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

1.
If $S\subseteq T$, then $\mathrm{Sp}(S)\subseteq \mathrm{Sp}(T)$. In particular, if $\mathrm{Sp}(S)=V$, every superset^{} of $S$ spans (generates) $V$.
Proof.
If $v\in \mathrm{Sp}(S)$, then $v={r}_{1}{v}_{1}+\mathrm{\cdots}+{r}_{n}{v}_{n}$ for ${v}_{i}\in S$. But ${v}_{i}\in T$ by assumption^{}. So $v\in \mathrm{Sp}(T)$ as well. If $\mathrm{Sp}(S)=V$, and $S\subseteq T$, then $V=\mathrm{Sp}(S)\subseteq \mathrm{Sp}(T)\subseteq V$. ∎

2.
If $S$ contains $0$, then $\mathrm{Sp}(S\{0\})=\mathrm{Sp}(S)$.
Proof.
Let $T=S\{0\}$. So $\mathrm{Sp}(T)\subseteq \mathrm{Sp}(S)$ by 1 above. If $v\in \mathrm{Sp}(S)$, then $v={r}_{1}{v}_{1}+\mathrm{\cdots}+{r}_{n}{v}_{n}$. If one of the ${v}_{i}$’s, say ${v}_{i}$, is $0$, then $v={r}_{2}{v}_{2}+\mathrm{\cdots}+{r}_{n}{v}_{n}\in \mathrm{Sp}(T)$. ∎

3.
It is not true that if ${S}_{1}\supseteq {S}_{2}\supseteq \mathrm{\cdots}$ 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 $i=1,2,\mathrm{\dots}$, let ${S}_{i}$ be the closed ball centered at the origin, with radius $1/i$. Then $\mathrm{Sp}({S}_{i})=V$. But the intersection of these ${S}_{i}$’s is just the origin, whose span is itself, not $V$. ∎

4.
$S$ 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 $0={r}_{1}{v}_{1}+\mathrm{\cdots}{r}_{n}{v}_{n}$, with ${r}_{1}\ne 0$. Then
$${v}_{1}={s}_{2}{v}_{2}+\mathrm{\cdots}+{s}_{n}{v}_{n},$$ (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 $\mathrm{Sp}(S)=\mathrm{Sp}(S\{v\})$. 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 onesided: basis implying minimal spanning set.
Title  properties of spanning sets 

Canonical name  PropertiesOfSpanningSets 
Date of creation  20130322 18:05:40 
Last modified on  20130322 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 