Prove $T$ commutes with $S^*$ 
Given $T,S\in\mathcal{L}(V)$ such that $T$ is normal and commutes with $S$.
Prove $T$ commutes with $S^*$.

I tried two ways to approach the problem, but got stuck:
First, I tried to prove something like $TS^*T^*=S^*TT^*.$
Second, I tried working with inner-products, but as I said, that didn't get me far.
Any help appreciated.
 A: I'm assuming that you are working over $\mathbb C$.
By the spectral theorem, there is an orthonormal basis $b$ of $V$ such that the matrix of $T$ with respect to $b$ is diagonal. For simplicity, suppose that the dimension of the space is $5$ and that $T$ hast three distinct eigenvalues $\lambda_1$, $\lambda_2$ and $\lambda_3$, the first two with multiplicity $2$. Then the matrix of $T$ will be$$\begin{pmatrix}\lambda_1&0&0&0&0\\0&\lambda_1&0&0&0\\0&0&\lambda_2&0&0\\0&0&0&\lambda_2&0\\0&0&0&0&\lambda_3\end{pmatrix}.$$You know that the matrix of $S$ with respect to the same basis commutes with this one. Therefore, it will be a block matrix:$$\begin{pmatrix}a_{11}&a_{12}&0&0&0\\a_{21}&a_{22}&0&0&0\\0&0&a_{33}&a_{34}&0\\0&0&a_{43}&a_{44}&0\\0&0&0&0&a_{55}\end{pmatrix}.$$
But then the matrix of $S^*$ (with respect to the same basis) will be$$\begin{pmatrix}\overline{a_{11}}&\overline{a_{21}}&0&0&0\\\overline{a_{12}}&\overline{a_{22}}&0&0&0\\0&0&\overline{a_{33}}&\overline{a_{43}}&0\\0&0&\overline{a_{34}}&\overline{a_{44}}&0\\0&0&0&0&\overline{a_{55}}\end{pmatrix}\text,$$which commutes with the matrix of $T$. Therefore, $T$ commutes with $S^*$.
