Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to prove that similarity $f:\mathbb R^n \to \mathbb R^n$, i.e there is $0<r<1$ such that $d(f(x),f(y))=r \space d(x,y)$ for any $x,y\in\mathbb R^n$, is surjection? Intuitively this is the case, but i can't prove it rigorously or give any counterexample. Thanks.

share|cite|improve this question
up vote 4 down vote accepted

You can compose a similarity of $\mathbf R^n$ with scaling (a bijection) to obtain an isometry, so you can assume without loss of generality that $f$ is an isometry.

$\mathbf R^n$ as a metric space has the property that for any two points, there is a unique geodesic line connecting them: the straight line. Isometries preserve distances, so they must also preserve geodesics, and hence also straight lines, so they are affine.

Therefore, if you compose an isometry $f$ with translation by $-f(0)$ (again a bijection), you obtain a linear isometry of $\mathbf R^n$, so without loss of generality $f$ is a linear isometry.

To check that a linear isometry is bijective, you just need to check that it transforms the sequence of basis vectors to a linearly independent sequence. That's not hard to do: if you had some $\sum_{i<n}\alpha_if(e_i)=0$, then, because $f$ is a linear isometry, $\sum_{i<n}\alpha_ie_i=0$, so $\alpha_i$ are all zero, and we're done.

This also shows that any similiarity is a composition of a linear isometry, translation and scaling, which, with some more work, can be used to show that it is also a composition of several reflections and scaling.

share|cite|improve this answer
Your answer is very helpful, thanks. – Deco Aug 13 '12 at 18:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.