If $U \in \mathcal B(\mathcal H)$ where $\mathcal H$ is a Hilbert space. Say, $U$ is a surjective isometry. Then my textbook writes:
$$\langle U^*Uv, v\rangle = \langle Uv, Uv\rangle = ||Uv||^2 = ||v||^2 = \langle v, v\rangle$$ for all $v \in \mathcal H$. Thus, $\langle (U^*U - I)v, v\rangle = 0$ for all $v$ and it follows by polarization that $U^*U - I = 0$, i.e., $U^*U = I$.
Can someone explain to me how $U^*U - I = 0$ follows from the polarization identity? I haven't come across this concept before.
Let $M = U^*U - I$. We are given that $\langle Mv,v \rangle = 0$ holds for all $v \in \mathcal H$. On the other hand, the polarization identity allows us to express $\langle Mx,y \rangle$ in terms of expressions of the form $\langle v,v \rangle$. In particular, for any $x,y \in \mathcal H$, we have $$ \langle M x,y \rangle = \frac 14 \sum_{k=0}^3 i^k \langle M(x + i^ky),x + i^ky\rangle. $$ Thus, we can conclude that $\langle Mx,y \rangle = 0$ holds for all $x, y \in \mathcal H$. It follows that for any $x\in \mathcal H$, $Mx = 0$ (why?), which is to say that $M = 0$.