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 |
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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 |