$AX = XA$, then show that $A^*X=XA^*$ 
Let $A \in M_{n\times n}$ be a normal matrix. If $X \in M_{n\times n}$ is any matrix such that $AX = XA$,
  then show that $A^*X=XA^*$  . 
Is this true if A is any matrix?

Since $A$ is normal ,$AA^*=A^*A$.
Let $A=$ \begin{bmatrix} a_{11} & a_{12} & a_{13}\\a_{21} & a_{22} &a_{23}\\a_{31}& a_{32}& a_{33}\end{bmatrix}
I took as example $A,X$ to be $3\times 3$ matrix
$X=$  \begin{bmatrix} x_{11} & x_{12} & x_{13}\\x_{21} & x_{22} &x_{23}\\x_{31}& x_{32}& x_{33}\end{bmatrix}
But I could not find how to do it.
But how to show that $A^*X=XA^*$  .
Please help.
 A: By the spectral theorem,
$$
A=\sum_{k=1}^n\lambda_k P_k
$$
where $P_k$ is a projection matrix, that is $P_k^*=P_k=P_k^2$, and $P_kP_l=0$ for $k\ne l$. The numbers $\lambda_1,\dots,\lambda_n$ are the (pairwise distinct) nonzero eigenvalues of $A$.
Assuming $AX=XA$, we have
$$
\sum_{k=1}^n \lambda_kP_kX=\sum_{k=1}^n \lambda_kXP_k
$$
Multiplying by $P_l$, we obtain
$$
\lambda_l P_lX=\lambda_lXP_l
$$
for $l=1,2,\dots,n$, which implies, as $\lambda_l\ne0$, that $P_lX=XP_l$.
Since $A^*=\sum_{k=1}^n\overline{\lambda_l}P_k$, the result should now be clear.

A hint for the proof of the top statement. Let $\lambda_0=0$ and $\lambda_1,\dots,\lambda_n$ the (pairwise distinct) eigenvalues of $A$. Then you can build the unitary matrix $P$ diagonalizing $A$ by taking $m_k$ normalized eigenvectors relative to $\lambda_k$ (for $k=0,1,\dots,n$), where $m_k$ is the multiplicity of $\lambda_k$. Of course, if $0$ is not an eigenvalue, we'll have $m_0=0$.
Now just realize that the projection matrices $P_k$ on the eigenspaces do as required, using the fact that eigenvectors relative to distinct eigenvalues are orthogonal.
To make an example, suppose the eigenvalues are $\lambda$ and $\mu$, the former having multiplicity $2$ and the latter multiplicity $1$. Then
$$
P=[u_1 \ u_2 \ u_3]
$$
where $Au_1=\lambda u_1$, $Au_2=\lambda u_2$, $Au_3=\mu u_3$. Define
$$
P_1=u_1u_1^*+u_2u_2^*
\qquad
P_2=u_3u_3^*
$$
Note that
$$
P=P_1+P_2
$$
(why?) and that $P_1P_2=0=P_2P_1$. Then
$$
A=PDP^*=\lambda P_1+\mu P_2
$$
just by computing.
