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.

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

b) Try to use the result proved here.

  • $\begingroup$ 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. $\endgroup$ – rackne May 6 '15 at 7:34
  • $\begingroup$ You know something about the trace. What is the trace of a matrix of the form $AC-CA$? $\endgroup$ – user37238 May 6 '15 at 7:44
  • $\begingroup$ ${Tr}({AC-CA}) = {Tr}(AC) - Tr({CA}) = 0$ but how do I know that ${Tr}((AC-CA)^{k})=0$? $\endgroup$ – rackne May 6 '15 at 8:10
  • $\begingroup$ Actually what you want is $\text{tr}((AB-BA)^k)=0$ to prove that $AB-BA$ is nilpotent. $\endgroup$ – user37238 May 6 '15 at 9:10
  • $\begingroup$ Ok I now proved it by induction again. Thanks. $\endgroup$ – rackne May 6 '15 at 9:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.