# Clarifying proof of Kunze’s theorem 3 in chapter 2

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.

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.