Given an endomorphism $\alpha$ acting on an inner product space s.t. $\alpha$ is normal, i.e. $\alpha \alpha ^* = \alpha ^* \alpha$, then show there is an eigenvector shared by both $\alpha$ and $\alpha ^*$. Show that the corresponding eigenvalues are conjugate.
I have tried to do this by showing there is a real eigenvector, which is easy enough, and then taking the conjugate of the eigenvector/value equation. Now show that the subspace $\langle v\rangle^\perp$ is invariant under both $\alpha$ and $\alpha^*$, where $v$ is a shared eigenvector of $\alpha$ and $\alpha^*$. I've been playing with inner products but all I can get so far is that mapping elements in this subspace gives another element of the subspace.
Let $\lambda$ an eigenvalue for $\alpha$. Since $\alpha$ and $\alpha^*$ commute, the eigenspace $E_\lambda(\alpha)$ of $\alpha$ associated to $\lambda$ is invariant by $\alpha^*$. By taking the restriction of $\alpha^*$ on $E_\lambda(\alpha)$ we see that there's an eigenvector $v$ of $\alpha^*$ (and then of $\alpha$ also) associated to an eigenvalue $\mu$.
Moreover, we have
$$\lambda||v||^2=\langle \alpha v,v\rangle=\langle v,\alpha ^*v\rangle=\overline \mu||v||^2\implies \lambda=\overline \mu$$