# Suppose $S$ is angle preserving, show that there is a constant such that $S=cT$ for some orthogonal map $T$

Let $$V,W$$ be finite dimensional inner product spaces, $$S$$ be an injective, linear map with $$\angle u,v=\angle S(u)S(v)$$. Prove there exists a nonzero constant c, such that $$S=cT$$. Where T is a linear orthogonal transformation.

So I'm not sure how to prove this. I can show $$=\vert S(u)\vert\vert S(v)\vert cos(\theta)=\vert S(u)\vert\vert S(v)\vert (\frac{}{\vert u \vert\vert v\vert})$$ but I don't think that helps at all.

Any hints would be preferred as right now I have no clue how to even start.

I have inadvertantly switched the roles of $$T$$ and $$S$$.
Let $$\{e_1,..,e_n\}$$ be an orthonormal basis. Then $$\langle e_i, e_j \rangle=0$$ for $$i\neq j$$. Also $$\langle (e_i+e_j), (e_i-e_j) \rangle=0$$. Hence $$\langle Te_i, Te_j \rangle=0$$ and $$\langle (Te_i+Te_j), (Te_i-Te_j) \rangle=0$$. Expanding this we get $$\|Te_i\|=\|Te_j\|$$. Thus $$c=\|Te_i\|$$ is indpendent of $$i$$. Any vector $$x$$ can be written as $$x=\sum a_ie_i$$ and we get $$\|Tx\|^{2}=c^{2}\|x\|^{2}$$ for all $$x$$. If $$S=\frac 1 {c^{2}} T$$ then $$S$$ is orthogonal.
• So I can see why this $\frac{1}{c^2}T$ is angle preserving but why does this prove every angle preserving transformation can be written in this way? – AColoredReptile Feb 18 '19 at 6:38
• In the answer $T$ is assumed to be angle preserving. In the question $S$ was assumed to have this property. So change the roles of $S$ and $T$ and everything is fine. – Jens Schwaiger Feb 18 '19 at 7:06