TITLE: $S\subseteq V \Rightarrow \text{span}(S)\cong S^{00}$
QUESTION [5 upvotes]: Let $S$ be a subset of a finite dimensional vector space $V$. How to show that
$$\text{span}(S)\cong S^{00}?$$
Here, $^0$ denotes the annihilator. That is,
$$S^0:=\{f\in V^*:f(s)=0,\forall s\in S\}$$
where $V^*$ is the dual space. So
$$S^{00}=\{F\in V^{**}:F(f)=0,\forall f\in S^0\}$$

REPLY [2 votes]: $\DeclareMathOperator{\span}{Span}$This actually does not require $V$ to be finite dimensional. First, $\span(S) \subset S^{00}$, since if $\alpha \in V^*$ vanishes on all of $S$, it must also vanish on $\span(S)$.
Conversely, suppose that $v \notin \span (S)$. Now, if $\mathcal{B}_0$ is any basis of $\span(S)$, $\mathcal{B}_0 \cup \{v\}$ is linearly independent. So extend this set to a basis $\mathcal{B}$ of $V$. Now, define an element $\alpha \in V^*$ as follows: for $w \in V$, express $w$ (uniquely) as a linear combination of elements in the basis $\mathcal{B}$. Then $v$ appears with some coefficient (possibly $0$) in this expression. Let $\alpha(w)$ be this coefficient. One can check that this defines a linear functional on $V$. But $\alpha(v) = 1$, and $\alpha(w) = 0$ for any $w \in \span(S)$. So $\alpha \in S^0$ and $v \notin S^{00}$.