Given matrices $A$ and $B$ such that $AB-BA=I$, show that $A^nB-BA^n=nA^{n-1}$.

Asked May 03 '22 at 21:47

Active May 03 '22 at 21:47

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I've among other things tried to multiply both sides of $AB=I+BA$ with $A^{n-1}$ in order to get $A^nB=A^{n-1}(I+BA)$. However from here I end up with trivial statements such as $A^nB=A^nB$.

asked May 03 '22 at 21:47

Tom

See my answer to this question https://math.stackexchange.com/a/2512124/399263. – zwim May 03 '22 at 21:51
Would it be cheating to say that because no such matrices $A$ and $B$ exist, the statement is trivially true? – eyeballfrog May 03 '22 at 21:57

Given matrices $A$ and $B$ such that $AB-BA=I$, show that $A^nB-BA^n=nA^{n-1}$.

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