$\phi_1, \dots , \phi_m$ a basis of $\left(\left(\operatorname{null}\phi_1 \right)\cap \cdots \cap \left(\operatorname{null}\phi_m\right)\right)^0$ The following is a problem from Axler's Linear Algebra Done Right. 

Suppose $V$ is a finite dimensional vector space and $\phi_1,\dots,\phi_m$ is a linearly independent list in $V^{'}.$ Prove that $$\operatorname{dim} \left(\left(\operatorname{null}\phi_1 \right)\cap \cdots \cap \left(\operatorname{null}\phi_m\right)\right)=(\operatorname{dim} V)-m$$

If we denote $U=\left(\operatorname{null}\phi_1 \right)\cap \cdots \cap \left(\operatorname{null}\phi_m\right)$, we know that $\operatorname{dim} U+ \operatorname{dim} U^0=\operatorname{dim} V$. Hence, we need only show that $\operatorname{dim} U^0=m$. Since the list $\phi_1,\dots,\phi_m \in U^0$ and is linearly independent, if $\operatorname{dim} U^0=m$, then $\phi_1,\dots,\phi_m$ must be a basis of $\left(\operatorname{null}\phi_1 \right)\cap \cdots \cap \left(\operatorname{null}\phi_m\right)$ if the statement that we are asked to prove is true. However, I have been unable to prove this statement. Does anyone have any hints as to how to go about proving this? Or am I going in the completely wrong direction to prove this? Thanks in advance.
 A: You wish to prove that $B = \{\phi_1, \ldots, \phi_m\}$ is a basis for $U^o$. Since $B$ is linearly independent, there exists a basis $B' = \{\phi_1, \ldots, \phi_{m}, \phi_{m+1}, \ldots, \phi_n\}$ for $V^*$ which contains $B$.
Find a basis $\{x_1, \ldots, x_n\}$ for $V$ such that $B'$ is its dual basis (do you know why such exists?). Now notice that for $k \in \{m+1, \ldots, n\}$ we have $x_k \in U$ because $\phi_i(x_k) = 0$, for every $i \in \{1, \ldots, m\}$.
Take $f \in U^o$. Then $f = \sum_{j = 1}^n \alpha_j\phi_j$, for some $\alpha_1, \ldots, \alpha_n \in \mathbb{F}$. Now for $k \in \{m+1, \ldots, n\}$ apply $f$ to $x_k$:
$$0 = f(x_k) = \sum_{j = 1}^n \alpha_j\phi_j(x_k) = \sum_{j = 1}^n \alpha_j\delta_{jk} = \alpha_k$$
We have $\alpha_k = 0$, $\forall k \in \{m+1, \ldots, n\}$ so $f = \sum_{j = 1}^m \alpha_j\phi_j$. In other words, $B$ spans $U^o$.
Since $B$ is linearly independent, it is a basis for $U^o$.
The proof of existence of the basis $\{x_1, \ldots, x_n\}$:
You can use the canonical isomorphism $F : V \to V^{**}$ which maps $x\in V$ to $\hat{x}\in V^{**}$ where $\hat{x}$ is a functional on $V^*$ defined as $\hat{x}(f) = f(x), \forall f\in V^*$.
Now let $\{{\phi_1}^*, \ldots, {\phi_n}^*\}\subseteq V^{**}$ be the dual basis of $\{\phi_1, \ldots, \phi_n\}$. Consider the set $S = \{x_1, \ldots, x_n\}\subseteq V$ where $x_j = F^{-1}({\phi_j}^*)$, or equivalently $\hat{x_j}={\phi_j}^*$. Since $S$ is actually the image of a linearly independent set $\{{\phi_1}^*, \ldots, {\phi_n}^*\}$ under the isomorphism $F^{-1}$, we get that $S$ is also linearly independent. Its cardinality equals $\dim V$ so $S$ is a basis for $V$. In fact, $\{\phi_1, \ldots, \phi_n\}$ is the dual basis of $S$:
$$\phi_i(x_j) = \hat{x_j}(\phi_i) = {\phi_j}^*(\phi_i) = \delta_{ij}, \quad\forall i, j \in \{1, \ldots, n\}$$ 
