Undergrad A = norm-preserving linear map (+other conditions) => A = lin isometry

[Shirish]

I'm studying "Semi-Riemannian Geometry: The Mathematical Langauge 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:

1. If $A$ is a linear isometry, then $\|A(v)\|=\|v\|\ \forall\ v\in V$
2. 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 $\langle A(v),A(v)\rangle=\langle v,v\rangle$.

Seems a bit incomplete. 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 (regardless of dimensions of $V$ and $W$)?

 

[PeroK]

The book proof is just a sketch. The point is that without the second condition you could have$$\langle Av, Aw\rangle = -\langle v,w\rangle$$Your proof needs to take this into account.

 

[Shirish]

The book proof is just a sketch. The point is that without the second condition you could have$$\langle Av, Aw\rangle = -\langle v,w\rangle$$Your proof needs to take this into account.

Yep that I understood already (I mean about why the additional conditions are needed apart from norm preservation), but I was still wondering if the whole theorem holds in more general conditions

 

[PeroK]

Yep that I understood already (I mean about why the additional conditions are needed apart from norm preservation), but I was still wondering if the whole theorem holds in more general conditions

Yes, it must. The proof is not dependent on the image space being $V$.

 

[martinbn]

From
$\langle v+w,v+w\rangle=\langle v,v\rangle+\langle w,w\rangle+2\langle v,w\rangle$

you have

$\langle v,w\rangle = \frac12 \left[\langle v+w,v+w\rangle-\langle v,v\rangle-\langle w,w\rangle \right]$

Thus the inner product of any pair of vectors is determined by inner products of the form $\langle a,a\rangle$. That is why they didn't write any more.