Why is $operatorname{Tr}([A,B]^{m}) = operatorname{Tr}([AB, [A,B]^{m-1}])$, if [A,[A,B]]=0$?
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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?
linear-algebra matrices
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edited Nov 28 '18 at 19:29
Ber