Given $|| u + v || = || u - v ||$ show $\langle u , v\rangle = 0$

Question

Given the norm of $u+v$ is equal to the norm of $u-v$, show $u$ and $v$ must be orthogonal.

Assuming we're working with the real inner product is this a valid way of solving this problem?

$|| u + v || = || u - v ||$

$\sqrt{\langle u+v, u+v\rangle}$ = $\sqrt{\langle u-v, u-v\rangle}$

$\langle u+v, u+v\rangle = \langle u-v, u-v\rangle$

$\langle u, u+v\rangle + \langle v, u+v\rangle = \langle u, u-v\rangle + \langle -v, u-v\rangle$

$\langle u, u\rangle + \langle u, v\rangle + \langle v, u\rangle + \langle v, v\rangle = \langle u, u\rangle + \langle u, -v\rangle + \langle -v, u\rangle + \langle -v, -v\rangle$

$\langle u, u\rangle + \langle u, v\rangle + \langle v, u\rangle + \langle v, v\rangle = \langle u, u\rangle - \langle u, v\rangle - \langle v, u\rangle + \langle v, v\rangle$

Cancelling like terms and setting it equal to zero we end up with

$4\langle u, v\rangle = 0$

Therefore,

$\langle u, v\rangle = 0$

Answer 1

Yes and by the way, this is called polarization identity in real Hilbert space: $$ (x,y)=\frac{1}{4}(\|x+y\|^2-\|x-y\|^2) $$