# On certain decomposition of unitary symmetric matrices

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!

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At least over the complex numbers, where I assume you are working, the result is best understood in terms of inner products and orthonormal bases. (By the way, you probably mean to use ${\bar A}^{T}UA$ in the complex case and use Hermitian matrices in place of symmetric ones). A complex square matrix $A$ is called normal if $A$ commutes with ${\bar A}^{T}.$ This includes unitary and Hermitian and skew-Hermitian. The relevant result is that a normal matrix has an orthonormal basis of eigenvectors. The change of basis matrix from the standard basis is then unitary – Geoff Robinson Jul 20 '12 at 12:02
any normal matrix can be unitarily diagonalized. – chaohuang Jul 20 '12 at 13:35

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.

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