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Linear isometry

Let f:\mathbb{R}^2 \rightarrow \mathbb{R}^2. If:

  • f(\mathbf{0})=\mathbf{0}
  • \left| {f\left( {\bf{u}} \right) - f\left( {\bf{v}} \right)} \right| = \left| {{\bf{u}} - {\bf{v}}} \right| for all {{\bf{u}},{\bf{v}}}

then prove that f is linear.


For convenience, identify \mathbb{R}^2 with \mathbb{C} here. Then note that for any such function f:\mathbb{C} \to \mathbb{C}, also z_1 \cdot f(z) a solution for any point z_1 on the unit circle. Also \overline{f(z)} is a solution. Note that \vert f(1)\vert=1 and hence we can wlog assume that f(1)=1. So f(i) is a point on the unit circle with distance \sqrt{2} to 1. Hence f(i) =\pm i, so w.l.o.g. assume that f(i)=i. But then for any z \in \mathbb{C}, both z and f(z) have the same distance to 0,1 and i. So supposing z \ne f(z), all 0,1,i lie on the perpendicular bisector between these points and in particular 0,1 and i are collinear which clearly is absurd. Hence f(z)=z for all z which proves the claim.

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