Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 4 / 15G (Linear Mathematics)

The zero-composition subspaces of a linear map are subspaces because the defining equations are homogeneous linear equations in $\beta$: they hold for the zero map and are preserved by addition and scalar multiplication. A map with $\beta \alpha =0$ vanishes on $\mathrm{im}\alpha$, so it factors uniquely through the quotient vector space $V/\mathrm{im}\alpha$. Conversely every map from that quotient to $U$ gives such a $\beta$. ThereforeThe condition $\alpha \beta =0$ says exactly that the image of $\beta$ lies in $\ker \alpha$. ConsequentlyBoth dimension formulas use the fact that the space of maps between spaces of dimensions $p,q$ has dimension $pq$: matrix units in any chosen bases supply a basis.

The dual map is defined by ${\alpha }^{\ast }(\lambda )=\lambda \circ \alpha$. Suppose ${\alpha }^{\ast }=0$. Then $\lambda (\alpha u)=0$ for every $u\in U$ and every $\lambda \in V^{\ast }$. The dual space separates vectors: a nonzero vector can be included in a basis and a coordinate functional takes value one on it. Thus $\alpha u=0$ for every $u$, and $\alpha =0$. This proves injectivity of $\Phi$.

Its source and target have the same finite dimension:An injective linear map between equal finite dimensions is surjective by the rank-nullity theorem. Therefore $\Phi$ is an isomorphism. The inverse is also described by taking the dual again and identifying each space canonically with its double dual.

Here $\Phi$ acts on $\beta :V\to U$, so it is the same dualization construction with the source and target interchanged: ${\Phi }_{V,U}:L(V,U)\to L(U^{\ast },V^{\ast })$. This is the natural family interpretation of the symbol used in part (b).

If $\beta \alpha =0$, composition reversal gives ${\alpha }^{\ast }{\beta }^{\ast }=0$, so ${\beta }^{\ast }\in M^r({\alpha }^{\ast })$. Conversely, if $\delta \in M^r({\alpha }^{\ast })$, the isomorphism in part (b) writes $\delta ={\beta }^{\ast }$ uniquely. Then ${\alpha }^{\ast }{\beta }^{\ast }=0$ implies $(\beta \alpha {)}^{\ast }=0$; injectivity of dualization gives $\beta \alpha =0$. ThusSimilarly, $\alpha \beta =0$ is equivalent to ${\beta }^{\ast }{\alpha }^{\ast }=0$, givingEquivalently, the reverse implications follow by taking double duals and using the canonical double-dual identification. The equalities are of subspaces under an isomorphism, not only inclusions.

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.