It is well known that an isometry $h \in \mathrm{Isom}(\Bbb R^2)$ can be written in the form $$h(v) = Av + w$$ for some orthogonal matrix $A\in \mathrm O_2(\Bbb R)$ and $w\in \Bbb R^2$. We also know that, depending on $\det A \in \{\pm1\}$, the matrix $A$ is a rotation or reflection around the origin $0\in \Bbb R^2$.
Assume now that $\det A=1$. Then, $A=\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin\theta & \cos \theta\end{pmatrix}$ for some angle $\theta\in [0,2\pi)$. Assume that $\theta\neq 0$.
In order to find the fixed points, we need to solve the equation $h(v)=v$ for $v\in \Bbb R^2$. We have $$h(v)=v\iff Av+w=v \iff w = (I_2 − A)v \iff v=(I_2-A)^{-1}w,$$ since $\det(I_2-A)=2(1 − \cos θ)\neq 0$. So, the fixed point of $h$ is $$P:=(I_2-A)w\in \Bbb R^2.$$ So, \begin{eqnarray} h(v) = Av+(P−AP)=A(v-P)+P \tag{*} \end{eqnarray}
Question. Why $(*)$ is a rotation around the fixed point $P$? How could we explain this geometrically?
Thanks!
If $v_0$ is a fixed point of $h(v) = Av+w$ then $h(v)$ can be represented as $h(v) = A(v-v_0) + Av_0 + w = A(v-v_0) + v_0$ where $A$ is a rotation matrix. But expression $A(v-v_0)+v_0$ is a rotation by matrix $A$ around $v_0$ because it represents translation of $v_0$ to the origin, rotation of the translated $v$ (which is $v-v_0$) by $A$, then translating it back to $v_0$.