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- 1. Lesson 14 Eigenvalues and Eigenvectors Math 20 October 22, 2007 Announcements Midterm almost done Problem Set 5 is on the WS. Due October 24 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. Geometric eﬀect of a diagonal linear transformation Example 20 Let A = . Draw the eﬀect of the linear transformation 0 −1 which is multiplication by A.
- 3. Geometric eﬀect of a diagonal linear transformation Example 20 Let A = . Draw the eﬀect of the linear transformation 0 −1 which is multiplication by A. y x
- 4. Geometric eﬀect of a diagonal linear transformation Example 20 Let A = . Draw the eﬀect of the linear transformation 0 −1 which is multiplication by A. y x e1
- 5. Geometric eﬀect of a diagonal linear transformation Example 20 Let A = . Draw the eﬀect of the linear transformation 0 −1 which is multiplication by A. y x e1 Ae1
- 6. Geometric eﬀect of a diagonal linear transformation Example 20 Let A = . Draw the eﬀect of the linear transformation 0 −1 which is multiplication by A. y e2 x e1 Ae1
- 7. Geometric eﬀect of a diagonal linear transformation Example 20 Let A = . Draw the eﬀect of the linear transformation 0 −1 which is multiplication by A. y e2 x e1 Ae1 Ae2
- 8. Geometric eﬀect of a diagonal linear transformation Example 20 Let A = . Draw the eﬀect of the linear transformation 0 −1 which is multiplication by A. y e2 v x e1 Ae1 Ae2
- 9. Geometric eﬀect of a diagonal linear transformation Example 20 Let A = . Draw the eﬀect of the linear transformation 0 −1 which is multiplication by A. y e2 v x e1 Ae1 Ae2
- 10. Geometric eﬀect of a diagonal linear transformation Example 20 Let A = . Draw the eﬀect of the linear transformation 0 −1 which is multiplication by A. y e2 v x e1 Ae1 Ae2 Av
- 11. Geometric eﬀect of a diagonal linear transformation Example 20 Let A = . Draw the eﬀect of the linear transformation 0 −1 which is multiplication by A. y e2 v x e1 Ae1 Ae2 Av
- 12. Geometric eﬀect of a diagonal linear transformation Example 20 Let A = . Draw the eﬀect of the linear transformation 0 −1 which is multiplication by A. y e2 v x e1 Ae1 Ae2 Av
- 13. Example Draw the eﬀect of the linear transformation which is multiplication by A2 . y x
- 14. Example Draw the eﬀect of the linear transformation which is multiplication by A2 . y x
- 15. Example Draw the eﬀect of the linear transformation which is multiplication by A2 . y x
- 16. Geometric eﬀect of a non-diagonal linear transformation Example 1/2 3/2 Let B = . Draw the eﬀect of the linear transformation 3/2 1/2 which is multiplication by B. y x
- 17. Geometric eﬀect of a non-diagonal linear transformation Example 1/2 3/2 Let B = . Draw the eﬀect of the linear transformation 3/2 1/2 which is multiplication by B. y x e1
- 18. Geometric eﬀect of a non-diagonal linear transformation Example 1/2 3/2 Let B = . Draw the eﬀect of the linear transformation 3/2 1/2 which is multiplication by B. y Be1 x e1
- 19. Geometric eﬀect of a non-diagonal linear transformation Example 1/2 3/2 Let B = . Draw the eﬀect of the linear transformation 3/2 1/2 which is multiplication by B. y e2 Be1 x e1
- 20. Geometric eﬀect of a non-diagonal linear transformation Example 1/2 3/2 Let B = . Draw the eﬀect of the linear transformation 3/2 1/2 which is multiplication by B. y e2 Be1 Be2 x e1
- 21. Geometric eﬀect of a non-diagonal linear transformation Example 1/2 3/2 Let B = . Draw the eﬀect of the linear transformation 3/2 1/2 which is multiplication by B. y e2 Be1 v Be2 x e1
- 22. Geometric eﬀect of a non-diagonal linear transformation Example 1/2 3/2 Let B = . Draw the eﬀect of the linear transformation 3/2 1/2 which is multiplication by B. y e2 Be1 v Be2 x e1
- 23. Geometric eﬀect of a non-diagonal linear transformation Example 1/2 3/2 Let B = . Draw the eﬀect of the linear transformation 3/2 1/2 which is multiplication by B. y e2 Be1 v Be2 x e1 Bv
- 24. Geometric eﬀect of a non-diagonal linear transformation Example 1/2 3/2 Let B = . Draw the eﬀect of the linear transformation 3/2 1/2 which is multiplication by B. y e2 Be1 v Be2 x e1 Bv
