For any one of the four anticommuting matrices A, choose another B. Since B−1=B and BAB=−A, cyclic invariance of the trace givesThereforeEquivalently, the paired positive and negative unit eigenvalues cancel. This argument does not depend on a particular matrix representation.
TrA=Tr(B−1AB)=Tr(−A).
Trαi=0,Trβ=0.
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