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 $(e_n)_n$?
closed as offtopic by András Bátkai, Alain Valette, Christian Remling, paul garrett, Myshkin Mar 5 at 22:15This question appears to be offtopic. The users who voted to close gave this specific reason:



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 (nonzero) eigenvectors. Generally speaking operators on Hilbert space, even the unitary ones, need not have any (nonzero) eigenvectors. An instructive example is the operator $M: L^2({\bf T}) \to L^2({\bf T})$ given by $Mf(e^{it}) = e^{it}f(e^{it})$ 

