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It is said that any finite-dimensional Hermitian matrix can be diagonalized by a unitary matrix. But isn't a special unitary matrix sufficient?

Am I making a mistake when I say that the phase that makes unitary matrices non-special doesn't do anything when diagonalizing Hermitian matrices?

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You mean how if $\lambda$ is a complex number with modulus $1$ and $U$ is unitary, then $UAU^*=\lambda U A (\lambda U)^*$, so it can be assumed that $\det U=1$ if that were ever more convenient? – Jonas Meyer Feb 4 '13 at 0:33
Yes, that's what I mean. Is that correct? Or are there cases when $\lambda$ matters? – QuantumDot Feb 4 '13 at 0:56
up vote 3 down vote accepted

Yes, one can always take the unitary matrix diagonalizing a Hermitian matrix to have determinant $1$, by multiplying by a scalar of modulus $1$.

So no, you are not making a mistake. I guess that because in many contexts such a normalization is not useful, and because it is easy to make the adjustment as needed once the arbitrary unitary case is known, this fact is not said as often.

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