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Linear Algebra Chapter 5: Eigenvalues and Eigenvectors Section 5.1. Eigenvalues and EigenvectorsProofs of Theorems April 10, 2020 () Linear Algebra April 10, 2020 1 / 23 Table of contents Page 300 Number 8 1 Page 300 Number 14 2

  1. Linear Algebra Chapter 5: Eigenvalues and Eigenvectors Section 5.1. Eigenvalues and Eigenvectors—Proofs of Theorems April 10, 2020 () Linear Algebra April 10, 2020 1 / 23

  2. Table of contents Page 300 Number 8 1 Page 300 Number 14 2 Theorem 5.1. Properties of Eigenvalues and Eigenvectors 3 Page 298 Example 8 4 Page 300 Number 18 5 Page 301 Number 30 6 Page 301 Number 32 7 Page 302 Number 38 8 Page 302 Number 40 9 () Linear Algebra April 10, 2020 2 / 23

  3. Page 300 Number 8 Page 300 Number 8 Page 300 Number 8. Find the characteristic polynomial, the real  − 1 0 0   . eigenvalues, and the corresponding eigenvectors for A = − 4 2 − 1  4 0 3 Solution. We have  − 1 0 0   1 0 0   − 1 − λ 0 0   =  . A − λ I = − 4 2 − 1  − λ 0 1 0 − 4 2 − λ − 1    4 0 3 0 0 1 4 0 3 − λ () Linear Algebra April 10, 2020 3 / 23

  4. Page 300 Number 8 Page 300 Number 8 Page 300 Number 8. Find the characteristic polynomial, the real  − 1 0 0   . eigenvalues, and the corresponding eigenvectors for A = − 4 2 − 1  4 0 3 Solution. We have  − 1 0 0   1 0 0   − 1 − λ 0 0   =  . A − λ I = − 4 2 − 1  − λ 0 1 0 − 4 2 − λ − 1    4 0 3 0 0 1 4 0 3 − λ So the characteristic polynomial is � � − 1 − λ 0 0 � � � � p ( λ ) = det( A − λ I ) = − 4 2 − λ − 1 � � � � 4 0 3 − λ � � � � 2 − λ − 1 � � = ( − 1 − λ ) � − (0) + (0) � � 0 3 − λ � = ( − 1 − λ ) ((2 − λ )(3 − λ ) − ( − 1)(0)) = ( − 1 − λ )(2 − λ )(3 − λ ). () Linear Algebra April 10, 2020 3 / 23

  5. Page 300 Number 8 Page 300 Number 8 Page 300 Number 8. Find the characteristic polynomial, the real  − 1 0 0   . eigenvalues, and the corresponding eigenvectors for A = − 4 2 − 1  4 0 3 Solution. We have  − 1 0 0   1 0 0   − 1 − λ 0 0   =  . A − λ I = − 4 2 − 1  − λ 0 1 0 − 4 2 − λ − 1    4 0 3 0 0 1 4 0 3 − λ So the characteristic polynomial is � � − 1 − λ 0 0 � � � � p ( λ ) = det( A − λ I ) = − 4 2 − λ − 1 � � � � 4 0 3 − λ � � � � 2 − λ − 1 � � = ( − 1 − λ ) � − (0) + (0) � � 0 3 − λ � = ( − 1 − λ ) ((2 − λ )(3 − λ ) − ( − 1)(0)) = ( − 1 − λ )(2 − λ )(3 − λ ). () Linear Algebra April 10, 2020 3 / 23

  6. Page 300 Number 8 Page 300 Number 8 (continued 1) Solution (continued). So the eigenvalues are λ 1 = − 1, λ 2 = 2, λ 3 = 3. To find the eigenvectors corresponding to each eigenvalue, we consider the v = � formula A � v = λ� v or ( A − λ I ) � 0 (see Note 5.1.A): () Linear Algebra April 10, 2020 4 / 23

