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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?

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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 .$

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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.

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  • $\begingroup$ Why/how do you know to try this? $\endgroup$ – CodeKingPlusPlus Dec 7 '12 at 4:16
  • $\begingroup$ 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. $\endgroup$ – Cameron Buie Dec 7 '12 at 4:26

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