Two-by-two anticommutator characteristic polynomial (${\chi }_{B\mapsto AB+BA}(t)=(t-\mathrm{tr}A{)}^2(t^2-2t\mathrm{tr}A+4\det A)$)

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.