Proposition 4
Tag 0AQK, permanent. Cite it as [Bourbaki, Tag 0AQK].
Lie Groups and Lie Algebras, chapter V, § 3. Groups of displacements generated by reflections
Let be a chamber and a set of walls of . Let be the subgroup of generated by the orthogonal reflections with respect to the elements of . For , denote by the unit vector orthogonal to on the same side of as . The following conditions are equivalent: (i) The group is finite. (ii) There exists a point of invariant under every element of . (iii) The hyperplanes belonging to have non-empty intersection. (iv) The family of vectors is free in .
By property (D2) at the beginning of § 3, the stabiliser in of every point of is finite, so (ii) implies (i).
Since the group is generated by the set of reflections with respect to the hyperplanes belonging to , the fixed points of are the points of belonging to every hyperplane , hence the equivalence of (ii) and (iii).
Assume that there exists a point of such that for all and let be such that . Since for such that (Prop. 3), and since for all , Lemma 3 (ii) implies that the for are linearly independent. Consequently, (iii) implies (iv).
Suppose finally that the family is free. Let be a point of . For any hyperplane , there exists a real number such that consists of the points of with . Since the family is free, there exists such that for all , and the point of belongs to all the hyperplanes . Thus, (iv) implies (iii).
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.