Inverse of a Rotation matrix

Question

If $R $ is a rotation matrix (determinant 1,orthonormal) can we say that $R^{-1}$ is also a rotation matrix?

If yes how do we prove it?

Answer 1

Hints:

You should be easily able to conclude that the determinant of $R^{-1}$ is $1$ from the identity $RR^{-1}=I$.

The orthonormality part is a bit trickier if you try to do it directly-but becomes trivial once you realise that $R^{-1}=R^t$. (I am assuming that the matrix has real entries).

Answer 2

Let $y = R(\alpha)x$ for two vectors $y$ and $x$ and $\alpha \in [0,2\pi)$. Then it must hold that $x = R(-\alpha)y$. From these two equalities you get $y = R(\alpha)R(-\alpha)y$. Since this relation must hold for all vectors $y$, you can conclude $$R(\alpha)R(-\alpha) = I$$

Then $R(-\alpha) = R^{-1}(\alpha)$. Since $R(-\alpha)$ is a rotation matrix by definition, so is $R^{-1}(\alpha)$. These arguments hold for the planar case, $x,y \in \mathbb{R}^2$. You can extend it to $\mathbb{R}^3$ by introducing the other rotational angles.