$T$ is an isometry if and only if $\langle Tx, Ty \rangle = \langle x, y \rangle$

Question

I want to prove:

A linear mapping $T:X \to Y$ between two pre-Hilbert spaces is an isometry if and only if the inner products $\langle Tx, Ty \rangle = \langle x, y \rangle$ for all $x, y \in X$.

=>

$4 \langle Tx,Ty \rangle = \|Tx+Ty\|^2-\|Tx-Ty\|^2 = \|T(x+y)\|^2-\|T(x-y)\|^2 = \|x+y\|^2-\|x-y\|^2 = 4 \langle x,y \rangle $ Therefore $\langle Tx, Ty \rangle = \langle x, y \rangle$

<=

How to show that?

Answer 1

Note that\begin{align}\bigl\lVert Tx\bigr\rVert^2&=\bigl\langle Tx,Tx\bigr\rangle\\&=\langle x,x\rangle\\&=\lVert x\rVert^2.\end{align}

Answer 2

Simply $$ \|Tx\| = \sqrt{\langle Tx, Tx\rangle} = \sqrt{\langle x, x \rangle} = \|x\|. $$