Inversion of rotation matrix

Question

For example, I have a two-dimensional rotation matrix $$ \begin{bmatrix} 0.5091 & -0.8607 \\ 0.8607 & \phantom{-}0.5091 \end{bmatrix} $$ and I have a vector I'd like to rotate, e.g. $(1, -0.5)$.

My problem is to find an inverse of the rotation matrix so that I can later “undo” the rotation performed on the vector so that I get back the original vector.

The rotation matrix is not parametric, created via eigendecomposition, I can't use angles to easily create an inverse matrix.

Answer 1

Recall that rotation matrices are orthogonal therefore

$$A^{-1}=A^T$$

indeed note that

$$A^{-1}=\begin{bmatrix}\cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha\end{bmatrix}^{-1} =\begin{bmatrix}\cos(-\alpha) & -\sin(-\alpha)\\ \sin(-\alpha) & \cos(-\alpha)\end{bmatrix}=\begin{bmatrix}\cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha\end{bmatrix}=A^T$$

Answer 2

That's easy: $$\begin{pmatrix}\cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha\end{pmatrix}^{-1} =\begin{pmatrix}\cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha\end{pmatrix}$$