Is the Phase of a Hermitian Matrix Hermitian? Let $A\in M_n(\mathbb{C})$ be a Hermitian matrix, $A^*=A$.
$A$ has a polar decomposition $A=U|A|$ with $U$ unitary.
I can show that
$$U|A|=U^*|A|=|A|U=|A|U^*,$$
and I am wondering is it the case that $U^*=U$.
Context: I have a purported alternative solution to this question. I am hoping to show that 
$$|\operatorname{Tr}(A\theta)|$$
is maximised over $\|\theta\|_{\infty}\leq$ at $\theta=U$, the phase of $A$. If $U$ is Hermitian, then 
$$\operatorname{Tr}(AU)=\operatorname{Tr}(UA)=\operatorname{Tr}(U^*A)=\operatorname{Tr}(|A|),$$
and I am away.
 A: If the matrix $A$ is invertible, then the polar decomposition is unique. You have already noted that $U^{*}|A|$ is a polar decomposition of $A$, hence $U = U^{*}$.
In the case that $A$ is not invertible, then the polar decomposition is no longer unique. I claim that in any case there exists a $U'$ such that $A = U'|A|$ and such that $U'^{*} = U'$.
First, let $A = U|A|$ be any polar decomposition, not neccesarily one with self-adjoint $U$.
Since $U$ is unitary, we have that the complement of the range of $|A|$ is not trivial. Note, furthermore that $\text{ker}(A) = \text{ran}(A)^{\perp}$ (because $A$ is self-adjoint). Now we show that the decomposition $\mathbb{C}^{n} = \ker(A) \oplus \text{ran}(A)$ is stable under $U$. Suppose that $v \in \ker(A)$, then we compute
\begin{equation}
 AUv = U|A|Uv = UAv = 0.
\end{equation}
Counting dimensions, we see that the decomposition $\mathbb{C}^{n} = \ker(A) \oplus \text{ran}(A)$ is stable under $U$. For the decomposition $A = U|A|$ it is irrelevant what $U$ does on $\text{ran}(A)^{\perp} = \ker(A)$, so we can define $U' = U|_{\ker(A)^{\perp}} \oplus \text{Id}_{\ker(A)}$, which is still unitary, and satisfies $U'^{*} = U'$. This is because $U|_{\ker(A)^{\perp}} |A||_{\ker{A}^{\perp}} = A|_{\ker{A}^{\perp}}$ is a polar decomposition of an invertible self-adjoint operator.
(Note that this same argument can be used with $i \text{Id}|_{\ker(A)}$ instead of $\text{Id}|_{\ker(A)}$ to show that there is a polar decomposition where $U^{*} \neq U$.)
