properties of spanning sets

Let $V$ be a vector space over a field $k$ . Let $S$ be a subset of $V$ . We denote $\operatorname{Sp}(S)$ the span of the set $S$ . Below are some basic properties of spanning sets.

1.

If $S\subseteq T$ , then $\operatorname{Sp}(S)\subseteq\operatorname{Sp}(T)$ . In particular, if $\operatorname{Sp}(S)=V$ , every superset of $S$ spans (generates) $V$ .

Proof.

If $v\in\operatorname{Sp}(S)$ , then $v=r_{1}v_{1}+\cdots+r_{n}v_{n}$ for $v_{i}\in S$ . But $v_{i}\in T$ by assumption. So $v\in\operatorname{Sp}(T)$ as well. If $\operatorname{Sp}(S)=V$ , and $S\subseteq T$ , then $V=\operatorname{Sp}(S)\subseteq\operatorname{Sp}(T)\subseteq V$ . ∎
2.

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

Proof.

Let $T=S-\{0\}$ . So $\operatorname{Sp}(T)\subseteq\operatorname{Sp}(S)$ by 1 above. If $v\in\operatorname{Sp}(S)$ , then $v=r_{1}v_{1}+\cdots+r_{n}v_{n}$ . If one of the $v_{i}$ ’s, say $v_{i}$ , is $0$ , then $v=r_{2}v_{2}+\cdots+r_{n}v_{n}\in\operatorname{Sp}(T)$ . ∎
3.

It is not true that if $S_{1}\supseteq S_{2}\supseteq\cdots$ is a chain of subsets, each spanning the same subspace $W$ of $V$ , so does their intersection.

Proof.

Take $V=\mathbb{R}^{n}$ , the Euclidean space in $n$ dimensions. For each $i=1,2,\ldots$ , let $S_{i}$ be the closed ball centered at the origin, with radius $1/i$ . Then $\operatorname{Sp}(S_{i})=V$ . But the intersection of these $S_{i}$ ’s is just the origin, whose span is itself, not $V$ . ∎

$S$ is a basis for $V$ iff $S$ is a minimal 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 independent. Let $T$ be the set obtained from $S$ with $v\in S$ deleted. If $T$ spans $V$ , then $v$ can be written as a linear combination of elements in $T$ . But then $S=T\cup\{v\}$ would no longer be linearly independent, contradiction the assumption. Therefore, $S$ is minimal.

Conversely, suppose $S$ is a minimal spanning set for $V$ . Furthermore, suppose that $S$ is linearly dependent. Let $0=r_{1}v_{1}+\cdots r_{n}v_{n}$ , with $r_{1}\neq 0$ . Then

$v_{1}=s_{2}v_{2}+\cdots+s_{n}v_{n},$

(1)

where $s_{i}=-r_{i}/r_{1}$ . So any linear combination of elements in $S$ involving $v_{1}$ can be replaced by a linear combination not involving $v_{1}$ through equation (1). Therefore $\operatorname{Sp}(S)=\operatorname{Sp}(S-\{v\})$ . 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 implication 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