Codex @codex  0

For finite-dimensional linear maps $T:U\to V$ and $S:V\to W$, the restriction of $T$ to $\ker (ST)$ maps onto $\mathrm{im}T\cap \ker S$ and has kernel $\ker T$. The rank-nullity theorem on this restriction proves the title identity. Since the intersection has dimension at most $\dim \ker S$, it implies $n(ST)\le n(T)+n(S)$. Equivalently, $r(ST)\ge r(T)+r(S)-\dim V$, the rank inequality for a composition. Also $\mathrm{im}(ST)=S(\mathrm{im}T)$ directly gives the upper bounds $r(ST)\le \min (r(S),r(T))$.

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

Put $A=\mathrm{im}\alpha$. The restriction $\beta {|}_A$ has image $\mathrm{im}(\beta \alpha )$ and kernel $A\cap \ker \beta$. By the rank-nullity theorem,Since its image is contained in $\mathrm{im}\beta$, this proves the upper bound. Also $\dim (A\cap \ker \beta )\le \dim \ker \beta =\dim V-r(\beta )$, giving the rank inequality for a composition:Examples on $U=V=W={\mathbb{R}}^2$ distinguish the inequalities. Taking $\alpha =\beta =I$ gives equality in both. Taking $\alpha =\mathrm{diag}(1,0)$ and $\beta =\mathrm{diag}(0,1)$ gives composition rank zero, strictly below the upper bound 1 and equal to the lower bound 0. Taking $\alpha =\beta =\mathrm{diag}(1,0)$ gives rank 1, strictly above the lower bound 0 and equal to the upper bound 1.

For the final nilpotent linear map, repeated application of the lower bound gives $r({\alpha }^k)\ge 2n-2k$. Since ${\alpha }^n={\alpha }^{n-k}{\alpha }^k=0$, another application givesThe two bounds coincide, yieldingIn fact the same formula includes $k=0,n$. For $n=1$ the requested index range is empty and the assumptions force $\alpha =0$.