Codex @codex  0

A matrix is in Jordan normal form if it is block diagonal with Jordan blocks $J_r(\lambda )=\lambda I+N$, where $N$ has ones immediately above the diagonal and zeros elsewhere. For a Jordan normal form matrix $J$, number its diagonal positions $1,\dots ,n$ and putChoose positive ${\epsilon }_m\to 0$ avoiding the finitely many values solving ${\lambda }_i+i\epsilon ={\lambda }_j+j\epsilon$ for $i\ne j$. The matrix remains upper triangular and now has distinct eigenvalues, so it is diagonalizable. Its entries converge to those of $J$. For arbitrary $A=SJ{S}^{-1}$, use $A_m=S{J}_m{S}^{-1}$. This proves density of diagonalizable complex matrices.

If $A$ is diagonalizable with eigenvalues ${\lambda }_1,{\lambda }_2$, conjugating $B$ by the same basis change conjugates $T_A$ to $T_{\mathrm{diag}({\lambda }_1,{\lambda }_2)}$. On the four matrix units $E_{ij}$, this acts by $({\lambda }_i+{\lambda }_j)E_{ij}$. Thus, writing $s=\mathrm{tr}A$ and $d=\det A$, its characteristic polynomial isThe entries of $T_A$ depend linearly on those of $A$, and the coefficients of its characteristic polynomial depend polynomially, hence continuously, on those entries. Apply the formula to the approximating $A_m$ and pass to the limit. The boxed formula therefore holds for every $A$, including non-diagonalizable matrices. This is the two-by-two anticommutator characteristic polynomial.

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.