Are isometries always linear?

Question

Let $E$ be a finite dimensional vector space (over a field of characteristic zero) and $f : E \rightarrow E$ be an isometry fixing 0.

Must $f$ be linear in this case ?

Note : I am NOT assuming that the norm of $E$ comes from a quadratic form (otherwise I know the answer is yes, as per Should isometries be linear?). I expect the answer to my question should be no, but I don't have any counter example.

Here's another related post: http://math.stackexchange.com/questions/866445/rigid-motion-on-mathbbr2-which-fixes-the-origin-is-linear/866471#866471 — Cameron Williams, Jul 19 '14 at 23:30
I am sorry myself, because the post I had linked (http://math.stackexchange.com/questions/59374/example-of-a-non-linear-isometry) was considering isometries between possibly different spaces. — PhoemueX, Jul 19 '14 at 23:44

score 2 · Accepted Answer · edited Apr 13 '17 at 12:20

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By the Mazur-Ulam theorem every bijective isometry between normed spaces, in particular between identical normed spaces, is affine. The linked paper also contains an example of injective non-affine isometry. See also Should isometries be linear?

edited Apr 13 '17 at 12:20

Community

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answered Jul 19 '14 at 23:28

Conifold

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Is it obvious that every isometry between the same (finite dimensional) normed space is always surjective? EDIT: Ok, it is open by invariance of domain. The image is complete, thus closed. Surjectivity then follows using the connectivity of $E$. – PhoemueX Jul 19 '14 at 23:46

Are isometries always linear?

1 Answers1

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