Proof of a modified Perron-Frobenius theorem using complex analysis

Question

For context, the question I'm trying to solve (Q1 of this) asks us to prove a modified version of the Perron-Frobenius theorem using complex analysis. Let $A \neq 0$ be a $n \times n$ real matrix with non-negative entries. Let $\lambda_1, \dots, \lambda_n$ in $\mathbb C$ be the (possibly repeated) eigenvalues of $A$ and $M = \max\{ |\lambda_1|, \dots, |\lambda_n|\}$. The goal is to show that $M$ is an eigenvalue of $A$.

The question asks us to consider the functions $F(z) = \sum_{j=1}^n \frac{M}{M - \lambda_j z}$ and its Taylor series $f(z) = \sum_{k=0}^\infty c_k z^k$ around $0$. I've shown that $$ c_k = \frac 1{M^k}(\lambda_1^k + \cdots + \lambda_n^k) = \frac 1{M^k}\mathrm{Tr}(A^k), $$ and the radius of convergence of $f$ is $1$ (the trace bit tells us that $c_k$ is real and non-negative). The plan now is to show that $\sum_{k=0}^\infty c_k = \infty$, i.e. $f$ blows up at $z = 1$, which necessitates $F$ blows up at $z = 1$ (since $f$ and $F$ agree for $x$ real $\nearrow 1$), which means $M$ is an eigenvalue. Since $c_k\geq 0$, it suffices to show that $\sum_{k=0}^\infty c_k$ does not converge. Terms with eigenvalues of smaller modulus than $M$ clearly die when $k \to \infty$ due to the $M^k$ in the denominator, so let $\lambda_{j_1}, \dots, \lambda_{j_m}$ be the eigenvalues of modulus $M$, so let them be $Me^{i\theta_1}, \dots, Me^{i\theta_m}$. Then my intuition is that we can find infinitely many $n$ making $$ |e^{in\theta_1} + \cdots + e^{in \theta_m}| > \frac12, \quad \text{say}, $$ by "simultaneously aligning the angles near an integer multiple of $2\pi$ using large $n$", then we'll have the result by the divergence test.

However, I have no idea how to show this. It feels somewhat similar in spirit to the density of integer multiples of an irrational number mod $1$ in $[0, 1)$ (e.g. this post), but the different terms interacting makes it trickier.

Answer 1

(This is a long comment.) In my opinion, showing that $1$ is a singular point of $f$ is an overkill. (By the way, this follows from Pringsheim’s theorem.) We only need to show that $1$ is a singularity of $F$. This can be done in a fairly elementary way. First, we have the following Tauberian lemma.

Lemma. Let $F:[0,1]\to\mathbb R$ be continuous. If $F(x)$ agrees on $[0,1)$ with a power series $\sum_{k=0}^\infty c_kx^k$ with nonnegative coefficients, then $\sum_{k=0}^\infty c_k$ converges to $F(1)$.

Proof. For any fixed $l\ge0$, we have, on $[0,1)$, $$ \left(\sum_{k=0}^lc_k\right)x^l\le\sum_{k=0}^lc_kx^k\le\sum_{k=0}^\infty c_kx^k=F(x). $$ Since the LHS and the RHS are continuous functions of $x$ on $[0,1]$, by taking $x\to1^-$, we obtain $\sum_{k=0}^lc_k\le F(1)$. Hence $\sum_{k=0}^\infty c_k\le F(1)$, because $\left\{\sum_{k=0}^lc_k\right\}_{l\in\mathbb N}$ is an increasing sequence that is bounded above, and equality must hold because $\sum_{k=0}^\infty c_k\ge \sum_{k=0}^\infty c_kx^k=F(x)$ on $[0,1)$. $\square$

In relation to Q1(ii) in your question, suppose $A$ is not nilpotent (so that $M>0$). If $M$ is not an eigenvalue of $A$, then $F$ is holomorphic in the open unit disc. Since $A$ is real, the restriction of $F$ on $[0,1]$ is real-valued. By the Tauberian lemma above, we have $F(1)=f(1)$. But then $f(z)$ will be absolutely convergent in the closed unit disc. Now $F=f$ in the open unit disc and $f$ is continuous in the closed unit disc. Hence $F$ is continuous on the unit circle. Yet, this is impossible, because $z=M\lambda_i^{-1}$ is a pole of $F$ on the unit circle whenever $|\lambda_i|=M$.