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2 2 0 1 2 3 4 1 6 6 a i 1 v

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Unformatted text preview: λ)(−λ)(−λ) = (1 − λ)λ2 . (?? pts) Solution: det(A − λI ) = ￿ 0 ￿ ￿ ￿0 0 −λ ￿ We have two eigenvalues, λ = 1 and λ = 0, the second having (algebraic) multiplicity 2. with λ = 1 we need to solve (A − λI )￿ = 0. As A − I = To find eigenvector associated x 0 3 4 0 −1 1 ￿3 r 0 −1 1 0 −1 1 → 0 0 −1 → 0 −1 → 0 0 0 −1 0 3 4 + r 1 + 7 r2 0 0 0 x1 1 indicates one free variable, x1 , and x2 = x3 = 0, we get a null space x2 = x1 0 . x3 0 1 Thus, eigenvalue λ = 1 is associated with eigenvector 0 . 0 To find eigenvector(s) associated with λ = 0 we need to solve (A − λI )￿ = 0. As x 130 134 −4￿2 r → 0 0 1 indicates only one free variable, x2 , A − λI = A = 0 0 1 000 0 0 0 x1 −3 and x1 = −3x2 , x3 = 0, we get a null space x2 = x2 1 . Thus, eigenvalue λ = 0 x3 0 −3 (of algebraic multiplicity 2) is associated with a single eigenvector 1 . 0 This means, in particular, that A has only two eigenvectors, so A is not diagonalizable. 2 Math 441 Exam 4 5 !6 1. (12 pts) Given A ! 3 !4 a) Find the eigenvalues a...
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This document was uploaded on 02/19/2014 for the course MATH 441 at WVU.

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