# Plane isometries/Section
## Theorem

Let
    $\varphi \colon \mathbb {R} ^{2}\longrightarrow \mathbb {R} ^{2}$
be a proper linear isometry. Then $\varphi$ is a rotation, and the describing matrix with respect to the standard basis has the form
    $D(\theta )={\begin{pmatrix}\operatorname {cos} \,\theta &-\operatorname {sin} \,\theta \\\operatorname {sin} \,\theta &\operatorname {cos} \,\theta \end{pmatrix}}\,$
with a uniquely determined rotation angle
$\theta \in [0,2\pi [$

### Proof

Let $(x,y)$ and $(u,v)$ be the images of the standard vectors $(1,0)$ and $(0,1)$. An isometry preserves the length of a vector; therefore,
    $\Vert {\begin{pmatrix}x\\y\end{pmatrix}}\Vert ={\sqrt {x^{2}+y^{2}}}=1\,.$
Hence, $x$ is a real number between $-1$ and $+1$, and $y=\pm {\sqrt {1-x^{2}}}$, that is, $(x,y)$ is a point on the real unit circle. The unit circle is parametrized by the trigonometric functions, that is, there exists a uniquely determined angle $\theta$, $0\leq \theta <2\pi$, such that
    $(x,y)=(\operatorname {cos} \,\theta ,\operatorname {sin} \,\theta )\,.$
Since an isometry preserves orthogonality, we have
    $\left\langle {\begin{pmatrix}x\\y\end{pmatrix}},{\begin{pmatrix}u\\v\end{pmatrix}}\right\rangle =xu+yv=0\,.$
In case $y=0$, this implies (because of $x=\pm 1$) $u=0$. Then $v=\pm 1$, and, due to the properness assumption, the sign must be that of $x$.
So let $y\neq 0$. Then
    ${\begin{pmatrix}-v\\u\end{pmatrix}}={\frac {u}{y}}{\begin{pmatrix}x\\y\end{pmatrix}}\,.$
Since both vectors have the length $1$, the modulus of the scalar factor $u/y$ is $1$. In case $u=y$, we had $v=-x$ and the determinant were $-1$. Therefore, $u=-y$ and $v=x$, yielding the claim.
$\Box$
The composition of two rotations ${\begin{pmatrix}\operatorname {cos} \,\alpha &-\operatorname {sin} \,\alpha \\\operatorname {sin} \,\alpha &\operatorname {cos} \,\alpha \end{pmatrix}}$ and ${\begin{pmatrix}\operatorname {cos} \,\beta &-\operatorname {sin} \,\beta \\\operatorname {sin} \,\beta &\operatorname {cos} \,\beta \end{pmatrix}}$ is ${\begin{pmatrix}\operatorname {cos} \,(\alpha +\beta )&-\operatorname {sin} \,(\alpha +\beta )\\\operatorname {sin} \,(\alpha +\beta )&\operatorname {cos} \,(\alpha +\beta )\end{pmatrix}}$. This property is intuitively clear, using intuitive properties of plane rotations. Using the addition theorems for the trigonometric functions, we can proof them. Moreover, the addition theorems follow from this property, see exercise. Therefore, the group of plane rotations is commutative.

## Theorem

Let
    $\varphi \colon \mathbb {R} ^{2}\longrightarrow \mathbb {R} ^{2}$
be an improper linear isometry. Then $\varphi$ is an axis reflection, and its describing matrix, with respect to the standard basis, has the form
    ${\begin{pmatrix}\cos \alpha &\sin \alpha \\\sin \alpha &-\cos \alpha \end{pmatrix}}$
with a uniquely determined angle
$\alpha \in [0,2\pi [$

### Proof

We consider
    $\varphi \circ {\begin{pmatrix}1&0\\0&-1\end{pmatrix}},$
which is by the multiplication theorem for the determinant a proper isometry. Because of fact, there exists a uniquely determined angle $\alpha \in [0,\pi [$ such that
    $\varphi \circ {\begin{pmatrix}1&0\\0&-1\end{pmatrix}}={\begin{pmatrix}\operatorname {cos} \,\alpha &-\operatorname {sin} \,\alpha \\\operatorname {sin} \,\alpha &\operatorname {cos} \,\alpha \end{pmatrix}}\,.$
Therefore,
    $\varphi ={\begin{pmatrix}\operatorname {cos} \,\alpha &-\operatorname {sin} \,\alpha \\\operatorname {sin} \,\alpha &\operatorname {cos} \,\alpha \end{pmatrix}}\circ {\begin{pmatrix}1&0\\0&-1\end{pmatrix}}={\begin{pmatrix}\cos \alpha &\sin \alpha \\\sin \alpha &-\cos \alpha \end{pmatrix}}\,.$
$\Box$
If such an axis rotation is given, then ${\begin{pmatrix}-\sin \alpha \\\cos \alpha -1\end{pmatrix}}$ is an eigenvector for the eigenvalue $1$, and the axis of reflection is $\mathbb {R} {\begin{pmatrix}-\sin \alpha \\\cos \alpha -1\end{pmatrix}}$, see exercise. An axis reflection is described, with respect to the basis consisting of a vector $\neq 0$ of the reflecting axis and a corresponding orthogonal vector $\neq 0$, by the matrix ${\begin{pmatrix}1&0\\0&-1\end{pmatrix}}$. This means that the description given in fact (with respect to the standard basis) can be improved drastically.