Recently encountered a matrix that looks similar to a circulant, and in case of a $5\times 5$ matrix, we have: $$ A = \begin{pmatrix} \omega^0 c_0 & \omega^1 c_1 & \omega^2 c_2 & \omega^3 c_2 & \omega^4 c_1 \\ \omega^1 c_1 & \omega^2 c_0 & \omega^3 c_1 & \omega^4 c_2 & \omega^5 c_2 \\ \omega^2 c_2 & \omega^3 c_1 & \omega^4 c_0 & \omega^5 c_1 & \omega^6 c_2 \\ \omega^3 c_2 & \omega^4 c_2 & \omega^5 c_1 & \omega^6 c_0 & \omega^7 c_1 \\ \omega^4 c_1 & \omega^5 c_2 & \omega^6 c_2 & \omega^7 c_1 & \omega^8 c_0 \end{pmatrix},$$ where $\omega = e^{2 \pi /N}$ is the $N^{\text{th}}$ root of unity, where $N$ is the dimension on which the matrix operates, and $c_{l}$ are some complex valued constants. So, this matrix closely resembles a complex symmetric circulant, except for these additional phase factors.
Are there any ideas on how to diagonalize? If no, at least may be there is a way to represent it as a product of 2-3 matrices, and that will give a hint on how to diagonalize it.
What I noticed up to now is that $P^{-2} \cdot \text{DFT}_{N}^{\dagger} \cdot A \cdot \text{DFT}_{N}$ produces a diagonal matrix, where $P$ is the circular shift matrix, and $\text{DFT}_{N}$ is the DFT matrix that diagonalizes any circulants. However, I must admit, that the diagonal elements are not the eigenvalues of $A$, which is understandable, as the transformation is not a similarity transformation, and this fact might be useless at all in the diagonalisation procedure :). However, it tells that a regular DFT matrix transforms $A$ to the following form: $$ \text{DFT}_{N}^{\dagger} \cdot A \cdot \text{DFT}_{N} = \begin{pmatrix} 0^{(2 \times N-2)} & D^{(2 \times 2)} \\ D^{(N-2 \times N-2)} & 0^{(N-2 \times 2)} \end{pmatrix}, $$ where $0^{(j \times k)}$ is a matrix of zeros of the corresponding size, and $D^{(j \times k)}$ is some diagonal matrix. So, it is already pretty close to diagonal.