Theorem   Let $X$ be a vector space and be functionals belonging to the dual space. A linear functional $f \in X^*$ belongs to the linear span of if and only if $\ker f \supseteq \bigcap_{i=1}^n \ker \phi_i$ .

Proof. The ``only if'' part is easy: if $f = \sum_{i=1}^n \lambda_i \phi_i$ for some scalars $\lambda_i$ , and $x \in X$ is such that $\phi_i(x) = 0$ for all $i$ , then clearly $f(x) = 0$ too.

The ``if'' part will be proved by induction on $n$ .

Suppose $\ker f \supseteq \ker \phi_1$ . If $f = 0$ , then the result is trivial. Otherwise, there exists $y \in X$ such that $f(y) \neq 0$ . By hypothesis, we also have $\phi_1(y) \neq 0$ . Every $z \in X$ can be decomposed into $z = x+ ty$ where $x \in \ker \phi_1 \subseteq \ker f$ , and $t$ is a scalar. Indeed, just set $t = \phi_1(z)/\phi_1(y)$ , and $x = z-ty$ . Then we propose that$$ f(z) = \frac{f(y)}{\phi_1(y)} \phi_1(z)\,, \text{ for all $z \in X$.}$$ To check this equation, simply evaluate both sides using the decomposition $z = x+ty$ .

Now suppose we have $\ker f \supseteq \bigcap_{i=1}^n \ker \phi_i$ for $n > 1$ . Restrict each of the functionals to the subspace $W = \ker \phi_n$ , so that $\ker f|_W \supseteq \bigcap_{i=1}^{n-1} \ker \phi_i|_W$ . By the induction hypothesis, there exist scalars such that $f|_W = \sum_{i=1}^{n-1} \lambda_i \phi_i|_W$ . Then $\ker ( f - \sum_{i=1}^{n-1} \lambda_i \phi_i ) \supseteq W = \ker \phi_n$ , and the argument for the case $n=1$ can be applied anew, to obtain the final $\lambda_n$ .