Science talk:Math Exam Resources/Courses/MATH152/April 2015/Question B 6 (a)

ZIMINGYIN:

The solution is correct, but excessively complicated and ignores the geometry of the question.

One should first determine the angle of rotation $\theta$ about the axis. The eigenvalues are then given by $1,e^{\pm i\theta }.$ (Note that this implies that the trace of the matrix is $1+2\cos \theta .$)

But an even more elegant solution for this particular problem is as follows (sketched below; please fill in the details):

A rotation matrix is orthogonal, and therefore has determinant $\pm 1$; it is easy to see in this case that the determinant is 1. One of the eigenvalues must be 1 (it's a rotation in an odd-dimensional space); let the two others be $z_{1},z_{2}$. Since the sum of the eigenvalues is the trace of the matrix, and since their product is the determinant, we have $1+z_{1}+z_{2}=1$ and $z_{1}z_{2}=1$, whence $\{z_{1},z_{2}\}=\{i,-i\}.$