Suppose $V$ is finite-dimensional and $\phi_1, \cdots, \phi_m \in V'$. Prove that the following three sets are equal to each other.
(a) $A = \text{span}(\phi_1, \cdots, \phi_m)$
(b) $B = \left(\bigcap_{i=1}^m \text{null}(\phi_i)\right)^0$
(c) $C = \{\phi \in V' ∶ \bigcap_{i=1}^m \text{null}(\phi_i) \subseteq \text{null}(\phi)\}$
Attempt:
Proving $A \subset B$
If $\phi \in \text{span}(\phi_1, \cdots, \phi_m)$, then for any arbitrary $x \in \bigcap_{i=1}^m \text{null}(\phi_i)$, we have
$\phi(x) = \sum_{i=1}^m c_i \phi_i (x)$ for some $c_i$. As $x \in \text{null}(\phi_i)~\forall 1 \le i \le m$, We have $\phi_i(x) = 0~ \forall 1 \le i \le m$ which gives us $\phi(x) = 0$. Thus $\phi$ annihilates $\bigcap_{i=1}^m \text{null}(\phi_i)$. Thus $\phi \in B$. Which gives us $A \subset B$.
Now lets say $\phi \in B$. Then we need to construct a linear combination of $\phi_i$ which agrees with $\phi$ on all basis elements of $V$. Here's where I'm lost. How to construct this linear combination. I can take a basis $v_1 , \cdots , v_k$ of $\bigcap_{i=1}^m \text{null}(\phi_i)$ and try to define $\phi$ as $\sum \phi_i$ and it would agree on the intersection of null space but I'm not sure how I get it to agree on the extended basis of $V$.