Isometries in $\Bbb R^2$ are linear

Question

I'm trying to prove that isometries $T:\Bbb R^2\to\Bbb R^2$ are linear (usual Euclidean metric). It is assumed that $T(0)=0$.

I think that I have proved that $T(\alpha v)=\alpha T(v)$, but I'm having problems with sum.

Context:

I have defined $\Bbb R^2$ as a set of points and a set of free vectors.The dot product is like that $(a,b)\cdot(c,d)=ac+bd$ and the module of a vector $v$ is $\sqrt{v\cdot v}$. I have defined lines as set of points $\{P+t\vec v:t\in\Bbb R\}$. I have managed to prove triangular inequality. Just that.

My idea is proving that $$\|Tu+Tv-T(u+v)\|^2=0$$ but I get dot products like $Tu\cdot T(u+v)$. I have tried other expresions like $$\|T(u+v)-Tu\|^2,$$ getting the same problem.

On the other hand, the trick $$u\cdot v=\frac12\left(\|u+v\|^2-\|u\|^2-\|v\|^2\right)$$ leads me to a circular reasoning, bacause I have to compute $T(u+v)$.

Any ideas?

Note: I can't deal with continuity in $\Bbb R^2$. This is for a High School student.

The problem that proving that isometries preserves inner productos is still there. And I haven't proved yet that the are surjective. In fact, this is the goal... — ajotatxe, Apr 16 '21 at 10:02
You can write the inner product in terms of norms using the polarization identity, and so that implies that isometries preserve the inner product. — memerson, Apr 16 '21 at 10:08
@memerson So... $$Tu\cdot Tv=\frac12\left(|Tu+Tv|^2-|Tu|^2-|Tv|^2\right)$$ and what should I do with $|Tu+Tv|^2$? — ajotatxe, Apr 16 '21 at 10:18
@big_GolfUniformIndia, yes, but that's not elementary proof, which what the OP is after. — Oskar Limka, Jan 21 '25 at 09:20

Michael Hoppe · Answer 1 · 2021-04-16T16:07:30.123

1

From $T(0)=0$ and $\|Tu-Tv\|=\|u-v\|$ we have $\|Tu\|=\|u\|$ and $\langle Tu,Tv\rangle=\langle u,v\rangle$, that is, $T$ is orthogonal.

Let $(a_1,a_2)$ an orthonormal basis for $\mathbb R^2$, then $ (Ta_1, Ta_2)$ is an orthonormal basis as well. Now $$\begin{align} Tp&=\langle Ta_1,Tp\rangle Ta_1+\langle Ta_2,Tp\rangle Ta_2\\ &=\langle a_1,p\rangle Ta_1+\langle a_2,p\rangle Ta_2. \end{align} $$

edited Apr 16 '21 at 16:07

answered Apr 16 '21 at 16:02

Michael Hoppe

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Isometries in $\Bbb R^2$ are linear

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