Let $(W, S) $ be a coxeter system where $S=\{s_1,…, s_n \} $ and the order of $s_is_j$ is denoted $m_{ij}$.
Let $V $ be a real vector space with basis $u_1,…, u_n $.
Let $B$ be the bilinear form defined by $$B(u_i, u_j )=-\cos(\pi / m_{ij} ) .$$
Let $\sigma_i (x)=x-2B(u_i, x)u_i $, then $\sigma_i $ is a linear map.
I am trying to show that $$H_i = \{v \in V : B(u_i, v)=0 \} $$ is a hyperplane and that $\sigma_i $ is the reflection of $V$ with respect to this hyperplane.
Outline of what I’ve done
The map $\phi_i :V \rightarrow \mathbb{R} $, $ \phi_i(v)=B(u_i, v) $ is clearly a surjective linear map and so by the rank nullity theorem we have $\dim (\ker (\phi_i ))=n-1 $ and so since $H_i = \ker (\phi_i) $ this means that $H_i $ is indeed a hyperplane.
Now we have $V=H_i \oplus U_i $ where $U_i $ is some one-dimensional subspace of $V$. Since $\phi_i(u_i)= B(u_i , u_i )=1 $ and so $u_i \not\in H_i $ and so $U_i = \langle u_i \rangle _{\mathbb{R}}$.
A problem arises from this because this then clearly means that for $j\neq i $ that $u_j \not\in U_i $ and so $u_j \in H_i = \ker(\phi_i ) $. But when it comes to calculating it, $\phi_i(u_j)= B(u_i, u_j )=-\cos(\pi / m_{ij}) $ but $m_{ij} $ is not necessarily 2 and so it doesn’t make sense.
What’s wrong here?
There are infinitely many one-dimensional subspaces $U$ such that $V = H_i\oplus U$. It is absolutely not true that a vector has to be in $H_i$ or $U$. Just picture a plane and a line in 3D, or two lines in 2D.