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.

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 $H$.
