Editing Rotation (section)

===Rotation plane===
{{main|Rotation plane}}

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.

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 +1s and −1s in the diagonal entries. Therefore, we do not have a proper rotation, but either the identity or the result of a sequence of reflections.

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, <math> \bar{v} </math> is also an eigenvector, and <math> v + \bar{v} </math> and <math> i(v - \bar{v}) </math> are such that their scalar product vanishes:

:<math> i (v^\text{T} + \bar{v}^\text{T})(v - \bar{v}) = i (v^\text{T} v - \bar{v}^\text{T} \bar{v} + \bar{v}^\text{T} v - v^\text{T} \bar{v} ) = 0 </math>

because, since <math> \bar{v}^\text{T} \bar{v} </math> is real, it equals its complex conjugate <math> v^\text{T} v </math>, and <math> \bar{v}^\text{T} v </math> and <math> v^\text{T} \bar{v} </math> are both representations of the same scalar product between <math> v </math> and <math> \bar{v} </math>.

This means <math> v + \bar{v} </math> and <math> i(v - \bar{v}) </math> are orthogonal vectors. Also, they are both real vectors by construction. These vectors span the same subspace as <math> v </math> and <math> \bar{v} </math>, which is an invariant subspace under the application of ''A''. Therefore, they span an invariant plane.

This plane is orthogonal to the invariant axis, which corresponds to the remaining eigenvector of ''A'', with eigenvalue 1, because of the orthogonality of the eigenvectors of ''A''.