Let R be a rotation matrix $R^TR=I$ and $\mathrm{det} R=1$. Usually we don’t talk about rotation matrix in higher space, though the definition of the rotation matrix is not constrained on 2D or 3D.

1. Eigenvalues and Eigenvectors:

Consider the rotation axis of R is x, then Rx=x. So R must have an eigenvalue as 1. The corresponding eigenvector is the rotation axis.

The other eigenvalues are all complex. And $|\lambda_i|=1$. That means the eigenvalues is the root of $\lambda^d=1$. For 3D, the eigenvalues are

$\{1, e^{-\theta}, e^{+\theta}\}$

2. Singular values and Singular vectors:

An SVD of A is $A=U\Sigma V^T$. The singular values of A is the square root of the eigenvalues of $A^TA$ and $AA^T$. If A is a rotation matrix, then $A^TA=AA^T=I$. So the eigenvalues of $I$ is 1. So the singular values of $A$ is all 1. So the condition number of a rotation matrix is

$\kappa=\frac{\lambda_{max}}{\lambda_{min}}=1$

So the numerical stability of R is the best!

The singular vectors of A is respectively the eigenvectors of $A^TA=I$ and $AA^T=I$.

Questions remained:

(1) why does the eigenvalues of R be the root of x^d=1?

(2) the eigenvector of R associating its complex eigenvalue is real or complex?