How to prove that two subspaces are equal if and only if their annihilators are equal By definition, $W_1=W_2 \rightarrow W_1^0=W_2^0$, $W_1, W_2 \subseteq V$ are subspaces of $V$. But how does one show the reverse implication? $V$ is not assumed to be finite dimensional.
The annihilator is defined as $S^0= \{f\in V^*| f(s)=0 \forall s\in S\}$, where $S\subseteq V, S$ is a subset of vector space V. 
 A: To prove $ W_1^0=W_2^0 \Rightarrow W_1=W_2 $ we proceed by contraposition; suppose $ W_1 \neq W_2$ then there is some vector $\alpha \in W_1 \setminus W_2$. Now suppose $B_{W_2} = \{w_1,...,w_k\}$ be a basis of $W_2$. Then $ B_{W_2}\cup \{ \alpha \}$ is linearly independent since $\alpha\notin W_2$. Extend $ B_{W_2}\cup \{ \alpha \}$ to a basis of $V$ by $B_V=\{w_1,...,w_k,\alpha,w_{k+2},...,w_n\}.$ 
Now define a (unique) linear functional $f:V\rightarrow \Bbb F $ such that $f(w_i)=0$ for $ i=1,...,k \quad; f(\alpha)=1_{\Bbb F} \; $and $f(w_j)=1_{\Bbb F}$ for $j=k+2,...,n$. Then clearly $f(w)=0 \; \forall  w\in W_2 $ (since $\forall w \in W_2 \;, w=a_1w_1+...+a_kw_k \; $for some $a_1,...,a_k \in \Bbb F$) i.e $f \in W_2^0$ but $f\notin W_1^0$ since $\alpha \in W_1 $ and $f(\alpha) \neq 0$. Hence $ W_1^0 \neq W_2^0$.
(We assume that $V$ is finite dimensional with $dim\;V=n$)
A: I assume that here we are dealing with a finite dimensional vector space $V$. In this case, there exists some basis $B=v_1,...,v_n$ for $V$. We define the dual basis to be $B'=\phi_1,...,\phi_n$, where the bases satisfy the Kroenecker Delta condition. That is
$$ \phi_i(v_j)=\delta_{ij}.$$
This means that $\phi_i(v_j)=1$ if and only if $i=j$. So consider then that we can express $W_1,W_2$ in terms of the basis vectors for $V$. So let
$$ W_1=span(v_k,...,v_{\ell})=W_2$$
Then, by definition
$$ W_1^0=span(\phi_1,...\phi_{k-1},\phi_{\ell+1},...,\phi_n)=W_2^0.$$
So, if we are given that $W_1^0=W_2^0$, we know that they are spanned by the same elements of the dual basis. And so:
$$ W_1^0=span(\phi_1,...\phi_{k-1},\phi_{\ell+1},...,\phi_n)=W_2^0 \implies W_1=span(v_k,...,v_{\ell})=W_2.$$
This completes the proof.
