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Suppose you have a unitary matrix $U$ such that $\overline{U}U=D$ for some diagonal unitary matrix $D$ (everything is taking place over $\mathbb{C}$). This is equivalent to a couple of other conditions:

1) $DU=U^T$

2) $DUD=U$

Furthermore, this implies that $U^2$ is symmetric. So most unitary matrices do not satisfy this property. But when they do, is it true that $D^2=I$, which is equivalent to saying that $D$ commutes with $U$?

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    $\begingroup$ What is the question exactly? $\endgroup$ Aug 11 '16 at 13:44
  • $\begingroup$ Is $D^2$ the identity? $\endgroup$ Aug 11 '16 at 13:57
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    $\begingroup$ Easy counterexamples are obtained by taking a $2\times 2$ matrix $U$ with zero diagonal. $\endgroup$ Aug 11 '16 at 19:07
  • $\begingroup$ Erg, yes. This is true. In fact it gives a recipe for counterexamples in every dimension. Hopefully it's some sort of degenerate case for what I need it for. But I'll have to think on it more. $\endgroup$ Aug 11 '16 at 22:49
  • $\begingroup$ @batconjurer: In general, you will need diagonal elements equal to zero to produce counterexamples: your condition implies that each row is a multiple of the corresponding column, and if the diagonal element is non-zero, then comparing those elements shows that the constant is $1$. $\endgroup$ Aug 12 '16 at 0:24
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From the fact that $U^2$ is symmetric you can get that:

$UU=U^TU^T \rightarrow U=U^TU^TU^*$

$UU=U^TU^T \rightarrow U=U^*U^TU^T$

so:

$U^TU^TU^*=U^*U^TU^T\rightarrow \bar{U}UU=UU\bar{U} \rightarrow DU=U\bar{D}$

So from that it can be seen that D commutes with only if D is a real matrix.

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  • $\begingroup$ Well, as I mentioned in the question, it's not hard to see that $D^2=I $ iff $D$ commutes with U. But any diagonal unitary satisfying $D^2=I $ can only have $\pm1$ on the diagonal and so is real. $\endgroup$ Aug 11 '16 at 22:42

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