On certain decomposition of unitary symmetric matrices This is by any means elementary, but since I have asked this question on Stark Exchange but received no satisfactory answers I decide to post it here.
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!
 A: If I parsed the original question correctly, $U$ is unitary as well as symmetric. In this case, we can indeed show $A^TUA=I$. Here's how.
Since $U$ is unitary, we can write it as $U=e^{iH}$ for some Hermitian matrix $H$. But, since $U^T=U$ by assumption, this shows that $U^T=(e^{iH})^T=e^{iH^T}=e^{i\overline{H}}=e^{iH}=U$, which implies that $H$ is actually real, symmetric. Now, simply define $A=e^{-iH/2}$; this matrix is unitary, and with this choice $A^TUA=I$.
A: Here is a way to answer the second question. I am assuming complex
matrices and using '$X^{*}=\overline{X}^{\mathrm{T}}$.  The first question (already answered by Suvrit ) I discuss at the end.
Assume '$U^{*}U=I$ and $U^{\mathrm{T}}=-U$. If $\mathbf{v}$ is nonzero and $U\mathbf{v}=\lambda\mathbf{v}$ then $|\lambda|=1$ and the spectral theorem for normal matrices implies $U^{*}\mathbf{v}=\overline{\lambda}\mathbf{v}$.  Take the
conjugate to find $U^{\mathrm{T}}\overline{\mathbf{v}}=\lambda\overline{\mathbf{v}}$.  Since $U^{\mathrm{T}}=-U$ we find $U\overline{\mathbf{v}}=-\lambda\overline{\mathbf{v}}$. This mean $\overline{\mathbf{v}}$ and $\mathbf{v}$ are orthogonal unit vectors. We could have started with an orthonormal basis for the whole eigenspace of $\lambda$ and so can find an orthonormal basis $\mathbf{v}_{1},\overline{\mathbf{v}_{1}},\dots,\mathbf{v}_{N},\overline{\mathbf{v}_{N}}$ so we see we are on an even dimensional space. Let the corresponding eigenvalues be $\lambda_{1},-\lambda_{1},\dots,\lambda_{N},-\lambda_{N}$.
We get a new real orthonormal basis $\mathbf{w}_{1},\dots,\mathbf{w}_{2N}$
by defining
$$
\mathbf{w}_{2j}=\frac{1}{\sqrt{2}}\mathbf{v}_{j}+\frac{1}{\sqrt{2}}\overline{\mathbf{v}_{j}}
$$
and
$$
\mathbf{w}_{2j+1}=\frac{i}{\sqrt{2}}\mathbf{v}_{j}-\frac{i}{\sqrt{2}}\overline{\mathbf{v}_{j}}
$$
Since 
$$
U\mathbf{w}_{2j}=\lambda_{j}\frac{1}{\sqrt{2}}\mathbf{v}_{j}-\frac{1}{\sqrt{2}}\lambda_{j}\overline{\mathbf{v}_{j}}=-i\lambda_{j}\mathbf{w}_{2j+1}
$$
and 
$$
U\mathbf{w}_{2j+1}=\lambda_{j}\frac{i}{\sqrt{2}}\mathbf{v}_{j}+\lambda_{j}\frac{i}{\sqrt{2}}\overline{\mathbf{v}_{j}}=i\lambda_{j}\mathbf{w}_{2j}
$$
we see there is a real orthogonal matrix $O$ so that
$$
U=O^{*}DO=O^{\mathrm{T}}DO
$$
where $D$ is block diagonal with blocks 
$$
\left[\begin{array}{cc}
0 & i\lambda_{j}\\
-i\lambda_{j} & 0
\end{array}\right].
$$
Let $Q$ be block diagonal with blocks
$$
\sqrt{i\lambda_{j}}\left[\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}\right]
$$
and $S$ be block diagonal with blocks
$$
\left[\begin{array}{cc}
0 & 1\\
-1 & 0
\end{array}\right].
$$
Then $D=QSQ=Q^{\mathrm{T}}SQ$ and 
$$
U=O^{\mathrm{T}}Q^{\mathrm{T}}SQO.
$$
The desired unitary is $QO$.
If $S$ is not what you consider the standard symplectic matrix, a permutation matrix will fix that.
In the case of $U$  unitary and $U^\mathrm{T} = U$ the proof starts the same, but we get $U\overline{\mathbf{v}}=\lambda\overline{\mathbf{v}}$  This leads to real eigenvectors, and so a real orthogonal matrix $O$  with $U=O^{\mathrm{T}}DO$ and $D$  diagonal.  So we have reduced to the case of a diagonal unitary, which is easy.  Any diagonal square root will do.
A: The key point is that unitary matrices have orthogonal eigenvectors, thus you can form an orthonormal basis of eigenvectors, which is the same thing as a unitary matrix $A$ with the property you describe.
Let $v$ and $w$ be two eigenvectors with different eigenvalues $\lambda_v$ and $\lambda_w$, then $(v \cdot w) = (Uv \cdot Uw)=(\lambda_v v \cdot \lambda_w w)= \lambda_v \bar{\lambda}_w (v \cdot w)$. Since $|\lambda_w|=1$, $\lambda_v \bar{\lambda}_w=\lambda_v \lambda_w^{-1}\neq 1$ so $v\cdot w=0$.
So choose an orthonormal basis for each eigenspace and take the union, then choose the unitary matrix mapping $e_n$ to the $n$th basis vector.
Edit: This is assuming the transpose in $A^T U A$ is the conjugate-transpose. If it is the normal transpose, Suvrit's answer is correct.
