 Solution

The algebraic dual space is $V^{\ast }={\mathrm{Hom}}_k(V,k)$, the vector space of linear functionals. For an ordered basis $v_1,\dots ,v_n$, define the dual basis by $v^i(v_j)={\delta }_{ij}$, equivalently $v^i({\sum }_j{x}_j{v}_j)=x_i$. 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 ${\alpha }^{\ast }:W^{\ast }\to V^{\ast }$, ${\alpha }^{\ast }(\ell )=\ell \circ \alpha$. Suppose the matrix of $\alpha$ satisfies $\alpha (v_i)={\sum }_j{A}_{ji}{w}_j$. ThenThese are the $i$th coordinates of ${\alpha }^{\ast }(w^j)$ in the dual basis, so 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, $\ker {\alpha }^{\ast }$ consists precisely of functionals on $W$ vanishing on $\mathrm{im}\alpha$, 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 $\dim W-\mathrm{rank}\alpha$ vectors. Thus $\dim \ker {\alpha }^{\ast }=\dim W-\mathrm{rank}\alpha$. The rank-nullity theorem givesFinally, if $\alpha$ is invertible, for every $\ell \in V^{\ast }$ 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