In some lecture notes about Reflection Groups, the writer constructs a vector space based on a Coxeter-system. For every $s \in S$ where $(W|S)$ is a Coxeter-system, he calls the related basis vector $\alpha_s$. He constructs the vector space $V$ as follows: $V = \bigoplus_{s \in S} \mathbb{R} \alpha_s$ and assigns the symmetrical bilinear form $B: V \times V \to \mathbb{R}, \;B(\alpha_s,\alpha_t) \mapsto -\cos \frac{\pi}{m_{st}}$ to it, where $m_{st}$ is the entry in the Coxeter-matrix belonging to $s, t \in S$ and $\frac{\pi}{\infty} := 0$. Furthermore, he defines a map $\sigma: S \to \mathrm{GL}(V) = \mathrm{Aut}(V), \;\; s \mapsto \sigma_s: \lambda \mapsto \lambda - 2B(\lambda,\alpha_s) \alpha_s$ which, if I understood correctly, assigns to every $s \in S$ a mirroring $\sigma_s$ at the "hyperplane" (in the sense of the bilinear form $B(\cdot, \cdot)$) perpendicular to the basis vector $\alpha_s$. By plugging in, we can verify that indeed $\sigma_s(\lambda)=\lambda$ for $\lambda \in \{\lambda \in V | \; B(\lambda,\alpha_s) = 0\}$ and $\sigma_s(\alpha_s) = -\alpha_s$.
Did I understand the motivation behind these definitions/constructions correctly so far?
Now, suddenly, the writer states that $\sigma_s$ is diagonalisable. However, in the following he doesn't seem to actually use this fact.
My question is: How can I easily see that this map is diagonalisable? Most statements similar to this in the lecture notes are explained at least briefly, whereas this is not, which is why I think it has to be "trivial". And also, why is it important that this map is diagonalisble? In the following, only one proposition is being proved, stating that for $(W,S)$ and the corresponding vector space with the previous bilinear form, there exists a unique group homomorphism $\sigma: W \to O(B)$ with $\sigma(s) = \sigma_s$ for all $s \in S$.
Thank you very much in advance for any answers.