= Invertible anticommuting matrices have even dimension and zero traces
{title2=$AB=-BA\ \Longrightarrow\ n\in2\mathbb Z,\ \operatorname{Tr}A=\operatorname{Tr}B=0$}
For complex <invertible matrices> $A,B$, $AB=-BA$ implies $B^{-1}AB=-A$ and $A^{-1}BA=-B$. Invariance of <matrix trace> under <matrix similarity> gives both zero traces. Invariance of <determinant> gives $\det A=(-1)^n\det A$, and invertibility forces $n$ even. Equivalently, <eigenvalues> occur in pairs $\lambda,-\lambda$ with equal <algebraic multiplicity>, and zero is excluded. Both invertibility assumptions matter: $A=\operatorname{diag}(1,0,0)$ and $B=\operatorname{diag}(0,1,0)$ anticommute in dimension three but are singular and have nonzero traces.
Back to article page
