define axis of rotation
Answers
Answer:
Explanation:
Axis of 2 dimensional rotations
2 dimensional rotations, unlike the 3 dimensional ones, possess no axis of rotation. This is equivalent, for linear transformations, with saying that there is no direction in the place which is kept unchanged by a 2 dimensional rotation, except, of course, the identity.
The question of the existence of such a direction is the question of existence of an eigenvector for the matrix A representing the rotation. Every 2D rotation around the origin through an angle {\displaystyle \theta }\theta in counterclockwise direction can be quite simply represented by the following matrix:
{\displaystyle A={\begin{bmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \end{bmatrix}}}{\displaystyle A={\begin{bmatrix}\cos \theta &-\sin \theta \\\sin \theta &\cos \theta \end{bmatrix}}}
A standard eigenvalue determination leads to the characteristic equation
{\displaystyle \lambda ^{2}-2\lambda \cos \theta +1=0}{\displaystyle \lambda ^{2}-2\lambda \cos \theta +1=0},
which has
{\displaystyle \cos \theta \pm i\sin \theta }{\displaystyle \cos \theta \pm i\sin \theta }
as its eigenvalues. Therefore, there is no real eigenvalue whenever {\displaystyle \cos \theta \neq \pm 1}{\displaystyle \cos \theta \neq \pm 1}, meaning that no real vector in the plane is kept unchanged by A.
Rotation angle and axis in 3 dimensions
Knowing that the trace is an invariant, the rotation angle {\displaystyle \alpha }\alpha for a proper orthogonal 3x3 rotation matrix {\displaystyle A}A is found by
{\displaystyle \alpha =\cos ^{-1}\left({\frac {A_{11}+A_{22}+A_{33}-1}{2}}\right)}{\displaystyle \alpha =\cos ^{-1}\left({\frac {A_{11}+A_{22}+A_{33}-1}{2}}\right)}
Using the principal arc-cosine, this formula gives a rotation angle satisfying {\displaystyle 0\leq \alpha \leq 180^{\circ }}{\displaystyle 0\leq \alpha \leq 180^{\circ }}. The corresponding rotation axis must be defined to point in a direction that limits the rotation angle to not exceed 180 degrees. (This can always be done because any rotation of more than 180 degrees about an axis {\displaystyle m}m can always be written as a rotation having {\displaystyle 0\leq \alpha \leq 180^{\circ }}{\displaystyle 0\leq \alpha \leq 180^{\circ }} if the axis is replaced with {\displaystyle n=-m}{\displaystyle n=-m}.)
Every proper rotation {\displaystyle A}A in 3D space has an axis of rotation, which is defined such that any vector {\displaystyle v}v that is aligned with the rotation axis will not be affected by rotation. Accordingly, {\displaystyle Av=v}{\displaystyle Av=v}, and the rotation axis therefore corresponds to an eigenvector of the rotation matrix associated with an eigenvalue of 1. As long as the rotation angle {\displaystyle \alpha }\alpha is nonzero (i.e., the rotation is not the identity tensor), there is one and only one such direction. Because A has only real components, there is at least one real eigenvalue, and the remaining two eigenvalues must be complex conjugates of each other (see Eigenvalues and eigenvectors#Eigenvalues and the characteristic polynomial). Knowing that 1 is an eigenvalue, it follows that the remaining two eigenvalues are complex conjugates of each other, but this does not imply that they are complex—they could be real with double multiplicity. In the degenerate case of a rotation angle {\displaystyle \alpha =180^{\circ }}{\displaystyle \alpha =180^{\circ }}, the remaining two eigenvalues are both equal to -1. In the degenerate case of a zero rotation angle, the rotation matrix is the identity, and all three eigenvalues are 1 (which is the only case for which the rotation axis is arbitrary).
A spectral analysis is not required to find the rotation axis. If {\displaystyle n}n denotes the unit eigenvector aligned with the rotation axis, and if {\displaystyle \alpha }\alpha denotes the rotation angle, then it can be shown that {\displaystyle 2\sin(\alpha )n=\{A_{32}-A_{23},A_{13}-A_{31},A_{21}-A_{12}\}}{\displaystyle 2\sin(\alpha )n=\{A_{32}-A_{23},A_{13}-A_{31},A_{21}-A_{12}\}}. Consequently, the expense of an eigenvalue analysis can be avoided by simply normalizing this vector if it has a nonzero magnitude. On the other hand, if this vector has a zero magnitude, it means that {\displaystyle \sin(\alpha )=0}{\displaystyle \sin(\alpha )=0}. In other words, this vector will be zero if and only if the rotation angle is 0 or 180 degrees, and the rotation axis may be assigned in this case by normalizing any column of {\displaystyle A+I}{\displaystyle A+I} that has a nonzero magnitude.[2]