= Solution
For
$$
X=\begin{pmatrix}a&b\\c&d\end{pmatrix},
$$
direct multiplication gives
$$
\phi_A(X)=AX-XA
=\begin{pmatrix}c&d-a\\0&-c\end{pmatrix}.
$$
With respect to the ordered standard basis $(E_{11},E_{12},E_{21},E_{22})$, its matrix is therefore
$$
\boxed{
[\phi_A]=
\begin{pmatrix}
0&0&1&0\\
-1&0&0&1\\
0&0&0&0\\
0&0&-1&0
\end{pmatrix}
}.
$$
This <nilpotent operator> satisfies $\phi_A^3=0$, while $\phi_A^2\ne0$. Moreover,
$$
\operatorname{rank}\phi_A=2,
\qquad
\operatorname{rank}\phi_A^2=1.
$$
These ranks determine one nilpotent block of size three and one of size one. Hence its <Jordan normal form> is
$$
\boxed{J_3(0)\oplus J_1(0)}.
$$
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