  7. Page 300 Number 8 Page 300 Number 8 (continued 1) Solution (continued). So the eigenvalues are λ 1 = − 1, λ 2 = 2, λ 3 = 3. To find the eigenvectors corresponding to each eigenvalue, we consider the v = � formula A � v = λ� v or ( A − λ I ) � 0 (see Note 5.1.A): v 1 = [ v 1 , v 2 , v 3 ] T an eigenvector corresponding to the λ 1 = − 1. With � v 1 = � eigenvalue λ 1 = − 1 we need ( A − λ 1 I ) � 0. () Linear Algebra April 10, 2020 4 / 23

  8. Page 300 Number 8 Page 300 Number 8 (continued 1) Solution (continued). So the eigenvalues are λ 1 = − 1, λ 2 = 2, λ 3 = 3. To find the eigenvectors corresponding to each eigenvalue, we consider the v = � formula A � v = λ� v or ( A − λ I ) � 0 (see Note 5.1.A): v 1 = [ v 1 , v 2 , v 3 ] T an eigenvector corresponding to the λ 1 = − 1. With � v 1 = � eigenvalue λ 1 = − 1 we need ( A − λ 1 I ) � 0. So we consider the augmented matrix     − 1 − ( − 1) 0 0 0 0 0 0 0  = − 4 2 − ( − 1) − 1 0 − 4 3 − 1 0    4 0 3 − ( − 1) 0 4 0 4 0  0 0 0 0   0 0 0 0  R 3 → R 3 + R 2 R 2 → R 2 − R 3 � � − 4 3 − 1 0 − 4 0 − 4 0     0 3 3 0 0 3 3 0  0 0 0 0   1 0 1 0  R 2 → R 2 / ( − 4) R 1 ↔ R 2 � � 1 0 1 0 0 0 0 0 R 3 → R 3 / 3     0 1 1 0 0 1 1 0 () Linear Algebra April 10, 2020 4 / 23

  9. Page 300 Number 8 Page 300 Number 8 (continued 1) Solution (continued). So the eigenvalues are λ 1 = − 1, λ 2 = 2, λ 3 = 3. To find the eigenvectors corresponding to each eigenvalue, we consider the v = � formula A � v = λ� v or ( A − λ I ) � 0 (see Note 5.1.A): v 1 = [ v 1 , v 2 , v 3 ] T an eigenvector corresponding to the λ 1 = − 1. With � v 1 = � eigenvalue λ 1 = − 1 we need ( A − λ 1 I ) � 0. So we consider the augmented matrix     − 1 − ( − 1) 0 0 0 0 0 0 0  = − 4 2 − ( − 1) − 1 0 − 4 3 − 1 0    4 0 3 − ( − 1) 0 4 0 4 0  0 0 0 0   0 0 0 0  R 3 → R 3 + R 2 R 2 → R 2 − R 3 � � − 4 3 − 1 0 − 4 0 − 4 0     0 3 3 0 0 3 3 0  0 0 0 0   1 0 1 0  R 2 → R 2 / ( − 4) R 1 ↔ R 2 � � 1 0 1 0 0 0 0 0 R 3 → R 3 / 3     0 1 1 0 0 1 1 0 () Linear Algebra April 10, 2020 4 / 23

  10. Page 300 Number 8 Page 300 Number 8 (continued 2) Solution (continued).     1 0 1 0 1 0 1 0 R 2 ↔ R 3  . � 0 0 0 0 0 1 1 0    0 1 1 0 0 0 0 0 + = 0 = − v 3 v 1 v 3 v 1 So we need v 2 + v 3 = 0 , or v 2 = − v 3 or, with r = v 3 0 = 0 = v 3 v 3 v 1 = − r as a free variable, = − r . v 2 v 3 = r () Linear Algebra April 10, 2020 5 / 23

