About equivalences of linear functionals on a real vector space $X$ Suppose we have a real vector space $X$ and $\varphi_1,\varphi_2,...,\varphi_n$ and $\psi$ linear functionals from $X$ to $\mathbb R$. I'm struggling with the following implication:
$$ "ker(\psi) \supset \cap_{1\le k \le n}ker(\varphi_k)" \implies " \exists \lambda_1,\lambda_2,...,\lambda_n \in \mathbb R: \psi=\sum_{1 \le k \le n}\lambda_k \varphi_k" $$
What I've tried so far was passing to the quotient $X/\cap_{1 \le k \le n}ker(\varphi_k)$ and then defining the unique linear functional $$\varphi: X/\cap_{1\le k \le n}ker(\varphi_k) \to \mathbb R, \space [x] \mapsto \psi (x)$$
I don't know how to proceed and to be honest I don't even know if this is the right approach but the kernel intersection immidiately triggered the idea to pass to the quotient.
Any help?
 A: Let $X$ be an $\mathbf{R}$-vector-space and $\phi_1, ..., \phi_n, \psi : X \rightarrow \mathbf{R}$ non-zero linear functionals. Assume $\psi \notin \langle \phi_1, ..., \phi_n \rangle$. We may assume w.l.o.g. that $\phi_1, ..., \phi_n$ are linearly independent, and hence $\psi, \phi_1, ..., \phi_n$ are linearly independent. We prove the following: 
Proposition: Let $f_1, ..., f_n$ be non-zero functionals on a vector-space $X$. If they are linearly independent, then $\ker f_n \nsupseteq \bigcap_{i = 1}^{n - 1} \ker f_i$. 
Proof: By induction. 
For $n = 1$ there's nothing to show. 
Assume the claim holds for $n \geq 1$. Let $x_1, ..., x_n \in X$ be such that $x_i \in \bigcap_{j \neq i} \ker f_j$ and $f_i(x_i) = 1$. Then for all $x \in X$ we have $$x - \sum_{i = 1}^n f_i(x) x_i \in \bigcap_{i = 1}^n \ker f_i$$
If $\ker f_{n + 1} \supset \bigcap_{i = 1}^n \ker f_i$, then 
$$f_{n + 1} \left(x - \sum_{i = 1}^n f_i(x) x_i \right) = 0$$
and so $f_{n + 1}(x) = \sum_{i = 1}^n f_{n + 1}(x_i) f_i(x)$, hence $f_{n + 1} \in \langle f_1, ..., f_n \rangle$, which is a contradiction. $\square$
