spanning sets of dual space


Theorem.

Let X be a vector spaceMathworldPlanetmath and ϕ1,…,ϕn∈X* be functionalsPlanetmathPlanetmathPlanetmath belonging to the dual spacePlanetmathPlanetmath. A linear functionalPlanetmathPlanetmath f∈X* belongs to the linear span of ϕ1,…,ϕn if and only if ker⁡f⊇⋂i=1nker⁡ϕi.

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

Proof.

The “only if” part is easy: if f=∑i=1nλi⁢ϕi 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 inductionMathworldPlanetmath on n.

Suppose ker⁡f⊇ker⁡ϕ1. If f=0, then the result is trivial. Otherwise, there exists y∈X such that f⁢(y)≠0. By hypothesisMathworldPlanetmathPlanetmath, we also have ϕ1⁢(y)≠0. Every z∈X can be decomposed into z=x+t⁢y where x∈ker⁡ϕ1⊆ker⁡f, and t is a scalar. Indeed, just set t=ϕ1⁢(z)/ϕ1⁢(y), and x=z-t⁢y. 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+t⁢y.

Now suppose we have ker⁡f⊇⋂i=1nker⁡ϕi for n>1. Restrict each of the functionals to the subspacePlanetmathPlanetmath W=ker⁡ϕn, so that ker⁡f|W⊇⋂i=1n-1ker⁡ϕi|W. By the induction hypothesis, there exist scalars λ1,…,λn-1 such that f|W=∑i=1n-1λi⁢ϕi|W. Then ker⁡(f-∑i=1n-1λi⁢ϕi)⊇W=ker⁡ϕn, 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 TheoremMathworldPlanetmath
Classification msc 15A99