2130_Chapter7_Notes

1 1 2 first nd the eigenvalues by solving the

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Unformatted text preview: −R1 1 −1 20 0 −3 30 1 2 −1 0 10 10 R −R2 R −2R2 −3− → 0 1 −1 0 −1 − → 0 1 −1 0 −− −− − 1R −3 2 00 00 00 00 6 / 13 Exa 2 (slide 2) 1 2 −1 0 10 10 1 1 0 −→ 0 1 −1 0 A | 0 = 2 1 −1 20 00 00 =⇒ x1 + x3 = 0 x2 − x3 = 0 =⇒ x1 = − x3 x2 = x3 Label x3 = c, then x1 −c −1 x = x2 = c = c 1 x3 c 1 for any constant c. 7 / 13 4 Eigenvalues & Eigenvectors For the matrix A, a scalar λ is an eigenvalue with eigenvector v if Av = λv, v = 0. Note that 0 is never an eigenvector. Eigenvalues and their eigenvectors can be found by solving a homogeneous equation: Av = λv ⇐⇒ Av − λv = 0 ⇐⇒ Av − λIv = 0 ⇐⇒ (A − λI)v = 0. This homogeneous equation either has: (1) the unique solution v = 0 ⇐⇒ det(A − λI) = 0, or (2) infinitely many solutions ⇐⇒ det(A − λI) = 0. Theorem λ is an eigenvalue for A ⇐⇒ det(A − λI) = 0. 8 / 13 Example 3 Find the eigenvalues and eigenvectors for 1 2 −1 A = 2 1 1 . 1 −1 2 First find the eigenvalues by solving the equation det(A − λI) = 0. 1−λ 2 −1 det(A − λI) = 2 1−λ 1 1 −1 2 λ 1−λ 1 2 −1 2 −1 = (1 − λ) −2 + −1 2 − λ −1 2 − λ 1−λ 1 = (1 − λ) (1 − λ)(2 − λ) + 1 − 2 (4 − 2λ) − 1 + 2 + (1 − λ) = (1 − λ)(3 − 3λ + λ2 ) − 2(3 − 2λ) + (3 − λ) = (1 − λ)(3 − 3λ + λ2 ) − 3 + 3λ = (1 − λ)(3 − 3λ + λ2 − 3) = (1 − λ)(λ2...
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This note was uploaded on 03/19/2014 for the course APMA 2130 taught by Professor Bernardfulgham during the Fall '09 term at UVA.

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