= Solution
The algebraic <dual space> is $V^*=\operatorname{Hom}_{\mathbb R}(V,\mathbb R)$, the <vector space> of all real <linear functionals>. For a finite <basis> $e_1,\ldots,e_n$, define the <dual basis> by $e_i^*(\sum_j a_je_j)=a_i$, so $e_i^*(e_j)=\delta_{ij}$. Every <linear functional> satisfies
$$
\lambda=\sum_{i=1}^n\lambda(e_i)e_i^*.
$$
This equality follows by evaluating both sides on each vector of the <basis>. It proves spanning directly. If $\sum_i b_ie_i^*=0$, evaluation on $e_j$ gives $b_j=0$, proving linear independence without assuming a theorem on dual dimensions.
For a <vector subspace> $U$, its <annihilator of a vector subspace> is
$$
U^\circ=\{\lambda\in V^*: \lambda(u)=0\text{ for all }u\in U\}.
$$
Extend a <basis> $e_1,\ldots,e_r$ of $U$ to a <basis> of $V$. Then $U^\circ$ has <basis> $e_{r+1}^*,\ldots,e_n^*$, so $\boxed{\dim U^\circ=n-\dim U}$.
The <dual map> of $\alpha:V\to W$ is $\alpha^*:W^*\to V^*$, $\lambda\mapsto\lambda\circ\alpha$. In the corresponding <dual bases> its <matrix> is the transpose of the <matrix> for $\alpha$. A <matrix> and its transpose have the same <rank>, giving $\operatorname{rank}\alpha^*=\operatorname{rank}\alpha$. Directly,
$$
\alpha^*\lambda=0\ \Longleftrightarrow\ \lambda(\alpha v)=0\text{ for all }v
\ \Longleftrightarrow\ \lambda\in(\operatorname{im}\alpha)^\circ.
$$
Thus $\boxed{\ker\alpha^*=(\operatorname{im}\alpha)^\circ}$. Also every $\lambda\circ\alpha$ vanishes on $\ker\alpha$, so $\operatorname{im}\alpha^*\subseteq(\ker\alpha)^\circ$. Both spaces have dimension $\operatorname{rank}\alpha$ by the preceding <annihilator of a vector subspace> formula and <rank-nullity theorem>, proving
$$
\boxed{\operatorname{im}\alpha^*=(\ker\alpha)^\circ.}
$$
Alternatively, a <linear functional> vanishing on the kernel descends to the image of $\alpha$ and can be extended to $W$ by extending a <basis>.
For the infinite-dimensional <polynomial ring>, the coefficient functionals $L_i$ are linearly independent: evaluate a finite relation on each monomial. However, they do not span the algebraic <dual space>. The <linear functional> $p\mapsto p(1)$ takes value one on every monomial, while any finite combination of the $L_i$ vanishes on every sufficiently high-degree monomial. Thus \b[the $L_i$ do not form a basis of $V^*$]. An infinite formal sum of coefficient functionals is not a finite linear combination.
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