Let $V$ be a vector space and $f,g_1,g_2,\ldots,g_m\in V^*$ its dual, prove that $f$ is a linear combination of $g_i$'s iff $\cap \text{ker} g_i \subseteq \text{ker}f$
I already understand that if the intersection of $g_i$'s is empty, then it is a basis for the dual, and therefore $f$ is a linear combination of $g_i$'s. And I also managed to prove the foward direction (if $f$ is a linear combination of $g_i$, then $\cap \text{ker} g_i \subseteq \text{ker} f$, since that for any $v \in \cap ker g_i$ $f(v)=0=a_1g_1(v)=g_2(v)=\ldots=g_m(v)=0$). But I'm not finding a way to prove the reverse direction. Tried chatgpt but got only nonsense. Can someone give me a hint or post a duplicate (really didn't find it out here).
Thanks in advance.
In a comment it says we can assume $V$ is finite dimensional, so choose a basis $v_1, ... v_n$ of $V$, and $v_1^*, ... v_n^*$ is the dual basis. Now, both $V$ and $V^*$ are isomorphic to $F^n$, and the evaluation map $V^* \times V \rightarrow F$ is the dot product.
We now have a system of linear equations $f = \sum_{i=1}^m a_i g_i$ with $n$ equations and $m$ variables. Gaussian elimination either gives a solution to this equation (which shows that $f$ is a linear combination of the $g_i$), or outputs some vector whose dot product with $f$ is nonzero, but whose dot product with $g_i$ are all zero. This is an element in $\cap ker(g_i)$ but not in $ker(f)$.