Let’s consider the following $ n \times n$ cyclic stochastic matrix

$ $ M= \begin{pmatrix} 0 & a_2 & & & &b_n \\ b_1 & 0& a_3& &&& \\ & b_2 & 0& \ddots & & \\ & &\ddots&\ddots &a_{n-1} & \\ & && &0 &a_n \\ a_1 & & & &b_{n-1} &0 \end{pmatrix} $ $

such that $ \forall i$ , $ a_i,\,b_i$ are positive real number, $ a_i+b_i = 1$ and all other component of the matrix are zeros. This is a cyclic matrix in the sense that the associated graph is cyclic.

From the Perron-Frobenius theorem, the eigenvalues $ \lambda$ of such matrix all belong to the unit circle. $ $ (\Re \lambda )^2 + (\Im \lambda )^2 \leq 1 $ $

From numerical explorations, I believe that all eigenvalues of $ M$ belong to the ellipse $ $ (\Re \lambda )^2 + \frac{(\Im \lambda )^2}{(\tanh p)^2} \leq 1 $ $

where $ p$ denote $ p = \frac{1}{2}\ln \frac{\sqrt[n]{\prod_i a_i}}{\sqrt[n]{\prod_i b_i}}$ , assumed to be positive, otherwise inverse $ a_i$ and $ b_i$ .

One of the extremal case is the symmetric case $ a_i=b_i$ where $ p=0$ and all eigenvalues are real. The equality is reached in the uniform case of all $ a_i$ to being equal to some value and all $ b_i$ being equal to another value, the matrix being then a circulant matrix.

I can already prove that the imaginary part of the eigenvalue is bounded by $ \tanh p$ (see below), but I am unable to extend the prove to include the real part. I also try to play with the Brauer theorem about oval of Cassini exposed into [Horn & Johnson, Matrix Analysis], but it did not get me anywhere

Do you have any hints or suggestions to prove the inclusion of the eigenvalue into the ellipse?

Proof for the imaginary part:

Denote $ z$ the left eigenvector associated with eigenvalue $ \lambda$ , we have from the eigenvalue equation $ \lambda z = z M $ , $ $ \forall i,\quad \lambda = a_i \frac{z_{i-1}}{z_i} + b_i\frac{z_{i+1}}{z_i} = \frac{a_i}{a_i+b_i} \frac{z_{i-1}}{z_i} + \frac{b_i}{a_i+b_i} \frac{z_{i+1}}{z_i} $ $ , where $ i+1$ and $ i-1$ ar evaluated modulo $ n$ , ad the second equality follow from $ a_i+b_i=1$ .

By taking the product of the imaginary part of all previous equation and denoting $ p_i= \ln \sqrt{\frac{a_i}{bi}}$ , we get $ $ \Im \lambda = \sqrt{\prod_i \,a_i b_i \Im \frac{z_{i-1}}{z_i} \Im \frac{z_{i+1}}{z_i} }\prod_i \frac{\sinh (p_i+\frac{1}{2}\ln \Im\frac{ z_{i+1} }{z_{i}} \Im\frac{z_{i} }{z_{i-1}} )}{\cosh p_i} \leq \prod_i \frac{\sinh (p_i+\frac{1}{2}\ln \Im\frac{ z_{i+1} }{z_{i}} \Im\frac{z_{i} }{z_{i-1}} )}{\cosh p_i}$ $ The inequality use that $ \prod_i \Im \frac{z_{i-1}}{z_i}\leq 1$ . The concavity of $ \ln \sinh$ and the convexity of $ \ln \cosh$ , give the result $ $ \Im \lambda \leq \tanh p.$ $