Question about subspaces of vector spaces From Rotman's Introduction to Algebraic Topology:

If $A$ is the affine set in $R^n$ spanned by an affine independent set
  $\{ p_0, ..., p_m\}$, then $A$ is a translate of an $m$-dimensional sub-vector-space V of $R^n$, namely
  $A = V + x_0$ for some $x_0 \in R^n$.
PROOF: Let $V$ be the sub-vector-space with basis $\{p_1-p_0, ..., p_m-p_0\}$ and set $x_0=p_0$.

What exactly is the sub-vector-space a subspace of?  
Because for 
$A =\{\sum_{i=0}^{m} t_i p_i : \sum t_i = 1\} = V + p_0 = \{v + p_0 : v \in V\} = \{ \sum_{i=1}^{m} j_i(p_i-p_0) + p_0\}$ 
wouldn't we have to have a similar summation restriction on the coefficients?
 A: Consider first a special case with only two points $p_0$ and $p_1$. The affine hull $A$ of those is the set of all affine combinations
$$
A=\{t_0p_0+t_1p_1\colon\ t_0+t_1=1\}.
$$

The condition $t_0+t_1=1$ means that we can choose $t_1$ as we like, it is a free variable, and then take $t_0=1-t_1$. Let's set this $t_0$ into the expression
$$
t_0p_0+t_1p_1=(1-t_1)p_0+t_1p_1=p_0+t_1(p_1-p_0).
$$
We see now that the set $A$ can be represented as a given translation by the vector $p_0$ plus the free movement along the vector $p_1-p_0$. Note that there is no restriction on $t_1$, it is a free variable.
In general, we do the same: since $\sum_{i=0}^nt_i=1$ we can write
$$
\sum_{i=0}^nt_ip_i=\sum_{i=0}^nt_ip_i-\underbrace{\sum_{i=0}^nt_i}_{=1}p_0+p_0=\sum_{i=0}^nt_i(p_i-p_0)+p_0=\sum_{i=1}^nt_i(p_i-p_0)+p_0.
$$
The variables $t_1,\ldots,t_n$ can be chosen freely, and the only dependent $t_0$ vanishes as $t_0(p_0-p_0)=0$. That's why we do not have any restrictions on the linear combinations of $p_i-p_0$, $i=1,\ldots,n$.
