Proposition 5

Suppose the vector space $\mathbf{E}$ to be of finite dimension n. The orthogonal group $\mathbf{O}(Q)$ associated with $Q$ is then generated by the symmetries with respect to the non-isotropic hyperplanes of $\mathbf{E}$.

The proposition being obvious for $n = 0$, we reason by induction on $n$. Let $u$ be an orthogonal transformation of $E$, and let $x$ be a non-isotropic vector of $E$ (Lemma 1); distinguish three cases:

a) First suppose that $u(x) = x$. Then the hyperplane $H$ orthogonal to $x$ is non-isotropic, and we have $u(H) = H$. The restriction $u'$ of $u$ to $H$ therefore belongs to the orthogonal group $O(Q')$ associated with the restriction $Q'$ of $Q$ to $H$. The induction hypothesis entails, since $Q'$ is non-degenerate, that we have $u' = v_1' \ldots v_m'$, where $v_i'$ is a symmetry with respect to a hyperplane $L_i$ of $H$. The endomorphism $v_i$ of $E$ which extends $v_i'$ and is such that $v_i(x) = x$ is then the symmetry with respect to the hyperplane $Ax + L_i$ of $E$. Obviously $u = v_1 v_2 \ldots v_m$.

b) In the second place, suppose that $u(x) = -x$. If $s$ denotes the symmetry with respect to the hyperplane $H$ orthogonal to $x$, and if we put $v = su$, we have $v(x) = x$, and we are reduced to case a).

c) Finally, let us pass to the general case, and put $y = u(x)$, so that $Q(y) = Q(x)$. Under these conditions, the vectors $x - y$ and $x + y$ cannot both be isotropic, for, from the relations $Q(x - y) = 0$ and $Q(x + y) = 0$, we would deduce, on adding member to member, $2(Q(x) + Q(y)) = 0$ ($§ 3$, no 4, déf. 2), whence $4Q(x) = 0$, contrary to the assumption. Suppose, for example, that $a = x - y$ is not isotropic; we then have

consequently, if $s$ denotes the symmetry with respect to the hyperplane orthogonal to $a$, formula (6) proves that $s(y) = y + a = x$; on putting $v = su$, we have $v(x) = x$, and we are reduced to case a). If $a = x - y$ is isotropic and $b = x + y$ is not isotropic, we see analogously that we are reduced to case b).