Euclidean plan/Proper isometry/Rotation/Fact/Proof

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.

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