Prove that the polar decomposition of normal matrices, $A=SU$, is such that $SU=US$ Assume $A$ is a normal matrix. Suppose $A=SU$ is a polar decomposition of $A$. Prove that $SU=US$.
I have no idea to prove this.
$A$ is normal then $AA^*=A^*A$. And then we have
$$
SS^*=U^*S^*SU.
$$
But I don't know how to continue.
 A: Let $\,A=U|A|$, then $\,A^*=|A|U^*$. By normality one obtains 
$$U|A|^2U^* = AA^* = A^*A = |A|^2,$$
an equality of positive-semidefinite matrices.
"Positive square-rooting" yields $\,U|A|U^* = |A|\;\Longleftrightarrow\; U|A| = |A|U$.
A: Hint: Note that, since $A = SU$ is a polar decomposition, $S$ is (Hermitian and) positive semidefinite.
So, as you noted, we have
$$
SS^* = U^*S^*SU \implies\\
S^2 = U^* S^2U
$$
From here, note that each side is positive semidefinite and that the positive semidefinite square root of such a matrix is uniquely defined. As such, we can take the square root of both sides.
A: Let $\newcommand{\bs}[1]{\boldsymbol{#1}} A$ be a normal matrix. It is therefore unitarily diagonalisable, i.e. writable as
$$A = \sum_k c_k P_{\bs u_k}$$
for a set of orthogonal projectors $P_{\bs u_k}\equiv \bs u_k \bs u_k^*$, where $\langle \bs u_j,\bs u_k\rangle=\delta_{jk}$ (and thus $\operatorname{Tr}(P_{\bs u_j} P_{\bs u_k})=\delta_{jk}$) and complex coefficients $c_k\in\mathbb C$.
The corresponding SVD (or more generally one possible choice of SVD) reads $A=UD V^\dagger$ with
$$
U = \sum_k e^{i\varphi_k} \bs u_k \bs e_k^*,\qquad
D=\sum_k |c_k| \bs e_k\bs e_k^*, \qquad
V=\sum_k \bs u_k \bs e_k^*,
$$
where $c_k\equiv |c_k|e^{i\varphi_k}$, and $\bs e_k$ are the basis vectors for some reference basis used to represent the matrices.
The polar decomposition reads $A=|A| U_A$ with
$$|A|\equiv UDU^\dagger= \sum_k |c_k| \bs u_k \bs u_k^*,
\qquad
U_A \equiv UV^\dagger = \sum_k e^{i\varphi_k} \bs u_k \bs u_k^*.$$
From this, it's clear that $|A|$ and $U_A$ commute, being them diagonal in the same basis.

It's worth seeing where exactly the above argument breaks in the case of non-normal matrices. For a generic matrix $A$ the SVD reads
$$A = UDV^\dagger = \sum_k p_k \bs u_k \bs v_k^*, \\
U\equiv \sum_k \bs u_k\bs e_k^*,
\qquad V \equiv \sum_k \bs v_k\bs e_k^*,
\qquad D \equiv \sum_k p_k \bs e_k \bs e_k^*,$$
for some positive coefficients $p_k>0$ and orthonormal sets $\{\bs u_k\}_k$ and $\{\bs v_k\}_k$. Then $A=|A| U_A$ with
$$|A| \equiv UD U^\dagger = \sum_k p_k \bs u_k \bs u_k^*,
\qquad
U_A \equiv UV^\dagger = \sum_k \bs u_k \bs v_k^*.
$$
Now we have
$$U_A |A| = (UV^\dagger U)DU^\dagger = \sum_{jk} p_k \langle \bs v_j,\bs u_k\rangle \bs u_j \bs u_k^*,$$
which in general differs from $|A| U_A=A$.
