Prove that $f$ is a linear combination of $f_1,f_2,\dots,f_n$. Let $V$ be a vector space and let $f, f_1,f_2,\dots,f_n$ be linear maps from $V$ to $\mathbb{R}$. Suppose that $f(x)=0$ whenever $f_1(x)=f_2(x)=\cdots=f_n(x)=0$. Prove that $f$ is a linear combination of $f_1,f_2,\ldots,f_n$.
The solution can be found here (first problem).. 
but I disagree that $a_k$ is guaranteed to exist. What if the set containing all vectors $u\in V$ such that $f_1(u)=f_2(u)=\cdots =f_{k-1}(u)$ is empty?

You can do the following. Let $V$ be an $K$-vector space ($K=\Bbb R$) of dimension say $m$. Let $V^*$ be its dual space. Now take $\varphi_1,\ldots,\varphi_n$ to  be linear forms in $V^*$, and let $\varphi$ be another form in $V^*$. You're assuming that $$\tag 1 \bigcap_{i=1}^n \ker\varphi_i\subseteq \ker \varphi$$
We may assume that the $n$ forms $\varphi_i$ are linearly independent. Extend them to a basis $$\{\varphi_1,\ldots,\varphi_{n},\varphi_{n+1},\ldots,\varphi_m\}$$
and let $\{v_1,\ldots,v_n,v_{n+1},\ldots,v_m\}$ be its predual basis in $V$. Now write $$\varphi=\alpha_1\varphi_1+\cdots+\alpha_n\varphi_n+\alpha_{n+1}\varphi_{n+1}+\cdots+\alpha_m\varphi_m$$
By definition we have that $\varphi_j(v_i)=\delta_{ij}$. Use $(1)$ to show that $\alpha_{n+1}=\cdots=\alpha_m=0$.
