Proof that $gxg^T=x, \forall x \in \mathbb{R}^{2\times 2} \implies g=\pm I$ To give some context, perhaps irrelevant, define $s:SL(2,\mathbb{R}) \rightarrow O(1,2)$ where given $g\in SL(2,\mathbb{R})$, $s(g):x\mapsto gxg^T$, $x$ is a symmetric $\mathbb{R}^{2\times 2}$. I want to find the Kernel of s.
In other words, we want the set of $g$ s.t. $x \rightarrow gxg^T = x$, which should turn out to be ${\pm I}$.
In short, show that
$gxg^T=x, \forall x \in \mathbb{R}^{2\times 2} \implies g=\pm I$.
Is there a mistake in my logic, and if there isn't, to show this how should I start?
 A: Using $x=I$ you obtain $g\cdot g^{T}=I$, so $g\in O(2,\mathbb{R})$. Thus $\exists \alpha \in [0,2\pi]: g=\begin{bmatrix} \cos\alpha & \sin\alpha\\ -\sin\alpha & \cos\alpha\end{bmatrix}(*)$ or $g=\begin{bmatrix} \cos\alpha & \sin\alpha\\ \sin\alpha & -\cos\alpha\end{bmatrix}(**)$. 
Suppose $(*)$ then using $x=\begin{bmatrix} 1 & 0\\ 0 & 0\end{bmatrix}$ you have $gxg^{T}=\begin{bmatrix} \cos^2\alpha & -\sin\alpha\cos\alpha\\ -\sin\alpha\cos\alpha & \sin^2\alpha\end{bmatrix}$. From the imposition $gxg^{T}=x$ we obtain $\sin\alpha =0\iff \alpha =0$ or $\alpha = \pi$ from which the solutions $g=I$ and $g=-I$.
Supposing $(**)$ and using same $x$ we obtain the condition $\begin{bmatrix} \cos^2\alpha & \sin\alpha\cos\alpha\\ \sin\alpha\cos\alpha & \sin^2\alpha\end{bmatrix}=\begin{bmatrix} 1&0\\0&0\end{bmatrix}$ so we have the possible values of $g=\pm\begin{bmatrix} 1&0\\0&-1\end{bmatrix}$. Instead this two values of $g$ aren’t solution: you can demonstrate it by parametrising $x=\begin{bmatrix} a&b\\ c& d\end{bmatrix}$ and seeing that $gxg^{T}=\begin{bmatrix} a&-b\\ -c& d\end{bmatrix}$ that is in general different by $x$.
