$S\subseteq V \Rightarrow \text{span}(S)\cong S^{00}$ 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
$$S^{00}=\{F\in V^{**}:F(f)=0,\forall f\in S^0\}$$
 A: First note that if $S$ is a subset of $V$, then $S^{00}=(\operatorname{span}(S))^{00}$, so we need only consider the case where $S$ is a subspace. 
To see this equality, observe that since $V$ is finite dimensional, the canonical mapping $\tau\colon V\to V^{**}$ is injective, hence surjective. Thus every element of $V^{**}$ has form $\bar{v}$ for $v\in V$. So take $\bar{v}\in S^{00}$. Thus $\bar{v}(f)=f(v)=0$ for all $f\in S^0$. But recall that taking the annihilator is order reversing, so since $S\subseteq\operatorname{span}(S)$, $\operatorname{span}(S)^0\subseteq S^0$. So for any $g\in\operatorname{span}(S)^0$, $g\in S^0$ as well. Thus for any such $g$, $\bar{v}(g)=g(v)=0$. Thus $S^{00}\subseteq\operatorname{span}(S)^{00}$. 
Conversely, take $\bar{v}\in\operatorname{span}(S)^{00}$. So $\bar{v}(f)=f(v)=0$ for all $f\in\operatorname{span}(S)^0$. But $\operatorname{span}(S)$ is a subspace, so necessarily $v\in\operatorname{span}(S)$. If not, there exists a linear functional annihilating $\operatorname{span}(S)$ but not $v$, contrary to the fact that $\bar{v}\in\operatorname{span}(S)^{00}$.
To be explicit, if $v=0$, clearly $v\in\operatorname{span}(S)$. So suppose $v\neq 0$, but $v\not\in\operatorname{span}(S)$. Let $\{s_1,\dots,s_k\}$ be a basis for $\operatorname{span}(S)$. Then $\{s_1,\dots,s_k,v\}$ is linearly independent. For suppose there is a nontrivial linear combination such that
$$
c_1s_1+\cdots+c_ks_k+cv=0.
$$
Necessarily $c\neq 0$, else we would have a nontrivial linear combination of $\{s_1,\dots,s_k\}$ equal to $0$. But then
$$
v=c^{-1}(-c_1s_1-\cdots-c_ks_k)\in\operatorname{span}(S)
$$
a contradiction. So $\{s_1,\dots,s_k,v\}$ is linearly independent. Extend it to a basis of $V$. Then define a linear functional sending $v$ to $1$, and all other basis vectors to $0$. This functional then annihilates $\operatorname{span}(S)$ but not $v$.
 So $v\in\operatorname{span}(S)$, so write $v=\sum c_is_i$ for some $s_i\in S$. Now take any $g\in S^0$. We have
$$
\bar{v}(g)=g(v)=g\left(\sum c_is_i\right)=\sum c_ig(s_i)=0
$$
since $g\in S^0$ annihilates every $s_i$. Thus $\operatorname{span}(S)^{00}\subseteq S^{00}$. 
Now let $\tau$ be the canonical map $\tau(v)=\bar{v}$. We need to show $\tau\colon S\to S^{00}$ is an isomorphism, and since $\tau$ is a monomorphism, all you need to do is show $\tau(S)=S^{00}$. Take $s\in S$, so $\tau(s)=\bar{s}$ is such that for all $f\in S^0$,
$$
\bar{s}(f)=f(s)=0
$$
whence $\tau(s)=\bar{s}\in S^{00}$. This implies $\tau(S)\subseteq S^{00}$. Also, if $\bar{v}\in S^{00}$, then for any $f\in S^0$, it follows that
$$
f(v)=\bar{v}(f)=0
$$
and thus every linear functional annihilating $S$ also annihilates $v$. Recall that if $v\not\in S$, then there exists a linear functional $g\in V^*$ such that $g(S)=\{0\}$ but $g(v)\neq 0$. Thus $v\in S$, and $\bar{v}=\tau(v)\in\tau(S)$, so $S^{00}\subseteq \tau(S)$, as required.
A: $\DeclareMathOperator{\span}{Span}$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 $v \notin S^{00}$. 
