Equality in the rank-sum and nullity-product inequalities

For endomorphisms $S,T$ of a finite-dimensional space of dimension $d$, $r(S+T)\le r(S)+r(T)$ and $n(ST)\le n(S)+n(T)$. If both equalities hold, the first gives $r(S)+r(T)\le d$, while the second gives $2d-r(S)-r(T)\le d$. Thus the rank sum is $d$, $S+T$ is an isomorphism, and $ST=0$. Conversely, $ST=0$ implies $\mathrm{im}T\subseteq \ker S$ and hence rank sum at most $d$, whereas invertibility of $S+T$ gives rank sum at least $d$. This proves both equalities.