spanning sets of dual space


Let X be a vector spaceMathworldPlanetmath and ϕ1,,ϕnX* be functionalsPlanetmathPlanetmathPlanetmath belonging to the dual spacePlanetmathPlanetmath. A linear functionalPlanetmathPlanetmath fX* belongs to the linear span of ϕ1,,ϕn if and only if kerfi=1nkerϕi.

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


The “only if” part is easy: if f=i=1nλiϕi for some scalars λi, and xX 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 kerfkerϕ1. If f=0, then the result is trivial. Otherwise, there exists yX such that f(y)0. By hypothesisMathworldPlanetmathPlanetmath, we also have ϕ1(y)0. Every zX can be decomposed into z=x+ty where xkerϕ1kerf, 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 zX.

To check this equation, simply evaluate both sides using the decomposition z=x+ty.

Now suppose we have kerfi=1nkerϕi for n>1. Restrict each of the functionals to the subspacePlanetmathPlanetmath W=kerϕn, so that kerf|Wi=1n-1kerϕi|W. By the induction hypothesis, there exist scalars λ1,,λn-1 such that f|W=i=1n-1λiϕi|W. Then ker(f-i=1n-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 TheoremMathworldPlanetmath
Classification msc 15A99