I'm studying "Semi-Riemannian Geometry: The Mathematical Language of General Relativity" by Stephen Newman. Theorem 4.4.4 in that book:
Let $(V,g)$ be a scalar product space, and let $A:V\to V$ be a linear map. Then:
- If $A$ is a linear isometry, then $\|A(v)\|=\|v\|\ \forall\ v\in V$
- If $\|A(v)\|=\|v\|\ \forall\ v\in V$, and if $A$ maps spacelike (resp. timelike and lightlike) vectors to spacelike (resp. timelike and lightlike) , then $A$ is a linear isometry.
The proof of part 2 is given like this:
Since $\|A(v)\|=\|v\|$ is equivalent to $|\langle A(v),A(v)\rangle|=|\langle v,v\rangle|$, the assumption regarding the way $A$ maps vectors yields $\langle A(v),A(v)\rangle=\langle v,v\rangle$.
I'm guessing the proof is incomplete? In that case, I'd like to know if my approach is correct:
$$\langle A(v+tw),A(v+tw)\rangle=\langle A(v)+tA(w),A(v)+tA(w)\rangle \\ =\langle A(v),A(v)\rangle+t^2\langle A(w),A(w)\rangle+2t\langle A(v),A(w)\rangle=\langle v,v\rangle+t^2\langle w,w\rangle+2t\langle A(v),A(w)\rangle$$
But $$\langle A(v+tw),A(v+tw)\rangle=\langle v+tw,v+tw\rangle=\langle v,v\rangle+t^2\langle w,w\rangle+2t\langle v,w\rangle$$
(I'm not even sure if the $t$ coefficient matters) The above shows that $\langle v,w\rangle=\langle A(v),A(w)\rangle$.
Is this fine? Secondly, wouldn't this same theorem more generally hold for a linear map $A:V\to W$, where $V,W$ are both scalar product spaces?