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Eigenvalues and Eigenvectors Artem Los (arteml@kth.se) February - PowerPoint PPT Presentation

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Eigenvalues and Eigenvectors Artem Los (arteml@kth.se) February 9th, 2017 Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 1 / 16 Overview What is an Eigenvalue? 1 Finding eigenvalues and eigenvectors 2 Exam

  1. Eigenvalues and Eigenvectors Artem Los (arteml@kth.se) February 9th, 2017 Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 1 / 16

  2. Overview What is an Eigenvalue? 1 Finding eigenvalues and eigenvectors 2 Exam question 3 Martin’s method 4 Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 2 / 16

  3. What is an Eigenvalue? Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 3 / 16

  4. Introduction problem � 3 / 4 � 1 / 4 find A 1000 (2 , 1) using A (1 , 1) = (1 , 1) Given A = Problem. 1 / 4 3 / 4 and A (1 , − 1) = 1 2 (1 , − 1). Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 4 / 16

  5. Introduction problem � 3 / 4 � 1 / 4 find A 1000 (2 , 1) using A (1 , 1) = (1 , 1) Given A = Problem. 1 / 4 3 / 4 and A (1 , − 1) = 1 2 (1 , − 1). Step 1: Find x 1 , x 2 so that (2 , 1) = x 1 (1 , 1) + x 2 (1 , − 1). x 1 = 3 2 and x 2 = 1 2 Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 4 / 16

  6. Introduction problem � 3 / 4 � 1 / 4 find A 1000 (2 , 1) using A (1 , 1) = (1 , 1) Given A = Problem. 1 / 4 3 / 4 and A (1 , − 1) = 1 2 (1 , − 1). Step 1: Find x 1 , x 2 so that (2 , 1) = x 1 (1 , 1) + x 2 (1 , − 1). x 1 = 3 2 and x 2 = 1 2 Step 2: A 1000 (2 , 1) = x 1 A 1000 (1 , 1) + x 2 A 1000 (1 , − 1) = Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 4 / 16

  7. Introduction problem � 3 / 4 � 1 / 4 find A 1000 (2 , 1) using A (1 , 1) = (1 , 1) Given A = Problem. 1 / 4 3 / 4 and A (1 , − 1) = 1 2 (1 , − 1). Step 1: Find x 1 , x 2 so that (2 , 1) = x 1 (1 , 1) + x 2 (1 , − 1). x 1 = 3 2 and x 2 = 1 2 Step 2: A 1000 (2 , 1) = x 1 A 1000 (1 , 1) + x 2 A 1000 (1 , − 1) = � 1 � 1000 = x 1 (1 , 1) + x 2 (1 , − 1) = 2 Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 4 / 16

  8. Introduction problem � 3 / 4 � 1 / 4 find A 1000 (2 , 1) using A (1 , 1) = (1 , 1) Given A = Problem. 1 / 4 3 / 4 and A (1 , − 1) = 1 2 (1 , − 1). Step 1: Find x 1 , x 2 so that (2 , 1) = x 1 (1 , 1) + x 2 (1 , − 1). x 1 = 3 2 and x 2 = 1 2 Step 2: A 1000 (2 , 1) = x 1 A 1000 (1 , 1) + x 2 A 1000 (1 , − 1) = � 1 � 1000 = x 1 (1 , 1) + x 2 (1 , − 1) = 2 = x 1 (1 , 1) = (3 / 2 , 3 / 2) Based on lecture notes from Olof Heden’s lecture in SF1604 Linear Algebra (in 2014) . Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 4 / 16

  9. Definition Eigenvalue The number λ is an eigenvalue of matrix A if A � x = λ� x for some vector � x � = 0. � x is an eigenvector to A belonging to the eigenvalue λ . Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 5 / 16

  10. Definition Eigenvalue The number λ is an eigenvalue of matrix A if A � x = λ� x for some vector � x � = 0. � x is an eigenvector to A belonging to the eigenvalue λ . We can use it to: find A 1000 given A . perform diagonalisations Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 5 / 16

  11. Finding eigenvalues and eigenvectors Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 6 / 16

  12. Finding eigenvalues and their corresponding eigenvectors Using the definition, we can deduce the following: Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 7 / 16

  13. Finding eigenvalues and their corresponding eigenvectors Using the definition, we can deduce the following: To find eigenvalues. Solve for λ (in the characteristic equation below): det ( A − λ I ) = 0 Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 7 / 16

  14. Finding eigenvalues and their corresponding eigenvectors Using the definition, we can deduce the following: To find eigenvalues. Solve for λ (in the characteristic equation below): det ( A − λ I ) = 0 Eg. we may get (2 − λ )(3 − λ ) 2 = 0, so the eigenvalues are λ 2 and λ 3 . Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 7 / 16

  15. Finding eigenvalues and their corresponding eigenvectors Using the definition, we can deduce the following: To find eigenvalues. Solve for λ (in the characteristic equation below): det ( A − λ I ) = 0 Eg. we may get (2 − λ )(3 − λ ) 2 = 0, so the eigenvalues are λ 2 and λ 3 . The algebraic multiplicity of λ 2 and λ 3 are 1 and 2, respectively. Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 7 / 16

