Anticommutator ($\{A,B\}$)

The anticommutator of two operators is $\{A,B\}=AB+BA$. They anticommute when this operator vanishes, equivalently when $AB=-BA$.

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: $A=\mathrm{diag}(1,0,0)$ and $B=\mathrm{diag}(0,1,0)$ anticommute in dimension three but are singular and have nonzero traces.

If $A$ is diagonalizable, the anticommutator map $B\mapsto AB+BA$ has eigenvalues $2{\lambda }_1$, ${\lambda }_1+{\lambda }_2$ twice, and $2{\lambda }_2$. Their product gives the displayed characteristic polynomial. The density of diagonalizable complex matrices and continuity of polynomial coefficients prove the formula also for defective matrices.