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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 $<S(u),S(v)>=\vert S(u)\vert\vert S(v)\vert cos(\theta)=\vert S(u)\vert\vert S(v)\vert (\frac{<Tu,Tv>}{\vert u \vert\vert v\vert})$ 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.

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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.

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  • $\begingroup$ 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? $\endgroup$ – AColoredReptile Feb 18 '19 at 6:38
  • $\begingroup$ 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. $\endgroup$ – Jens Schwaiger Feb 18 '19 at 7:06
  • $\begingroup$ @JensSchwaiger Thanks for the comment. $\endgroup$ – Kavi Rama Murthy Feb 18 '19 at 7:22
  • $\begingroup$ Ah I see thanks. $\endgroup$ – AColoredReptile Feb 18 '19 at 7:57

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