Proposition 3

Let s be a reflection in V.

(i) A subspace V' of V is stable under s if and only if $V_s^- \subset V'$ or $V' \subset V_s^+$.

(ii) An endomorphism u of V commutes with s if and only if $V_s^+$ and $V_s^-$ are stable under u.

(i) If $V' \subset V_s^+$, then $s(x) = x$ for all $x$ in $V'$, so $s(V') \subset V'$. Assume that $V_s^- \subset V'$; then, for any $x$ in $V'$, $s(x) - x \in V_s^- \subset V'$, hence $s(x) \in V'$; thus $s(V') \subset V'$. Conversely, assume that $s(V') \subset V'$; if $V' \not\subset V_s^+$, there exists $x$ in $V'$ with $s(x) \neq x$; the non-zero vector $a = s(x) - x$ belongs to the line $V_s^-$, and hence generates this space; since $a \in V'$, we have $V_s^- \subset V'$.

(ii) Assume first that $u$ commutes with $s$. If $x$ is a vector such that $s(x) = \varepsilon.x$ (where $\varepsilon = \pm 1$), then $s(u(x)) = u(s(x)) = \varepsilon.u(x)$, so $V_s^+$ and $V_s^-$ are stable under $u$. Conversely, assume that $V_s^+$ and $V_s^-$ are stable under $u$; it is clear that $us - su$ vanishes on $V_s^+$ and on $V_s^-$, and since V is the direct sum of $V_s^+$ and $V_s^-$, we have $us - su = 0$.