 Solution

Let $m=\dim U$, $n=\dim V$, $r=\mathrm{rank}\alpha$ and $r_{\ast }=\mathrm{rank}{\alpha }^{\ast }$. The isomorphism of part (c) preserves dimensions, so the formulas of part (a), applied also to ${\alpha }^{\ast }:V^{\ast }\to U^{\ast }$, giveIf $m>0$, cancellation proves $\mathrm{rank}\alpha =\mathrm{rank}{\alpha }^{\ast }$. If $m=0$, both maps are zero and the equality remains true. This argument derives dual-rank equality from the requested subspace correspondence without assuming it beforehand.