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
−42
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.
 Spring '08
 STAFF
 Math, Linear Algebra, Algebra

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