A function $\mathbb{R}^n\to\mathbb{R}^n$ that preserves distances must be a linear map followed by a translation

Asked May 16 '12 at 05:50

Active May 16 '12 at 05:59

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Are isometric normed linear spaces isomorphic?
$ f: \mathbb{R}^n \to \mathbb{R}^m $ preserving distances

Consider the set of all functions $\varphi : \mathbb{R}^n \rightarrow \mathbb{R}^n$ that preserve distance.

That is $\forall a,b : \| a-b\| = \|\varphi(a) - \varphi(b)\| $

Is the following statement true or false?

$\forall \varphi: \exists v \in \mathbb{R}^n, M \in \mathbb{R}^{n\times n}: \varphi(x) = Mx + v$

Why or why not? (ie rough sketch of proof)

edited Apr 13 '17 at 12:20

Community

asked May 16 '12 at 05:50

Andrew Tomazos

1

This is a repeat of this question – Arturo Magidin May 16 '12 at 05:55
And indeed, $M$ must be orthogonal. Presumably the distance is the $2$-norm. – copper.hat May 16 '12 at 05:59
See also this question – t.b. May 16 '12 at 07:28

0 Answers0