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.
 A: 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 $H$
A: The book does not say what you are claiming that it says. 
This particular $H$ can be written as a polynomial in its square $H^2$, a fact which is true for positive semi-definite matrices (as the book says on the page that you link).
Thus $HU = g(H^2)U = Ug(H^2) = UH.$ The result is not true for arbitrary Hermitian $H$.
A: Counterexample for part 2 of the question. Let
$$
A=UH=\left(
\begin{array}{cc}
 -i & 0 \\
 0 & i \\
\end{array}
\right)=
\left(
\begin{array}{cc}
 0 & -i \\
 i & 0 \\
\end{array}
\right)
\left(
\begin{array}{cc}
 0 & 1 \\
 1 & 0 \\
\end{array}
\right)
$$
then $A A^\dagger = A^\dagger A = 1$ but $H$ and $U$ still do not commute.
May be the theorem is that $H$ and $U$ can be chosen to commute, as there could be some freedom in what is $H$ and what is $U$. 
