TITLE: Representation of a linear functional in vector space
QUESTION [11 upvotes]: In the book Functional Analysis, Sobolev Spaces and Partial Differential Equations
of Haim Brezis we have the following lemma:
Lemma. Let $X$ be a vector space and let
$\varphi, \varphi_1, \varphi_2, \ldots, \varphi_k$ be $(k + 1)$ linear functionals on $X$ such that
$$
[\varphi_i(v) = 0 \quad \forall\; i \in \{1, 2, \ldots , k\}] \Rightarrow [\varphi(v) = 0].
$$
Then there exist constants $\lambda_1, \lambda_2, \ldots, \lambda_k\in\mathbb{R}$ such that $\varphi=\lambda_1\varphi_1+\lambda_2\varphi_2+\ldots+\lambda_k\varphi_k$.
In this book, the author used separation theorem to prove this lemma. I would like ask whether we can use only knowledge of linear algebra to prove this lemma.
Thank you for all helping.

REPLY [2 votes]: Here is a repackaging of linearalgebraist's answer:
Let 
$$
L: X\to\mathbb{R}^k\\
Lx:=(\phi_1(x),\dots, \phi_k(x))
$$
Then
$$
{L}^*: (\mathbb{R}^k)^* \to X^* \\
{L}^*f:=f\circ {L}= f_1\phi_1+\dots+f_k\phi_k ,
$$
so the conclusion you seek is equivalent to the general algebraic fact that 
$$
\text{im} L^* = (\ker L)^{\bot}:=\{\phi \in X^* : \phi (x) = 0\; \forall x \in \ker L\}
$$
More intuitively,
$L$ is injective on $X/\ker L$, which means that knowing $Lx$ amounts to knowing $x$ up to an element of $\ker L$. Hence, knowing $Lx$ implies knowing  $\phi(x)$, because the element of $\ker L$ doesn't affect that. Clearly, all of this "knowing things" is linear, so you can write $\phi(x)$ as a linear combination of $\phi_1(x),\dots,\phi_k(x)$.