spanning sets of dual space

Theorem.

Let $X$ be a vector space and ${\varphi }_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{\varphi }_n\mathrm{\in }{X}^{\mathrm{*}}$ befunctionals belonging to the dual space.A linear functional $f\mathrm{\in }{X}^{\mathrm{*}}$ belongs to the linear span of${\varphi }_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{\varphi }_n$ if and only if$\ker \mathit{}f\mathrm{\supseteq }{\mathrm{\bigcap }}_{i\mathrm{=}\mathrm{1}}^n\ker \mathit{}{\varphi }_i$.

$\ker$ refers to the kernel.Note that the domain $X$ need not be finite-dimensional.

Proof.

The “only if” part is easy: if $f={\sum }_{i=1}^n{\lambda }_i{\varphi }_i$for some scalars ${\lambda }_i$, and $x\in X$is such that ${\varphi }_i(x)=0$ for all $i$, then clearly $f(x)=0$ too.

The “if” part will be proved by induction on $n$.

Suppose $\ker f\supseteq \ker {\varphi }_1$.If $f=0$, then the result is trivial.Otherwise, there exists $y\in X$ such that $f(y)\ne 0$.By hypothesis, we also have ${\varphi }_1(y)\ne 0$.Every $z\in X$ can be decomposed into $z=x+ty$where $x\in \ker {\varphi }_1\subseteq \ker f$, and $t$ is a scalar.Indeed, just set $t={\varphi }_1(z)/{\varphi }_1(y)$, and $x=z-ty$.Then we propose that

$$f(z)=\frac{f(y)}{{\varphi }_1(y)}{\varphi }_1(z),\text{ for all }z\in X\text{.}$$

To check this equation, simply evaluate both sides using the decomposition$z=x+ty$.

Now suppose we have $\ker f\supseteq {\bigcap }_{i=1}^n\ker {\varphi }_i$for $n>1$.Restrict each of the functionalsto the subspace $W=\ker {\varphi }_n$, so that${\ker f|}_W\supseteq {{\bigcap }_{i=1}^{n-1}\ker {\varphi }_i|}_W$.By the induction hypothesis, there exist scalars ${\lambda }_1,\mathrm{\dots },{\lambda }_{n-1}$such that ${f|}_W={{\sum }_{i=1}^{n-1}{\lambda }_i{\varphi }_i|}_W$.Then $\ker (f-{\sum }_{i=1}^{n-1}{\lambda }_i{\varphi }_i)\supseteq W=\ker {\varphi }_n$, and the argument for the case $n=1$can be applied anew, to obtain the final ${\lambda }_n$.∎