# spanning sets of dual space

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

###### Proof.

The “only if” part is easy: if $f=\sum_{i=1}^{n}\lambda_{i}\phi_{i}$ 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.

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 $\lambda_{1},\ldots,\lambda_{n-1}$ 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 $\lambda_{n}$. ∎

Title spanning sets of dual space SpanningSetsOfDualSpace 2013-03-22 17:17:28 2013-03-22 17:17:28 stevecheng (10074) stevecheng (10074) 6 stevecheng (10074) Theorem  msc 15A99