Apply a fraction of a rotation matrix without extracting axis-angle In 3D, you can take any pure rotation matrix and find an axis-angle representation of the same transformation (although not necessarily unique).  From that representation, you could create a new matrix that represents a fractional part of the original transformation by taking a fraction of the original angle, keeping the original axis, and creating a new matrix.
In high-dimensional space, axis-angle doesn't make sense since it seems that rotation actually happens within a plane and not necessarily around an axis per se.  Is it possible to find a rotation matrix that rotates a space by some fraction of the angle by which some other rotation matrix rotates without first finding the plane and angle of rotation?
 A: Essentially you are asking to raise the rotation matrix to an arbitrary power. To do this you can use the fact that $X^a=\exp(a \log X)$ for any matrix $X$. To compute the matrix logarithm we use
$$\log X = \log (I - (I-X)) = -\sum_{n=1}^\infty \frac{(I-X)^n}{n}$$
Now if you can diagonalize $I-X$ (perhaps there is a proof that this is always possible for $X$ a rotation matrix?) to give $I-X=SDS^{-1}$ then you have
$$\log X = -S \left(\sum_{n=1}^\infty \frac{D^n}{n}\right) S^{-1}$$
which is fast to compute (again you'd need a proof that this converges when $X$ is a rotation matrix). Now you compute the matrix exponential in the same way. Letting $\log X=\hat{S}\hat{D}\hat{S}^{-1}$ for $\hat{D}$ diagonal,
$$X^a = \exp(a\log X) = \hat{S} \left(\sum_{n=0}^\infty \frac{a^n \hat{D}^n}{n!}\right) \hat{S}^{-1}$$
which again is fast to compute.

Thinking off the top of my head now, it seems that since all rotations are in a plane, the eigenvalues of a rotation X in $\mathbb{R}^n$ must be $e^{\pm \mathrm{i}\theta}$ for some $\theta$ (once each) and $1$ ($n - 2$ times). Therefore the matrix $I-X$ has eigenvalues $1-e^{\pm\mathrm{i}\theta}$ and $0$ ($n - 2$ times), so its diagonalisation D has a particularly simple form. 
