Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $A$ has an eigenvector $x$ with eigenvalue $\lambda$, find an eigenvector for the matrix $B = S^{-1}AS$. Find the corresponding eigenvalue as well.

Do: $(B-\lambda I)x = 0$

($S^{-1}AS - \lambda I)x = 0$. I am kind of stuck here...

I was thinking about an expression for $A^{-1}$ and minimal polynomials. Is that a good place to start?

share|improve this question
add comment

2 Answers 2

up vote 1 down vote accepted

Given $A=SBS^{-1}$ and

$$ (\lambda I-A)x=0 \implies (\lambda I- SBS^{-1} )x=0 \implies (\lambda S - SB )S^{-1}x=0 $$

$$ \implies S(\lambda I - B )S^{-1}x=0 \implies (\lambda I - B )S^{-1}x=S^{-1}0=0 .$$

The last equation implies that $B$ has the same eigenvalue as $A$ (which is a fact for the similarity matrices) with the eigenvector $ z=S^{-1}x .$

share|improve this answer
add comment

Try $y=S^{-1}x$ for an eigenvector of $B$. (Do you see why this is a non-$0$ vector?) The corresponding eigenvalue should be easy enough to figure out.

share|improve this answer
Why/how do you know to try this? –  CodeKingPlusPlus Dec 7 '12 at 4:16
Well, we know that $(CA)x=C(Ax)=C(\lambda x)=\lambda Cx$ for any (appropriately sized) matrix $C$, yes? The problem is, we still have that $C$ floating around, so we can't say that $x$ is an eigenvector for $CA$. What if we wanted $Cx$ to be an eigenvector of some matrix, instead of just $x$? Well, if $C$ is invertible, then we know that $\lambda Cx=(CA)x=(CA)Ix=(CAC^{-1})Cx$, so it was just a matter of figuring out how to get a $Cx$ factor on the far right of the scalar-free expression, somehow. In this particular case, we're just letting $C=S^{-1}$, but the approach is the same. –  Cameron Buie Dec 7 '12 at 4:26
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.