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

For a linear map $T:V\to V$, its kernel of a linear map and image of a linear map areIts nullity of a linear map is $n(T)=\dim \ker T$, and its rank of a linear map is $r(T)=\dim \mathrm{im}T$.

Choose a basis $u_1,\dots ,u_k$ of $\ker T$ and extend it to a basis $u_1,\dots ,u_k,v_1,\dots ,v_m$ of $V$. The vectors $T(v_1),\dots ,T(v_m)$ span $\mathrm{im}T$, because applying $T$ to an arbitrary linear combination of the full basis eliminates the $u_i$ terms. They are also linearly independent: if ${\sum }_j{b}_{j}T(v_j)=0$, then ${\sum }_j{b}_j{v}_j\in \ker T$, so ${\sum }_j{b}_j{v}_j={\sum }_i{a}_i{u}_i$ for some coefficients. Independence of the extended basis forces all $a_i,b_j$ to vanish.

Consequently $r(T)=m$, $n(T)=k$ and $\dim V=k+m$. This proves the rank-nullity theorem:

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.