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.