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 $V={ℝ}^n$, 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 one-sided: basis implying minimal spanning set.