Prove that Tr$[x,y]^m=0$ for $m \ge 1$

Question

$V$ is a complex vector space and $x,y \in gl(V)$ are linear maps such that $[x,y]$ commutes with both $x$ and $y$. Let $z=[x,y]$. Show that tr$z^m=0$ for all $m\ge 1$.

I have thought about this question for a long time, but I don't know how to use the condition that $z$ commutes with $x$, $y$. Can somebody give me some hints? Thank you.By the way, we haven't learnt Engel's Theorem or Lie's Thm at this point, so I cannot use that.

Answer 1

We have$\def\tr{\mathop{\rm tr}}$ \begin{align*} \tr([x,y]^{m}) &= \tr\bigl((xy-yx)[x,y]^{m-1}\bigr)\\ &= \tr(xy[x,y]^{m-1}) - \tr(y[x,y]^{m-1}x) & \text{as $x[x,y]^{m-1}=[x,y]^{m-1} x$}\\ &= \tr(xy[x,y]^{m-1}) - \tr(xy[x,y]^{m-1}) & \text{property of the trace}\\ &= 0 \end{align*}