I'm currently pretty stuck at showing this. This is the task:
Let an automorphism f: $V\buildrel\over\rightarrow V$ display orthogonal vectors to orthogonal vectors which means:
f(v) $\bot$ f(w) if v $\bot$ w. (for v,w in V)
Show there exists an orthogonal map g and λ $\epsilon$ $R_{>0}$ so that f=λg.
In a first task we had to show that (v+w) $\bot$(v-w) $\leftrightarrow$ ||v||=||w||, what I've done and it's a hint to use this. However I did not come up with anything special so far. I thought of the identity, but how do I know that f(v)=x=λv=λg(v) which I concluded wrong. I also thought of g=$\frac{1}{λ}$f. For orthogonal vectors it seems to work as the definition of an orthogonal map would be:
$<$g(v),g(w)$>$=$<$$\frac{1}{λ}$f(v),$\frac{1}{λ}$f(w)$>$=0=$<$v,w$>$ and obviously f=λg (maybe this is a wrong thought?)
but now for any vector v $\epsilon$ V it is not sure on what f displays v so that g is really orthogonal. I don't know how to make use of the hint here. Maybe work with the bijectivity? Nevertheless I would appreciate some help!
Being an automorphism means that $f:V\to V$ is linear and invertible.
Fix an arbitrary unit vector $v$ (so that $\|v\|=1$, i.e. $\langle v, v\rangle=1^2=1$), and let $\lambda:=\|f(v) \|$. Since $f$ is invertible, $\lambda\ne0$. We have to show $g:=\frac1\lambda f$ is orthogonal.
Here is the key point: let $w$ be another unit vector, then $\|w\|=1=\|v\|$ so $v+w\perp v-w$ thus $f(v) +f(w) \perp f(v) -f(w) $ implying $\|f(w) \|=\|f(v) \|=\lambda$.
We get that $g$ maps unit vectors to unit vectors, hence preserves the norm, and preserves orthogonality, as $f$ did.
For any unit vector $w$, use that the orthogonal decomposition of an arbitrary vector $x$ along $w$ as $x=\alpha w+x_\perp$ is unique with $\alpha=\langle x, w\rangle$.
Apply $g$ to the decomposition and deduce $\langle g(w), g(x) \rangle=\langle w, x\rangle$.