Suppose we have an aperiodic matrix $A_t$ that has entries that are either $0$ or are positive integer powers of $t$, i.e. we could have $$A_t = \begin{pmatrix} 0 & t & t^2\\ t & t^2 & 0\\ t & 0 & t \end{pmatrix}$$ for example.

Suppose $t>0$ and let $\Lambda(t)$ denote the unique, real, simple maximal eigenvalue of $A_t$ guaranteed by the Perron-Frobenius Theorem. If we consider the function $$f(t) = \log\Lambda(e^t)$$ then it is possible to show using a variational principle and perturbation theory that $f(t)$ is increasing, convex and analytic (this is non-trivial!) with uniformly bounded (for $t\in\mathbb{R}$) first derivative. In particular the limits $$\lim_{t\to\infty} \frac{f(t)}{t} = \alpha_1 \ \ \text{and} \ \ \lim_{t\to - \infty} \frac{f(t)}{t} = \alpha_2$$ both exist and are finite. My question is the following: can we calculate the error term associated to these limits? That is, can we find $g(t)$ such that $$f(t) = \alpha_1 t + O(g(t))$$ as $t\to\infty$ for example?

Edit: I would ideally like to show that the error is $O(1)$. This is equivalent to the fact that $\Lambda(e^t)$ grows purely exponentially, i.e. there exists $C, \lambda \ge 1$ such that $$\frac{1}{C} \lambda^t \le \Lambda(e^t) \le C \lambda^t$$ for all $t$ sufficiently large. An equivalent inequality should also hold for $-t$ sufficiently large.

Any thoughts/insights would be greatly appreciated - thanks!

  • $\begingroup$ We have $A_t=tB+O(t^2)$. Then we have $\Lambda(A_{e^t})=e^t\Lambda(B)+O(e^{2t})$. Hence $log \Lambda(A_{e^t})=t+log(\Lambda(B))+O(e^{t})$. Hence if $B\neq 0$, $\alpha_2=1$ and error is $log(\Lambda(B))$. $\endgroup$ – user35593 Nov 30 '20 at 13:09
  • $\begingroup$ Thanks for your reply, but I'm confused by your comment - the expression $\Lambda(A_{e^t})=e^t\Lambda(B) + O(e^{2t})$ is an asymptotic expression where the "error term" $e^{2t}$ is larger than the "lead term" $e^t$ as $t \to\infty$. $\endgroup$ – Zestylemonzi Nov 30 '20 at 13:39
  • $\begingroup$ yes it only works for $t\rightarrow -\infty$ sorry $\endgroup$ – user35593 Nov 30 '20 at 14:57

Denote by $\mu(A_t)$ the max spectral radius of $A_t$ which is defined as the maximal cycle geometric mean $$ \mu(A_t) := \max \{ (a_{i_1i_2}a_{i_2i_3}\dots a_{i_ki_1})^{1/k}\}$$ where the maximum is taken over all cycles in the matrix $A$, $k$ is the length of the cycle, and for each cycle the indices $i_1,\ldots,i_k$ are distinct (so that it is really a cycle). Here this means that $$ \mu(e^t) := \mu(A_{e^t}) = \exp\left( \frac{\ell}{k} t\right)$$ for some maximal cycle of length $k$ and the powers along this cycle sum to $\ell$. In particular, the maximum is attained in the same cycle for all $t\geq0$, namely one in which sum of the exponents divided by length of the cycle is maximized. For $t<0$ on the other hand we need to minimize the factor $\ell/k$ over all cycles.

If $n$ is the size of the matrix, then it is known for any nonnegative matrix $B$ that $$ \mu(B) \leq \Lambda(B) \leq n \mu(B), $$ (equation (6.10) in Elsner, Johnson, Dias da Silva, The Perron root of a weighted geometric mean of nonnegative matrices. Linear and Multilinear Algebra 24 (1988) 1-13.)

So here this means $$ \exp\left( \frac{\ell}{k} t\right) \leq \Lambda(e^t) \leq n\exp\left( \frac{\ell}{k} t\right),\quad t\geq0$$ where the $\ell/k$ is actually easy to get from the matrix $A_t$. Similarly, but with a different constant (in general) for $t<0$.


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