Find the eigenvalues of the following matrix: $$\begin{bmatrix} 3&2&3&0&0&0 \\ 0&4&2&0&0&0 \\ 0&0&5&0&0&0 \\ 0&0&0&0&1&0 \\ 0&0&0&0&0&1 \\ 0&0&0&1&0&0 \end{bmatrix}$$
I understand that you can partition this matrix and the way to get the eigenvalues of 3,4,5 are pretty straightforward. However, I'm confused about how to get the complex eigenvalues, which are (1$\pm \sqrt{3}i$)/2. Can anyone help me out here? I have the solutions, so just need a walkthrough about how to get there.
There’s the tried-and-true general method: find the roots of the characteristic polynomial of $$\begin{bmatrix}0&1&0\\0&0&1\\1&0&0\end{bmatrix},$$ which is $\lambda^3-1$, so its eigenvalues are the three cube roots of unity: $1$ and $e^{\pm 2/3\pi i}$.
You might also recognize this as a 3-d rotation matrix, so you know that one of its eigenvalues will be $1$ and the other two are $e^{\pm i\theta}$, where $\theta$ is the rotation angle. It’s not too hard to visualize this particular rotation: it’s 120° about the axis $(1,1,1)^T$. If you don’t happen to spot that, you can extract the angle from the trace of the matrix. Since the trace is equal to the sum of the eigenvalues, the trace of a 3-d rotation matrix is equal to $1+e^{i\theta}+e^{-i\theta}=1+2\cos\theta$. The sign ambiguity isn’t a problem here since were going to use both $\theta$ and $-\theta$, anyway.
Finally, you might recognize this a circulant matrix, in particular a cyclic permutation matrix whose eigenvalues are the cube roots of unity.