I can't figure out:
On one hand: Yes, they are defective, because rotations have no eigenvalues/vectors and hence can't have an n-dimensional eigenspace for n-x-n matrix.
On the other:
No, because:
1. Rotations are necessarily orthogonal - they follow the form (in 2d):
2. Orthogonal matrices are necessarily normal (as per definition here)
3. Normal matrices are necessarily diagonalizable (as per spectral theorem), which is to say non-defective.
Which of course is a contradiction. Where is my logic flawed?
Excluding the cases when $\theta = n\pi$, where $n = 0, 1,2,...$, rotation matrices don't have real eigenvalues. They do have complex eigenvalues. You need to extend to the complex field.
I'll refer you to this document: https://mathcs.clarku.edu/~ma130/complexeigen.pdf
When $\theta = n\pi$, there is a single eigenvalue, but you'll find that the eigenspace is two-dimensional as the matrices are just the identity matrix and its negative. Any vector multiplied by the matrix will either be itself or its negative. Thus, every vector in the space is an eigenvector. And the matrix is obviously diagonal already.
For the other $\theta$, there are two distinct complex eigenvalues with an eigenvector each.