= Solution
Let $m=\dim U$, $n=\dim V$, $r=\operatorname{rank}\alpha$ and $r_* =\operatorname{rank}\alpha^*$. The <isomorphism> of part (c) preserves dimensions, so the formulas of part (a), applied also to $\alpha^*:V^*\to U^*$, give
$$
(n-r)m=\dim M^l(\alpha)=\dim M^r(\alpha^*)=(n-r_*)m.
$$
If $m>0$, cancellation proves \b[$\operatorname{rank}\alpha=\operatorname{rank}\alpha^*$]. 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.
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