= Solution
Direct <matrix multiplication> gives $A^2=0$. Denote the columns by $c_1,\ldots,c_4$; they satisfy
$$
c_3=c_1-2c_2,\qquad c_4=-c_1+c_2.
$$
Thus the <matrix rank> is at most two. Over the <real numbers> or <complex numbers>, the upper-left two-by-two <minor> is $-3\ne0$, so the <matrix rank> is exactly two. Every <Jordan block> has size at most two because $A^2=0$, and each size-two <Nilpotent Jordan block> contributes one to the <matrix rank>. Consequently
$$
\boxed{J=J_2(0)\oplus J_2(0)=
\begin{pmatrix}0&1&0&0\\0&0&0&0\\0&0&0&1\\0&0&0&0\end{pmatrix}}.
$$
Equivalently the <nullities> are $(2,4)$ and the <minimal polynomial> is $x^2$.
There is a <field> qualification in the source: Question 2 initially allows an arbitrary <field>, and part (c) does not specify a new one. The preceding answer holds over every <field> whose <characteristic> is not three. In <characteristic> three, the first row vanishes and the other three rows all become $(2,1,0,2)$, giving <matrix rank> one while $A^2=0$. Under that reading the answer is instead
$$
\boxed{J=J_2(0)\oplus J_1(0)\oplus J_1(0)\qquad(\operatorname{char}F=3)}.
$$
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