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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!

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It is simply the spectral theorem --- unitary matrices are normal! – Federico Poloni Jul 29 '12 at 14:32
It seems you do not read my statement carefully:) – Zhang Xiao Jul 29 '12 at 14:41
What do you mean by a unitary matrix over an arbitrary field? Are you fixing an involution on $F$? – Qiaochu Yuan Jul 29 '12 at 15:40
Yuan: the formal statement is like this: for any unitary symmetry matrix (over $\Bbb C$), there exists some unitary matrix $A$ such that $A^TUA=I$. And for unitary skew matrix (again over $\Bbb C$), some unitary $A$ such that $A^TUA=S$, $S$ is the standard symplectic matrix. The crucial point is to make every matrix unitary. – Zhang Xiao Jul 29 '12 at 15:53
So this is a question over $\mathbb{C}$, not over an arbitrary field $F$? – Qiaochu Yuan Jul 29 '12 at 22:21
up vote 2 down vote accepted

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.

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I like this down-to-ground proof, many thanks! – Zhang Xiao Jul 30 '12 at 5:33

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 $A^TUA=I$.

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Thanks a lot Suvrit! I find your answer quite simple and illuminating. – Zhang Xiao Jul 30 '12 at 5:29
A technical point here: one needs to select $H$ with spectrum in $[0,2\pi)$ so that $\overline{H}$ has the same spectrum and we can conclude form the equality of exponentials the equality of $H$ and $\overline{H}$. – Terry Loring Aug 2 '12 at 2:53

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.

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@Will Sawin, if the questioner intends literally what was written, there is not a hermitian "inner product", but complex-bilinear, no? – paul garrett Jul 29 '12 at 19:18
How do you define a unitary matrix without a hermitian inner product? – Will Sawin Jul 29 '12 at 20:11
@Will Sawin, I don't know what the questioner really wants... Of course, with "matrices", there're "transpose", and "transpose-conjugate", but, as you suggest, it's hard to know what point this might have. The questioner's mention that the outcome could be skew under some (?) hypothesis is a clue, but ... :) – paul garrett Jul 29 '12 at 20:21
There's two definitions of transpose, but only one definition of unitary, as far as I'm aware. I think it's safe to guess that Zhang Xiao wants that one. – Will Sawin Jul 29 '12 at 21:23
@Will Sawin, your guess is as good as mine, but the questioner's remark about "skew-unitary" makes me think the questioner uses "unitary" as synonym for (what would more often nowadays be) "orthogonal". So "skew-unitary" means "skew-symmetric", maybe, and then the question is about "normal form" for skew-symmetric things? Would be mildly compatible with asking about canonical forms for symmetric things. Or maybe not... :) – paul garrett Jul 29 '12 at 21:56

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