Exact form of distance preserving map $T:\mathbb R^2 \to \mathbb R^2.$

Asked Feb 14 '18 at 15:46

Active Feb 14 '18 at 15:46

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Let $T:\mathbb R^2 \to \mathbb R^2$ be distance preserving, i.e., $|T(a)-T(b)|=|a-b|$ for all $a,b\in \mathbb R^2.$ From here can we say the exact form of $T$?

If $T(x)=Ax+b,$ where $A^TA=I$ and $x,b\in \mathbb R^2$ and $b$ is fixed, then $T$ is distance preserving.

But my question is that 'Can we say that the above mentioned mappings are only distance preserving map?'

asked Feb 14 '18 at 15:46

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Reduce to the case that $T(0) = 0$ by replacing $T$ with $S(x) = T(x) - T(0)$. – fred goodman Feb 14 '18 at 18:04
Closely related: Classification of Euclidean plane isometries – hardmath Feb 16 '18 at 20:26

Exact form of distance preserving map $T:\mathbb R^2 \to \mathbb R^2.$

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