annihilator is the intersection of sets If $W$ is a subspace of a finite dimensional vector space $V$ and $\{g_{1},g_{2},\cdots, g_{r}\}$ is a basis of the annihilator $W^{\circ}=\{f \in V^{\ast}| f(a)=0, \forall a \in W\}$, then $W=\cap_{i=1}^{r} N_{g_{i}}$, where for $f \in V^{\ast}$, $N_{f}=\{a \in V| f(a)=0\}$ 
How shall I prove this?
 A: A hint:
To show $W\subset\bigcap^r_{i=1}N_{g_i}$, you just need to unpack definitions.
For the other inclusion, you should use the definition of the linear structure on $V^*$ to show that if $g_i(a)=0$ for all $i$, then in fact $f(a)=0$ for all $f\in W^\circ$. Then you either need to use (if you know it) or prove (if you don't) that for any $v\in V$, if $f(v)=0$ for all $f\in W^\circ$, then $v\in W$.
A: We wish to prove that 
$$W = \bigcap_{i=1}^{r} N_{g_{i}}$$

Step $1$: Proving $W \subset \bigcap_{i=1}^{r} N_{g_{i}}$
Let $w \in W$. We know that the annihilator $W^{o}$ is the set of linear functionals that vanish on $W$. If $g_{i}$ is in the basis for $W^{o}$, it is certainly in $W^{o}$. Thus, the $g_{i}$ all vanish on $W$, so that $W \subset N_{g_i}$. We thus see that $w \in N_{g_{i}}$ for all $1 \leq i \leq r$. Hence 
$$w \in \bigcap_{i=1}^{r} N_{g_{i}}$$
But $w$ was arbitrary, so 
$$W \subset \bigcap_{i=1}^{r} N_{g_{i}}$$

Step $2$: Proving $W \supset \bigcap_{i=1}^{r} N_{g_{i}}$
Let $\{\alpha_{1}, \cdots, \alpha_{s}\}$ be a basis for $W$, and extend it to a basis for $V$, $\{ \alpha_{1} ,\cdots, \alpha_{n}\}$. Likewise, extend $\{g_{1}, \cdots, g_{r}\}$ to a basis for $V^{*}$, $\{g_{1}, \cdots, g_{n}\}$, noting that $\mbox{dim}(V) = \mbox{dim} (V^{*})$. It is easily seen that one can choose these two bases to be dual to each other, as the dual basis to $\{\alpha_{1}, \cdots, \alpha_{s}\}$ is not contained in $W^{o}$, and the dual basis for $V \setminus W$ must be contained in $W^{o}$, just by the nature of the dual basis. 
(To make it easier to choose the correct basis for $V \setminus W$, try looking at the double dual $V^{**}$, which is naturally isomorphic to $V$, and looking at the dual basis for $W^{o}$ in $V^{**}$)
We get
$$ g_{i}(\alpha_{j}) = \left\{ \begin{array}{cc}
1 & i= j\\
0 & i \neq j
\end{array}
 \right.$$
Let $v \in \bigcap_{i=1}^{r} N_{g_{i}}$, so that $g_{i}(v) = 0$ for all $1 \leq i \leq r$, and write
$$v = c_{1}\alpha_{1} + \cdots + \cdots c_{n}\alpha_{n}$$
Taking $g_{i}$ of both sides, where $i$ ranges from $1$ through $r$,
$$g_{i}(v) = c_{1}g_{i}(\alpha_{1}) + \cdots + c_{n}g_{i}(\alpha_{n}) = 0$$
However, these $\{ g_{i}: 1 \leq i \leq r\}$ form a dual basis to $V \setminus W$. Hence, if $v$ had a nonzero component $c_{k}\alpha_{k}$ in $V \setminus W$, it would not vanish on $g_{k}$, $1 \leq k \leq r$. Then $g_{k}(r) \neq 0$, contrary to what we have shown. Thus $v$ must be contained in $W$. 
