# Show that $\mathrm{O}(q)$ is the union of a set of vector symmetries and the set of applications $\gamma_a$

We fix $$q$$ the quadratic form on $$\mathbb{R}^2$$ given by $$q\left(x_1, x_2\right)=x_1 x_2$$. Let

$$\mathrm{O}(q)=\left\{f \in \mathrm{GL}_2(\mathbb{R}) \mid q(x)=q(f(x)) \text { pour tout } x \in \mathbb{R}^2\right\}$$

Show that $$\mathrm{O}(q)$$ is the union of a set of vector symmetries and the set of applications $$\gamma_a$$ defined by $$\gamma_a\left(e_1\right)=a e_1$$ and $$\gamma_a\left(e_2\right)=a^{-1} e_2$$ (for $$a \in \mathbb{R}^*$$ ).

I know that the matrix of $$q$$ in the canonical basis is $$M_q = \left(\begin{array}{ll} 0 & 1/2 \\ 1/2 & 0 \end{array}\right)$$ and the relation $$q(x) = q(f(x))$$ gives that if $$P$$ is the matrix of $$f \in \mathrm{GL}_2(\mathbb{R})$$ then $$M_q = P^t M_q P.$$ What do I have to do next?

Let $$P= \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix}$$ be an element of $$O(q)$$,then $$\forall x_{1},x_{2}\in \mathbb{R},acx_{1}+(ad+bc)x_{1}x_{2}+bd x_{2} =x_{1}x_{2}$$ so,we must have $$ac=bd=0$$ and $$ad+bc=1$$,if $$a=0$$, then $$c\neq 0 ,b\neq 0$$, so $$d=0$$ and $$bc=1$$, in particular, $$P$$ must be of the form: $$P_{b}= \begin{pmatrix} 0 & b \\ b^{-1} & 0 \\ \end{pmatrix}$$ if $$a\neq 0$$, then necesarly $$c=b=0,ad=1$$,So $$P$$ must be of the form: $$P= \begin{pmatrix} a & 0\\ 0 & a^{-1} \\ \end{pmatrix}$$ which correspond to $$\gamma_{\alpha}$$,finally remark that $$P_{b}=R\circ\gamma_{b^{-1}}$$ where $$R$$ is a reflection along the line having equation $$x_{2}=x_{1}$$