If a subspace is $T$-invariant, prove that it is $T^*$-invariant.

The problem is as follows.

Let $T:V \longrightarrow V$ be a normal ($T^*T=TT^*$) linear operator on a space $V$ over the field $\mathbb{C}$ such that $\dim(V)<\infty$. Let $W$ be a subspace of $V$.

Prove that if $W$ is $T$-invariant (that is, $T(W)\subset W$) then $W$ is $T^*$-invariant.

What I've already tried: I know that if $T:V \longrightarrow V$, then $\ker(T^*(V))=[T(V)]^\perp$. If this holds using the restriction $T|_W$ instead of $T$ and $W$ instead of $V$ then I can prove it, but in the end I couldn't prove that this property holds in that case, so I'm afraid that it isn't valid at all.

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This is proved in the answer here. Note that in general $W$ is $T$ invariant iff $W^\perp$ is $T^*$ invariant. So what you want to prove says that every invariant subspace of a normal operator is reducing (in finite dimension), which makes this close to a duplicate. – 1015 Jul 18 '13 at 23:15
Thanks for pointing that, I'd never figure it by myself. So in this case... should I close the question? – Wheepy Jul 18 '13 at 23:51
You can also answer your question, that's ok. – 1015 Jul 18 '13 at 23:58
Hm, so should I show that this is equivalent to the question in the link you've mentioned? – Wheepy Jul 19 '13 at 0:27
Either that, or directly answer your question. Up to you. Maybe the second option is better. – 1015 Jul 19 '13 at 0:29