If $T:\mathbb{R^2}\to \mathbb{R^2}$ is a linear transformation such that $\langle x,y \rangle =0 \implies \langle T(x),T(y) \rangle =0 $ for each $x,y \in \mathbb{R^2} $, show that $T=aS$ ,where $S:\mathbb{R^2}\to \mathbb{R^2}$,is an orthogonal transformation.
My attempt.
Instead of showing $T$ orthogonal I have decided to show $S=\frac{1}{a}T$ orthogonal.
For this we need to show that $\langle S(x),S(x) \rangle = \langle x,x \rangle $ .So I let $x=x_1e_1+x_2e_2$.But I am unable to get rid of $a$.Is my approach right?
First of all, notice that if $x_1y_1 + x_2y_2 =0$, then by assumption $x_1y_1\langle T(e_1),T(e_1) \rangle + x_2y_2\langle T(e_2),T(e_2) \rangle =0.$ In particular, for $x= \left( \begin{array} \ 1 \\ -1 \end{array} \right) $ and $y= \left( \begin{array} \ 1 \\ 1 \end{array} \right) $ we get $\langle T(e_1),T(e_1) \rangle = \langle T(e_2),T(e_2) \rangle$. Now set $a:= \sqrt{\langle T(e_1),T(e_1) \rangle} = ||T(e_1)||$ and verify that $\frac{1}{a}T$ is indeed orthogonal.