## A short alternative to Gordan

The nicely named “theorem of the alternative” of Paul Gordan states the following

If $V \subset \mathbb{R}^n$ is a linear subspace then either $V$ contains a vector with positive coordinates or the orthogonal complement $V^{\perp}$ contains a non-zero vector with non-negative coordinates.”

These options are mutually exclusive, hence the name of the theorem. Gordan gave a proof of this in 1873 which involved a very clever inductive argument. I just thought of a self contained very short proof of this theorem. In fact I will prove an equivalent statement

If $v_1, \dotsc, v_n \in \mathbb{R}^k$ then either there is a vector $v \in \mathbb{R}^k$ such that the inner product $\langle v, v_j \rangle$ is positive for all $j \in \{1, \dotsc, n\}$ or the convex hull $C$ of $v_1, \dotsc, v_n$ contains the origin.”

The equivalence can be seen as follows. Let $M$ be the $n \times k$ matrix with $v_j$ as its rows. Then the columns of $M$ span a subspace $V \subset \mathbb{R}^n$ and this transforms the second statement into the first and vice versa.

Now let $v \in C$ have minimal norm. If $|v| = 0$ then $C$ contains the origin. If $|v| > 0$ and $w$ is any point in the convex hull then the line segment from $v$ to $w$ lies in $C$ and contains no point with smaller norm than $v$. So for all $t \in (0,1]$ we have

$\dfrac{|(1-t) v + t w|^2 - |v|^2}{2t} = \langle v, w \rangle - |v|^2+ \dfrac{t}{2}|w-v|^2 \geq 0$.

This implies that $\langle v, w \rangle \geq |v|^2 > 0$ and in particular this holds for all $w \in \{v_1, \dotsc, v_n \}$. Done!