= Solution
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 $1,\ldots,n$ and put
$$
J_m=J+\varepsilon_m\operatorname{diag}(1,\ldots,n).
$$
Choose positive $\varepsilon_m\to0$ avoiding the finitely many values solving $\lambda_i+i\varepsilon=\lambda_j+j\varepsilon$ 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 $A=SJS^{-1}$, use $A_m=SJ_mS^{-1}$. 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 $T_{\operatorname{diag}(\lambda_1,\lambda_2)}$. On the four matrix units $E_{ij}$, this acts by $(\lambda_i+\lambda_j)E_{ij}$. Thus, writing $s=\operatorname{tr}A$ and $d=\det A$, its <characteristic polynomial> is
$$
\begin{aligned}
\chi_{T_A}(t)&=(t-2\lambda_1)(t-2\lambda_2)(t-\lambda_1-\lambda_2)^2\\
&=\boxed{(t-s)^2(t^2-2st+4d)}.
\end{aligned}
$$
The 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>.
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