 Solution

Write $V^{\ast }=V^{\prime }$ for the dual space. Define $\theta (v)(f)=f(v)$. For fixed $v$, this is a linear functional on $V^{\ast }$; evaluation is also linear in $v$, so $\theta :V\to V^{\ast \ast }$ is a linear map. If $v\ne 0$, extend $v$ to a basis and take its first coordinate linear functional $f$; then $f(v)=1$. Thus $\theta$ is injective. A dual basis has the same number of elements as the original basis, so $\dim V=\dim V^{\ast }=\dim V^{\ast \ast }$. The rank-nullity theorem now makes $\theta$ an linear isomorphism.

The annihilator of a vector subspace of a linear subspace $U\le V$ isExtend a basis $v_1,\dots ,v_k$ of $U$ to a basis $v_1,\dots ,v_n$ of $V$, with dual basis $f_1,\dots ,f_n$. A linear functional belongs to $U^{\circ }$ exactly when its first $k$ coordinates in this dual basis vanish. Hence $U^{\circ }$ has basis $f_{k+1},\dots ,f_n$, provingThe dual map is ${\alpha }^{\ast }(h)=h\circ \alpha$; here this is the algebraic dual map, denoted ${\alpha }^{\prime }$ in the source. If $v\in \ker \alpha$, then $({\alpha }^{\prime }h)(v)=h(\alpha v)=0$. Consequently $\mathrm{im}{\alpha }^{\prime }\subseteq (\ker \alpha {)}^{\circ }$. Moreover,because coordinate linear functionals separate vectors. This proves $(\mathrm{im}{\alpha }^{\prime }{)}^{\circ }=\theta (\ker \alpha )$, including the requested second inclusion. Taking dimensions givesThe first inclusion has equal dimensions on both sides, soFor a matrix $A=(a_{ij})$ representing $\alpha$, the matrix of ${\alpha }^{\prime }$ in the corresponding dual bases is $A^{\mathsf{T}}$: its $ji$ entry is $h_i(\alpha v_j)=a_{ij}$. Thus the column rank of $A^{\mathsf{T}}$ is the row rank of $A$, while the column rank of $A$ is $\dim \mathrm{im}\alpha$. Applying the established equality to these two linear maps proves row rank equals column rank over every field.