TITLE: On certain decomposition of unitary symmetric matrices
QUESTION [9 upvotes]: It is well known that a symmetric matrix over field $\Bbb F$ is congruent to a diagonal matrix, i.e., there exists some A s.t. $A^TUA=D$ with $U$ symmetric and $D$ diagonal. If $\Bbb F=\Bbb C$ then we can make $D=I$.
Recently I learned that if $U$ is unitary that we can do one step further by requiring $A$ to be unitary too. A similar result holds for unitary skew matrices. But I fail to figure out a proof myself.
Can anyone provide a proof of this or at least help me to locate some references? Many thanks!

REPLY [5 votes]: Let $U$ be complex symmetric unitary, i.e., $U^T = U$ and $U^* U = I$. From Takagi factorization we know that $U = V D V^T$, i.e.,
$$D = V^{-1} U V^{-T},$$
for some unitary $V$ and real diagonal $D$. Here, we do not have $D = I$, but we do have the unitarity of $V$.
Using the unitarity of $U$ and $V$, we see that
\begin{align*}
D^* D &= (V^{-T})^* U^* V^{-*} V^{-1} U V^{-T} = \overline{V^{-1}} U^* (V V^*)^{-1} U V^{-T} = \overline{V^{-1}} U^* U V^{-T} \\
&= \overline{V^{-1}} V^{-T} = \overline{(V^* V)^{-1}} = {\rm I}.
\end{align*}
In other words, $D$ is real diagonal unitary, which means that it has diagonal elements $-1$ and $1$. Define a complex diagonal matrix $X = \operatorname{diag}(x_1, x_2, \dots, x_n)$:
$$x_k = \begin{cases}
1, & D_{kk} = 1, \\
{\rm i}, & D_{kk} = -1.
\end{cases}$$
Obviously, $D = X X^T$. Defining $A = VX$ we get
$$U = V D V^T = (VX) (VX)^T = A {\rm I} A^T,$$
which is the required form. Notice that $X$ is unitary, so $A$ is unitary as well.