While the statement seems intuitively plausible, I am currently struggling to see a proof that any convex hull "spanned" by a set of $n+1$ affinely independent $n$-vectors, $\{x_1,x_2,…,x_{n+1}\}$, is uniquely characterized by this set of vectors.
That is to say, how can I see that $$\left\{\sum_{j=1}^{n+1}\lambda_j\,x_j:\sum_{j=1}^{n+1}\lambda_j=1\text{ and }\lambda_j\geq0\,\,\forall\,\,j\right\}=\left\{\sum_{j=1}^{k+1}\gamma_j\,z_j:\sum_{j=1}^{k+1}\gamma_j=1\text{ and }\gamma_j\geq0\,\,\forall\,\,j\right\}$$implies that $\{x_1,x_2,…,x_{n+1}\}=\{z_1,z_2,…,z_{k+1}\}.$
Thank you very much.
Best Answer
Let use now use the barycentric coordinates associated with the simplex $\Delta$. For simplicity, let us write $s_j,t_j$ instead of $x_j-O,z_j-O$ for the vectors. Note that, thanks to the affine linear independence of the two sets, these new sets of vectors are linearly independent.
Since the two set have the same convex hull we already know that every element of one set can be written as a linear combination of elements from the other. Thus our problem is reduced to prove that such a linear combination consists of only one vector.
As a proof by contradiction, let $$t_l=\sum \lambda_j s_j \\ \exists_{p\neq q}\lambda_p,\lambda_q≠0$$
This implies (together with the fact that the two set have the same convex hull) $$s_p=\sum_{j=0}^{k+1}\gamma_jt_j\ (1) \\ s_q=\sum_{j=0}^{k+1}\hat{\gamma}_jt_j\ (2)$$ Combining with the precedent requirement, we get
$$s_l=\lambda_p\sum_{j=0}^{k+1}\gamma_js_j+\lambda_q\sum_{j=0}^{k+1}\hat{\gamma}_js_j+...$$
By linear independence (and the fact that $\lambda\ge 0$) we obtain
$$\gamma_{j\neq l}=0\\ \hat{\gamma}_{j\neq l}=0$$
This yeld (thanks to (1),(2) )
$$s_p=\gamma_l t_l\\ s_q=\hat{\gamma}_l t_l$$
This contradicts the linear independence of $s_p, s_q$.