TITLE: Show matrix is nilpotent
QUESTION [6 upvotes]: I have matrices $A,B$ of dimension $n$ with real coefficients which satisfy the following: $A^2-B^2=c(AB-BA)$ where $c$ is a real number. If $c\neq0$ , prove that $(AB-BA)^n = 0$. 
So far, I've been able to show that $AB-BA$ is singular. Can someone help?

REPLY [3 votes]: As you mentioned in your another question, this is just a variant of that question.
Let $X=A-B$ and $Y=A+B$. Then
\begin{align}
XY&=(A-B)(A+B)=(c+1)(AB-BA),\\
YX&=(A+B)(A-B)=(c-1)(AB-BA).
\end{align}
Hence $XY-YX = 2(AB-BA)$ and $XY+YX = 2c(AB-BA)$, i.e. $XY+YX=c(XY-YX)$ for some $c\ne0$. So, the result from the aforementioned question shows that $(XY-YX)^n=0$. Hence $(AB-BA)^n=0$.