A root system of vectors in R^n admits a simple system; but does every finite set of vectors

Question

I'm reading about root systems in the context of finite reflection groups. As I understand it, every root system (a set $\Phi$ of vectors in $R^n$ with some nice properties) admits a simple system, i.e. a subset of $\Phi$ such that

1. the vectors in $\Phi$ are linearly independent;
2. every vector in $\Phi$ expressed as linear sum of the simple vectors has all non-negative or all non-positive coefficients.

My question is, forgetting about root systems, does every arbitary finite set of vectors in $R^n$ admit a simple system: ie some subset satisfying the two conditions above? I'm struggling to prove it, or to find a counter-example!

I'm not insisting that the coefficients be integers: partly because I can see this wouldn't be true generally, and partly because the book I'm following (Humphrey's) doesn't insist on it, and if I understand correctly we get some root systems where the integer condition isn't satisfied (eg Dihedral groups).

Thoughts I've had so far:

• If my intuition serves me right, then what we are trying to prove is that from any finite set of vectors, we can pick a linearly independent set such that all the other vectors fall into the "double cone" of the chosen vectors. To this end, we would want to pick nicely spread out vectors with obtuse angles so the cone is really wide.
• Hence, a counter example might use a set of vectors with lots of acute angles.
• A proof, if it exists, might follow the proof for root systems, and take a smallest subset of vectors satisfying condition 2) above. We would then have to show linear independence.
• We require the set of vectors to be finite, as some infinite sets of vectors won't have simple systems (eg, all of them in $R^n$).

Thank you for reading my question. Any thoughts appreciated!

Answer 1

First of all, you should have required the subset $S$ of simple roots to be linearly independent, not $\Phi$ (otherwise, the solution is quite simple: Take $S=\Phi$).

With this in mind, there is a counter-example already when $n=3$. Consider a pyramid $P$ in $R^3$ with tip at the origin and such that $P$ has quadrilateral cross-section with vertices $v_1,...,v_4$. Now, take your set $\Phi$ of vectors to be $\{v_1,...,v_4\}$.