  11. Page 300 Number 8 Page 300 Number 8 (continued 2) Solution (continued).     1 0 1 0 1 0 1 0 R 2 ↔ R 3  . � 0 0 0 0 0 1 1 0    0 1 1 0 0 0 0 0 + = 0 = − v 3 v 1 v 3 v 1 So we need v 2 + v 3 = 0 , or v 2 = − v 3 or, with r = v 3 0 = 0 = v 3 v 3 v 1 = − r as a free variable, = − r . So the collection of all eigenvectors of v 2 v 3 = r   − 1  where r ∈ R , r � = 0. λ 1 = − 1 is � v 1 = r − 1  1 () Linear Algebra April 10, 2020 5 / 23

  12. Page 300 Number 8 Page 300 Number 8 (continued 2) Solution (continued).     1 0 1 0 1 0 1 0 R 2 ↔ R 3  . � 0 0 0 0 0 1 1 0    0 1 1 0 0 0 0 0 + = 0 = − v 3 v 1 v 3 v 1 So we need v 2 + v 3 = 0 , or v 2 = − v 3 or, with r = v 3 0 = 0 = v 3 v 3 v 1 = − r as a free variable, = − r . So the collection of all eigenvectors of v 2 v 3 = r   − 1  where r ∈ R , r � = 0. λ 1 = − 1 is � v 1 = r − 1  1 () Linear Algebra April 10, 2020 5 / 23

  13. Page 300 Number 8 Page 300 Number 8 (continued 3) Solution (continued). v 2 = � λ 2 = 2. As above, we consider ( A − 2 I ) � 0 and consider the augmented matrix     − 1 − (2) 0 0 0 − 3 0 0 0  = − 4 2 − (2) − 1 0 − 4 0 − 1 0    4 0 3 − (2) 0 4 0 1 0  1 0 0 0   1 0 0 0  R 1 → R 1 / ( − 3) R 2 → R 2 +4 R 1 � � − 4 0 − 1 0 0 0 − 1 0 R 3 → R 3 − 4 R 1     4 0 1 0 0 0 1 0  1 0 0 0   1 0 0 0  R 3 → R 3 + R 2 R 2 →− R 2 � �  . 0 0 − 1 0 0 0 1 0    0 0 0 0 0 0 0 0 () Linear Algebra April 10, 2020 6 / 23

  14. Page 300 Number 8 Page 300 Number 8 (continued 4) = 0 = 0 v 1 v 1 Solution (continued). So we need v 3 = 0 , or v 2 = v 2 or, with 0 = 0 = 0 v 3 v 1 = 0 s = v 2 as a free variable, = . So the collection of all eigenvectors v 2 s v 3 = 0   0  where s ∈ R , s � = 0. of λ 2 = 2 is � v 2 = s 1  0 () Linear Algebra April 10, 2020 7 / 23

  15. Page 300 Number 8 Page 300 Number 8 (continued 4) = 0 = 0 v 1 v 1 Solution (continued). So we need v 3 = 0 , or v 2 = v 2 or, with 0 = 0 = 0 v 3 v 1 = 0 s = v 2 as a free variable, = . So the collection of all eigenvectors v 2 s v 3 = 0   0  where s ∈ R , s � = 0. of λ 2 = 2 is � v 2 = s 1  0 () Linear Algebra April 10, 2020 7 / 23

  16. Page 300 Number 8 Page 300 Number 8 (continued 5) Solution (continued). v 3 = � λ 3 = 3. As above, we consider ( A − 3 I ) � 0 and consider the augmented matrix  − 1 − (3) 0 0 0   − 4 0 0 0   = − 4 2 − (3) − 1 0 − 4 − 1 − 1 0    4 0 3 − (3) 0 4 0 0 0     − 4 0 0 0 1 0 0 0 R 1 → R 1 / ( − 4) R 2 → R 2 − R 1 � �  . 0 − 1 − 1 0 0 1 1 0 R 3 → R 3 + R 1 R 2 → − R 2    0 0 0 0 0 0 0 0 v 1 = 0 v 1 = 0 So we need v 2 + v 3 = 0 , or v 2 = − v 3 . . . 0 = 0 v 3 = v 3 () Linear Algebra April 10, 2020 8 / 23

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