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Why is $\operatorname{Tr}([A,B]^{m}) = \operatorname{Tr}([AB, [A,B]^{m-1}])$, where $[A,B] = AB-BA$ for two quadratic matrices $A,B$ with$ [A,[A,B]]=0$ and $Tr$ is the trace of a matrix?

I tried to rewrite this and reduce it to $\operatorname{Tr}([A,B]^{m-1} AB) = \operatorname{Tr}(BA [A,B]^{m-1} ) $ the following way:

$$[A,B]^m = (AB-BA) [A,B]^{m-1} $$ $$[AB,[A,B]^{m-1}] = AB [A,B]^{m-1} - [A,B]^{m-1} AB$$

But now I do not see how to use $[A,[A,B]] = 0$.

Does anyone has hints for this or a hint how to advance?

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From $$[A,[A,B]]=0$$ we have that $$A[A,B]=[A,B]A$$ And $$A[A,B]^n=[A,B]^nA$$ So using your work: $$\text{Tr}([A,B]^{m-1}AB)=\text{Tr}(B[A,B]^{m-1}A)=\text{Tr}(BA[A,B]^{m-1})$$ which is what you wanted to prove.

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  • $\begingroup$ Thanks, I looked for such an identity and somehow did not see it. :) $\endgroup$ – MPB94 Nov 28 '18 at 19:37
  • $\begingroup$ @MPB94 which one was new? The one from the commutator? I didn't see that for the first time either, just when I wrote everything down :) $\endgroup$ – Botond Nov 28 '18 at 19:43
  • $\begingroup$ $A[A,B] = [A,B] A$. I somehow wanted something like that and did not see it haha. $\endgroup$ – MPB94 Nov 28 '18 at 19:47

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