 Solution

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 $\alpha =0$.