# 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. 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. 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. 3.
###### 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 $V$. ∎

4. 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 $0=r_{1}v_{1}+\cdots r_{n}v_{n}$, 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 PropertiesOfSpanningSets 2013-03-22 18:05:40 2013-03-22 18:05:40 CWoo (3771) CWoo (3771) 7 CWoo (3771) Result msc 15A03 msc 16D10