= Solution
Write $V^*=V'$ for the <dual space>. Define $\theta(v)(f)=f(v)$. For fixed $v$, this is a <linear functional> on $V^*$; evaluation is also <linear> in $v$, so $\theta:V\to V^{**}$ is a <linear map>. If $v\ne0$, 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 $\dim V=\dim V^*=\dim V^{**}$. The <rank-nullity theorem> now makes $\theta$ an <linear isomorphism>.
The <annihilator of a vector subspace> of a <linear subspace> $U\le V$ is
$$
U^\circ=\{f\in V^*: f(u)=0\text{ for every }u\in U\}.
$$
Extend a <basis> $v_1,\ldots,v_k$ of $U$ to a <basis> $v_1,\ldots,v_n$ of $V$, with <dual basis> $f_1,\ldots,f_n$. A <linear functional> belongs to $U^\circ$ exactly when its first $k$ coordinates in this <dual basis> vanish. Hence $U^\circ$ has <basis> $f_{k+1},\ldots,f_n$, proving
$$
\boxed{\dim U+\dim U^\circ=\dim V.}
$$
The <dual map> is $\alpha^*(h)=h\circ\alpha$; here this is the algebraic <dual map>, denoted $\alpha'$ in the source. If $v\in\ker\alpha$, then $(\alpha'h)(v)=h(\alpha v)=0$. Consequently $\operatorname{im}\alpha'\subseteq(\ker\alpha)^\circ$. Moreover,
$$
\theta(v)\in(\operatorname{im}\alpha')^\circ
\iff h(\alpha v)=0\text{ for every }h\in W^*
\iff\alpha v=0,
$$
because coordinate <linear functionals> separate <vectors>. This proves $(\operatorname{im}\alpha')^\circ=\theta(\ker\alpha)$, including the requested second inclusion. Taking <dimensions> gives
$$
\dim\operatorname{im}\alpha'=\dim V-\dim\ker\alpha=\dim\operatorname{im}\alpha.
$$
The first inclusion has equal <dimensions> on both sides, so
$$
\boxed{\operatorname{im}\alpha'=(\ker\alpha)^\circ,\qquad
(\operatorname{im}\alpha')^\circ=\theta(\ker\alpha).}
$$
For a <matrix> $A=(a_{ij})$ representing $\alpha$, the matrix of $\alpha'$ 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 $\dim\operatorname{im}\alpha$. Applying the established equality to these two <linear maps> proves \b[row rank equals column rank] over every <field>.
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