f linear with $\|f(x)\| = \|x\|$ then $\langle f(x),f(y)\rangle = \langle x,y\rangle$

Question

On a vector Space $V$ with scalarproduct $\langle\cdot,\cdot\rangle$ and $f : V \to V$ linear with $\|f(x)\| = \|x\|$ for all $x$ in $V$ then prove that $\langle f(x),f(y)\rangle= \langle x,y\rangle $ for all $x,y$ in $V$

I tried this : $\langle x,y\rangle\leq \|x\|\cdot\|y\| = \|f(x)\|\cdot\|f(y)\|$
and same $\langle f(x),f(y)\rangle \leq \|f(x)\|\cdot\|f(y)\| = \|x\|\cdot\|y\|$
I don't know how to get the eqaulity though. Any help? Thank you

Answer 1

Here we can use the polarization identity of the Euclidean norm (i.e. $\|x-y\|_2^2 = \|x\|_2^2 + \|y\|_2^2 - 2\langle x, y \rangle$)

Since $f$ is linear we note that

$$\|x-y\| = \|f(x-y)\| = \|f(x) - f(y)\|.$$

The square both sides and apply polarization to both sides to get

$$\|x\|^2 + \|y\|^2 - 2\langle x, y \rangle = \|f(x)\|^2 + \|f(y)\|^2 - 2\langle f(x), f(y) \rangle. $$

Use the fact that $f$ preserves the norms, we can simplify and get

$$\langle x, y \rangle = \langle f(x), f(y) \rangle. $$

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Edit: This proof is only true when working over $\mathbb{R}$ as pointed out in the comment. (Can be modified for $\mathbb{C}$ by looking $x-iy$ as well.)