Let $S$ and $W$ be subsets of a vector space $V$. Show that if $S$ is a subset of $W$, then $\mathrm{span}(S)$ is a subspace of $\mathrm{span}(W)$

Question

Let $S$ and $W$ be subsets of a vector space $V$. Show that if $S$ is a subset of $W$, then $\mathrm{span}(S)$ is a subspace of $\mathrm{span}(W)$.

Ok I'm finally understanding what each of these things mean.. but I'm running out of ideas on how to actually show it without using arbitrary vector space examples like Rn. Since sets and subspaces are kind of new its hard to figure out how to write correct proofs with them.

I can see how if $V = \mathbb{R}^3$ and $S = \{(1,1,1)\}$ and $W = \{(1,1,1), (1,1,2)\}$ then $\mathrm{span}(S) = R1(line), span(W) = R2(plane)$ and that R1 is a subspace of R2... im just having trouble showing this officially when all vector spaces and sets are completely in general....

I could get as far as writing out $S=(u_1,u_2,\dots,u_n)$, $u_i \in V$ and same with $W$, but I can't really figure out what else to do in this sort of proof.

Answer 1

Well, we know the span of a set of vectors is, by definition, the set of all finite linear combinations. So, if we have a set $\{v_{1}, v_{2}, \dots, v_{n} \}$, then:

span$\{v_{1}, v_{2}, \dots, v_{n} \} = \{c_{1}v_{1} + c_{2}v_{2} + \dots + c_{n}v_{n} \mid c_{i} \in \mathbb{R} \}$ (all possible linear combinations where the coefficients are real numbers).

Note: In my example set above, I only have finitely many elements. But there is no reason to believe that when we discuss the span of a set, the underlying set is finite. It could be infinite. But the span, by definition, consists of all possible finite linear combinations of elements in our underlying set. So you only have finite combinations -- no infinite sums.

Ok, so then if $S$, a set, is a subset of $W$, another set, then every element of $S$ is in $W$, right? So that means any linear combination of elements of $S$ is also a linear combination of elements of $W$ since elements of $S$ are elements of $W$. Then that means all possible linear combinations of elements of $S$ are linear combinations of elements of $W$. So, that means the span of $S$ is contained in the span of $W$. Does this make sense?

Answer 2

Let $V$ form a vector space over the field $F$

For any set $S \subseteq V$, $span(S)$ always forms a subspace of $V$.(check).Let $w\in span(S)$ then $w=c_1s_1+c_2s_2+......+c_ns_n$ where $c_i\in F $ and $s_i \in S$.Since $S\subseteq W$ so $s_i \in W$ and hence $w \in span(W)$.

Answer 3

I am not sure if I'm reading your question correctly, but did you mean to say "subspace" instead of subset? If so, vector spaces are closed with respect to scalar multiplication and addition; therefore, $\mathrm{span}(S)$ is the set of all linear combinations of vectors in $S$, which is a subspace of $W$ (i.e. $\mathrm{span}(S)$ consists of linear combinations of elements of $W$, which lie in $W$). Now show that $\mathrm{span}(S)$ is closed with respect to scalar multiplication and addition.

Also, $\mathbb{R}^n$ is not an arbitrary vector space. It is actually a specific one (or a family of vector spaces). In your question, $V$ would be the arbitrary vector space. But it is still good to build intuition with examples as you have.