# Trying to understand a theorem: why is a spanned subset, a set of linear combinations?

I'm trying to understand a theorem from the subspaces section of the Hoffman/Kunze's book. The theorem and proof are the following:

Theorem The subspace spanned by a non-empty subset $S$ of a vector space $V$ is the set of all linear combinations of vectors in $S$.

Proof

My problem is that I don't see how the conclusion of the proof follows from the premises.

For example, wouldn't it be possible for $L \subset W$ and for a vector $\alpha \in W$ be such that $\alpha \notin L$, so that $\alpha$ is not a linear combination of $S$?

I tried to reach a contradiction by developing this idea but couldn't find any, so I still can't my wrap my head completely around this proof.

For reference, the book uses this definition of spanned subspace:

Definition Let $S$ be a set of vectors in a vector space $V$. The subspace spanned by $S$ is defined to be the intersection $W$ of all subspaces of $V$ which contain $S$.

• Try not to be daunted by the abstract terseness of Hoffman & Kunze's presentation. Essentially, we're talking about 2 differing ways of looking at $W = \text{span}(S)$-one is from the "outside", we compare $W$ to other subspaces of $V$, without knowing much about what is IN it (besides $S$, of course). The other view is from the "inside"-which vectors, specifically, is $W$ made of. Commented Jun 27, 2015 at 15:14

## 1 Answer

No. As soon as we have that $L$ is a subspace with $L \subseteq W$, then from the definition of $W$ we have that $L$ is a subspace containing $S$, and $W$ is a subspace containing $S$, so:

$W \subseteq W \cap L = L$ (since $W$ is in all such intersections).

A more intuitive way to phrase the definition of $W$ is to say $W$ is the smallest subspace of $V$ containing $S$, in the sense that if any subspace of $V$ contains $S$, it necessarily contains $W$ (note the important distinction here between subSET and subSPACE).

• I got the Aha! moment. Once you have proved L contains S and is a subspace of V, it becomes one of the sets in the intersection which produces the subspace spanned by S.
– enzo
Commented Jun 27, 2015 at 15:11
• Exactly! And then $L \subset W$, and $W \subset L$, so $W = L$. Commented Jun 27, 2015 at 15:16