Clarifying Proof that Span(v$_1, $v$_2,...,$v$_n$) is a Subspace of $V$

Question

Theorem: If v$_1, $v$_2,...,$v$_n$ are elements of a vector space $V$ then Span(v$_1, $v$_2,...,$v$_n$) is a subspace of $V$.

I understand the proof of the closure properties under addition and scalar multiplication, but what I am confused about is why Span(v$_1, $v$_2,...,$v$_n$) is a subset to begin with.

A subset will contain elements of $V$ and the theorem states that v$_1, $v$_2,...,$v$_n$ are elements of a vector space $V$. So then why is Span(v$_1, $v$_2,...,$v$_n$) a subset when it doesn't contain elements of $V$, since it is the set of all linear combinations of v$_1, $v$_2,...,$v$_n$?

Answer 1

First keep in mind (as you have said) that $v_1,v_2,\ldots,v_n$ are in $V$.

Since $V$ is a vector space it is closed under scalar multiplication, so all possible elements $\alpha_1v_1,\alpha_2v_2,\ldots,\alpha_nv_n$ are in $V$.

Since $V$ is a vector space it is also closed under addition. So all possible elements $\alpha_1v_1+\alpha_2v_2+\cdots+\alpha_nv_n$ (that is, linear combinations) are in $V$.

But such expressions are exactly the elements of ${\rm span}\{v_1,v_2,\ldots,v_n\}$. So all elements of ${\rm span}\{v_1,v_2,\ldots,v_n\}$ are in $V$.

Answer 2

$V$, being a vector space, is closed under scalar multiplication and addition. Thus any linear combination of elements of $\{v_1, ... , v_n\}$ is in $V$, so Span($\{v_1, ... , v_n\}$) is a subset of $V$.

Answer 3

It sounds like you are confusing the language "v$_1, $v$_2,...,$v$_n$ are elements of a vector space $V$" with the language "v$_1, $v$_2,...,$v$_n$ are all of the elements of a vector space $V$". The two statements are different. The former only means that v$_1, $v$_2,...,$v$_n$ are among the elements of a vector space $V$, but there could be others (and there are -- the linear combinations of v$_1, $v$_2,...,$v$_n$, and even more if these vectors don't span $V$).