Clarifying proof of Kunze’s theorem 3 in chapter 2

Question

I am an undergraduate (took one course in Linear Algebra) working through the 2nd edition of Hoffman and Kunze's Linear Algebra.

Theorem 3 in chapter 2 states: "The subspace W spanned by a non-empty subset S of a vector space V is the set of all linear combinations of vectors in S."

I am having trouble understanding the proof of this theorem. I have written up what I think is a sufficient proof (it is basically Kunze's with the "left out" steps written fully out).

To my understanding, this is approached as a set equality proof. The fact that $$\textbf{L} \subset \textbf{W}$$ is clear to me.

What's less clear is how $$\textbf{W} \subset \textbf{L}$$

Any help would be much appreciated. As previously mentioned, I have written up what I think is a sufficient proof and will provide it as an answer.

Answer 1

Note: I defer to Kunze for the proof that $\textbf{L}$ is a subspace because it was not the point of confusion and thus importance to me.

My proof: Suppose $\textbf{W}$ is a subspace spanned by the non-empty set $\textbf{S}$. Let $\textbf{v}, \textbf{u} \in \textbf{W}$ be vectors.

$\textbf{W}$ is a subspace containing $\textbf{S}$

​$\implies \textbf{W}$ is closed under scalar multiplication

$\implies$ if $\textbf{v} \in \textbf{W}$, then $c\textbf{v} \in \textbf{W}$ where $c \in \textbf{F}$ and if $ \textbf{u} \in \textbf{W}$, then $d\textbf{u} \in \textbf{W}$ where $d \in \textbf{F}$

Being a subspace also $\implies \textbf{W}$ is closed under vector addition

$\implies c\textbf{v} + d\textbf{u} \in \textbf{W}$

$\implies \textbf{W}$ contains all linear combinations of vectors in $\textbf{S}$.

Thus, the set of all linear combinations of vectors in $\textbf{S}$ denoted $\textbf{L}$ is contained in $\textbf{W}$. More concisely, $\textbf{L} \subset \textbf{W}$.

Now, $\textbf{L}$ obviously contains $\textbf{S}$. It turns out that $\textbf{L}$ is also a subspace of $\textbf{V}$. This is proven nicely in (Kunze, 37). Thus, $\textbf{L}$ is a subspace of $\textbf{V}$ containing $\textbf{S}$. In other words, it is one of the intersected sets which "contributes" to the subspace $\textbf{W}$ spanned by $\textbf{S}$.

Now, to reiterate, $\textbf{W}$ and $\textbf{L}$ are subspaces of $\textbf{V}$ containing $\textbf{S}$. And, $\textbf{W}$ is by definition the intersection of all subspaces containing $\textbf{S}$. Thus, $\textbf{W} \subset (\textbf{W} \cap \textbf{L})$. $\textbf{W} \cap \textbf{L} = \textbf{L}$ because $\textbf{L} \subset \textbf{W}$. Thus, $\textbf{W} \subset \textbf{L}$.

We have proven that $\textbf{L} \subset \textbf{W}$ and $\textbf{W} \subset \textbf{L}$​. Thus, $\textbf{W} = \textbf{L}$, i.e. the statement of the theorem.