Why $H^2U=UH^2$ implies $H$ and $U$ commutes? Why does $H^2U=UH^2$ imply $H$ and $U$ commutes, where $H$ is a Hermitian matrix and $U$ is a unitary matrix?
This comes from the book 'Theory of Matrices' on p277 http://www.maths.ed.ac.uk/~aar/papers/gantmacher1.pdf
Now I know the implication is false. How to prove the following:
If a matrix $A$ is normal, i.e $AA^*=A^*A$, then the polar and unitary factor of the polar decomposition of $A$, $A=UH$, commute.
PS:Actually the book was right. I forgot to mention H is not only Hermitian, H is also positive semi-definite.

That cannot be true. Take 
$$
H=\left(
\begin{array}{cc}
 0 & 1 \\
 1 & 0 \\
\end{array}
\right)
$$
then $H^2=1$ it commutes with anything, however 
$$U=\left(
\begin{array}{cc}
 0 & -i \\
 i & 0 \\
\end{array}
\right)$$
is unitary and does not commute with $H$
