Frobenius theorem experts,
Why for an irreducible, non-negative, and non-singular matrix $A \in M_n$ (where $n$ is a prime number) is either
$\rho\left(A\right)$ is the only eigenvalue of $A$ of maximum modulus?
or all the eigenvalues of $A$ have maximum modulus?
(how the prime number size of a matrix is playing a role?)
Thank you so much in advance
If $A$ is aperiodic, the Frobenius theorem tells us there is only one eigenvalue of maximum modulus. So suppose $A$ is periodic with period $q > 1$. Consider the directed graph corresponding to $A$, with vertices $\{1,\ldots,n\}$ and an arc $i \to j$ iff the matrix entry $a_{ij} \ne 0$. The vertices are partitioned into $q$ classes $C_1, \ldots, C_q$ where arcs from $C_i$ all go into $C_{i+1}$ (identifying $C_{q+1}$ as $C_1$). Now if $C_i$ has more vertices than $C_{i+1}$, the matrix rows corresponding to $C_i$ must be linearly dependent, implying that $A$ is singular. So if $A$ is nonsingular, all $C_i$ have the same number $m$ of vertices, and $n = m q$. But if $n$ is prime, we must have $m = 1$ and $q = n$. Thus the directed graph is a cycle, $A^q$ is a constant multiple of $I$, and the eigenvalues of $A$ are the $q$'th roots of that constant, which all have the same modulus.