spanning sets of dual space
Theorem.
Let X be a vector space and ϕ1,…,ϕn∈X* be
functionals
belonging to the dual space
.
A linear functional
f∈X* belongs to the linear span of
ϕ1,…,ϕn if and only if
kerf⊇⋂ni=1kerϕi.
ker refers to the kernel. Note that the domain X need not be finite-dimensional.
Proof.
The “only if” part is easy: if f=∑ni=1λiϕi for some scalars λi, and x∈X is such that ϕi(x)=0 for all i, then clearly f(x)=0 too.
The “if” part will be proved by induction on n.
Suppose kerf⊇kerϕ1.
If f=0, then the result is trivial.
Otherwise, there exists y∈X such that f(y)≠0.
By hypothesis, we also have ϕ1(y)≠0.
Every z∈X can be decomposed into z=x+ty
where x∈kerϕ1⊆kerf, and t is a scalar.
Indeed, just set t=ϕ1(z)/ϕ1(y), and x=z-ty.
Then we propose that
f(z)=f(y)ϕ1(y)ϕ1(z), for all z∈X. |
To check this equation, simply evaluate both sides using the decomposition z=x+ty.
Now suppose we have kerf⊇⋂ni=1kerϕi
for n>1.
Restrict each of the functionals
to the subspace W=kerϕn, so that
kerf|W⊇⋂n-1i=1kerϕi|W.
By the induction hypothesis, there exist scalars λ1,…,λn-1
such that f|W=∑n-1i=1λiϕi|W.
Then ker(f-∑n-1i=1λiϕi)⊇W=kerϕn, and the argument for the case n=1
can be applied anew, to obtain the final λn.
∎
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 |