= Solution
The algebraic <dual space> is $V^*=\operatorname{Hom}_k(V,k)$, the vector space of linear functionals. For an ordered <basis> $v_1,\ldots,v_n$, define the <dual basis> by $v^i(v_j)=\delta_{ij}$, equivalently $v^i(\sum_jx_jv_j)=x_i$. Every functional satisfies $\ell=\sum_i\ell(v_i)v^i$, proving spanning. Evaluating a zero linear combination on each $v_j$ forces every coefficient to vanish, proving independence.
For a <linear map> $\alpha:V\to W$, its <dual map> is $\alpha^*:W^*\to V^*$, $\alpha^*(\ell)=\ell\circ\alpha$. Suppose the matrix of $\alpha$ satisfies $\alpha(v_i)=\sum_jA_{ji}w_j$. Then
$$
\alpha^*(w^j)(v_i)=w^j(\alpha(v_i))=A_{ji}.
$$
These are the $i$th coordinates of $\alpha^*(w^j)$ in the dual basis, so \b[the dual matrix is] $\boxed{A^T}$. For complex vector spaces this is still an ordinary transpose, because the algebraic dual uses linear functionals, not conjugate-linear ones.
To prove equality of ranks directly, $\ker\alpha^*$ consists precisely of functionals on $W$ vanishing on $\operatorname{im}\alpha$, its <annihilator>. A basis of the image extends to a basis of $W$; such functionals have zero values on the image basis and arbitrary values on the remaining $\dim W-\operatorname{rank}\alpha$ vectors. Thus $\dim\ker\alpha^*=\dim W-\operatorname{rank}\alpha$. The <rank-nullity theorem> gives
$$
\boxed{\operatorname{rank}\alpha^*=\operatorname{rank}\alpha.}
$$
Finally, if $\alpha$ is invertible, for every $\ell\in V^*$ the two compositions give $(\ell\circ\alpha^{-1})\circ\alpha=\ell$ and $(\ell\circ\alpha)\circ\alpha^{-1}=\ell$. Hence the two dual maps are mutual inverses and
$$
\boxed{(\alpha^*)^{-1}=(\alpha^{-1})^*.}
$$
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