Matrix Representation of Rotation in $\mathbb{R^3}$

Question

I am reading the article: "The Banach-Tarski Paradox" by Karl Stromberg. At page no 153, author give a $3 \times 3$ matrix $\phi$ as follows:

$$ \phi= \begin{pmatrix} -\cos\theta & 0 & \sin\theta \\ 0 & -1 & 0 \\ \sin\theta & 0 & \cos \theta\\ \end{pmatrix} $$ (here $\theta$ is a fixed real number such that $\cos \theta$ is transcendental number.)

And author states that:

"Geometrically, $\phi$ rotates $\mathbb{R^3}$ by $180^{\circ}$ about the line in the $xz$ plane whose equation is $x\cos \frac{1}{2}\theta=z\sin \frac{1}{2}\theta.$"

But I can't understand:

• why does $\phi$ rotate $\mathbb{R^3}$ by "$180^{\circ}$" ?
• And how to find the axis of this rotation? (Axis is the subspace $\{x \in \mathbb{R^3} | \phi(x)=x\}$.)

I know $\phi$ is orthogonal and of determinant $1$ so it is a rotation. If I denote elementary rotation about the co-ordinate axis as following: $$ R_x(\theta)= \begin{pmatrix} 1& 0 & 0 \\ 0 & \cos \theta & -\sin\theta \\ 0& \sin\theta & \cos\theta\\ \end{pmatrix} $$ $$ R_y(\theta)= \begin{pmatrix} \cos \theta& 0 & \sin\theta \\ 0 & 1 & 0\\ -\sin\theta& 0 & \cos\theta\\ \end{pmatrix} $$ and $$ R_z(\theta)= \begin{pmatrix} \cos \theta& -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0\\ 0 & 0 & 1\\ \end{pmatrix} $$ Here $R_x(\theta)$ represents the rotation through $x$ axis about an angle $\theta$ in anticlockwise direction and similar for $R_y(\theta)$ and $R_z(\theta)$. So $\phi$ can be obtained from these elementary matrices.But still I can't conclude the author's statements.

And if possible,please suggest a good resource for the Group of isometries of $\mathbb{R^3}$ that include the matrix representation of an arbitrary member in $SO_3(\mathbb{R})$.

Please help. Thank you.

Answer 1

As you mentioned, the matrix is orthogonal with determinant 1 and is real. Hence it must be a rotation. To find the axis I think the best way will be to find the eigenvector corresponding to eigenvalue 1. To find the angle, you may want to try to find two other eigenvalues which will be of the form $e^{\pm i\psi}$ ($\psi$ being the angle of rotation.) But a quicker way will be to check the trace: $$\operatorname{tr}(R)=1+e^{+i\psi}+e^{-i\psi}=1+2\operatorname{cos}(\psi)$$ In your case, $\cos(\psi)=-1$ and it is a $\pi$ rotation for sure. In general you will find $$\psi=\pm\operatorname{Arccos}(\frac{\operatorname{tr}(R)-1}{2})$$ This sign ambiguity comes from the fact that any $+\psi$ rotation around some direction $\hat{n}$ is equivalent to a $-\psi$ direction around $-\hat{n}$.

Answer 2

The matrix $\begin{bmatrix} -\cos \theta & \sin \theta \\ \sin \theta &\cos \theta \end{bmatrix}$ is the matrix of a reflection in the plane and you can figure out the line of reflection. Let's leave it for later to figure out the line.

The matrix in the problem statement implements such a reflection in the $x$-$z$ plane and sends $y$ to $-y$. Therefore it is the rotation by angle $\pi$ in the line in the $x$-$z$ plane.

Now to figure out the line: In general, $\begin{bmatrix} \cos \eta & \sin \eta \\ \sin \eta &-\cos \eta \end{bmatrix}$ is the reflection in the line at angle $\eta/2$ to the $x$-axis, so $\begin{bmatrix} -\cos \theta & \sin \theta \\ \sin \theta &\cos \theta \end{bmatrix}$ is the reflection in the line at angle $\pi/2 - \theta/2$ to the $x$-axis.