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I understand that similar matrices have the same eigenvalues and corresponding eigenvectors, I get the following notation:

$$(P^{-1} A P)\cdot v=\lambda v\rightarrow(P^{-1} A P)\cdot P^{-1}v=\lambda P^{-1}v$$

Why do we multiple both sides with $P^{-1}$?

gbox
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1 Answers1

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The statement you write is either wrong, or uses $v$ for two different vectors, which is probably what is confusing you. Here's what it should say:

Let $\lambda$ be an eigenvalue of $A$. Then let $v \ne 0$ be some corresponding eigenvector. We have that $$ Av = \lambda v $$ Now consider the nonzero vector $w = P^{-1} v$. Notice that $$ (P^{-1} A P) w = P^{-1} A P P^{-1} v = P^{-1} A v = P^{-1} (\lambda v) = \lambda (P^{-1} v) = \lambda w. $$ Therefore, $\lambda$ is an eigenvalue of $P^{-1} A P$ as well (with an example eigenvector given by $w$).

This works for any $P$, so if $A$ and $B$ are similar, then every eigenvalue of $A$ is an eigenvalue of $B$, and every eigenvalue of $B$ is an eigenvalue of $A$. Therefore, $A$ and $B$ have the same eigenvalues.

Caleb Stanford
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  • Thanks. So we can conclude that if $v\neq 0$ is an eigenvector of $A$ (and $B$) so is $P^{-1}\cdot v$? – gbox Aug 25 '15 at 14:58
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    @gbox No--$A$ and $B$ do not have the same eigenvectors. They only have the same eigenvalues. If $B = P^{-1} A P$ and $v$ is an eigenvector of $A$ then $P^{-1}v$ is an eigenvalue of $B$. – Caleb Stanford Aug 25 '15 at 15:28