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

Write $K_i=\ker T^i$, with $K_0=\{0\}$. The minimal polynomial $x^m$ means $T^m=0$ and $T^{m-1}\ne 0$. Choose $w$ with $T^{m-1}w\ne 0$. For each $0\le i<m$, set $w_i=T^{m-1-i}w$. ThenThus $w_i\in K_{i+1}\setminus K_i$. Also $K_i\subseteq K_{i+1}$ because applying $T$ to zero still gives zero. Therefore the kernel dimensions of powers of a nilpotent map increase strictly until the nilpotency index:The same reasoning covers $m=1$, when the nilpotent linear map is zero on a nonzero vector space.

Put $d_i=\dim K_i$, including $d_0=0$. The restriction of $T^i$ to $K_{i+1}$ has kernel $K_i$, so the rank-nullity theorem givesLikewise $T^{i-1}{|}_{K_i}$ has kernel $K_{i-1}$ and hence image dimension $d_i-d_{i-1}$. But $T(K_{i+1})\subseteq K_i$, which impliesComparing dimensions yields the required concavity of the kernel dimensions of powers of a nilpotent map:

The increments ${\Delta }_i=d_i-d_{i-1}$ are positive integers by part (a)(i), are nonincreasing by part (a)(ii), and sum to $d_m=4$. Thus they form an integer partition of $4$. The complete list is $(4)$, $(3,1)$, $(2,2)$, $(2,1,1)$, and $(1,1,1,1)$.

To prove all five occur over every field, let $J_s(0)$ be a Nilpotent Jordan block of size $s$. It sends each standard basis vector to its predecessor, with the first sent to zero, soFor a block diagonal matrix, the nullities add. The following explicit nilpotent matrices therefore realize every possible sequence:In each row the largest Jordan block size is exactly the nilpotency index $m$, so the displayed sequence stops at the correct place.

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