I’m looking at $ N\times N$ matrices $ M_N$ with elements $ $ M_N=\left( \rho^{|i-j|} \right)_{i,j=1}^N,$ $ where $ \rho$ is a complex number of unit modulus. These matrices with $ \rho\in\mathbb R$ and $ |\rho|<1$ have been studied in detail before, with a nice exposition to be found here, which includes some more references.

In the cited article, there is an implicit form of the eigenvalues, given through $ $ \lambda_j = \frac{1-\rho^2}{1-2\rho\cos\theta_j+\rho^2}, $ $ where $ \theta_j$ are roots of the following function $ $ G(\theta) = \sin[(n+1)\theta]-2\rho\sin[n\theta]+\rho^2\sin[(n-1)\theta]. $ $ (Even though only real $ \rho$ were considered in the article, this works for complex $ \rho$ also.)

Intriguingly, if $ \rho=e^{i\phi}$ , the largest eigenvalue of $ M_N$ seems to be essentially independent of $ \phi$ (as long as $ \phi$ is sufficiently different from 0 (i.e., $ \mathcal O(1/N)$ ). Since the inverse of $ M_N$ is almost tridiagonal (see the article for its form), it can be efficiently diagonalized numerically. I’ve checked up to $ N=10^6$ and the largest eigenvalue (mostly real) seems to *almost* follow a power law (roughly $ N^{0.85}$ ), but not quite. In fact, it looks slightly curved, so perhaps it is approaching $ N$ .

Another important thing I’ve noticed is that the $ \theta$ corresponding to the largest eigenvalue is very close to $ \phi$ , and seems to approach $ \phi$ as $ N\to\infty$ . Indeed, it is the same scaling as the actual eigenvalue $ \lambda$ , which follows from the fact that $ \theta=\phi$ makes the denominator of the formula for $ \lambda$ vanish. Expanding $ \theta=\phi+\delta\phi$ it becomes clear that $ $ \ln\lambda\to -\ln(\theta-\phi) + \text{const.} $ $

So I’ve been trying for a long time to extract how $ \theta$ approaches $ \phi$ from $ G(\theta)$ , but failed. I’d appreciate any pointers to a solution.