IMC 1998 · Day 2 · P7

We use induction on $k$. By passing to a subset, we may assume that $f_1, \dots, f_k$ are linearly independent.

Since $f_k$ is independent of $f_1, \dots, f_{k-1}$, by induction there exists a vector $a_k \in V$ such that $f_1(a_k) = \dots = f_{k-1}(a_k) = 0$ and $f_k(a_k) \ne 0$. After normalising, we may assume that $f_k(a_k) = 1$. The vectors $a_1, \dots, a_{k-1}$ are defined similarly to get $$f_i(a_j) = \begin{cases} 1 & \text{if } i = j \\ 0 & \text{if } i \ne j. \end{cases}$$ For an arbitrary $x \in V$ and $1 \le i \le k$, $f_i(x - f_1(x) a_1 - f_2(x) a_2 - \dots - f_k(x) a_k) = f_i(x) - \sum\limits_{j=1}^{k} f_j(x) f_i(a_j) = f_i(x) - f_i(x) f_i(a_i) = 0$, thus $f(x - f_1(x) a_1 - \dots - f_k(x) a_k) = 0$. By the linearity of $f$ this implies $f(x) = f_1(x) f(a_1) + \dots + f_k(x) f(a_k)$, which gives $f(x)$ as a linear combination of $f_1(x), \dots, f_k(x)$.