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Let $u\colon H\to H$ be a unitary operator on a separable Hilbert space $H$ and let $(e_n)_n$ be a fixed orthonormal basis in $H$. Is it possible to decompose $u$ as $u=v^*dv$ where $v$ is a unitary and $d$ is a diagonal operator with respect to $(e_n)_n$?

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closed as off-topic by András Bátkai, Alain Valette, Christian Remling, paul garrett, Myshkin Mar 5 at 22:15

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are there non-diagonalizable unitaries? – Carlo Beenakker Mar 4 at 13:00
I am not vey familiar with this topic. I thoght infinite dimensionality of $H$ might be a problem. If not, I would be grateful for a reference. – Krzysztof Mar 4 at 13:05

This is a bit basic for MO, but since the OP may not be familiar with operator theory: if an operator is diagonalizable it would have to possess lots of (non-zero) eigenvectors. Generally speaking operators on Hilbert space, even the unitary ones, need not have any (non-zero) eigenvectors. An instructive example is the operator $M: L^2({\bf T}) \to L^2({\bf T})$ given by $Mf(e^{it}) = e^{it}f(e^{it})$

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