Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 4 / 10F / a / Solution

The algebraic dual map is $\delta ={\alpha }^{\vee }:W^{\ast }\to V^{\ast }$, defined by $\delta (\phi )=\phi \circ \alpha$. The annihilator of a vector subspace $U\subseteq V$ isEvery functional $\phi \circ \alpha$ vanishes on $\ker \alpha$, so $\mathrm{im}\delta \subseteq (\ker \alpha {)}^{\circ }$. Conversely, let $\lambda \in (\ker \alpha {)}^{\circ }$. Define a functional on $\mathrm{im}\alpha$ by $\tilde{\lambda }(\alpha v)=\lambda (v)$. It is well defined because two preimages differ by an element of $\ker \alpha$. Extend a basis of $\mathrm{im}\alpha$ to one of $W$ and extend the functional arbitrarily, for instance by zero on the extra basis vectors. The resulting $\phi \in W^{\ast }$ satisfies $\delta \phi =\lambda$.

This is the dual image and kernel annihilator identity. ThereforeThe basis extension theorem also gives $\dim U^{\circ }=\dim V-\dim U$. By the rank-nullity theorem,Apply rank-nullity theorem to both maps, using $\dim W^{\ast }=\dim W$, to obtainOver $\mathbb{C}$, this is the algebraic dual, without complex conjugation; the adjoint operator in part (b) uses the inner products instead.