Why does $A \in$ SO(2) and $A \ne I \implies $ker$(A-I) = \{0\}$ I've come across this in proofs for Euclidean geometry a few times. Can someone tell me why this is true?
$$A \in \text{SO}(2) \text{ and }  A \ne I \implies \text{ker}(A-I) = \{0\}$$

Let $A\in\textrm{SO}(2,\mathbb{R})\setminus\{I_2\}$, there exists $\theta\in\mathbb{R}\setminus 2\pi\mathbb{Z}$ such that: $$A=\begin{pmatrix}\cos(\theta)&-\sin(\theta)\\\sin(\theta)&\cos(\theta)\end{pmatrix}.$$
Therefore, one gets: $$\det(A-I_2)=(\cos(\theta)-1)^2+\sin^2(\theta)=2(1-\cos(\theta))\neq 0.$$
Finally, $A-I_2\in\textrm{GL}(2,\mathbb{R})$ and $\textrm{ker}(A-I_2)=\{0\}$.
