How do I prove that for $\|T(x)\|=\|x\|$ for all $x$ in a vector space iff $\left<T(x),T(y)\right>=\left<x,y\right>$?

Question

Cleaner version: $$\left \| T(x) \right \|=\left \| x \right \|,\forall x\in V\Longleftrightarrow \left \langle T(x),T(y) \right \rangle=\left \langle x,y \right \rangle$$

I know that $\left \langle x,y \right \rangle=\frac{1}{4}\left \|x+y \right \|^2-\frac{1}{4}\left \|x-y \right \|^2$, for converting between inner products and an expression of their magnitudes, but I'm lost on how to clean it up so I can use the fact that $\left \|x \right \|=\left\|T(x)\right \|$

Answer 1

If $\langle x,y \rangle = \langle Tx,Ty \rangle$, then taking $y=x$ gives $\|x\| = \|Tx\|$.

If $\|x\| = \|Tx\|$ for all $x$, then the polarisation identity gives: $\left \langle Tx,Ty \right \rangle=\frac{1}{4}\left \|Tx+Ty \right \|^2-\frac{1}{4}\left \|Tx-Ty \right \|^2 =\frac{1}{4}\left \|T(x+y) \right \|^2-\frac{1}{4}\left \|T(x-y) \right \|^2 =\frac{1}{4}\left \|x+y \right \|^2-\frac{1}{4}\left \|x-y \right \|^2 = \langle x,y \rangle$