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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
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any normal matrix can be unitarily diagonalized. – chaohuang Jul 20 '12 at 13:35

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This is called the spectral theorem for self-adjoint and normal operators. It can be found here: http://www.math.brown.edu/%7Etreil/papers/LADW/LADW.pdf (page 159).

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