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


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


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.

  • 1
    $\begingroup$ I have generalized my answer to take non-invertible $A$ into account. (I'm posting this comment because I'm not sure you get an update when I edit my answer.) $\endgroup$ – Peter Nov 2 '17 at 15:44

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 $U^{*} \neq U$.)

  • $\begingroup$ Thank you Peter. $A$ is not in general invertible. $\endgroup$ – JP McCarthy Nov 2 '17 at 14:47
  • $\begingroup$ Ah, your question stated $A \in \text{GL}(\mathbb{C}^{n})$, which I took to mean that $A$ is invertible... $\endgroup$ – Peter Nov 2 '17 at 14:47
  • $\begingroup$ Peter, sorry my bad. I meant just $A\in M_n(\mathbb{C})_{\text{sa}}$. $\endgroup$ – JP McCarthy Nov 2 '17 at 14:47

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