= Solution
The algebraic <dual map> is $\delta=\alpha^\vee:W^*\to V^*$, defined by $\delta(\varphi)=\varphi\circ\alpha$. The <annihilator of a vector subspace> $U\subseteq V$ is
$$
U^\circ=\{\lambda\in V^*: \lambda(u)=0\text{ for all }u\in U\}.
$$
Every functional $\varphi\circ\alpha$ vanishes on $\ker\alpha$, so $\operatorname{im}\delta\subseteq(\ker\alpha)^\circ$. Conversely, let $\lambda\in(\ker\alpha)^\circ$. Define a functional on $\operatorname{im}\alpha$ by $\widetilde\lambda(\alpha v)=\lambda(v)$. It is well defined because two preimages differ by an element of $\ker\alpha$. Extend a <basis> of $\operatorname{im}\alpha$ to one of $W$ and extend the functional arbitrarily, for instance by zero on the extra basis vectors. The resulting $\varphi\in W^*$ satisfies $\delta\varphi=\lambda$.
This is the <dual image and kernel annihilator identity>. Therefore
$$
\boxed{\operatorname{im}\delta=(\ker\alpha)^\circ.}
$$
The <basis extension> theorem also gives $\dim U^\circ=\dim V-\dim U$. By the <rank-nullity theorem>,
$$
\boxed{\dim\operatorname{im}\delta=\dim V-\dim\ker\alpha=\dim\operatorname{im}\alpha.}
$$
Apply <rank-nullity theorem> to both maps, using $\dim W^*=\dim W$, to obtain
$$
\boxed{\dim\ker\alpha-\dim\ker\delta=\dim V-\dim W.}
$$
Over $\mathbb C$, this is the algebraic dual, without complex conjugation; the <adjoint operator> in part (b) uses the inner products instead.
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