$X$ is normal matrix and $AX=XB$ and $XA=BX$.why $A{X^} = {X^}B$ and ${X^}A = B{X^}$?

Question

Let $A,B,X \in {M_n}$ and $X$ is normal matrix and

$AX=XB$

$XA=BX$

Why $A{X^*} = {X^*}B$ and ${X^*}A = B{X^*}$?

Closely related: Show that if $T_1, T_2$ are normal operators that commutes then $T_1+T_2$ and $T_1T_2$ are normal. — user1551, May 07 '15 at 09:33

score 2 · Answer 1 · answered May 07 '15 at 08:12

Hints:

Set $$ X' = \begin{pmatrix} 0 & 0 \\ X & 0 \end{pmatrix} \text{ and } A' = \begin{pmatrix} A & 0 \\ 0 & B \end{pmatrix} $$ Then $X'$ is normal and commutes with $A'$. Thus you have reduced the problem to the statement that if $XA = AX$ then $X^{\ast}A = AX^{\ast}$.
Now use the fact that $X$ is diagonalizable, and now you are reduced to the case where $X$ is diagonal.
Prove it when $X$ is diagonal.

BTW, this is (a special case of) the Fuglede-Putnam theorem.

$X$ is normal matrix and $AX=XB$ and $XA=BX$.why $A{X^*} = {X^*}B$ and ${X^*}A = B{X^*}$?