# Showing that an Isometry on the Euclidean Plane fixing the origin is Linear

Suppose $f$ is an isometric (i.e., distance preserving) function on $\mathbb{E}^2$ such that $f(0,0) = (0,0)$. Then I want to show that $f$ is necessarily linear. Now $f$ is linear iff $f$ is both additive and homogenous. The following is an attempted proof for the homogeneity of f (missing the last step); still more, I have no idea how to argue for the additivity of f. Any ideas?

Let $x \in \mathbb{E}^2$ and $\alpha \in \mathbb{R}$

We know that $\forall x \in \mathbb{E}^2$, $||x - 0|| = ||f(x) - f(0)|| = ||f(x) - 0||$ so that $||x|| = ||f(x)||$.

From this we immediately have the following facts:

$||x|| = ||f(x)||$

$||\alpha x|| = ||f(\alpha x)||$

We can then argue that since $||\alpha x|| = |\alpha| ||x|| = |\alpha| ||f(x)|| = ||\alpha f(x)||$, we also have that $||f(\alpha x)|| = || \alpha f(x)||$.

Finally, we have that

$||\alpha x - x|| = ||\alpha f(x) - f(x)||$

iff $||(\alpha - 1)x|| = ||(\alpha - 1) f(x)||$

iff $|\alpha - 1| ||x|| = |\alpha - 1| ||f(x)||$

Since the last of these statements is in fact true, we now have $||\alpha x - x|| = ||\alpha f(x) - f(x)||$ as desired.

Now at this point it seems like I have all of the facts required to assert that $f(\alpha x) = \alpha f(x)$, but I can't figure out how to formally state why without illegally appealing to visual intuition. Any ideas?

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Any line $g\subset{\mathbb E}^2$ can be viewed as being the median of two suitably chosen points. It follows that an isometry $f:\ {\mathbb E}^2\to{\mathbb E}^2$ maps lines onto lines.

Given an $x\ne0$ and an $\alpha\ne1$ the three different points $0$, $x$, and $y:=\alpha x$ are collinear; therefore their images $0$, $x'$, and $y'$ are collinear as well. As $x'\ne0$ we necessarily have $y'=\beta x'$ for some $\beta\in{\mathbb R}$.

Since $f$ is an isometry it follows that $$|\beta|\ \|x'\|=\|\beta x'\|=\|y'\|=\|y\|=\|\alpha x\|=|\alpha|\ \|x\|\ ,$$ whence $|\beta|=|\alpha|$, or $\beta=\pm\alpha$. Assume that $\beta=-\alpha$. Then $$|\alpha+1|\ \|x'\|=|1-\beta|\ \|x'\|=d(x',y')=d(y,x)=|\alpha-1|\ \|x\|\ .$$ This says that $\alpha$ is equidistant from $-1$ and $1$; whence $\alpha=0$. It follows that $\beta=\alpha$ in all cases.

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It suffices to prove that $||f(\alpha x) - \alpha f(x) || = 0$. Now

$$\begin{eqnarray*} ||f(\alpha x) - \alpha f(x) || &=& ||f(\alpha x)||^2-2\alpha\langle f(\alpha x),f(x) \rangle + \alpha^2 ||f(x)||^2 \\ &=& 2\alpha^2 ||f(x)||^2 - 2\alpha\langle f(\alpha x),f(x) \rangle \end{eqnarray*}$$

where $\langle \cdot,\cdot \rangle$ denotes the usual Euclidean inner product. From tbe last line in your question and using the fact that $f$ is an isometry, we get that $$||f(\alpha x) - f(x) || = ||\alpha f(x) - f(x)||.$$

Expanding the left and right out, this is saying that $\alpha||f(x)||^2 = \langle f(\alpha x),f(x) \rangle$. It now follows that

$$\begin{eqnarray*} ||f(\alpha x) - \alpha f(x)|| &=& 2\alpha^2 ||f(x)||^2 - 2\alpha\langle f(\alpha x),f(x) \rangle \\ &=& 2\alpha^2 ||f(x)||^2 - 2\alpha ( \alpha||f(x)||^2 )\\ &=& 2\alpha^2 ||f(x)||^2 - 2\alpha^2 ||f(x)||^2 \\ &=& 0\end{eqnarray*}$$

from which we conclude that $f(\alpha x) - \alpha f(x) = 0$, i.e.

$$f(\alpha x) = \alpha f(x).$$

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Looks fine now. – Brian M. Scott Sep 12 '12 at 7:24
Looks like something wrong. Why is this true $α||f(x)||^2=⟨f(αx),f(x)⟩$?! – Yong Yang May 5 '14 at 2:20
Please refer to link – Yong Yang May 5 '14 at 2:44