Prove span is the smallest containing subspace In Linear Algebra Done Right, it proved that the span of a list of vectors in $V$ is the smallest subspace of $V$ containing all the vectors in the list.
I followed the proof that $span(v_1,...,v_m)$ is a subspace of $V$. But I don't follow the proof of smallest subspace.
Each $v_j$ is a linear combination of $v_1,...,v_m$ (to show this, set $a_j = 1$ and let the other $a$'s in $a_1v_1+...+a_mv_m$ equal $0$) [I don't understand the sentence in the parenthesis. What is the message?]. Thus $span(v_1,...v_m)$ contains each $v_j$.
Conversely, because subspaces are closed under scalar multiplication and addition, every subspace of $V$ containing each $v_j$ contains $span(v_1,...v_m)$ [I don't understand this line].
Thus $span(v_1,...v_m)$ is the smallest subspace of $V$ containing all vectors $v_1,...v_m$ [how to reach this conclusion?]
 A: 1) We wish to express $v_j$ as a linear combination of the vectors $v_1, \dots, v_m$. So suppose
$$v_j = a_1v_1 + \dots + a_mv_m.$$
Then taking $a_j = 1$ and $a_i = 0$ for $i \neq j$ does the trick.
2) Recall that subspaces are closed under scalar multiplication and addition (its the definition). So if you take a subspace of $V$ containing ALL of $v_1, \dots, v_m$, then the subspace in question must also contain all linear combinations of $v_1, \dots, v_m$. That is, it contains their span.
3) We saw in 1) that the span contains all $v_j$. We saw in 2) that any subspace containing $v_j$ contains the span, so the span is indeed the smallest such subspace. 
A: Let $M$ be the smallest vector subspace of $V$ containing $v_1,\dotsc, v_n$. We claim that $M=\text{span}(v_1,\dotsc,v_n)$. Indeed, note that
$$
v_i=0\cdot v_1+0\cdot v_2+\dotsb+1\cdot v_i+\dotsb+0\cdot v_n
$$
for all $1\leq i\leq n$. So by the definition of the span as the collection of all linear combinations of $v_1,\dotsc, v_n$, we have that
$$
v_i\in\text{span}(v_1,\dotsc, v_n)
$$
for all $1\leq i\leq n$. But $M$ is the smallest vector subspace containing $v_1,\dots, v_n$, so
$$
M\subseteq \text{span}(v_1,\dotsc, v_n).
$$
For the reverse inclusion, since $v_i\in M$ for $1\leq i\leq n$ and $M$ is a vector subspace (closed under addition and scalar multiplication) it follows that
$$
a_1v_1+\dotsb+a_nv_n\in M
$$
for all $(a_1,\dotsc, a_n)\in\mathbb{F}^n$ (i.e. $M$ contains all linear combinations of $v_1,\dotsc, v_n$). So
$$
\text{span}(v_1,\dotsc, v_n)\subseteq M.
$$
