 −1
  41. 41. Computing An with diagonalization Example (Worksheet Problem 4) 1/2 3/2 Let A = 3/2 1/2 . Find A100 . Solution 1 1 2 0 We know A = PDP−1 , where P = and D = . 1 −1 0 −1 Now An = (PDP−1 )n = (PDP−1 )(PDP−1 ) · · · (PDP−1 ) n −1 −1 −1 = PD(P P)D(P P) · · · D(P P)DP−1 = PDn P−1 And Dn is easy!
  42. 42. Computing An with diagonalization Example (Worksheet Problem 4) 1/2 3/2 Let A = 3/2 1/2 . Find A100 . Solution 1 1 2 0 We know A = PDP−1 , where P = and D = . 1 −1 0 −1 Now An = (PDP−1 )n = (PDP−1 )(PDP−1 ) · · · (PDP−1 ) n −1 −1 −1 = PD(P P)D(P P) · · · D(P P)DP−1 = PDn P−1 And Dn is easy! So 1 1 1 2100 0 1 1 1 2100 + 1 2100 − 1 A100 = = 2 1 −1 0 1 1 −1 2 2100 − 1 2100 + 1

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