  16. Finding eigenvalues and their corresponding eigenvectors Using the definition, we can deduce the following: To find eigenvalues. Solve for λ (in the characteristic equation below): det ( A − λ I ) = 0 Eg. we may get (2 − λ )(3 − λ ) 2 = 0, so the eigenvalues are λ 2 and λ 3 . The algebraic multiplicity of λ 2 and λ 3 are 1 and 2, respectively. Solve for � v (for λ 2 and λ 3 ): To find corresponding eigenvectors. v = � ( A − λ I ) � 0 The dimension of the solution space is the geometric multiplicity. Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 7 / 16

  17. Example Problem. Find the eigenvalues and eigenvectors of � 2 � − 2 A = 0 3 Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 8 / 16

  18. Example Problem. Find the eigenvalues and eigenvectors of � 2 � − 2 A = 0 3 Step 1: Solve for λ in det( A − λ I ) = 0 Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 8 / 16

  19. Example Problem. Find the eigenvalues and eigenvectors of � 2 � − 2 A = 0 3 Step 1: Solve for λ in det( A − λ I ) = 0 � � 2 − λ − 2 � � � = � � 0 3 − λ � Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 8 / 16

  20. Example Problem. Find the eigenvalues and eigenvectors of � 2 � − 2 A = 0 3 Step 1: Solve for λ in det( A − λ I ) = 0 � � 2 − λ − 2 � � � = (2 − λ )(3 − λ ) = 0 � � 0 3 − λ � The roots are λ 1 = 2 and λ 2 = 3, which are also the eigenvalues of A . Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 8 / 16

  21. Example Problem. Find the eigenvalues and eigenvectors of � 2 � − 2 A = 0 3 Step 1: Solve for λ in det( A − λ I ) = 0 � � 2 − λ − 2 � � � = (2 − λ )(3 − λ ) = 0 � � 0 3 − λ � The roots are λ 1 = 2 and λ 2 = 3, which are also the eigenvalues of A . Both of them have algebraic multiplicity of 1, since we don’t have any double roots, etc. Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 8 / 16

  22. Example (continued) Find the eigenvalues and eigenvectors of Problem. � 2 � − 2 A = 0 3 v = � Step 2: To find the corresponding eigenvectors, solve ( A − λ I ) � 0 Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 9 / 16

  23. Example (continued) Find the eigenvalues and eigenvectors of Problem. � 2 � − 2 A = 0 3 v = � Step 2: To find the corresponding eigenvectors, solve ( A − λ I ) � 0 Case λ = 2 � 0 � − 2 0 0 1 0 Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 9 / 16

  24. Example (continued) Find the eigenvalues and eigenvectors of Problem. � 2 � − 2 A = 0 3 v = � Step 2: To find the corresponding eigenvectors, solve ( A − λ I ) � 0 Case λ = 2 � 0 � − 2 0 0 1 0 x 2 has to be 0 and x 1 can be any, let’s say 1. � 1 � � v = 0 Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 9 / 16

  25. Example (continued) Find the eigenvalues and eigenvectors of Problem. � 2 � − 2 A = 0 3 v = � Step 2: To find the corresponding eigenvectors, solve ( A − λ I ) � 0 Case λ = 2 Case λ = 3 � 0 � − 1 � � − 2 0 − 2 0 0 1 0 0 0 0 x 2 has to be 0 and x 1 can be any, let’s say 1. � 1 � � v = 0 Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 9 / 16

  26. Example (continued) Find the eigenvalues and eigenvectors of Problem. � 2 � − 2 A = 0 3 v = � Step 2: To find the corresponding eigenvectors, solve ( A − λ I ) � 0 Case λ = 2 Case λ = 3 � 0 � − 1 � � − 2 0 − 2 0 0 1 0 0 0 0 x 2 has to be 0 and x 1 can By parametrization of be any, let’s say 1. − x 1 − 2 x 2 = 0, we get: � 2 � 1 � � � � v = v = 0 − 1 Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 9 / 16

  27. Example (continued) Find the eigenvalues and eigenvectors of Problem. � 2 � − 2 A = 0 3 v = � Step 2: To find the corresponding eigenvectors, solve ( A − λ I ) � 0 Case λ = 2 Case λ = 3 � 0 � − 1 � � − 2 0 − 2 0 0 1 0 0 0 0 x 2 has to be 0 and x 1 can By parametrization of be any, let’s say 1. − x 1 − 2 x 2 = 0, we get: � 2 � 1 � � � � v = v = 0 − 1 Geometric multiplicity is 1 in both cases, since the dimension is 1. Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 9 / 16

  28. Exam question Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 10 / 16

  29. Exam question From https://www.kth.se/social/files/5858ed94f276542bb5877d62/tentor.pdf , p. 2 Artem Los (arteml@kth.se) Eigenvalues and Eigenvectors February 9th, 2017 11 / 16

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