Let $x^1,\dotsc,x^n$ be vectors in $\mathbb{R}^n$ such that no $x^i$ can be written as a convex combination of the other vectors. I'm trying to show that there exists a unique hyperplane that passes through these points.
As it is well known that there exists a unique hyperplane through $n$ affinely independent points in $\mathbb{R}^n$, my approach was to show that the set of vectors is affinely independent. At first sight, this seemed trivial, but I was unable to do the formal proof.
This is correct for $n\leq 3$ but wrong for $n\geq 4$
$n=2$ is trivial. For $n=3$ you only have to consider two cases: The three points lie on a line or they do not lie one one line. In the first case, it is clear that one of them is the convex combination of the other two, and that the hyperplane is not unique. In the second case, it is also quite clear that none of the points is a convex combination of the other two, and that there is a unique hyperplane.
For $n=4$, a counterexample can easily be given. Take any four vectors with $x_1+x_2 = x_3+x_4$, and each three of them are linearly independent. The four points form a 2D rhomboid in 4D space. For example $$ x_1 = e_1\;,\;\; x_2 = e_2\;,\;\; x_3 = e_3\;,\;\; x_4 = e_1+e_2-e_3 $$ This approach can easily be extended to higher dimensions.
The deeper reason behind this is the following: Affine independence of $x_1,\ldots,x_n$ means that $$ \sum_{i=1}^{n} \alpha_i x_i = 0\;,\;\;\sum_{i=1}^{n} \alpha_i = 0 $$ has only one solution, which is $\alpha_i = \ldots = \alpha_n = 0$. If there is no affine independence and $n\leq 3,$ then one of the $\alpha_i$ of the non-trivial solution will have a different sign than the other ones, which allows us to construct the corresponding $x_i$ as a convex combinations of the other $x_j$. So affine dependence implies convex combination, from which we conclude "absence of convex combinations implies affine independence." using proof by contradiction.
If there is no affine independence and $n\geq 4,$ then it can happen that there is more than one positive $\alpha_i$ as well as more than one negative $\alpha_i$ in each non-trivial solution, which makes it impossible to construct one of the $x_i$ as a convex combination of the other $x_j$.