About two systems of linear equations defining the same linear subspace Assume that we have in a linear space $X$ a linear subspace $V$ over $F$ defined system of linear equations :
$$
f_1=0, \\
f_2=0, \\
\,\vdots \\
f_k=0,
$$
where linear functionals $f_1,\ldots,f_k \in X^*$ and are linearly independent.
Let $g\in X^*$ be such that system
$$
f_1=0, \\
f_2=0, \\
\,\vdots \\
f_k=0, \\
g=0,
$$
defines the same subspace $V$. 
How to prove that $f_1,\ldots,f_k,g$ are linearly dependent?
 A: The essential result here is that if a matrix $A$ has a trivial null space then the range of $A^T$ is the entire space. The problem is reduced to a
finite dimensional one using quotient space.
Define $F:X \to \mathbb{F}^k$ by $F(x) = (f_1(x),...,f_k(x))$. The hypotheses show that $V=\ker F \subset \ker g$.
Define $\overline{F}: {X \over V} \to \mathbb{F}^k$ in the usual way, that is
$\overline{F}([x]) = F(x)$. Note that $\overline{F}$ is injective.
Note that $p=\dim {X \over V} \le k$.
If $x_1-x_2 \in V$, we have $g(x_1) = g(x_2)$, hence we can unambiguously
define $\overline{g} :{X \over V} \to \mathbb{F}$ by $\overline{g}([x]) = g(x)$.
Let $[x_1],...,[x_p]$ form a basis for ${X \over V}$, and let
$\Phi, \gamma$ be the representations of $\overline{F},\overline{g}$ using this basis for the domain. Since $\ker \Phi = \{0\}$, we have ${\cal R} \Phi^T = \mathbb{F}^p$, hence $\gamma^T = \Phi^T \lambda$ for some
$\lambda \in \mathbb{F}^k$.
In particular, we have $\gamma = \lambda^T \Phi$ and so we have
$g = \sum_{i=1}^k \lambda_i f_i$.
(Note that linear independence of the $f_i$ is not needed.)
A: Your statement can be formulated as follows:
Let $g, f_1,...,f_k \in X^*$. If $$Ker f_1 \cap...\cap Ker f_k=Ker f_1 \cap...\cap Ker f_k \cap Ker g$$ then $g\in Span_F \{f_1,...,f_k \}$, that is there exist $\alpha_1,...,\alpha_k \in F$ such that $g=\alpha_1f_1+...+\alpha_k f_k$.
I used a similar idea as copper.hat. Let $F: X \rightarrow F^k$, $F(x)=(f_1(x),...,f_k(x))$. Since $Ker F \subset ker g$ the following linear mapping $g: Im F \rightarrow F$, $h(F(x))=g(x)$ for $x\in X$ is well defined.
We next extend $h$ to a linear mapping $H: F^k\rightarrow F$. Then for some scalars $a_1,...,a_k\in F$ we have $H(x_1,...,x_k)=\alpha_1x_1+...+\alpha_k x_k$.
Hence we obtain
$$
\alpha_1f_1+..+\alpha_kf_k=g,
$$
which proves in particular that $f_1,...,f_k,g$ are linearly dependent.
More generally: 
 Let $g, f_1,...,f_k \in X^*$ and $c,c_1,...,c_k \in F$.
If $$f_1^{-1}(c_1) \cap...\cap  f_k^{-1}(c_k)= f_1^{-1}(c_1) \cap...\cap f_k^{-1}(c_k) \cap  g^{-1}(c) \neq \emptyset$$ then there exist $\alpha_1,...,\alpha_k \in F$ such that $g=\alpha_1 f_1+...+\alpha_k f_k$ and $c=\alpha_1 c_1+...+\alpha_kc_k$.
Indeed, let $x_0 \in f_1^{-1}(c_1) \cap...\cap  f_k^{-1}(c_k)= f_1^{-1}(c_1) \cap...\cap  f_k^{-1}(c_k) \cap  g^{-1}(c)$.
Then 
$$
x_0+Ker f_1 \cap...\cap Ker f_k=x_0+Ker f_1 \cap...\cap Ker f_k \cap Ker g.
$$
Hence $Ker f_1 \cap...\cap Ker f_k=Ker f_1 \cap...\cap Ker f_k \cap Ker g$.
By the first part there are $\alpha_1,...,\alpha_k \in F$ such that $g=\alpha_1 f_1+...+\alpha_k f_k$. In particular $c=g(x_0)=\alpha_1 f_1(x_0)+...+\alpha_k f_k(x_0)= \alpha_1c_1+...+\alpha_kc_k$.
