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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

  1. $\rho\left(A\right)$ is the only eigenvalue of $A$ of maximum modulus?

  2. 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

Rodrigo de Azevedo
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user550103
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2 Answers2

2

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.

Robert Israel
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0

You can do this with Newton's Identities. The aperiodic case is the simplest form of Perron-Frobenius theory and there is exactly one max modulus eigenvalue $\rho(A)$, the Perron Root.

Now let $A$ have period $2\leq d\leq n$
(note $d\gt n \implies A$ is nilpotent, impossible). This tells you that $\text{trace}\big(A^k\big) =0$ when $k\%d \neq 0$ (where $\%$ is the modulo sign). Write out the characteristic polynomial for $A$:

$p(x) = x^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} +... + a_{1}x^{1}+ a_0$
where $a_n =1$ since we've defined the characteristic polynomial to be monic.
$a_{n-1} =0 = -\text{trace}\big(A\big)$ which is a Base Case for the inductive hypothesis on $r$, that $a_{n-r } = 0$ when $r\%d \neq 0$ for $r\leq n$. Applying Newton's Identities:

$a_{n-r} = -\frac{1}{r} \sum_{j=1}^{r} a_{n-r + j}\cdot \text{trace}\big(C^j\big) =0\text{ when r%d}\neq 0$
since $\text{trace}\big(C^j\big)\neq 0\implies j\%d = 0\implies a_{n-r + j}=0$ since $(-r+j)\%d = -r\%d \neq 0\implies a_{n-r + j}=0$ by (strong) induction hypothesis.

Conclude:
$\det\big(A\big)\neq 0 \implies a_{n-n}=a_0\neq 0\implies n\%d =0\implies d =n$ since $n$ is prime $\implies p\big(x\big) = x^n - c=\prod_{k=0}^{n-1} \Big(x-\omega^k\cdot\rho(A)\Big)$ for primitive $n$th root of unity $\omega$

user8675309
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