Eigenvalues of a block circulant matrix How does one calculate the eigenvalues of a unitary 3 by 3 block circulant matrix where
$$U = -\frac{i}{3} \begin{pmatrix}
    \Lambda_{1} & \Lambda_{2} & \Lambda_{3} \\
    \Lambda_{3} & \Lambda_{1} & \Lambda_{2} \\
    \Lambda_{2} & \Lambda_{3} & \Lambda_{1} \end{pmatrix}$$
where
$$ \Lambda_{1}  = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & -2\sqrt{2} \\ -2\sqrt{2} & 0 & 0 & -1 \\ 0 & 0 & -1 & 0 \end{pmatrix}, \quad \Lambda_{2} = \begin{pmatrix} 0 & 0 & -2\sqrt{2} & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} ,  \quad \Lambda_{3}  = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & -2\sqrt{2} & 0 & 0 \end{pmatrix} $$

This won't be a full answer, but here's some thoughts that (I think) should move you in the right direction.
Let $P$ denote the matrix
$$
P = \pmatrix{0&1&0\\0&0&1\\1&0&0}
$$
Then, in terms of Kronecker products, we can write your matrix as
$$
3iU = I \otimes \Lambda_1 + P \otimes \Lambda_2 + P^2 \otimes \Lambda_3
$$
It is actually very easy to calculate the eigenvalues of $(P \otimes \Lambda_2 + P^2 \otimes \Lambda_3)$: notably, $\Lambda_2,\Lambda_3$ commute, and are both nilpotent.  It follows that $P \otimes \Lambda_2$ and $P^2 \otimes \Lambda_3$ are also commuting nilpotent matrices.  It follows that $(P \otimes \Lambda_2 + P^2 \otimes \Lambda_3)$ is a nilpotent matrix, which means that its only eigenvalue is $0$.  Moreover, it satisfies $(P \otimes \Lambda_2 + P^2 \otimes \Lambda_3)^2 = 0$.  Its rank is at most $3$.
The $\Lambda_1$ term, however, makes things a bit tricky, and I'm not sure how to continue this line of analysis.
Here's a "brute force" approach that might help.


*

*Your matrix $\Lambda_1$ is diagonalizable.  To begin, compute an invertible $S$ (whose columns are eigenvectors) such that $D = S^{-1}\Lambda_1S$ is diagonal.  The eigenvalues of $\Lambda_1$ are the $4$ distinct (but non-real) roots of its characteristic polynomial, $p(x) = x^4 - 2x^2 + 9$

*Let $V = 3iU$ (so that I can be lazy and leave off the $-\frac i3$).  We note that
$$
(I \otimes S)^{-1} V (I \otimes S) = 
\pmatrix{S\Lambda_{1}S^{-1} & S\Lambda_{2}S^{-1} & S\Lambda_{3}S^{-1} \\
    S\Lambda_{3}S^{-1} & S\Lambda_{1}S^{-1} & S\Lambda_{2}S^{-1} \\
    S\Lambda_{2}S^{-1} & S\Lambda_{3}S^{-1} & S\Lambda_{1}S^{-1}} = \\
\pmatrix{D & M_2 & M_3 \\
    M_3 & D & M_2 \\
    M_2& M_3 & D} = I \otimes D + P \otimes M_2 + P^2 \otimes M_3
$$
The matrix $I \otimes D$ is diagonal, and the matrix $(P \otimes M_2 + P^2 \otimes M_3)$ is similar to the matrix $(P \otimes \Lambda_2 + P^2 \otimes \Lambda_3)$ analyzed above.  The matrices $M_2,M_3$ have rank $1$.


I hope you find these observations to be useful.
