properties of spanning sets


Let V be a vector spaceMathworldPlanetmath over a field k. Let S be a subset of V. We denote Sp(S) the span of the set S. Below are some basic properties of spanning sets.

  1. 1.

    If ST, then Sp(S)Sp(T). In particular, if Sp(S)=V, every supersetMathworldPlanetmath of S spans (generates) V.

    Proof.

    If vSp(S), then v=r1v1++rnvn for viS. But viT by assumptionPlanetmathPlanetmath. So vSp(T) as well. If Sp(S)=V, and ST, then V=Sp(S)Sp(T)V. ∎

  2. 2.

    If S contains 0, then Sp(S-{0})=Sp(S).

    Proof.

    Let T=S-{0}. So Sp(T)Sp(S) by 1 above. If vSp(S), then v=r1v1++rnvn. If one of the vi’s, say vi, is 0, then v=r2v2++rnvnSp(T). ∎

  3. 3.

    It is not true that if S1S2 is a chain of subsets, each spanning the same subspacePlanetmathPlanetmathPlanetmath W of V, so does their intersectionMathworldPlanetmath.

    Proof.

    Take V=n, the Euclidean space in n dimensionsPlanetmathPlanetmath. For each i=1,2,, let Si be the closed ball centered at the origin, with radius 1/i. Then Sp(Si)=V. But the intersection of these Si’s is just the origin, whose span is itself, not V. ∎

  4. 4.

    S is a basis for V iff S is a minimalPlanetmathPlanetmath spanning set of V. Here, minimal means that any deletion of an element of S is no longer a spanning set of V.

    Proof.

    If S is a basis for V, then S spans V and S is linearly independentMathworldPlanetmath. Let T be the set obtained from S with vS deleted. If T spans V, then v can be written as a linear combinationMathworldPlanetmath of elements in T. But then S=T{v} would no longer be linearly independent, contradictionMathworldPlanetmathPlanetmath the assumption. Therefore, S is minimal.

    Conversely, suppose S is a minimal spanning set for V. Furthermore, suppose that S is linearly dependent. Let 0=r1v1+rnvn, with r10. Then

    v1=s2v2++snvn, (1)

    where si=-ri/r1. So any linear combination of elements in S involving v1 can be replaced by a linear combination not involving v1 through equation (1). Therefore Sp(S)=Sp(S-{v}). But this means that S is not minimal, contrary to our assumption. Therefore, S must be linearly independent. ∎

Remark. All of the properties above can be generalized to modules over rings, except the last one, where the implicationMathworldPlanetmath is only one-sided: basis implying minimal spanning set.

Title properties of spanning sets
Canonical name PropertiesOfSpanningSets
Date of creation 2013-03-22 18:05:40
Last modified on 2013-03-22 18:05:40
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 7
Author CWoo (3771)
Entry type Result
Classification msc 15A03
Classification msc 16D10