From the definition in the previous section, we have \begin{equation} \SO(2) = \{ M\in\RR^{2\times 2} : M^T M = I, \det M = 1 \} \end{equation} It is easy to show that the most general element of $\SO(2)$ takes the form \begin{equation} M = \begin{pmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \\ \end{pmatrix} \label{2rot} \end{equation} representing a counterclockwise rotation by $\alpha$ in the $xy$ plane. It is in fact enough to check this on a basis, such as \begin{align} \ii &= \begin{pmatrix} 1\\ 0\end{pmatrix} \\ \jj &= \begin{pmatrix} 0\\ 1\end{pmatrix} \end{align} noting that \begin{align} M \ii &= \begin{pmatrix} \cos\alpha\\ \sin\alpha\end{pmatrix} \\ M \jj &= \begin{pmatrix} -\sin\alpha\\ \cos\alpha\end{pmatrix} \end{align} which indeed correspond to a counterclockwise rotation of the $x$ and $y$ axes through an angle $\alpha$, resulting in the rotated basis $\{M\ii,M\jj\}$ — which are just the columns of $M$.