Corollary

Two distinct reflections s and u commute if and only if $V_s^- \subset V_u^+$ and $V_u^- \subset V_s^+$.

If $V_s^- \subset V_u^+$ and $V_u^- \subset V_s^+$, Prop. 3, (i) shows that $V_u^+$ and $V_u^-$ are stable under $s$, hence $su = us$ by Prop. 3, (ii).

Conversely, if $su = us$, the subspace $V_s^-$ is stable under $u$ by (ii); by (i), there are two possible cases:

a) We have $V_u^- \subset V_s^-$: since they are both of dimension 1, these spaces are therefore equal, hence $V_s^- \not\subset V_u^+$; since $V_u^+$ is stable under $s$, we have $V_u^+ \subset V_s^+$ and so these two hyperplanes are equal. But then $s = u$, contrary to our assumptions.

b) We have $V_s^- \subset V_u^+$: the image of $1 - s$ is thus contained in the kernel of $1 - u$, so $(1 - u).(1 - s) = 0$. Since $s$ and $u$ commute, we have $(1 - s).(1 - u) = 0$, in other words $V_u^- \subset V_s^+$.