<P> This discussion applies to a proper rotation, and hence det A = 1 (\ displaystyle \ det A = 1). Any improper orthogonal 3x3 matrix B (\ displaystyle B) may be written as B = − A (\ displaystyle B = - A), in which A (\ displaystyle A) is proper orthogonal . That is, any improper orthogonal 3x3 matrix may be decomposed as a proper rotation (from which an axis of rotation can be found as described above) followed by an inversion (multiplication by - 1). It follows that the rotation axis of A (\ displaystyle A) is also the eigenvector of B (\ displaystyle B) corresponding to an eigenvalue of - 1 . </P> <P> As much as every tridimensional rotation has a rotation axis, also every tridimensional rotation has a plane, which is perpendicular to the rotation axis, and which is left invariant by the rotation . The rotation, restricted to this plane, is an ordinary 2D rotation . </P> <P> The proof proceeds similarly to the above discussion . First, suppose that all eigenvalues of the 3D rotation matrix A are real . This means that there is an orthogonal basis, made by the corresponding eigenvectors (which are necessarily orthogonal), over which the effect of the rotation matrix is just stretching it . If we write A in this basis, it is diagonal; but a diagonal orthogonal matrix is made of just + 1's and - 1's in the diagonal entries . Therefore, we don't have a proper rotation, but either the identity or the result of a sequence of reflections . </P> <P> It follows, then, that a proper rotation has some complex eigenvalue . Let v be the corresponding eigenvector . Then, as we showed in the previous topic, v _̄ (\ displaystyle (\ bar (v))) is also an eigenvector, and v + v _̄ (\ displaystyle v+ (\ bar (v))) and i (v − v _̄) (\ displaystyle i (v - (\ bar (v)))) are such that their scalar product vanishes: </P>

What is meant by rotation in computer graphics