TITLE: Proving that a linear isometry on $\mathbb{R}^{n}$ is an orthogonal matrix
QUESTION [8 upvotes]: 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!

REPLY [8 votes]: We know that $\,\langle\, x,y\,\rangle =0\,\,\,\forall\,y\Longleftrightarrow x=0\,$ , so
$$\forall x,y\,\,:\,\langle\,x,y\,\rangle=\langle\,Ax,Ay\,\rangle=\langle\,x,A^tAy\,\rangle\Longrightarrow \langle\,x,(A^tA-I)y\,\rangle=0$$ 
$$\Longrightarrow (A^tA-I)y=0\,\,\,\,\forall y\,\,\Longrightarrow A^tA-I=0\Longrightarrow A^tA=I$$