Let A and B be $n \times n$ complex matrices such that
$A(AB-BA) = (AB-BA)A$
a) Show that for every positive integer $k$, the matrix $(AB-BA)^k$ is of the form $AC-CA$, where $C$ is an $n \times n$ complex matrix.
b) Prove that $AB-BA$ is nilpotent.
I have tried the following.
$A^{-1}A(AB-BA) = A^{-1}(AB-BA)A \implies AB-BA = A^{-1}(AB-BA)A$
$A(AB-BA)A^{-1} = (AB-BA)AA^{-1} \implies AB-BA = A(AB-BA)A^{-1}$
Then $(AB-BA)^{k} = (A^{-1}(AB-BA)A)^k = A^{-1}(AB-BA)^kA$
and $(AB-BA)^{k} = A(AB-BA)^{k}A^{-1}$.
I don't know where do I go from here, thanks for your help.
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rackne
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You are not allowed to use $A^{-1}$: you don't know if $A$ is invertible. – Crostul May 06 '15 at 07:11
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@user1551 Correct:I misread and thought it was to be proved that $;A;$ itself is nilpotent. Thanks, deleting comment. – Timbuc May 06 '15 at 08:35
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Part (b) is a duplicate of $AB−BA$ is a nilpotent matrix if it commutes with $A$, but I will not cast a close vote because the OP seems to have trouble with part (a). – user1551 May 06 '15 at 08:35
1 Answers
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a) We prove this by induction. For $k=1$, just take $B=C$. Suppose that you can find a matrix $C_k$ such that $(AB-BA)^k=AC_k-C_kA$. Then you have $$(AB-BA)^{k+1} = (AC_k-C_kA)(AB-BA) = A(C_k(AB-BA)) - C_kA(AB-BA), $$ but, by hypothesis you know that $$A(AB-BA) = (AB-BA)A $$ Consequently, $$(AB-BA)^{k+1} = A(C_k(AB-BA)) - (C_k(AB-BA))A, $$ so you can take $C_{k+1} = C_k(AB-BA)$ to get that $(AB-BA)^{k+1} = AC_{k+1}-C_{k+1}A$.
b) Try to use the result proved here.
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Hey, thanks for part a. In the question you linked, they are trying to prove if trace is equal to zero then the matrix is nilpotent but we don't know anything about the trace here. – rackne May 06 '15 at 07:34
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You know something about the trace. What is the trace of a matrix of the form $AC-CA$? – user37238 May 06 '15 at 07:44
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${Tr}({AC-CA}) = {Tr}(AC) - Tr({CA}) = 0$ but how do I know that ${Tr}((AC-CA)^{k})=0$? – rackne May 06 '15 at 08:10
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Actually what you want is $\text{tr}((AB-BA)^k)=0$ to prove that $AB-BA$ is nilpotent. – user37238 May 06 '15 at 09:10
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