inverse of similarity transformation

Question

If $S$ is a similiarity transformation, i.e. there exists $c>0$, such that $$ \lvert S(x)-S(y)\rvert = c\lvert x-y\rvert. $$ Then, apparently, we have that $$ \big\lvert S^{-1}(x)-S^{-1}(y)\big\rvert = \frac{1}{c}\lvert x-y\rvert. $$ How do we prove this? As far as I can see S is not necessarily linear.

Answer 1

Let T(x) = S(x) / c, then it is an isometry in R^n. It is a famous theorem in linear algebra that the isometry in R^n is precisely a composition of an orthogonal transform( particularly, linear ) and a translation.

Answer 2

I think this works... \begin{align*} |x-y| &= \left|S\left(S^{-1}(x)\right) - S\left(S^{-1}(y)\right)\right|\\ &= c\left|S^{-1}(x) - S^{-1}(y)\right|. \end{align*}