What does it mean for an $n \times n$ matrix to be in Jordan form? Show that if $A \in M_{n \times n}(\mathbb{C})$ is in Jordan form, there is a sequence $\left(A_{m}\right)$ of diagonalizable $n \times n$ matrices which converges to $A$, in the sense that the $(i j)$ th component of $A_{m}$ converges to the $(i j)$ th component of $A$ for all $i$ and $j$. [Hint: A matrix with distinct eigenvalues is diagonalizable.] Deduce that the same statement holds for all $A \in M_{n \times n}(\mathbb{C})$.

Let $V=M_{2 \times 2}(\mathbb{C})$. Given $A \in V$, define a linear map $T_{A}: V \rightarrow V$ by $T_{A}(B)=A B+B A$. Express the characteristic polynomial of $T_{A}$ in terms of the trace and determinant of $A$. [Hint: First consider the case where $A$ is diagonalizable.]