Example 712 the matrix 3 1 3 5 has characteristic

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Unformatted text preview: s called the characteristic polynomial of the matrix A. For any root λ of (3), the space (4) is called the eigenspace corresponding to the eigenvalue λ. Example 7.1.2. The matrix 3 1 3 5 has characteristic polynomial (3 − λ)(5 − λ) − 3 = 0; in other words, λ2 − 8λ + 12 = 0. Hence the eigenvalues are λ1 = 2 and λ2 = 6, with corresponding eigenvectors v1 = 3 −1 and v2 = 1 1 respectively. The eigenspace corresponding to the eigenvalue 2 is v ∈ R2 : 1 1 3 3 v=0 = c 3 −1 :c∈R . The eigenspace corresponding to the eigenvalue 6 is v ∈ R2 : −3 1 3 −1 v=0 = c 1 1 :c∈R . Example 7.1.3. Consider the matrix −1 A= 0 0 6 −13 −9 −12 30 . 20 To find the eigenvalues of A, we need to find the roots of −1 − λ 6 −12 det 0 −13 − λ 30 = 0; 0 −9 20 − λ in other words, (λ + 1)(λ − 2)(λ − 5) = 0. The eigenvalues are therefore λ1 = −1, λ2 = 2 and λ3 = 5. An eigenvector corresponding to the eigenvalue −1 is a solution of the system 0 6 −12 1 (A + I)v = 0 −12 30 v = 0, with root v1 = 0 . 0 −9 21 0 An eigenvector corresponding to the eigenvalue 2 is a solution of the system −3 6 −12 0 (A − 2I)v = 0 −15 30 v = 0, with root v2 = 2 . 0 −9 18 1 An eigenvector corresponding to the eigenvalue 5 is a solution of the system −6 6 −12 1 (A − 5I)v = 0 −18 30 v = 0, with root v3 = −5 . 0 −9 15 −3 Note that the three eigenspaces are all lines through the origin. Note also that the eigenvectors v1 , v2 and v3 are linearly independent, and so form a basis for R3 . Chapter 7 : Eigenvalues and Eigenvectors page 3 of 12 Linear Algebra c W W L Chen, 1982, 2005 Example 7.1.4. Consider the matrix 17 A = 45 −30 −10 −28 20 −5 −15 . 12 To find the eigenvalues of A, we need to find the roots of 17 − λ −10 −5 det 45 −28 − λ −15 = 0; −30 20 12 − λ in other words, (λ + 3)(λ − 2)2 = 0. The eigenvalues are therefore λ1 = −3 and λ2 = 2. An eigenvector corresponding to the eigenvalue −3 is a solution of the system 20 −10 −5 1 (A + 3I)v = 45 −25 −15 v = 0, with root v1 = 3 . −30 20 15 −2 An eigenvector corresponding 15 −10 (A − 2I)v = 45 −30 −30 20 to the eigenvalue 2 is a solution of the system −5 1 −15 v = 0, with roots v2 = 0 10 3 2 v3 = 3 . 0 and Note that the eigenspace corresponding to the eigenvalue −3 is a line through the origin, while the eigenspace corresponding to the eigenvalue 2 is a plane through the origin. Note also that the eigenvectors v1 , v2 and v3 are linearly independent, and so form a basis for R3 . Example 7.1.5. Consider the matrix 0 0. 3 2 A = 1 0 −1 0 0 To find the eigenvalues of A, we need to find the roots of 2−λ −1 0 det 1 0−λ 0 = 0; 0 0 3−λ in other words, (λ − 3)(λ − 1)2 = 0. The eigenvalues are therefore λ1 = 3 and λ2 = 1. An eigenvector corresponding to the eigenvalue 3 is a solution of the system −1 −1 0 0 (A − 3I)v = 1 −3 0 v = 0, with root v1 = 0 . 0 0 0 1 An eigenvector corresponding to the 1 (A − I)v = 1 0 eigenvalue 1 is a solution of the system −1 0 1 −1 0 v = 0, with ro...
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