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Let $V$ be a nontrivial finitely-generated inner product space and let $\alpha$ and $\beta$ be selfadjoint endomorphisms of $V$ satisfying $\alpha\beta = \beta\alpha$. Show that $\alpha$ and $\beta$ have a common eigenvector?

I'm totally lost. I appreciate any help.

I know that if $v$ be an eigenvalue of $\alpha$ associated with eigenvalue $c$, we get $\alpha(v)=cv$. Applying $\beta$, we get

$$\beta\alpha(v)=\alpha(\beta(v))=c\beta(v). $$ Hence $\beta(v) $, is an eigenvector of $\alpha$ associated with eigenvalue $c$, so $v$ and $\beta(v)$ belong to the same eigenspace. I'm not sure if I can say the dimension of eigenspaces of $\alpha$ is $1$. If it's true, there exists an scalar $\lambda$ satisfying $\beta(v)=\lambda v$, and I'm done.

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Let $V_c$ denote the eigenspace of $\alpha$ with eigenvalue $c$, then your argument shows that $\beta$ restricts to a linear map $V_c\to V_c$. Furthermore this restriction is selfadjoint, so you can always diagonalize $\beta\mid_{V_c}$ to get an eigenvector of $\beta$ that lives in $V_c$.

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What you need to read from $\alpha\beta v=c\beta v $ is that the $\alpha $-eigenspace for $c $ (say, $K $) is invariant under $\beta $. Assuming your scalars are $\mathbb C $, there is an eigenvector $w $ for $\beta $ (likely with an eigenvalue other than $c $) in $K $.

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