Timeline for Commutation fo a self-adjoint operator with a unitary operator

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Dec 21, 2019 at 18:50	comment	added	Christian Remling		To sum up the discussion so far, this will work (for arbitrary $A$) if and only if $e^{ix}$ is injective on $x\in\sigma(M)$ (because then $M$ is a function of $e^{iM}$, and if you don't have this condition, then there's Jochen's counterexample).
Dec 21, 2019 at 16:09	comment	added	user131781		As is so often the case, the counterexample suggests what is going wrong here. The general problem would be the question: if $A$ and $f(M)$ commute, is the same true of $A$ and $M$? So what you need is that $f$ be invertible in a suitable sense. It is easy to formulate a precise version which I will leave as an exercise to the OP since I am typing this on a pad.
Dec 21, 2019 at 15:03	comment	added	Jochen Glueck		@AlexandreEremenko: Not quite: Let $M$ be the diagonal matrix with diagonal entries $2\pi$ und $-2\pi$, and let $A$ be any matrix that does not commute with $M$.
Dec 21, 2019 at 14:56	comment	added	Alexandre Eremenko		In finite dimension, two diagonalizable operators commute iff there is a basis in which they are both diagonal. So in finite dimensuion your (1) and (2) are equivalent.
Dec 21, 2019 at 14:46	history	asked	Bazin	CC BY-SA 4.0