Let $\phi: \mathbb{R}^2 \to \mathbb{R}^2$ be a lineair mapping. Prove that the following statements are equivalent:
- $\phi$ is isometric.
- For all $x \in \mathbb{R}^2$ the following holds: $|\phi(x)| = |x|$ (distance to (0,0))
- For all $x,y \in \mathbb{R}^2$ the following holds: $\langle \phi(x),\phi(y) \rangle = \langle x,y\rangle$ (scalar product)
To prove the equivalence I want to prove the following implications:
$1 \Rightarrow 2$, $2 \Rightarrow 3$ and $3 \Rightarrow 1$.
I have no difficulties proving the first implication. Because $\phi$ is a linear mapping $(0,0)$ stays at its place and because $\phi$ is isometric the distance from $(0,0)$ to $x$ and $\phi(x)$ must be the same.
The latter two I don't quite understand.
Is anybody willing to give me some tips to solve this problem?
If you always have $\langle\phi(x),\phi(x)\rangle=\langle x,x\rangle$, then, if $x,y\in\Bbb R^2$,$$\langle\phi(x+y),\phi(x+y)\rangle=\langle x+y,x+y\rangle.\tag1$$But\begin{align}(1)&\iff\langle\phi(x)+\phi(y),\phi(x)+\phi(y)\rangle=\langle x,x\rangle+2\langle x,y\rangle+\langle y,y\rangle\\&\iff\|\phi(x)\|^2+2\langle\phi(x),\phi(y)\rangle+\|\phi(y)\|^2=\|x\|^2+2\langle x,y\rangle+\|y\|^2\\&\iff\langle\phi(x),\phi(y)\rangle=\langle x,y\rangle.\end{align}
On the other hand if you always have $\langle\phi(x),\phi(y)\rangle=\langle x,y\rangle$, then, in particular, $\|\phi(x)\|^2=\|x\|^2$, and therefore $\|\phi(x)\|=\|x\|$. And so$$\|\phi(x)-\phi(y)\|=\|\phi(x-y)\|=\|x-y\|,$$and so $\phi$ is an isometry.