I wish to prove that if $T:\mathbb{R}^{n}\to\mathbb{R}^{n}$ is defined by $T(v)=Av$ (where $A\in M_{n}(\mathbb{R})$) is an isometry then $A$ is an orthogonal matrix.
I am familiar with many equivalent definition for $A\in M_{n}(\mathbb{R})$ to be orthogonal, and it doesn't matter to me which one to show. What I tried to do is the following: $||x-y||=||Ax-Ay||\implies\langle x-y,x-y\rangle=\langle Ax-Ay,Ax-Ay\rangle\implies\langle x-y,x-y\rangle=\langle x-y,A^{t}A(x-y)\rangle$,
from here I thought that I will be able to deduce $A^{t}A=I$ and complete the proof, but I was unable to do so.
How can I complete the proof, or prove this in another fashion ? Help is appreciated!