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I know that in 2-dimension, a rotation matrix has no non-zero eigenvector. But in 3-d space, I can only imagine the rotation axis to be an eigenvector, and it looks like that the rotation angle can't be represent through eigenvectors or eigenvalues. So I guess in 3-dimension, any rotation matrix has and only has one eigenvector. Am I right?

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What if the matrix rotates $180^{o}$? Consider 2D and 3D cases. –  Thomas E. Jan 26 '12 at 14:16

3 Answers 3

The general situation is: by the spectral theorem, after rotating coordinates your matrix has a basis of orthonormal eigenvectors $v_1... v_n$ with corresponding eigenvalues $\lambda_1 ... \lambda_n$. If $\lambda_i$ is not real then $\bar{\lambda}_i$ will be the eigenvalue corresponding to $\bar{v_i}$ (take complex conjugates of each component). Then $Re(v_i) = {v_i + \bar{v_i} \over 2}$ and $Im(v_i) = {v_i - \bar{v_i} \over 2i}$ will be a real basis for the space spanned by $v_i$ and $\bar{v_i}$, and $M$ acts as a rotation on that two-dimensional subspace which is neither 0 nor 180 degrees (since it has no real eigenvectors in that subspace).

So in general, after rotating coordinates $M$ is a direct sum of 1 by 1 and 2 by 2 matrices, where the 2 by 2's are all rotation matrices, not by 0 or 180 degrees, and the 1 by 1's have to be either 1 or -1 since the eigenvectors of a rotation have magnitude 1. In 3 dimensions we thus have the following possibilities after rotating coordinates, keeping in mind that a rotation has to have determinant 1:

1) M is the identity matrix.

2) M is a diagonal matrix with two -1's and one 1 along the diagonal.

3) M has an eigenvector $v$ (the axis of rotation) and is a rotation in the two perpendicular directions that is neither 0 nor 180 degrees.

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You are right, but to be more precise it only has one real eigenvector. Rotations in 2 dimensions have the eigenvectors $(-i,1)$ and $(i,1)$ associated with the eigenvalues $\cos(t) - i \sin(t)$ and $\cos(t) + i \sin(t)$.

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This is not quite correct in general; rotate the $xy$ plane by 180 degrees; then $(1,0,0)$ and $(0,1,0)$ are eigenvectors with eigenvalue $-1$, and $(0,0,1)$ is an eigenvector with eigenvalue $1$. –  Arturo Magidin Jan 26 '12 at 16:16

In two dimension the result you quoted is false as stated: the matrix $$ \begin{pmatrix} -1 & 0 \\ 0 & -1\end{pmatrix} = \begin{pmatrix} \cos \pi & \sin \pi \\ -\sin \pi & \cos \pi \end{pmatrix} $$ is a rotation matrix. And every vector is an eigenvector.

It is true, however, if you explicitly disallow this particular case.

In three dimensions, note that since rotations preserves vector norms, you have that if $v$ is an eigenvector of a rotation $A$ you must have $Av = \pm v$. Now supposing you have two linearly independent eigenvectors $v$ and $w$. Let $u$ be the unique vector orthogonal to $v$ and $w$. Then you have $$ (Au)^Tv = u^T A v = u^T (\pm v) = 0 $$ and similarly $$ (Au)^Tw = 0 $$ and hence you get that $Au$ is proportional to $u$, and hence you must have $u$ is an eigenvector also.

This means that if a rotation matrix has more than 1 eigendirections, it must have a set of three linearly independent eigendirections.

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In two dimensions, $I$ is also a rotation matrix. –  Stefan Smith Mar 28 '13 at 22:39

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