Anticommuting matrices

A variant on commutation of matrices is anticommutation. As before, consider $M$ and N as two matrices for which

$$\begin{eqnarray*} M N & = & - N M, \end{eqnarray*}$$

with X an eigenvector for M whereby $M X = \lambda X$. Then

$$\begin{eqnarray*} M N X & = & - N M X \\ M (N X) & = & (- \lambda) (N X). \end{eqnarray*}$$

Either $N$ is singular, $\lambda$ is zero (making $M$ singular), or $(N X)$ is another eigenvalue belonging to $-\lambda$. Evidently the relationship is a mutual one, $M$ mapping eigenvectors of $N$ into new ones with reversed eigenvalue sign. We need to consider four regions: vectors annihilated by $M$, vectors annihilated by $N$, the vectors with positive eigenvalue associated with $M$, and those with negative eigenvalues, conjugated from the first group by $N$.

Setting up the explicit format of the previous section and repeating the derivation leads to the conclusion that

$$\begin{eqnarray*} (\lambda_i + \lambda_j) k_{ij} & = & 0. \end{eqnarray*}$$

Various schematic representations of the situation are possible, but are probably best summarized by saying that M and N have the respective forms:

$$\begin{eqnarray*} \left[ \begin{array}{ll\vert l\vert l} A & . & . & . \\ .... .... & . & . & . \\ \hline . & . & . & G \end{array} \right] \end{eqnarray*}$$

The $F$ and $G$ parts can be discarded except for singularity (but then the dimension of $M$ and $N$ must be even), leaving the general impression that anticommuting matrices can be brought to a form with one of them diagonal, the other antidiagonal, and both with their nonzero eigenvalues arranged in negative pairs.