- 25. Geometric eﬀect of a non-diagonal linear transformation Example 1/2 3/2 Let B = . Draw the eﬀect of the linear transformation 3/2 1/2 which is multiplication by B. y e2 Be1 v Be2 x e1 Bv
- 26. Geometric eﬀect of a non-diagonal linear transformation Example 1/2 3/2 Let B = . Draw the eﬀect of the linear transformation 3/2 1/2 which is multiplication by B. y x
- 27. Geometric eﬀect of a non-diagonal linear transformation Example 1/2 3/2 Let B = . Draw the eﬀect of the linear transformation 3/2 1/2 which is multiplication by B. y v1 x
- 28. Geometric eﬀect of a non-diagonal linear transformation Example 1/2 3/2 Let B = . Draw the eﬀect of the linear transformation 3/2 1/2 which is multiplication by B. y Bv1 v1 x
- 29. Geometric eﬀect of a non-diagonal linear transformation Example 1/2 3/2 Let B = . Draw the eﬀect of the linear transformation 3/2 1/2 which is multiplication by B. y Bv1 v1 x v2
- 30. Geometric eﬀect of a non-diagonal linear transformation Example 1/2 3/2 Let B = . Draw the eﬀect of the linear transformation 3/2 1/2 which is multiplication by B. y Bv1 Bv2 v1 x v2
- 31. Geometric eﬀect of a non-diagonal linear transformation Example 1/2 3/2 Let B = . Draw the eﬀect of the linear transformation 3/2 1/2 which is multiplication by B. y Bv1 Bv2 v1 x 2e2 v2
- 32. Geometric eﬀect of a non-diagonal linear transformation Example 1/2 3/2 Let B = . Draw the eﬀect of the linear transformation 3/2 1/2 which is multiplication by B. y Bv1 Bv2 v1 x 2e2 v2
- 33. Geometric eﬀect of a non-diagonal linear transformation Example 1/2 3/2 Let B = . Draw the eﬀect of the linear transformation 3/2 1/2 which is multiplication by B. y Bv1 Bv2 v1 x 2e2 v2
- 34. Geometric eﬀect of a non-diagonal linear transformation Example 1/2 3/2 Let B = . Draw the eﬀect of the linear transformation 3/2 1/2 which is multiplication by B. y 2Be2 Bv1 Bv2 v1 x 2e2 v2
- 35. Geometric eﬀect of a non-diagonal linear transformation Example 1/2 3/2 Let B = . Draw the eﬀect of the linear transformation 3/2 1/2 which is multiplication by B. y 2Be2 Bv1 Bv2 v1 x 2e2 v2
- 36. Geometric eﬀect of a non-diagonal linear transformation Example 1/2 3/2 Let B = . Draw the eﬀect of the linear transformation 3/2 1/2 which is multiplication by B. y 2Be2 Bv1 Bv2 v1 x 2e2 v2
- 37. Geometric eﬀect of a non-diagonal linear transformation Example 1/2 3/2 Let B = . Draw the eﬀect of the linear transformation 3/2 1/2 which is multiplication by B. y 2Be2 Bv1 Bv2 v1 x 2e2 v2
- 38. The big concept 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.
- 39. Finding the eigenvector(s) corresponding to an eigenvalue Example (Worksheet Problem 1) 0 −2 Let A = . The number 3 is an eigenvalue for A. Find −3 1 an eigenvector corresponding to this eigenvalue.
- 40. Finding the eigenvector(s) corresponding to an eigenvalue Example (Worksheet Problem 1) 0 −2 Let A = . The number 3 is an eigenvalue for A. Find −3 1 an eigenvector corresponding to this eigenvalue. Solution We seek x such that Ax = 3x =⇒ (A − 3I)x = 0.
- 41. Finding the eigenvector(s) corresponding to an eigenvalue Example (Worksheet Problem 1) 0 −2 Let A = . The number 3 is an eigenvalue for A. Find −3 1 an eigenvector corresponding to this eigenvalue. Solution We seek x such that Ax = 3x =⇒ (A − 3I)x = 0. −3 −2 0 1 2/3 0 A − 3I 0 = −3 −2 0 0 00
- 42. Finding the eigenvector(s) corresponding to an eigenvalue Example (Worksheet Problem 1) 0 −2 Let A = . The number 3 is an eigenvalue for A. Find −3 1 an eigenvector corresponding to this eigenvalue. Solution We seek x such that Ax = 3x =⇒ (A − 3I)x = 0. −3 −2 0 1 2/3 0 A − 3I 0 = −3 −2 0 0 00 −2/3 −2 So , or , are possible eigenvectors. 1 3
- 43. Example (Worksheet Problem 2) The number −2 is also an eigenvalue for A. Find an eigenvector.
- 44. Example (Worksheet Problem 2) The number −2 is also an eigenvalue for A. Find an eigenvector. Solution We have 2 −2 1 −1 A + 2I = . −3 3 00 1 is an eigenvector for the eigenvalue −2. So 1
- 45. Summary To ﬁnd the eigenvector(s) of a matrix corresponding to an eigenvalue λ, do Gaussian Elimination on A − λI.
