Hilbert space: $\langle Ux, Uy \rangle = \langle x, y \rangle$ iff $\| Ux \| = \| x \|$

Question

I have to prove:

Let $(\mathbb{H}, \langle \cdot, \cdot\rangle)$ be a Hilbert space and $U \in L(\mathbb{H})$. Prove that $\langle Ux, Uy \rangle = \langle x, y \rangle$ for all $x, y \in \mathbb{H}$ iff $\| Ux \| = \| x \|$.

My idea:
If $\langle Ux, Uy \rangle = \langle x, y \rangle$, then for $x=y$ we have: $\|Ux \|^2= \langle Ux, Ux \rangle = \langle x, x \rangle = \|x\|^2$, hence $\| Ux \| = \| x \|$, because the norm is positive definite. How do I prove the statement for $x \neq y$?

If $\| Ux \| = \| x \|$, then $\langle Ux, Ux \rangle = \langle x, x \rangle$. The same question: what about $x \neq y$?

Answer 1

Consider $\|Ux + Uy\|^2 = \|U(x+y)\|^2 = \|x+y\|^2$. Expand both sides to find $$\|Ux\|^2 + 2 \langle Ux,Uy \rangle + \|Uy\|^2 = \|x\|^2 + 2 \langle x,y \rangle + \|y\|^2$$ which reduces to $$\langle Ux,Uy \rangle = \langle x,y \rangle.$$

Of course this assume that $H$ is a real Hilbert space. If it is a complex Hilbert space you should experiment with expressions like $\|Ux \pm i Uy\|^2$.