spanning sets of dual space
Theorem.
Let be a vector space and be
functionals
belonging to the dual space
.
A linear functional
belongs to the linear span of
if and only if
.
refers to the kernel. Note that the domain need not be finite-dimensional.
Proof.
The “only if” part is easy: if for some scalars , and is such that for all , then clearly too.
The “if” part will be proved by induction on .
Suppose .
If , then the result is trivial.
Otherwise, there exists such that .
By hypothesis, we also have .
Every can be decomposed into
where , and is a scalar.
Indeed, just set , and .
Then we propose that
To check this equation, simply evaluate both sides using the decomposition .
Now suppose we have
for .
Restrict each of the functionals
to the subspace , so that
.
By the induction hypothesis, there exist scalars
such that .
Then , and the argument for the case
can be applied anew, to obtain the final .
∎
Title | spanning sets of dual space |
---|---|
Canonical name | SpanningSetsOfDualSpace |
Date of creation | 2013-03-22 17:17:28 |
Last modified on | 2013-03-22 17:17:28 |
Owner | stevecheng (10074) |
Last modified by | stevecheng (10074) |
Numerical id | 6 |
Author | stevecheng (10074) |
Entry type | Theorem![]() |
Classification | msc 15A99 |