- 46. Find the eigenvalues of a matrix Example (Worksheet Problem 3) Determine the eigenvalues of −4 −3 A= . 3 6
- 47. Find the eigenvalues of a matrix Example (Worksheet Problem 3) Determine the eigenvalues of −4 −3 A= . 3 6 Solution Ax = λx has a nonzero solution if and only if (A − λI)x = 0 has a nonzero solution, if and only if A − λI is not invertible.
- 48. Find the eigenvalues of a matrix Example (Worksheet Problem 3) Determine the eigenvalues of −4 −3 A= . 3 6 Solution Ax = λx has a nonzero solution if and only if (A − λI)x = 0 has a nonzero solution, if and only if A − λI is not invertible. What magic number determines whether a matrix is invertible?
- 49. Find the eigenvalues of a matrix Example (Worksheet Problem 3) Determine the eigenvalues of −4 −3 A= . 3 6 Solution Ax = λx has a nonzero solution if and only if (A − λI)x = 0 has a nonzero solution, if and only if A − λI is not invertible. What magic number determines whether a matrix is invertible? The determinant!
- 50. Find the eigenvalues of a matrix Example (Worksheet Problem 3) Determine the eigenvalues of −4 −3 A= . 3 6 Solution Ax = λx has a nonzero solution if and only if (A − λI)x = 0 has a nonzero solution, if and only if A − λI is not invertible. What magic number determines whether a matrix is invertible? The determinant! So to ﬁnd the possible values λ for which this could be true, we need to ﬁnd the determinant of A − λI.
- 51. −4 − λ −3 det(A − λI) = 6−λ 3 = (−4 − λ)(6 − λ) − (−3)(3) = (−24 − 2λ + λ2 ) + 9 = λ2 − 2λ − 15 = (λ + 3)(λ − 5) So λ = −3 and λ = 5 are the eigenvalues for A.
- 52. We can ﬁnd the eigenvectors now, based on what we did before. We have A + 3I =
- 53. We can ﬁnd the eigenvectors now, based on what we did before. We have −1 −3 A + 3I = 3 9
- 54. We can ﬁnd the eigenvectors now, based on what we did before. We have −1 −3 13 A + 3I = 3 9 00
- 55. We can ﬁnd the eigenvectors now, based on what we did before. We have −1 −3 13 A + 3I = 3 9 00 −3 So is an eigenvector for this eigenvalue. 1
- 56. We can ﬁnd the eigenvectors now, based on what we did before. We have −1 −3 13 A + 3I = 3 9 00 −3 So is an eigenvector for this eigenvalue. Also, 1 A − 5I =
- 57. We can ﬁnd the eigenvectors now, based on what we did before. We have −1 −3 13 A + 3I = 3 9 00 −3 So is an eigenvector for this eigenvalue. Also, 1 −9 −3 A − 5I = 3 1
- 58. We can ﬁnd the eigenvectors now, based on what we did before. We have −1 −3 13 A + 3I = 3 9 00 −3 So is an eigenvector for this eigenvalue. Also, 1 −9 −3 1 1/3 A − 5I = 3 1 0 0
- 59. We can ﬁnd the eigenvectors now, based on what we did before. We have −1 −3 13 A + 3I = 3 9 00 −3 So is an eigenvector for this eigenvalue. Also, 1 −9 −3 1 1/3 A − 5I = 3 1 0 0 −1/3 −1 So or would be an eigenvector for this eigenvalue. 1 3
- 60. Summary To ﬁnd the eigenvalues of a matrix A, ﬁnd the determinant of A − λI. This will be a polynomial in λ, and its roots are the eigenvalues.
- 61. Example (Worksheet Problem 4) Find the eigenvalues of −10 6 A= . −15 9
- 62. Example (Worksheet Problem 4) Find the eigenvalues of −10 6 A= . −15 9 Solution The characteristic polynomial is −10 − λ 6 = (−10−λ)(9−λ)−(−15)(6) = λ2 +λ = λ(λ+1). −15 9−λ
- 63. Example (Worksheet Problem 4) Find the eigenvalues of −10 6 A= . −15 9 Solution The characteristic polynomial is −10 − λ 6 = (−10−λ)(9−λ)−(−15)(6) = λ2 +λ = λ(λ+1). −15 9−λ The eigenvalues are 0 and −1.
- 64. Example (Worksheet Problem 4) Find the eigenvalues of −10 6 A= . −15 9 Solution The characteristic polynomial is −10 − λ 6 = (−10−λ)(9−λ)−(−15)(6) = λ2 +λ = λ(λ+1). −15 9−λ 3 The eigenvalues are 0 and −1. A set of eigenvectors are and 5 2 . 3
- 65. Applications In a Markov Chain, the steady-state vector is an eigenvector corresponding to the eigenvalue 1.

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