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Problem

Let $A$ and $B$ be similar matrices. Prove that the geometric multiplicities of the eigenvalues of $A$ and $B$ are the same. [Hint: show that, if $B=P^{-1}AP$, then every eigenvector of $B$ is of the form $P^{-1}\vec{v}$ for some eigenvector $\vec{v}$ of A].

Attempted Proof

Write $A \vec{v} = \lambda \vec{v}$. Since $B=P^{-1}AP$, this is equivalent to saying
$$B (P^{-1}\vec{v})=P^{-1}APP^{-1}\vec{v} = P^{-1}A\vec{v} = \lambda (P^{-1} \vec{v})$$

Since $P^{-1} \vec{v}$ is an eigenvector of $B$ only for eigenvalues of $A$, the dimensions of the eigenspaces must be the same.

Is this proof complete?

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Anthony Peter
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1 Answers1

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You should also show that every eigenvector of $B$ is of this form (just for completeness):

If $Bv=\lambda v$ then $B(P^{-1}Pv)=\lambda P^{-1}Pv$. Set $w=Pv$ here. Observe that $w$ is an eigenvector of $A$. Indeed, $BP^{-1} = P^{-1}A$, so $$P^{-1}Aw = \lambda P^{-1}w$$

(Based on a comment by Blah)

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