Fuglede's theorem in finite-dimensional vector space Let $V$ be a finite dimensional vector space and $A$ be normal operator on $V$ and $B$ is an operator such that $AB=BA$. Show that $BA^*=A^*B$.
I guess that this problem should not be so difficult. I have tried different approaches and I got some identities which do not lead to desired equality. 
So I would be thankful if you show the solution to this problem, please!
 A: The way to think about this problem is when $B$ is diagonalizable, and $A$ being normal is diagonalizable (over $\mathbb C$) so we can call on simultaneous diagonalizability, recognize that being normal $A^*$ may also be simultaneously diagonalized with $B$ (via the same similarity transform that we'd use on $AB$)  which implies that $A^*B = BA^*$.  However it is  conceivable that $B$ might be defective-- so a more direct argument can be employed to compute the norm of the commutator
$\Big\Vert  A^*B - BA^*\big\Vert_F^2$
$=\text{trace}\Big(\big(A^*B - BA^*\big)^*\big(A^*B - BA^*\big)\Big)$
$=\text{trace}\Big(\big(B^*A - AB^*\big)\big(A^*B - BA^*\big)\Big)$
$=\text{trace}\Big(B^*AA^*B\Big) + \text{trace}\Big(AB^*BA^*\Big)- \text{trace}\Big(B^*ABA^*\Big) -\text{trace}\Big(AB^*A^*B\Big) $
$=\text{trace}\Big(AA^*BB^*\Big) + \text{trace}\Big(B^*BA^*A\Big)- \text{trace}\Big(B^*ABA^*\Big) -\text{trace}\Big(BAB^*A^*\Big) $
$=\text{trace}\Big(AA^*BB^*\Big) + \text{trace}\Big(B^*BA^*A\Big) - \text{trace}\Big(B^*BAA^*\Big) -\text{trace}\Big(ABB^*A^*\Big)$
$=\text{trace}\Big(AA^*BB^*\Big) + \text{trace}\Big(B^*BA^*A\Big) - \text{trace}\Big(B^*BAA^*\Big) -\text{trace}\Big(A^*ABB^*\Big)$
$=\text{trace}\Big(AA^*BB^*\Big) + \text{trace}\Big(B^*BA^*A\Big) - \text{trace}\Big(B^*BA^*A\Big) -\text{trace}\Big(AA^*BB^*\Big)$
$=0$
thus by positive definiteness of the (squared) Frobenius norm we have
$\Big\Vert  A^*B - BA^*\big\Vert_F^2 = 0 \longrightarrow A^*B - BA^* = \mathbf 0\longrightarrow A^*B = BA^*$
A: This is Fuglede's theorem. From the spectral theorem, $A$ can be written as
$$A=\sum_{i=1}^{n}\lambda_iP_i,$$
and $A^{*}$ can be expressed similarly by replacing $\lambda_i$ by its conjugate. Now, notice that for $p_i(x)=x\prod_{j : \lambda_j \neq \lambda_i}(x-\lambda_j)$
$$p_i(A)=\sum_{j=1}^{n}p_i(\lambda_j)P_j=\lambda_i\prod_{j : \lambda_j \neq \lambda_i}(\lambda_i-\lambda_j)\sum_{j : \lambda_j=\lambda_i}P_j.$$
(The first equality is a generalization of the fact that $A^k = \sum_{i=1}^{n}\lambda_i^kP_i$.) This implies that
$$\lambda_i\sum_{j : \lambda_j=\lambda_i}P_j=\frac{p_i(A)}{\prod_{j : \lambda_j \neq \lambda_i}(\lambda_i-\lambda_j)}.$$
Given that A and B commute, we have that $p_i(A)$ and $B$ also commute. So,
$$\lambda_i\sum_{j : \lambda_j=\lambda_i}BP_j=\lambda_i\sum_{j : \lambda_j=\lambda_i}P_jB,$$
and by replacing $\lambda_i$ with its conjugate and summing over all distinct eigenvalues we get that $BA^{*}=A^{*}B$.
