Matrix
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Find the characteristic polynomial and the eigenvalues
Find a basis for the eigenspace of each of the distinct eigenvalues
Factorize A
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Solution Summary
This provides an example of finding characteristic polynomial, eigenvalues, basis for eigenspace, factorization.
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Solution:
a)
Let A = [■(4&-2&2@1&1&1@0&0&2)]
The characteristic polynomial is given by
|A-λI| = 0
|■(4-λ&-2&2@1&1-λ&1@0&0&2-λ)| = 0
(4-λ)[(1-λ)(2-λ)-1*0]+2[1(2-λ)-0]+2[1*0-0(1-λ)]= 0
(4-λ)(2-λ)(1-λ)+2(2-λ)=0
(2-λ)(λ^2-5λ+4+2)=0
(2-λ)(λ^2-5λ+6)=0
(2-λ)(λ-2)(λ-3)=0
λ=2,2,3
Corresponding eigen vectors are
λ=2
Plugging λ=2 we get
[■(2&-2&2@1&-1&1@0&0&0)][■(x@y@z)]=0
2x-2y+2z=0
x-y+z=0
The above two equations are one same.
x+z=y
Assume x = -t
z = 2t and y =t
[■(x@y@z)]= [■(-t@t@2t)]=t[■(-1@1@2)]
The eigen vector is [■(-1@1@2)]
Also we can assume x = 0, z = t
Then y = t
Therefore the other vector is
[■(x@y@z)]= ...
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