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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.

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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.

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Your answer is very helpful, thanks. –  Deco Aug 13 '12 at 18:31
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