# 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

$$[[A,B]^2,C]=0.$$

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I think the "warmup" has been asked here a few times. Think traces. – Dylan Moreland Jun 12 '12 at 5:15
Ok, I deleted it so the discussion doesn't get cluttered. – Potato Jun 12 '12 at 5:30
Well, if it's a good hint for the problem then I think it's good to leave it, perhaps with a link to a relevant question if one of us can dig that up. – Dylan Moreland Jun 12 '12 at 5:30
It's actually not a good hint. I solved it using a similar method but there's a slicker way. – Potato Jun 12 '12 at 5:31
Here's a related question: Prove $[A,B]-[C,D]+[C,D]\cdot[A,B]$ also commutes with every $2\times2$ complex matrix. – user641 Jun 12 '12 at 13:53

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 $[[A,B]^2,C]=0$.

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Very nice. I will just mention for the reference of others that the Caley-Hamilton theorem can be used to get the basically the same results. – Potato Jun 12 '12 at 6:45
@Potato: Yes, I realized that this morning, but not last night at midnight. I'm posting the other argument separately. – Arturo Magidin Jun 12 '12 at 13:38

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 $C\in M_{2\times 2}(\mathbf{F})$.

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