If every eigenvector of $T$ is also an eigenvector of $T^{*}$ then $T$ is a normal operator 
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)
 A: If you consider matrices with respect to an orthonormal basis, what I meant in my comment about unitary triangularization is that for every $n$-by-$n$ complex matrix $A$ there is a unitary matrix $U$ such that $UAU^{-1}$ is upper triangular.  An algorithm to find $U$ appears in Hogben's Handbook of linear algebra.  If we forget about matrices, what this says is that given an operator $T$ on $V$, there exists an orthonormal basis $(e_1,\ldots,e_n)$ and constants $(a_{ij})_{i\leq j}$ such that $\displaystyle{Te_j=\sum_{i=1}^j a_{ij}e_i}$.  In particular, $e_1$ is an eigenvector for $T$.  Therefore it is also one for $T^*$, and this forces $a_{1j}=0$ for $j>1$.  Then $a_{12}=0$ implies that $e_2$ is an eigenvector for $T$, thus also one for $T^*$, thus $a_{2j}=0$ for $j>2$.  Since $a_{13}=a_{23}=0$, $e_3$ is an eigenvector for $T$. And so on (omitting the induction proof).  Eventually it comes out that $T$ and $T^*$ are simultaneously diagonalized, so they commute.
A: 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.
