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!



On certain decomposition of unitary symmetric matrices
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.


