Sufficient condition for an operator to be orthogonal Lets say I want to prove that an operator $T:V\rightarrow V$ is orthogonal, i.e. for all $u,v \in V: \langle \ f(u) \mid f(v)\ \rangle = \langle \ u \mid v\ \rangle$, Is it sufficient to prove that $\|v\|=\|f(v)\|$ for all $v\in V$?
because I am trying to understand a proof of a theorem that states the following:

$\forall v\in V$ s.t. $v$ is a unit vector $f(v)$ is also a unit vector then $f$ is orthogonal.

In the proof the start with a $v\neq 0$ and they get to $\|v\|=\|f(v)\|$. shouldn't we prove that for all $u,v \in V: \langle \ f(u) \mid f(v)\ \rangle = \langle \ u \mid v\ \rangle$?
 A: Yes, it is sufficient to prove that $f$ preserves norms (and by homogeneity, that is equivalent to $f$ mapping unit vectors to unit vectors). This follows from the polarisation identity.
Let $u, v \in V$ arbitrary. If $V$ is a real inner product space, we have
\begin{align}
\langle u \mathop{\vert} v\rangle &= \frac{1}{4}\lVert u+v\rVert^2 - \frac{1}{4}\lVert u - v\rVert^2 \\
&= \frac{1}{4}\lVert f(u+v)\rVert^2 - \frac{1}{4}\lVert f(u-v)\rVert^2 \\
&= \frac{1}{4}\lVert f(u) + f(v)\rVert^2 - \frac{1}{4}\lVert f(u) - f(v)\rVert^2 \\
&= \langle f(u)\mathop{\vert} f(v)\rangle
\end{align}
if $f\colon V \to V$ is a norm-preserving linear map.
If $V$ is a complex inner product space, the polarisation identity is
$$4\langle u\mathop{\vert} v\rangle = \lVert u + v\rVert^2 - \lVert u - v\rVert^2 + i\lVert u + iv\rVert^2 - i\lVert u - iv\rVert^2$$
(assuming the inner product is linear in the first argument and antilinear in the second; in the physicist's convention, the right hand side equals $4\langle v\mathop{\vert} u\rangle$ instead). The rest of the argument is the same as in the real case.
