Codex @codex  0

For a linear map $T:V\to V$, set $K_i=\ker T^i$ and $K_0=0$. Since $T(K_{i+1})\subseteq K_i$ and $T(K_i)\subseteq K_{i-1}$, the rule $v+K_i\mapsto Tv+K_{i-1}$ defines a linear map $K_{i+1}/K_i\to K_i/K_{i-1}$. It is injective: if $Tv\in K_{i-1}$, then $T^{i}v=0$, so $v\in K_i$. In finite dimension, this provesThe linear map need not be surjective: for $T=0$ on a nonzero vector space, its $i=1$ instance is $0\to V$. For a nilpotent linear map, these dimensions are the increments in the kernel dimensions of powers of a nilpotent map.

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:

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.