Euler's rotation theorem

Remarks

Proof idea: since $R^T R = I$ and $\det R = 1$, one gets $\det(R - I) = \det(R^T)\det(R - I) = \det(I - R^T) = \det(I - R) = -\det(R - I)$, because the size $3$ is odd. So $\det(R - I) = 0$, and $1$ is an eigenvalue with an eigenvector $u$. The plane orthogonal to $u$ is mapped to itself, and there $R$ acts as an element of $SO(2)$, that is, a plane rotation. The trace formula follows by computing $\tr R$ in an orthonormal basis that starts with $u$.

Example: the cyclic permutation matrix $R$ with $R e_1 = e_2$, $R e_2 = e_3$, $R e_3 = e_1$ lies in $SO(3)$. It fixes $u = (1, 1, 1)/\sqrt{3}$, and $\tr R = 0$ gives $\cos\theta = -1/2$, so it is the rotation by $2\pi/3$ about the line through $(1, 1, 1)$.

The hypothesis $\det R = 1$ is needed: the matrix $-I$ is orthogonal with $\det(-I) = -1$, and it fixes no nonzero vector. The odd dimension is also needed. In $\R^4$, take the rotations by $\pi/2$ in the plane of $e_1, e_2$ and in the plane of $e_3, e_4$. Their block matrix has determinant $1$ and fixes no nonzero vector.

Going deeper

For a student

The theorem says that $SO(3)$ is the set of rotations about axes through the origin, which justifies calling its elements rotations. It gives the axis-angle description of orientations, which Rodrigues' rotation formula turns into an explicit matrix and unit quaternions turn into an algebra. Its counterpart for rigid motions that also translate is Chasles' theorem. In rigid body kinematics it leads to the existence of an instantaneous axis of rotation and of the angular velocity vector.