Past exam of the linear algebra course of the University of Oxford / 2023 / 2 / c / Solution

Direct matrix multiplication gives $A^2=0$. Denote the columns by $c_1,\dots ,c_4$; they satisfyThus the matrix rank is at most two. Over the real numbers or complex numbers, the upper-left two-by-two minor is $-3\ne 0$, 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. ConsequentlyEquivalently 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