Proof help: The span of any list of vectors in V is a subspace of V. I'm having some trouble understanding this proof. 
Let $U = ()$, the empty set, and define span$(U) = 0 \subset V$. Now let $U = (v_1 \cdot \cdot \cdot v_n)$ be a list of vectors in $V$. This means $$span(U) = \sum ^n _{i=1} a_i v_i $$ if every $a_i = 0$, we have the empty set, $0 \in V$. 
The next step of the proof is to let $$u = \sum ^ n _{i=1} a_i v_i \ \ \ \ \ \ \ \ v = \sum ^n _{i=1} b_i v_i$$
Where $v,u$ are vectors. $u+v \in span(U)$. For every $u = (a_{1} v_{1} + \cdot \cdot \cdot+ a_{i} v_{i})$  there is $au = (a)a_{1} v_{1} + \cdot \cdot \cdot+ (a)a_{i} v_{i}) \in span(U)$
I'm pretty confused with this. How can $u,v$ be in the span of U, if the span of $U$ is the empty set? 
I also don't understand the last step, how does multiplying $u$ by a constant mean it is in the span of $U$?
 A: Let $U$ be a set of vectors: $U=\{v_1,v_2,\dots,v_n\}$ (we'll suppose that $U$ is finite for clarity, but these concepts carry over for infinite sets).  Then we define the span of $U$ to be the set of all vectors which can be written as:
$$
\sum_{i=1}^n a_iv_i
$$
where the $a_i$ are arbitrary scalars.  This representation need not be unique: for example, if $U=\left\{\begin{pmatrix}1\\0\end{pmatrix},\begin{pmatrix}2\\0\end{pmatrix}\right\}$ then $\begin{pmatrix}3\\0\end{pmatrix}$ is in the span of $U$ because: 
$$
\begin{pmatrix}3\\0\end{pmatrix}=\begin{pmatrix}1\\0\end{pmatrix}+\begin{pmatrix}2\\0\end{pmatrix}
$$
but also: 
$$
\begin{pmatrix}3\\0\end{pmatrix}=5\begin{pmatrix}1\\0\end{pmatrix}-\begin{pmatrix}2\\0\end{pmatrix}
$$
If $U$ is the empty set then we define the span of $U$ to be the set $\{0\}$, because it makes things easier.  Every other set has $0$ in its span (Why?) and it turns out that defining the span of $\{\}$ in this way means that various theorems hold for all possible $U$ that would otherwise only have held for all $U$ except the empty set.  But you don't ever need to worry about the span of $\{\}$ again.  
What this does mean, though, is that the span of a set $U$ is never empty: it must always contain the zero vector $0$.  
Now, in order to show that $\textrm{span} U$ is a subspace of $V$, we need to check that it's closed under addition and scalar multiplication.  This is easy.  Let $u$ and $v$ be two arbitrary vectors in the span of $U$, and let $a$ be an arbitrary constant.  Since $u$ and $v$ are in the span of $U$, they can be written as: 
$$
u=\sum_{i=1}^na_iv_i\hspace{24pt}v=\sum_{i=1}^nb_iv_i
$$
where the $a_i$ and the $b_i$ are some choice of scalars (which need not be unique).  Then $u+v$ is given by the sum: 
$$
u+v=\sum_{i=1}^na_iv_i+\sum_{i=1}^nb_iv_i=\sum_{i=1}^n(a_i+b_i)v_i
$$
Since this is of the desired form (Write $c_i=a_i+b_i$.  Then $u+v=\sum_{i=1}^nc_iv_i$.), it must be a member of the span of $U$.  So $\textrm{span} U$ is closed under addition.  In addition, $au$ is given by the following:
$$
au=a\sum_{i=1}^na_iv_i=\sum_{i=1}^na\times a_iv_i
$$
This is also of the desired form (Write $d_i=a\times a_i$.  Then $au=\sum_{i=1}^nd_iv_i$.), so it is a member of the span of $U$.  So $\textrm{span} U$ is closed under scalar multiplication.  
These two facts tell us that the span of $U$ is a subspace of $V$.  
A: 
Now let $U = (v_1 \cdot \cdot \cdot v_n)$ be a list of vectors in $V$

The case $U = \emptyset$ was already done on line 1. After that we are working with $U$ consisting of a finite number of vectors.
For the second question, if $u$ is in the span of $U$, then
\begin{align}
au &= a(a_1 v_1 + \cdots + a_n v_n) \\
   &= (aa_1) v_1 + \cdots + (aa_n) v_n
\end{align}
which means that $au$ can be written as a sum of "scalar times vector in $U$", i.e. that $au$ is also in the span of $U$.
If you are pedantic, the proof only deals with the case that $U$ is finite. Perhaps "lists" are assumed to be in your book. Also note that if all $a_i$ are $0$, we don't get the empty set, but the zero vector. 
