About the dimension of intersection of null spaces of linear functions In Page 115 of Axler's book Linear Algebra Done Right (third edition) there is an exercise (numbered 30). See blow:

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

I have tried for a long time, but failed. In the case of $m=1,$ the result is trivial, just a consequence of the Fundamental Theorem of Linear Maps. But in the case  of $m=2,$ I have tried as follows. Since $\phi_1, \phi_2$ is linearly independent, $$\dim\text{null}(\phi_1)=\dim\text{null}(\phi_2)=\dim V-1.$$ Thus 
\begin{align*}
&\dim \big(\text{null}(\phi_1)\cap \text{null}(\phi_2)\big)\\
=&\dim(\text{null}(\phi_1))+\dim(\text{null}(\phi_2))-\dim\big(\text{null}(\phi_1)+\text{null}(\phi_2)\big)\\
=&\dim V-1+\dim V-1-\dim\big(\text{null}(\phi_1)+\text{null}(\phi_2)\big)\\
\geq &\dim V-1+\dim V-1-\dim V\\
=&\dim -2.
\end{align*}
But how to show the reversed inequality, that is,
$\dim \big(\text{null}(\phi_1)\cap \text{null}(\phi_2)\big)\leq \dim -2?$
Moreover, it seems that the method is hardly valid for $m\geq 3.$
Can anyone help me?
 A: Proof.  $\quad$ Consider the mapping
\begin{gather*}
 f\colon V\to\mathbb{F}^m,
\end{gather*}
such that
\begin{gather*}
f(v)=\big(\phi_1(v),\dots,\phi_m(v)\big)
\end{gather*}
for all $v\in V.$  It is clear that $ \text{null} f=\text{null}(\phi_1)\cap\cdots \text{null}(\phi_m).$  Indeed, let $v\in  \text{null}f.$ Then $f(v)=0,$ which implies that 
\begin{gather*}
 \big(\phi_1(v),\dots,\phi_m(v)\big)=0,
\end{gather*}
and thus $\phi_j(v)=0$ for $j=1,\dots,m.$ So we have $v\in \cap_{j=1}^m \text{null}(\phi_j).$ Hence we have proved $ \text{null} f\subset \cap_{j=1}^n  \text{null} \phi_j.$  In the other direction of inclusion, suppose $v\in \cap_{j=1}^m \text{null}(\phi_j).$ Then $\phi_j(v)=0$ for all $j=1,\dots, m.$ Thus $f(v)=0,$ which implies that $v\in \text{null} f.$ Hence $\cap_{j=1}^m \text{null}(\phi_j)\subset  \text{null} f.$ Therefore $\cap_{j=1}^m \text{null}(\phi_j)= \text{null} f.$ 
We then show that $\dim \text{range} f=m.$  Let  $v_1,\dots, v_n$ be a basis of $V$ and $e_1,\dots, e_m$ be the standard basis of $\mathbb{F}^m.$ 
Then
\begin{gather*}
 f(v_k)= \big(\phi_1(v_k),\dots, \phi_m(v_k)\big) 
 = \sum_{j=1}^{m}\phi_j(v_k)e_j.
\end{gather*}
Thus the matrix of $f$ with respect to the given bases is $A:=\big(\phi_j(v_k)\big)_{m\times n}.$ Therefore, by 3.117 and 3.118, 
\begin{gather*}
 \dim \text{range} f=\text{column}\text{ rank }{\mathcal{M}(f)}=\text{row rank } \mathcal{M}(f).
\end{gather*}
We claim that the list of rows of $A$ is linearly independent. Indeed, let $a_1,\dots, a_m\in\mathbb{F}.$ Suppose 
$a_1A_{1,\cdot}+\cdots+a_mA_{m,\cdot}=0.$ That is, 
\begin{gather*}
 a_1\big(\phi_1(v_1),\dots, \phi_1(v_n)\big)+\cdots+a_m\big(\phi_m(v_1),\dots,\phi_m(v_n)\big)=0.
\end{gather*}
We have 
\begin{gather*}
 \big(a_1\phi_1(v_1)+\cdots+a_m\phi_m(v_1),\dots, a_1\phi_1(v_n)+\cdots+a_m\phi_m(v_n)\big)=0.
\end{gather*}
Thus 
\begin{gather*}
 a_1\phi_1(v_j)+\cdots+a_m\phi_m(v_j)=0,\qquad \text{for all $j=1,\dots, n.$}
\end{gather*}
Particularly, 
\begin{gather*}
 a_1\phi_1(v_1)+\cdots+a_m\phi_m(v_1)=0.\tag{1}
\end{gather*}
Because $\phi_1,\dots, \phi_m$ is linearly independent, from (1) it follows that  $a_1=\cdots=a_m=0.$ Thus 
the list of rows of $A$ is linearly independent. As a result, we have the row rank of $A$ is $m.$ Thus $\dim \text{range} f=m.$ Finally, by the Fundamental Theorem of Linear Maps, we have 
\begin{align*} 
 &\dim\big( \text{null}(\phi_1)\cap\cdots\cap \text{null}(\phi_m)\big)=\dim \text{null} f\\
 =&\dim V-\dim \text{range} f\\
 =&\dim V-m.
\end{align*}
A: Let $W = \text{null}\:\varphi_{1}\cap\dots\cap \text{nulll}\:\varphi_{m}$. We need to prove that $\text{dim}\:W^0 = m$. It is clear that $\{\varphi_{1},\dots,\varphi_{m}\}$ is a linearly independent subset of $W^{0}$, so $\text{dim}\:W^{0}\geq m$. It can be shown that for any subspaces $U_1$ and $U_2$ of a finite-dimensional vector space $V$ one has $(U_1\cap U_2)^{0} = U_1^{0}+U_2^{0}$. Therefore,
\begin{align}
\text{dim}\:W^0 &= \text{dim}\:(\text{null}\:\varphi_{1}^0+\dots+\text{null}\:\varphi_{m}^{0})\\
&\leq \text{dim}\:\text{null}\:\varphi_{1}^{0}+\text{dim}\:\text{null}\:\varphi_{m}^{0}\\
&= 1+\dots+1\\
&= m.
\end{align}
