Isometries of $\mathbb{R}^2$

Question

Show that if $A:\mathbb{R}^2\to \mathbb{R}^2$ is a proper rotation, then it may be represented by a matrix of the form $$\pmatrix{ \cos(\theta)& -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \\}.$$ Further, any improper rotation is given by $$\pmatrix{ 1 & 0 \\ 0 & -1 \\} \dot\ \pmatrix{ \cos(\theta)& -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \\}.$$ Conclude then that any isometry of $\mathbb{R}^2$ is a composition of a translation, a proper rotation and possibly a reflection with respect to the y-axis.

I do not know how to do this problem. Any help with be greatly appreciated.

Note: If $f:\mathbb{R}^n\to \mathbb{R}^n$ is an isometry, then $$f(p)=f(o)+A(p),$$ where $o$ is the origin of $\mathbb{R}^n$ and $A$ is an orthogonal transformation. So if $f:\mathbb{R}^n\to \mathbb{R}^n$ is an isometry with $f(o)=o$ we say that it is a rotation, and if $A=f-f(o)$ is identity we say that $f$ is a translation.

Answer 1

For rotations our goal is to find a matrix $R_\phi$ so that for any vector on the unit circle $u = (\cos \theta ,\sin \theta)^T$ we have $R_\phi u = (\cos (\theta + \phi), \sin(\theta + \phi))^T$. Towards this end we use the addition formulas for sine and cosine which gives us $$R_\phi u = (\cos(\theta) \cos (\phi) - \sin (\theta) \sin (\phi), \sin(\theta)\cos(\phi) + \cos(\theta)\sin(\phi))^T$$

and while it seems a bit daunting this is a linear combination of $u$ where the coefficients are given by the terms involving $\phi$. Sometimes students find it helpful to substitute $x=\cos(\theta)$ and $y=\sin(\theta)$ at this stage to make this more transparent. With a little work we can see this is $$R_\phi u =\pmatrix{ \cos(\phi)& -\sin(\phi) \\ \sin(\phi) & \cos(\phi) \\}\pmatrix{\cos(\theta) \\ \sin(\theta)}$$ and we've now recovered the matrix suggested in the problem. Seeing that $R_\phi$ is orthogonal is a routine application of the pythagorean theorem since $\det R_\phi = 1$. Seeing that this works for every vector in the plane can be done by considering $v=\lambda u$ for some scalar $\lambda > 0$. The improper rotation case will be when $\lambda < 0$.

We're basically done at this point since for any two isometries $f(x)$ and $g(x)$ the composition $f(g(x))$ will be an isometry. This means we can check the translations and reflections independently then argue the composition will also be an isometry. Making sure a rotation around an arbitrary point and a reflection across an arbitrary line also work is a matter of choosing an origin and the appropriate basis vectors. I'll leave those details to you.

Answer 2

Continuing from Lays's solution...

From the equation $a^2+b^2=1$, we may assume that $a=\cos\theta$ where $0\leq\theta<2\pi$. Then, $b=\pm\sin\theta$. But, since $(\cos\theta,\sin\theta)=(\cos(-\theta),-\sin(-\theta))$, we may assume that $(a,b)=(\cos\theta,-\sin\theta).$ With some similar ideas, from the equation $c^2+d^2=1$, we can get $(c,d)=(\sin\phi,\cos\phi)$. From the equation $ac+bd=0$, we get $\sin(\phi-\theta)=0$. Hence, $\phi=\theta$ or $\phi=\pi+\theta$. In the first case we have $(c,d)=(\sin\theta,\cos\theta)$ and in the second case, $(c,d)=(-\sin\theta,-\cos\theta)$. In conclusion, the group of orthogonal matrices is $$O(2)=SO(2)\bigcup\begin{bmatrix} 1&0\\0&-1 \end{bmatrix}SO(2)$$ where $SO(2)=\left\{\begin{bmatrix} \cos\theta&-\sin\theta\\\sin\theta&\cos\theta \end{bmatrix}\right|0\geq\theta<2\pi\}$ is the group of rotation matrices of $\Bbb R^2$. The identity $$\begin{bmatrix}\cos(\alpha+\theta)\\\sin(\alpha+\theta)\end{bmatrix}=\begin{bmatrix} \cos\theta&-\sin\theta\\\sin\theta&\cos\theta \end{bmatrix}\begin{bmatrix}\cos\alpha\\\sin\alpha\end{bmatrix}$$ explains the geometric meaning: The rotation matrix sends the point $P(r\cos\alpha,r\sin\alpha)$ given in polar coordinates to $P'=(r\cos(\alpha+\theta),r\sin(\alpha+\theta))$ in other words, it rotates the point about the origin through angle $\theta$.

There is a fundamental fact which can be proved synthetically: An isometry of $\Bbb R^2$ is created by at most three reflections in some lines. The composite of two reflections is a rotation when the reflection lines intersect and a translation when they are parallel. Hence, we conclude that the isometry group of $\Bbb R^2$ is $E(2)=O(2)⋊\Bbb R^2$.

Answer 3

I think I have come up with an answer (from some hints from my professor), but I am having difficulty finishing it.

Since a proper rotation is a type of rigid motion, we can use an orthogonal transformation to show this.

Let $A$ be an orthogonal transformation of $\mathbb{R}^2$ (and suppose $A$ is the identity matrix):

$$\pmatrix{ a& b \\ c & d \\}.$$

I know:

• $\text{det}A =\text{ad - bc}=1$
• $A^TA=\pmatrix{ a& b \\ c & d \\}\pmatrix{ a& c \\ b & d \\} = \pmatrix{ a^2+b^2& ac+bd \\ ac+bd & c^2+d^2\\}.$

Thus:

$\begin{cases} a^2+b^2=1 \\ c^2+d^2 = 1 \\ ac+bd=0 \\ \text{ad - bc}=1 \end{cases}$

I can note that $a^2+b^2=1$ and $c^2+d^2 = 1$ are just points on the unit circle which can be expressed as $\cos\theta$ and $\sin\theta$. Also, since $ac+bd=0 $ then $\pmatrix{ a \\b}\dot\ \pmatrix{ c \\d}=0$ which means they are orthogonal.

But this is where I am having difficulty. Why is $a=\cos\theta$, $b=-\sin\theta$, $c=\sin\theta$, and $d=\cos\theta$? Also, how does this matrix rotate by an angle theta?