Past exam of the mathematics course of the University of Cambridge / 2002 / ib / Paper 4 / 6F / Solution

For a linear map $T:U\to V$, its rank of a linear map is $\dim \mathrm{im}T$, and its nullity is $\dim \ker T$. The rank-nullity theorem states $\dim U=\mathrm{rank}T+\mathrm{nullity}T$. One obtains it by extending a basis of the kernel to a basis of $U$; the images of the added vectors form a basis of the image.

Restrict $\beta$ to $A=\mathrm{im}\alpha$. Its image is $\mathrm{im}(\beta \alpha )$ and its kernel is $A\cap \ker \beta$. ThusThe intersection dimension is at most $\dim \ker \beta =\dim V-\mathrm{rank}\beta$, proving the lower bound. It is nonnegative, proving the upper bound by $\mathrm{rank}\alpha$, while $\mathrm{im}(\beta \alpha )\subseteq \mathrm{im}\beta$ proves the other upper bound. Consequently the rank inequality for a composition is