If $A$ is normal and $A$ and $B$ commute, then $A^*$ and $B$ commute Let $A$ is a normal matrix: $A^*\! A = A A^*\!\!$,$\,$ and $AB = BA$. Prove that $A^*\!B=BA^*\!\!$.
I can prove that if $\det A\ne 0$ by multiplication $AB=BA$ by $A^*$ left and right and using some manipulation. But I have no idea what to do if $\det A = 0$.

In fact, if $|A|\neq 0$, then we can find a sequence $A_n$ such that $\lim_{n\rightarrow \infty}A_n=A$ and every term $A_n$ is inverse, that is, $|A_n|\neq 0$. 
