$S\subseteq V \Rightarrow \text{span}(S)\cong S^{00}$

Question

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 $$S^{00}=\{F\in V^{**}:F(f)=0,\forall f\in S^0\}$$

Answer 1

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.