 Solution

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.