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I numerically verified the case for the $2\times2$ matrix but don't know how to prove it theoretically. Let's say, if $\Sigma$ is an $n\times n$ matrix, and we have $A \Sigma A^{\dagger}=\Sigma$, where $A$ is a unitary matrix, and $\Sigma$ is a Hermitian matrix, then does it imply $\Sigma$ also to be a diagonal matrix?

For clarification, $\Sigma$ is real in my case, but I don't know if 'real' really matters to prove $\Sigma$ is diagonal.

Here is a simple numerical verification for $2\times2$ case: Let

$$A = \begin{bmatrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{bmatrix}, $$

which is a rotation matrix, let $\Sigma = \begin{bmatrix} a & b\\ c & d \end{bmatrix}$, from the equation $A \Sigma A^{\dagger}=\Sigma$ we get $a=d, c=-b$, now hermitian implies $c=b$, so we get $c=b=0$ and prove $\Sigma$ is a diagonal matrix.

lcv
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dtouro
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    Doesn't this always hold for $A = I$? – Mason Dec 06 '22 at 22:42
  • @Mason Yes when given $A = I$, we can't say $\Sigma$ is diagonal, probably I should add "for arbitrary unitary matrix $A$" in the description? I added a simple numerical verification for 2*2 case in my description. – dtouro Dec 06 '22 at 23:13
  • Your "verification" is just an example of unitary $A$ for which your implication holds, but this is not true of any unitary $A$: Mason gave you a counterexample – Anne Bauval Dec 06 '22 at 23:13
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    Do you mean you intend to modify your question by requiring that $\Sigma$ commutes with every unitary matrix? But then the answer is obviously yes. – Anne Bauval Dec 06 '22 at 23:18
  • By the way, your example would still be wrong. – lcv Dec 06 '22 at 23:23
  • @AnneBauval can you give a proof why commutes with every unitary matrix imply a diagonal matrix? Or any sources? – dtouro Dec 07 '22 at 00:15
  • @lcv Why you think it's wrong? – dtouro Dec 07 '22 at 00:19
  • @dtouro Use the SVD to write $\Sigma = VDV^*$, with $V$ unitary and $D$ diagonal. – Mason Dec 07 '22 at 01:03
  • @dtouro Even if you let $[A(\theta),\Sigma] = 0, ,, \forall \theta$ you only obtain $\Sigma = aI + b \sigma^y$ [as $A(\theta) = \exp(-i \theta \sigma^y)$]. Here $\sigma^y$ is the purely imaginary Pauli matrix. – lcv Dec 07 '22 at 15:05
  • @lcv Thanks I see your point, for $\Sigma$ in complex field, this example is wrong. – dtouro Dec 07 '22 at 17:28
  • Yes, since you spoke of unitary I assumed a complex field. – lcv Dec 07 '22 at 17:30

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Your initial conjecture is trivially false, as Mason's counterexample $A=I$ shows.

Therefore, so is your "numerical verification for 2*2 case" (take $\theta=0$).

An answer to (a strengthening of) your modified question (in the comments) is the following: if an arbitrary (not necessarily hermitian) matrix $\Sigma\in M_n(\Bbb C)$ commutes with every unitary matrix $A\in U_n(\Bbb C),$ then $\Sigma$ is a scalar matrix. This is because every complex $n\times n$ matrix is a linear combination of two unitary matrices, and if $\Sigma$ commutes with every complex $n\times n$ matrix, then it is scalar.

Anne Bauval
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