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?
$$A \in \text{SO}(2) \text{ and }  A \ne I \implies \text{ker}(A-I) = \{0\}$$

Alternatively, compute the eigenvalues of a matrix $A$ as given in C. Falcon's post to be $e^{\pm i\theta}$. If $ker(A-I)\neq\{0\}$, then there is some eigenvector $\vec{x}$ of $A$ with corresponding eigenvalue $1$. So, $e^{\pm i\theta}=1$ implies $\theta=2\pi k$ for some $k\in\mathbb{Z}$ implies $A=I$.
