# Prove that A(AB-BA) = (AB-BA)A implies AB-BA is nilpotent.

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.

• You are not allowed to use $A^{-1}$: you don't know if $A$ is invertible. – Crostul May 6 '15 at 7:11
• @user1551 Correct:I misread and thought it was to be proved that $\;A\;$ itself is nilpotent. Thanks, deleting comment. – Timbuc May 6 '15 at 8:35
• 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 6 '15 at 8:35

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 $(AB-BA)^{k+1} = AC_{k+1}-C_{k+1}A$.
• You know something about the trace. What is the trace of a matrix of the form $AC-CA$? – user37238 May 6 '15 at 7:44
• ${Tr}({AC-CA}) = {Tr}(AC) - Tr({CA}) = 0$ but how do I know that ${Tr}((AC-CA)^{k})=0$? – rackne May 6 '15 at 8:10
• Actually what you want is $\text{tr}((AB-BA)^k)=0$ to prove that $AB-BA$ is nilpotent. – user37238 May 6 '15 at 9:10