Let $\mathbf p_1, \ldots,\mathbf p_n$ be points in $\Re^d$ and let $k$ be the maximum number of affinely independent points among them. Let $\{I_1,\ldots,I_m\}$ be the set of all sets $I_i=\{j_1,\ldots,j_k\}$ of $k$ indexes such that the points $\mathbf p_{j_1},\ldots,\mathbf p_{j_k}$ are affinely independent.
Given a point $\mathbf p$ in the affine hull of $\mathbf p_1, \ldots,\mathbf p_n$, let $\mathbf x^{I_i}$ be the unique vector in $\Re^n$ such that $\mathbf p=x^{I_i}_1\mathbf p_1+\dots+x^{I_i}_n\mathbf p_n$ and $x^{I_i}_h=0$ for all $h\notin I_i$.
Equivalently, given a point $p$ in the affine hull of $\mathbf p_1, \ldots,\mathbf p_n$, the vector $\mathbf x^{I_i}$ (with $I_i=\{j_1,\ldots,j_k\}$) is the one in $\Re^n$ such that:
$$x^{I_i}_h=\begin{cases} 0& \text{if $h\notin I_i$}\\ y_l & \text{if $h=j_l$ for some $l$} \end{cases}$$
where $\mathbf y\in \Re^k$ and $\mathbf p=y_1\mathbf p_{j_1}+\ldots+y_k\mathbf p_{j_k}$.
Roughly speaking, $\mathbf x^{I_i}$ is the vector of the (normalized) barycentric coordinates of $\mathbf p$ with respect to the points whose indexes are in $I_i$, filled with zeros on the components whose indexes are not in $I_i$.
What I'm asking is whether the following statement is true for any $\mathbf p_1,...,\mathbf p_n$ and $\mathbf p$:
$$\{\mathbf x\in\Re^n:\mathbf p=x_1\mathbf p_1+\dots+x_n\mathbf p_n \mbox{ and } x_1+\dots+x_n=1\}=\{\alpha_1 \mathbf x^{I_1}+...+\alpha_m\mathbf x^{I_m}: \boldsymbol\alpha\in \Re^m\mbox{ and }\alpha_1+\dots+\alpha_m=1\}$$
$$?$$
Put another way: is the set of all possible barycentric coordinate vectors for $\mathbf p$ with respect to $\mathbf p_1, \ldots, \mathbf p_n$ equal to the affine hull of $ \mathbf x^{I_1},\ldots,\mathbf x^{I_m}$ ?
It is straightforward to see that any vector in the affine hull of $ \mathbf x^{I_1},\ldots,\mathbf x^{I_m}$ is a barycentric coordinate vector for $\mathbf p$ with respect to $\mathbf p_1, \ldots, \mathbf p_n$, but the converse is not, to me.
EXAMPLE
Consider the points
$\mathbf p_1=(-1,3)$
$\mathbf p_2=(1,5)$
$\mathbf p_3=(2,6)$
$\mathbf p_4=(10,5)$
$\mathbf p_1,\mathbf p_2,\mathbf p_3,\mathbf p_4$ are not affinely independent, and not even $\mathbf p_1,\mathbf p_2,\mathbf p_3$ are. In this case the maximum number of affinely independent points is $3$ and the triplets of such points are:
$\mathbf p_1,\mathbf p_2,\mathbf p_4$
$\mathbf p_1,\mathbf p_3,\mathbf p_4$
$\mathbf p_2,\mathbf p_3,\mathbf p_4$
There are $3$ triplets so $m=3$ and:
$I_1=\{1,2,4\}$
$I_2=\{1,3,4\}$
$I_3=\{2,3,4\}$
Letting $\mathbf p=(10,1)$ we have:
$\mathbf x^{\{1,2,4\}}=(2,-22/9,0,13/9)$ since $(10,1)=2*\mathbf p_1-22/9*\mathbf p_2+(0*\mathbf p_3)+13/9*\mathbf p_4$
$\mathbf x^{\{1,3,4\}}=(32/27,0,-44/27,13/9)$ since $(10,1)=32/27*\mathbf p_1+(0*\mathbf p_2)-44/27*\mathbf p_3+13/9*\mathbf p_4$
$\mathbf x^{\{2,3,4\}}=(0,32/9,-4,13/9)$ since $(10,1)=(0*\mathbf p_1)+32/9*\mathbf p_2-4*\mathbf p_3+13/9*\mathbf p_4$
The question in this particular case is whether the following statement is true:
$$\{\mathbf x\in\Re^4:\mathbf (10,1)=x_1*(-1,3)+x_2*(1,5)+x_3*(2,6)+x_4*(10,5) \mbox{ and } x_1+x_2+x_3+x_4=1\}=\{\alpha_1 (2,-22/9,0,13/9)+\alpha_2 (32/27,0,-44/27,13/9)+\alpha_3(0,32/9,-4,13/9): \boldsymbol \alpha\in \Re^3\mbox{ and }\alpha_1+\alpha_2+\alpha_3=1\}$$
$$?$$