4

The following is the related problem:
Improper Rotations in Even Dimensions

I want the simpler explanation.

An improper rotation is rotation, followed by reflection in the plane perpendicular to the axis of rotation.

Consider an improper rotation in $2D$:

$$R_I=\begin{bmatrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{bmatrix} $$

Obviously, $\text {det}( R_I)=-1$.

Consider the following simple case:

  1. $R_I\begin{bmatrix} 1\\ 0 \end{bmatrix}=\begin{bmatrix} \cos\theta \\ \sin\theta \end{bmatrix}$. However, this is the same as $R\begin{bmatrix} 1\\ 0 \end{bmatrix}$, where $R$ is the proper rotation matrix.
    $$R=\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$$
  2. $R_I\begin{bmatrix} 0\\ 1 \end{bmatrix}=\begin{bmatrix} \sin\theta \\ -\cos\theta \end{bmatrix}$. and $R\begin{bmatrix} 0\\ 1 \end{bmatrix}=\begin{bmatrix} -\sin\theta \\ \cos\theta \end{bmatrix}$. Both are $180^o$ out of phase.

I am confused about how to explain the definition of improper rotation by just two examples.

Community
  • 1
sleeve chen
  • 8,576
  • 1
    What is your definition of an improper rotation? If it is "An improper rotation is rotation, followed by reflection in the plane perpendicular to the axis of rotation" I have trouble parsing that. In 2D there is no axis of rotation. And there is only a single plane, the whole space, so how do you reflect w.r.t. it?? – Jyrki Lahtonen Jun 27 '16 at 07:40
  • @JyrkiLahtonen The definition is from https://en.wikipedia.org/wiki/Improper_rotation. Maybe you can think the rotation axis is the $z$-axis with the origin at $(0,0)$? – sleeve chen Jun 27 '16 at 07:41
  • @JyrkiLahtonen Or this definition does not hold for 2D case? – sleeve chen Jun 27 '16 at 07:42
  • Hmm. That (=Wikipedia) definition assumes 3D. What is your definition of an improper rotation in 2D or in general? – Jyrki Lahtonen Jun 27 '16 at 07:43
  • @JyrkiLahtonen So in 2D, no definition for improper rotation?. Maybe I can figure out the case in 3D. – sleeve chen Jun 27 '16 at 07:45
  • 1
    If you assume that an improper rotation has matrix $R$ such that $R^T=R^{-1}$, and $\det R=-1$. Then $$\det(R+I)=\det(R+RR^T)=\det R \det (I+R^T)=-\det(I+R^T)=-\det(R+I)$$ shows that $\det(R+I_2)=0$. In other words, $-1$ is an eigenvalue of $R$. In 2D this implies that $R$ is just a reflection. – Jyrki Lahtonen Jun 27 '16 at 07:47
  • 1
    Certainly your matrix $R_I$ is a reflection. In 2D a rotation followed by a reflection is another reflection. You may have seen this when studying the dihedral groups. – Jyrki Lahtonen Jun 27 '16 at 07:48
  • @JyrkiLahtonen But in the first case, it does not reflect? – sleeve chen Jun 27 '16 at 07:55
  • 1
    Yes it does. Unless I made a mistake your matrix $R_I$ is the reflection with respect to the line in the direction of $\vec{u}=(\cos(\theta/2),\sin(\theta/2))^T$. That is, $\vec{u}$ is an eigenvector of $R_I$ corresponding to eigenvalue $\lambda=+1$, and the 2D vector orthogonal to $\vec{u}$ is an eigenvector of $R_i$ belonging to $\lambda=-1$. – Jyrki Lahtonen Jun 27 '16 at 08:09
  • 1
    In both your cases, $\vec{v}=\binom 10$ and $\vec{v}=\binom 01$, the bisectors of the angles between $\vec{v}$ and $R_I\vec{v}$ are the same. – Jyrki Lahtonen Jun 27 '16 at 08:11

1 Answers1

3

I assume that an improper rotation means an element of the orthogonal group with determinant $=-1$. In other words, an $n$-dimensional improper rotation is represented by a matrix $R$ such that $RR^T=I_n$ and $\det R=-1$.

All such matrices have $\lambda=-1$ as an eigenvalue. This is because $$\det(R+I)=\det(R+RR^T)=\det R \det (I+R^T)=-\det(I+R^T)=-\det(R+I),$$ which implies that $\det(R+I)=0$.

This has the following corollary

A 2-dimensional improper rotation is just the orthogonal reflection w.r.t. a line through the origin.

Proof. We saw above that $\lambda_1=-1$ is an eigenvalue of $R$. If $\lambda_2$ is the other eigenvalue, then $\lambda_1\lambda_2=\det R=-1$, so we can conclude that $\lambda_2=1$. If $\vec{u}$ is an eigenvector belonging to $\lambda_2$, and $\vec{v}\perp\vec{u}$ is another unit vector, then (because $R$ preserves lengths and angles) we can conclude that $R\vec{v}\perp R\vec{u}$. But here $R\vec{u}=\vec{u}$, and in 2D the only unit vectors $\perp\vec{u}$ are $\pm\vec{v}$. So we can conclude that $R\vec{v}=\pm\vec{v}$. The plus sign cannot occur, for then we would have $R=I_2$. Therefore $R\vec{v}=-\vec{v}$. This implies that $R$ is the orthogonal reflection w.r.t. the line spanned by $\vec{u}$.

Jyrki Lahtonen
  • 140,891