Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics I / 3 / b

Since $S^{\prime }$ is the restriction of $S$ to $\mathrm{im}T$,This image of a linear map is contained in $\mathrm{im}S$, so $r_{ST}\le r_S$. Also, applying $S$ to a basis of $\mathrm{im}T$ gives a spanning set for $\mathrm{im}{S}^{\prime }$, of size $r_T$. Thus $r_{ST}\le r_T$, and

For $u\in U$, the vector $Tu$ already belongs to $\mathrm{im}T$, soTherefore $\ker (ST)=T^{-1}(\ker S^{\prime })$. Since $\ker S^{\prime }=\ker S\cap \mathrm{im}T$, its intersection with $\mathrm{im}T$ is itself. Applying a with $X=\ker S^{\prime }$ yieldsThe inequality follows from the inclusion $\ker S^{\prime }\subseteq \ker S$ and monotonicity of dimension for vector subspaces.

By i and ii,The domain of a function of $S^{\prime }$ is $\mathrm{im}T$. Its rank-nullity theorem gives $r_{S^{\prime }}+n_{S^{\prime }}=\dim \mathrm{im}T=r_T$. Applying the rank-nullity theorem for $T$ now giveswhich is exactly the rank-nullity theorem for the function composition $ST$.