## A lemma on a rather frequently encountered set of Hermitian matrices

I’m documenting a small ‘lemma’ that I think is worth mentioning. Suppose are 4 Hermitian matrices () satisfying

where denotes the identity matrix. Then these matrices have eigenvalues , are traceless and are necessarily of even dimension.

For , the anticommutator above gives . So, for any eigenvector and eigenvalue , we have

or equivalently .

Traceless-ness has a neat proof:

Suppose . Then

where the last equality follows from . But , so

.

Finally, suppose the numbers of +1 eigenvalues and -1 eigenvalues are and respectively. The dimension of the matrix is then . Since the trace equals the sum of eigenvalues, we have

So, the dimension is , which is clearly always an even number.

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