TITLE: If every eigenvector of $T$ is also an eigenvector of $T^{*}$ then $T$ is a normal operator
QUESTION [6 upvotes]: Let $V$ be a finite-dimensional inner product space over $\mathbb{C}$ and $T: V \to V$ a linear transformation.
  Show that if every eigenvector of $T$ is also an eigenvector of $T^{*}$ then $T$ is a normal.

We need to show that $\forall v \in V,\ TT^*v=T^*Tv$.
I started by picking $v$ to be an eigenvector of $T$, hence $Tv=\lambda_1 v$, and $T^*v=\lambda_2 v$. Therefore: $$TT^*v=T(\lambda_2 v)=\lambda_2 Tv=\lambda_1 \lambda_2 v=...=T^*Tv$$
But I'm stuck when $v$ is not an eigenvector.
(Also, this question is taken from a chapter about Jordan forms, but I can't see how it relates)

REPLY [2 votes]: Hint for a coordinate-free proof: Begin by proving the following 
${\bf Lemma.}$ If $U$ is an invariant subspace of $T\ $ then $U^\perp$ is an invariant subspace of $T^*$,
then use induction on the dimension.