$AB-BA$ is a nilpotent matrix if it commutes with $A$ I saw this in a MathOverflow post and am putting it here for posterity.
Problem:
Let $A$ and $B$ by square matrices and set $C=AB-BA$. If $AC=CA$, prove $C$ is nilpotent.
 A: Suppose $\mathcal A$ is a normed algebra and $\delta:\mathcal A\to\mathcal A$ is a bounded derivation.  If $x\in A$ and $\delta(\delta(x))=0$, then $\lim\limits_{n\to\infty}\|\delta(x)^n\|^{1/n}=0$.  This is proved as Theorem 2.2.1 in Sakai's Operator algebras in dynamical systems.
This applies to the case where $\mathcal A=M_n(\mathbb C)$, $\delta(X)= AX-XA$.  For an $n$-by-$n$ real or complex matrix $C$, $\lim\limits_{n\to\infty}\|C^n\|^{1/n}=0$ if and only if $C$ is nilpotent.
(I also remarked on this application in an answer to a different question where the result was applicable.)
A: I use this theorem: 

If $\forall i\ge1$, $\mathrm{tr}(C^i)=0$, then $C$ is nilpotent.

You can easily prove by  induction that $\mathrm{tr}(C^i)=0$ for all $i\ge1$. 
Theorem. $\forall i\ge1$, $\mathrm{tr}(C^i)=0$ iff $C$ is nilpotent.
Proof: $C$ is a real matrix but you may assume $C$ is a complex matrix and $$f(x)=(x-a_1)(x-a_2)\cdots(x-a_n)$$ is its characteristic polynomial in the complex field. You can prove that $\mathrm{tr}(C^k)=\sum_{i=1}^n a_i^k$ by induction, and if  $\forall k\in\mathbb N$, $\mathrm{tr}(C^k)=\sum_{i=1}^n a_i^k=0$  then $a_i=0$. Hence, $f(x)=x^n$, so $C^n=0$ and it is shown that $C$ is nilpotent.
