Can I always replace a translation together with a rotation with a rotation around some other point?

Question

Background / Context The last days I have been implementing some affine transformation motion compensation for video applications.

left : original non-motion estimated sequence. right : Index frame objects in black. Motion object in white.

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Question : If I have a set of points in 2D plane which are subjected to a known translation and a known rotation (note : no scaling here).

Will I always be able to find some point which I can find some angle which I can rotate around which accomplishes the same motion?

In other words, can I make the translational part $0$ by before some rotation (which we once again are free to choose) making an arbitrary translation and then back again after the rotation ?

My own thoughts have mostly been focused around matrix representations for this motion and how I possibly can do linear algebra and factor things out. But any approach or proof or counter-example would be of interest.

Answer 1

The answer is clearly...no.

Suppose that we have a non-zero translation composed with a rotation of angle zero. This (affine) map has no fixed point as it is a translation. If it would be equal to a rotation, it would have a fixed point. A contradiction.

Answer 2

Your question is unclear. Here is my interpretation. Given a non-empty set $S=\{v_0,v_1,v_2,\ldots,v_n\}$ of distinct labelled points and a rigid motion $f:v\mapsto R(v+t)$, where $R$ is a 2D rotation matrix and $t\in\mathbb R^2$ is the translation vector, do there exist a centre $c$ and a rotation matrix $Q$ such that each $f(v_i)$ is identical to the rotated image of $v_i$ by $Q$ about the centre $c$? That is, do there exist $c$ and $Q$ such that the following holds? $$ f(v_i)-c=Q(v_i-c)\quad\forall i\in\{0,1,\ldots,n\}.\tag{1} $$ We consider four cases. The answer to your question is “yes” in the first three cases and “no” in the last one.

1. $R\ne I$. Since the identity matrix is the only 2D rotation matrix with $1$ as an eigenvalue, $R^T-I$ is nonsingular in this case. Now condition $(1)$ is satisfied by $Q=R$ and $c=(R^T-I)^{-1}t$.
2. $f$ is a pure rotation ($t=0$). Clearly $(1)$ is satisfied by $Q=R$ and $c=0$.
3. $S=\{v_0\}$ is a singleton set. Then $(1)$ is satisfied by any rotation matrix $Q\ne I$ with $c=(I-Q)^{-1}\left(f(v_0)-Qv_0\right)$.
4. $f$ is a genuine translation (i.e., $R=I$ and $t\ne0$) and $S$ has at least two points. Since no rotation (by any angle about any origin) can map one directed straight line to a different parallel one with the same direction, the the answer to your question is negative here.

However, if the points are unlabelled, the problem becomes whether there exist $Q$ and $c$ such that $f(S)-c=Q(S-c)$. In cases 1-3 above, the answer remain positive, but in case 4, the answer now depends on the precise details of $f$ and $S$. For instance, when $f$ represents a translation to the right by two units, $f(S)$ is the rotated image of $S$ by 180 degrees about the origin if $S=\{(-1,1),(-1,-1)\}$, and $f(S)$ is not a rotated image of $S$ (by any angle about any point) if $S=\{(0,0),(0,1),(1,0)\}$.