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Let us consider the space $M_n(\mathbb{C})$. By a unitary matrix $U=(u_{ij})$ we mean that $U^{-1}=(\overline{u_{ji}})$.

Q. Let $U$ be a unitary matrix. I am looking for the pairs of matrices $(D,A)$ satisfying the following conditions: (1) $D$ is a diagonal matrix in $M_n(\mathbb{C})$, (2) $UAU^{-1}=D$, (3) $U$ and $A$ have the same eigen-vectors. In general case, does there exist such a pair? How can we find a formula to produce all such of the pairs?

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    $\begingroup$ $U = A = D$?... $\endgroup$
    – LSpice
    Commented Apr 24, 2022 at 14:38
  • $\begingroup$ $D$ should be a diagonal matrix not diagonalizable. $\endgroup$
    – ABB
    Commented Apr 24, 2022 at 14:42
  • $\begingroup$ Indeed; my comment was suggesting a unitary matrix $U$ and a diagonalisable matrix $A$ both of which are already diagonal as an obvious, silly example. My answer says that in fact this is (essentially; we don't actually need $U = A$) the only example. $\endgroup$
    – LSpice
    Commented Apr 24, 2022 at 14:46
  • $\begingroup$ $A$ is diagonalizable, so if $U,A$ have the same eigenvectors, then they commute, so $A=D$, and $U$ is any diagonal unitary matrix. $\endgroup$
    – Christian Remling
    Commented Apr 24, 2022 at 17:06

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I mentioned the silly example $U = A = D$ offhand, but, in fact, it's essentially the only example. In general, note that the columns of $U^{-1}$ are the eigenvectors of $A$. So we're asking for, at least, a unitary matrix $U$ that admits the columns of $U^{-1}$ as eigenvectors. But $U U^{-1} = I$, so this means that $U$ must be diagonal.

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