Past exam of the mathematics course of the University of Oxford / 2014 / Prelims / Mathematics I / 2 / b

Restrict $S$ to the vector subspace $W=\mathrm{im}T$. This restriction has kernel of a linear map $W\cap \ker S$ and image of a linear map $S(W)=\mathrm{im}(ST)$. The rank-nullity theorem on $W$ therefore givesApply the rank-nullity theorem to $T$ and $ST$ on $V$ to obtain the exact nullity of a composition identity:Since the intersection is a vector subspace of $\ker S$, its dimension is at most $n(S)$. Hence

The image of a linear map $\mathrm{im}(ST)=S(\mathrm{im}T)$ lies inside $\mathrm{im}S$. Therefore $r(ST)\le r(S)$. Also $S$ restricted to $\mathrm{im}T$ cannot have rank of a linear map greater than its domain's dimension, so $r(ST)\le \dim \mathrm{im}T=r(T)$. ThusEquivalently, apply the rank-nullity theorem to that restriction: $r(ST)=r(T)-\dim (\mathrm{im}T\cap \ker S)\le r(T)$. This proof keeps track of the intermediate vector subspace, rather than assuming an invertible factor.