2130_Chapter7_Notes

# 1 1 2 first nd the eigenvalues by solving the

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This note was uploaded on 03/19/2014 for the course APMA 2130 taught by Professor Bernardfulgham during the Fall '09 term at UVA.

Ask a homework question - tutors are online