Verifying Proof: The span of a list of vectors in V is the smallest subspace of V containing all vectors in the list. I’m working through Axler, and I believe I have a different proof than the one presented in Chapter 2. Unfortunately, I’m not confident enough in my understanding of the concepts to be sure my proof isn’t flawed.
I would appreciate any feedback. Thanks!
Claim: Let $V$ be a vector space over $\mathbb{F}$. $\forall v \in V$, $span(v_1,\ldots,v_m)$ is the smallest subspace of $V$ that contains $v_1,\ldots,v_m$.
Proof: Let $U_i$ be a subspace of $V$ such that $U_i = \{ \alpha v_i \mid \alpha \in \mathbb{F}\}$. Clearly $0 \in U_i$. Also, consider that $\beta v_i \in U_i$ by construction, and $\forall u,v \in U_i$ we have $u + v = \alpha_1 v_i + \alpha_2 v_i = (\alpha_1 + \alpha_2)v_i \in U_i$. So, $U_i$ is a subspace.
Now, consider $\sum^m U_i$. We have already proven that $U_1 + \ldots +U_m$ is the smallest subspace of $V$ which contains all $u \in \sum^m U_i$. Since $u \in \sum^m U_i = \{\alpha_1 v_1 + \ldots + \alpha_m v_m \mid \alpha_i \in \mathbb{F}, v_i \in U_i \}$, by definition $span(v_1,\ldots,v_m) = \sum^m U_i$.
$\therefore span(v_1,\ldots,v_m)$ is the smallest subspace containing $v_1,\ldots,v_m$.
 A: Trusting that you have a proof that $U_1 + \ldots + U_m$ is the smallest subspace containing $U_1, \ldots, U_m$, then this idea is solid, and the proof is presented relatively well. However, I do have some notes for you.

*

*When proving $U_i$ is a subspace, the order of your logical statements is slightly jumbled. The way you have written it, it's not 100% clear how $u$ and $v$ relate to $\alpha v_i$ and $\beta v_i$. I can very easily guess how they are related, but it's worth (for the practice) structuring your proof so that it is 100% clear. I would write it like so:Suppose $u, v \in U_i$. Then there exist $\alpha, \beta \in \Bbb{F}$ such that $u = \alpha v_i$ and $v = \beta v_i$. We then have$$u + v = \alpha v_i + \beta v_i = (\alpha + \beta)v_i.$$Since $\alpha, \beta \in \Bbb{F}$, a field, we have $\alpha + \beta \in \Bbb{F}$, and so $u + v \in U_i$.


*Your proof that $U_i$ is a subspace is missing closure under scalar multiplication. This is a hole in your proof that needs to be filled.


*I'm very much happy with your explanation that$$U_1 + \ldots + U_m = \operatorname{span}(v_1, \ldots, v_m).$$However, the left hand side is the smallest subspace containing each $U_i$, while the right hand side is the smallest subspace containing each $\{v_i\}$. There's a small logical gap there: could there be a smaller subspace that contains each $\{v_i\}$, but not the entirety of $U_i$? Of course, the answer is "no", since any subspace that contains $v_i$ contains $\alpha v_i$ for all $\alpha \in \Bbb{F}$ by closure under scalar homogeneity, and thus contains all of $U_i$. Ideally, this should be made clear.
A: Your proof has the right idea but it is not written properly. Let's take a look.
Let $v_1,\ldots,v_m \in V.$ Want to show that the span $span(v_1,\ldots,v_m)$ is the smallest subspace of $V$ that contains $v_1,\ldots,v_m$.
Proof:
Step 1:
Show that $span(v_1,\ldots,v_m)$ is a subspace of V. It's clear by the definition. Note that  $v_1,\ldots,v_m \in span(v_1,\ldots,v_m)$. You just check that the def of the subspace is satisfied.
Step 2.
Show that $span(v_1,\ldots,v_m)$ is the smallest such subspace. To do that let W be another subspace that contains $v_1,\ldots,v_m \in V.$ Since W is a subsapce of V, W contains any linear combinations of the elements of  $v_1,\ldots,v_m.$ That means W contains $span(v_1,\ldots,v_m)$.
This completes the proof.
