Consider a $(N-1)\times(N-1)$ matrix A with real entries
$$ A_{mn} = \begin{cases} \sum_{l=1}^{N-1} K_{ml} \cos{\left(\theta_l-\theta_m\right)}+2 K_{mN}\cos{\theta_m} & \text{if $m=n$}\\ -K_{mn}\cos{\left( \theta_n-\theta_m \right) }+K_{nN}\cos{\theta_n} & \text{if $m \neq n$} \end{cases} $$
composed of real angles $\theta_j\in[0,2\pi)$ for all $1\leqslant j \leqslant N-1$ and real positive symmetric coefficients $K_{jl}=K_{lj}$ for all $1\leqslant j,l \leqslant N$.
Is A diagonalizable over $\mathbb{R}$ (as numerical sampling suggests), or at least over $\mathbb{C}$?
If no general answer is possible: what if we additionally set $K_{jl}=1$ for all $1\leqslant j,l \leqslant N$?
Or can we at least say that almost all A are diagonalizable over $\mathbb{C}$ (or even $\mathbb{R}$), i.e., diagonalizable with probability 1 for randomly chosen angles and coefficients?
For sufficiently small $N$, analytically computing the eigenvectors would do the job, but I wonder if there is a general result for unspecified $N$ - or a counterexample. Indeed A is the sum of a symmetric matrix and a matrix whose $n$-th column vector reads as $K_{nN}\cos{\theta_n}\cdot(1,1,..,1)^T$, but this doesn't get me any further.