- #1

- 244

- 32

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##.

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##)?