Matrix commutator question Here's a nice question I heard on IRC, courtesy of "tmyklebu."
Let $A$, $B$, and $C$ be $2\times 2$ complex matrices. Define the commutator $[X,Y]=XY-YX$ for any matrices $X$ and $Y$. Prove
$$[[A,B]^2,C]=0.$$
 A: Since the trace is additive, and $\mathrm{trace}(XY)=\mathrm{trace}(YX)$, it follows that the trace of any commutator of matrices is zero.
Thus, the trace of $[A,B]$ is $0$, hence it has eigenvalues $\lambda$ and $-\lambda$.
Case 1. $\lambda\neq 0$. Then $[A,B]$ is diagonalizable, and hence so is $[A,B]^2$ (same conjugating matrix). But the eigenvalues of $[A,B]^2$ are $(\lambda)^2$ and $(-\lambda)^2$, so $[A,B]^2$ is diagonalizable with all eigenvalues equal; hence $[A,B]^2$ is actually a scalar multiple of the identity, and therefore commutes with every matrix. Therefore, $[[A,B]^2,C] = 0$.
Case 2. $\lambda=0$. If $[A,B]$ is diagonalizable, then it is the zero matrix, and therefore $[A,B]^2=0$. If $[A,B]$ is not diagonalizable, then its Jordan canonical form is
$$\left(\begin{array}{cc}0 &1\\
0 & 0
\end{array}\right),$$
and hence $[A,B]^2=0$. Either way, $[A,B]^2=0$, hence $[A,B]^2$ commutes with any matrix, so $[[A,B]^2,C]=0$.
A: Here's a better argument (not posted at midnight...) which shows that the result holds over any field: we don't need the matrices to be complex.
As in the other answer, the trace of $[A,B]$ is $0$. Therefore, the characteristic polynomial of $[A,B]$ is $x^2+\det[A,B]$. By the Cayley-Hamilton Theorem,
$$[A,B]^2 = -\det[A,B]I.$$
Therefore, $[A,B]^2$ is a scalar matrix, and therefore lies in the center of $M_{2\times 2}(\mathbf{F})$. We conclude that $[[A,B]^2,C]=0$ for any matrix $C\in M_{2\times 2}(\mathbf{F})$. 
