A linear mapping which multiplies the length of every vector by the same factor preserves, or reverses oriented angles.

Question

Let $T$ be a non-singular linear mapping from $\mathbb{R}^2$ to itself such that $\|Tu\|=r\|u\|$ for some $r>0$ and every $u\in \mathbb{R}^2$.

For $u,v\in\mathbb{R}^2$ let $\theta_{u,v}$ be the oriented angle between $u$ and $v$ (defined between $-\pi$ and $\pi$). I want to show that $\theta_{Tu,Tv}=\theta_{u,v}$ for all $u,v$ or $\theta_{Tu,Tv}=-\theta_{u,v}$ for all $u,v$.

I was able to prove that $\cos\theta_{Tu,Tv}=\cos\theta_{u,v}$ so that $\theta_{Tu,Tv}=\pm\theta_{u,v}$ for all $u,v$. I want to prove that in. this case, if $\theta_{Tu,Tv}=\theta_{u,v}$ for some nonparallel $u,v$ then the equation must hold for all $u,v$ and analogously if $\theta_{Tu,Tv}=-\theta_{u,v}$ for some nonparallel $u,v$. Any idea to do this last part?

Answer 1

There is this result:

Let $\phi : \mathbb{R}^n \to \mathbb{R}^n$ be a distance preserving map, i.e.

$$\|\phi(x) - \phi(y)\| = \|x - y\|, \quad\forall x,y \in \mathbb{R}^n$$ Then there exists an orthogonal matrix $A \in O(n)$ such that $$\phi(x) = Ax + \phi(0), \forall x\in \mathbb{R}^n$$

In your case, notice that $\frac1r T$ is distance preserving and linear so $\frac1r T = A$ for some $ A \in O(2)$.

Hence $T = rA$.

It is known that $O(2)$ consists of compositions of rotations around the origin and reflections with respect to lines through the origin. Scaling this by $r > 0$ does not change the orientation.

Therefore, it should be clear that $T$ either preserves the sign of the angle (if $\det T > 0$, so $A$ is a rotation) or it flips the sign of the angle (if $\det T < 0$ so $A$ consists of a rotation and a reflection).