Linear map that maps a circle to a circle

Question

Is it true that a linear map $f:\mathbb R^2\to\mathbb R^2$ that maps a circle to a circle is the composition of a rotation and scaling? I mean rotation around the origin and scaling centered at the origin.

Answer 1

Firstly, the answer to your question is no, because you may need to do a flip as well. If you add that, the answer becomes yes.

Here is an elementary argument (see below for a "scientific" one):

Step 1) Center goes to center: Consider a parallelogram inscribed into your circle. It is automatically a rectangle, and intersection of its diagonal is the center of the original circle, which we will call $C$. Its image under any linear map is a parallelogram inscribed in the image circle, hence also a rectangle. It's diagonals intersect at $C'$, the center of the image circle. The image of the intersection of diagonals is the intersection of diagonals of the image. So the center goes to the center, $C'=f(C)$.

Step 2) Case 1: If $C$, the center of original circle, is the origin, then after composing with a rescaling you have a map that sends original circle to itself. Such a map preserves lengths, and so must be a rotation or a composition of a flip and a rotation. Case 2: If $C$ is not the origin, then there exists a rotation and rescaling that takes $C'$ to $C$. Composing $f$ with them you have now a map that takes $C$ to $C$ and a circle around $C$ to a concentric circle. The vector from origin to $C$ is fixed; since the map is linear the whole line from origin to $C$ is fixed; so in fact the circle in question is also fixed. Consider tangent lines to the circle parallel to $OC=OC'$. These lines are either taken to themselves or exchanged. In the first case the composite map is identity, and original map is rotation composed with rescaling; in the second case it is a flip through the line $OC$ and the original map is composition of rescaling, rotation and this flip.

Now, in general if an affine map from $\mathbb{R}^n\to \mathbb{R}^n$ takes a sphere to a sphere then after pre and post composing with rescalings and translations it takes the standard sphere through the origin to itself. This means result is distance-preserving, so lies in the orthogonal group $O(n)$. This in turn means that the map is a composition of a rescaled element of orthogonal group and a translation (any initial translation can be commuted out to a translation acting after the linear map; this is a manifestation of the structure of affine group as a semidirect product of translations and linear maps). If the original map is actually linear, instead of just affine, then the translation bit is automatically trivial; thus proving that the map itself is a rescaled element of orthogonal group (aka a composition of a rotation with possibly a reflection and/or a rescaling).