Given a set $X \subset \mathbb R^n$ of $n+2$ points $x_1, x_2, \ldots x_{n+2}$, consider the following set of equations ($w_i$ are $n+2$ variables)

$\sum_{i=1}^{n+2} w_i x_i = 0$

$\sum_{i=1}^{n+2} w_i = 0$.

It is clear that solutions of $w_i$ exists since there are $n+2$ variables $w_i$ but only $n+1$ constraints. However, the question is for what $X$ then the solution of $w_i$ is unique up to scaling, in other words, all solutions have the form $w = \alpha \ w^*$ for any real $\alpha$?

I guess that the necessary and sufficient condition is that all points of $X$ are not on the same hyperplane, but not sure it's correct or how to get it.


Not all solutions need be of that form.

E.g. for $n = 2$ let us consider the set of points

$$x_1 = (1,1),\qquad x_2 = (-1,1),\qquad x_3 = (1,-1),\qquad x_4 = (-1,-1).$$

Then the solutions

$$w_1 = w_2 = w_3 = w_4 = 1,\qquad w'_1 = w'_4 = 1,\ w'_2 = w'_3 = 2$$

cannot be derived from one another via scaling. This example can be easily generalised to higher dimensions.

  • $\begingroup$ Sorry, I made a typo (already edited), the second line was the constraint $\sum_i w_i = 0$. $\endgroup$ – Vuong Bui Jul 7 '18 at 14:16
  • $\begingroup$ In that case you should check en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem It implies that the rank of the solutions is $(n+2) - (n+1) = 1$, i.e. all solutions come from scaling. $\endgroup$ – Darth Geek Jul 7 '18 at 17:01

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