Given transformation $f:\mathbb{R}^2 \to \mathbb{R}^2 $
such that $f(0,0)=(0,0)$
and $||f(x,y)||=||(x,y)||$ —which (if I'm not mistaken) implies that $f$ is an isometry
How can I prove that this transformation is linear?
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1It may be a duplicate in spirit, but is not in reality. As stated, the question incorrectly assumes that the condition on the norms is isometry. It is not. Isometry would be $\lVert f(x,y) - f(u,v)\rVert = \lVert (x,y) - (u,v)\rVert$. – Ingix May 08 '19 at 08:32
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It need not be linear actually. All you need is to define a family of function $(f_r)_{r>0}$ from the circle with radius $r$ to itself, and these functions would define a function $f$ defined on the whole plane that preserves norms but not necessarily distances. – Arnaud D. May 08 '19 at 09:03
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@ArnaudD. I know that. The point is to make the OP aware of the difference between the condition they actually stated and what they believe it to be. It may be an error in exactly coyping the condition, or an error in understanding. – Ingix May 08 '19 at 10:38
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@Ingix My comment was more adressed to the author of the question, to point out that a clarification is really needed : is the question about distance-preserving maps, in which case the question is indeed a duplicate, or about norm-preserving maps, in which case it is false? – Arnaud D. May 08 '19 at 10:45
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@Ingix So to be clear I agree with you, and I have upvoted your comment and voted to reopen your question (although I guess it could be argued that at the moment it falls under "unclear what you're asking" and should be closed for that reason...) Right now the Review is finished – Arnaud D. May 08 '19 at 10:48