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Let $Q: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be an operator that preserves all distances. Is this condition alone enough for us to say that $Q$ must be a linear operator?

If not, what are some counterexamples (whether simple or pathological), where an operator preserves all distances but is not a linear operator?

[Edit] Did I even give the correct interpretation of "preserves all distances"?

[Edit] I guess not! What I meant is that $Q$ is an isometry (i.e. $||Qx - Qy|| = ||x - y||$ for any $x$, $y$).

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2 Answers 2

up vote 8 down vote accepted

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So, to summarize: no, Q also has to fix the origin. –  Qiaochu Yuan Jan 10 '11 at 14:46

(EDIT: when I wrote this answer, the question asked if "for all $x$, $\|Qx\|=\|x\|$" was the right definition of "$Q$ preserves all distances").

No, your condition is not equivalent to preserving all distances. For example, the map which maps $(1,0,0)$ to $(-1,0,0)$ and fixes everything else satisfies your condition but does not preserve all distances, as it reduces the distance between $(1,0,0)$ and $(-1,0,0)$ from 2 to 0.

The natural notion of preserving all distances is for $Q$ to be an isometry: that is, $\|Q(x)-Q(y)\|=\|x-y\|$ for all $x$ and $y$. A map with this property need not be linear (for example, $Q(x)=x+(1,0,0)$ has this property), but it must be affine. Thus if you also require $Q$ to fix $0$, then it is linear.

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Why is it linear then? –  Rasmus Jan 10 '11 at 7:50
While I agree with your second paragraph, I don't understand the example of the first paragraph. In fact, the condition $\|Qv\| = \|v\|$ implies that $Q$ is linear. –  t.b. Jan 10 '11 at 7:52
@Theo: it's a pretty simple example, hopefully only obscured by my typo of "with" for "to". If $v$ is not $(1,0,0)$, then $Qv$=$v$, so certainly $\|Qv\|=\|v\|$. If $v=(1,0,0)$, then $\|Qv\|=\|(-1,0,0)\|=1=\|v\|$. –  Chris Eagle Jan 10 '11 at 7:57
@Rasmus: I can't actually remember how that proof goes. Hopefully someone will explain it. For the moment, it's on page 489 of this book. –  Chris Eagle Jan 10 '11 at 8:06
Ah, right. I forgot about the surjectivity part in the Mazur-Ulam theorem. –  t.b. Jan 10 '11 at 8:29

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