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Consider the vectors $r^1 = (0,2,-1)$, $r^2 = (-1,0,2)$, and $r^3 = (2,-1,0)$. Two properties of these vectors that interest us here are: 1) The $i$'th coordinate of $r^i$ is 0, and 2) The intersection of the convex hull of these three vectors and the nonnegative orthant in $R^3$ is the unit simplex.

Which are the triplets of vectors in $R^3$ that satisfy these two properties? Using a pencil and paper one can easily convince oneself that the answer is the following: either (in the description of the rays below, $x \geq 0$) a) $r^1$ lies on the ray $(0,-x,1+x)$, $r^2$ lies on the ray $(1+x,0,-x)$, and $r^3$ lies on the ray $(-x,1+x,0)$; or b) $r^1$ lies on the ray $(0,1+x,-x)$, $r^2$ lies on the ray $(-x,0,1+x)$, and $r^3$ lies on the ray $(1+x,-x,0)$.

Question: what happens in $R^4$, that is, what are the possible configurations for $r^1,\cdots,r^4$ so that the two properties above are satisfied? what happens in $R^n$?

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  • $\begingroup$ Note that in your example, the first coordinate of $r^1$ is not 0, while the first condition requires that $r^i_i=0$ for every $i$. $\endgroup$
    – Eilon
    Commented Apr 11, 2017 at 15:33
  • $\begingroup$ You're right. Will delete. $\endgroup$ Commented Apr 12, 2017 at 1:46

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