Every subspace of the dual of a finite-dimensional vector space is an annihilator Exercise 26 page 115 of Linear Algebra Done Right by Sheldon Axler is the following:

Suppose $V$ is finite-dimensional and $\Gamma$ is a subspace of $V'$. Show that $$\Gamma=\{v\in V:\varphi(v)=0\text{ for every }\varphi\in\Gamma\}^0$$

where $V'$ is the dual space of $V$ and, for any $S\subset V$, $S^0$ is the annihilator of $S$.
Attempt:
Let $S=\{v\in V:\varphi(v)=0\text{ for every }\varphi\in\Gamma\}$. Clearly, $\Gamma\subset S^0$.I tried to show that $S^0\subset\Gamma$ using some bases of $V$ and $V'$, but I failed.
I also tried to show that $\dim{\Gamma}=\dim{V}-\dim{S}=\dim{S^°}$. If $s_1,\dots,s_n$ is a basis of $S$ and $s_1',\dots,s_n'$ its dual, and if $\psi_1,\dots,\psi_p$ is a basis of $\Gamma$, it's easy to see that $s_1',\dots,s_n',\psi_1,\dots,\psi_p$ is a linearly independent list of $V'$. Also, if you extend $s_1,\dots,s_n$ to a basis $s_1,\dots,s_n,v_1,\dots,v_m$ of $V$, then its dual $s_1',\dots,s_n',v_1',\dots,v_m'$ is a basis of $V'$; in fact $S^0=\text{span}\{v_1',\dots,v_m'\}$. Two remarkable facts:$$\forall i\in[1,m],\,\exists j\in[1,p],\,\psi_j(v_i)\neq0$$ because $v_j\notin S$, and $$\forall i\in[1,p],\,\exists j\in[1,m],\,\psi_i(v_j)\neq 0$$ because $\psi_i\neq 0$; in other words the matrix of the inclusion map from $\Gamma$ to $V'$ with respect to the basis of $\Gamma$ and the dual base of the chosen basis of $V$ has no $0$ row nor $0$ column. But this doesn't seem to provide a way to prove that any linear comination of the $(v_i')_{1\le i\le m}$ is a linear combination of the elements of the basis of $\Gamma$. I believe that $v_1,\dots, v_m$ should be chosen more carefuly but I fail to.
Could you please help me? Thank you in advance!
 A: For a subspace $T\subseteq V'$ define $T^0 := \{v\in V\,|\, \varphi(v) = 0\text{ for all $\varphi\in T$}\}$ (which is the same as your definition under the identification $V = V''$). In your case, we have $\Gamma^0 = S$.
Then we have $\dim V = \dim U + \dim U^0$ for each subspace $U\subseteq V$: Take a basis $u_1,\dotsc,u_d$ of $U$ and complete it to a basis $u_1,\dotsc,u_n$ of $V$. Denoting by $u'_1,\dotsc,u'_n$ the dual basis, it is easy to see that $U^0$ is spanned by $u'_{d+1},\dotsc,u'_n$, which proves the claim.
Likewise, we have $\dim V' = \dim T + \dim T^0$ for any subspace $T\subseteq V'$. More precisely, let $\varphi_1,\dotsc,\varphi_d$ be a basis of $T$ and complete it to a basis $\varphi_1,\dotsc,\varphi_n$ of $V'$. Then it follows directly from the definitions that $T^0 = \bigcap_{i=1}^d \ker\varphi_i$. It remains to check that $\dim \bigcap_{i=1}^d \ker \varphi_i = n-d$. Indeed, putting $W_j = \bigcap_{i=1}^j \ker\varphi_i$, for $j=1,\dotsc,d$, we obtain an increasing chain
$$
0 = W_n \subset W_{n-1} \subset\dotsb \subset W_1 \subset V.
$$
Since obviously $\dim W_j \ge \dim W_{j+1} \ge \dim W_j-1$, for all $j$, and $\dim W_1 = n -1$, we must in fact have $\dim W_j = n-j$. In particular,
$$
\dim T^0 = \dim \bigcap_{i=1}^d\ker\varphi_i = \dim W_d = n-d = \dim V' - \dim T.
$$
Now, it follows that $T^{00} = T$ for $T\subseteq V'$: $T\subseteq T^{00}$ is clear and both have the same dimension, because of
$$\dim T = \dim V' - \dim T^0 = \dim V - \dim T^0 = \dim T^{00}.$$
With this, we conclude $\Gamma = \Gamma^{00} = S^0$ (using $\Gamma^0 = S$ for the last equality).
A: I really didn't understand the user218931's solution. But, I think I solved this problem in another way:
Let $S = \{v \in V: \varphi(v)=0\text{ for every }\varphi\in\Gamma \}$ and let $F$ be the field. Let $\phi_1, \phi_2, \cdots, \phi_m$ be a basis for $\Gamma$.
By the definition of annihilator, $\Gamma \subset S^{0}$. Now, to prove that $ S^{0} \subset \Gamma$, first it is necessary prove that S is a subspace of V:


*

*$\varphi(0)=0 \Rightarrow 0 \in S$

*If $v_1 \in S$ and $v_2 \in S$, then for every $\varphi \in \Gamma$ we have $\varphi(v_1 + v_2) = \varphi(v_1) + \varphi(v_2) =0$. So, $v_1+v_2 \in S$.

*if $v \in S$ and $\lambda \in F$, then for every $\varphi \in \Gamma$ we have $\varphi(\lambda v)= \lambda\varphi(v)=0$. Thus, $\lambda v \in S$.


Therefore, $S$ is a subspace of $V$. Therefore we can use:
$$\dim S + \dim S^{0}= \dim V$$
Now, notice that the definition of $S$ is equivalent to $S = null \space \phi_1 \space \cap null \space \phi_2 \space \cap \cdots \cap \space null \space \phi_m$, because $v \in S$ if and only if $\phi_i(v)=0 $ for all $i \in \{1,2,..., m\}$ .  Besides, it is known that 
$$\dim(null \space \phi_1 \space \cap null \space \phi_2 \space \cap \cdots \cap \space null \space \phi_m) = \dim V - m $$
Thus, $\dim S^{0} = m =\dim \Gamma$. Since the set $\Gamma \subset S^0$, then $\Gamma= S^0$.
A: Suppose there is some $s'\in S^0,s'\notin \Gamma.$  Then there is some $x\in V$ with $s'(x)\neq0$ while $\varphi(x)=0$ for all $\varphi\in\Gamma$. But then $x\in S$, so $s'(x)=0$, a contradiction.  
The * hard * part is showing that there exists such an $x$.  The only way I can think of showing there is by using the Hahn-Banach theorem, but there is definitely a simpler way.  I'll edit once I can find one.
