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!



Growth of eigenvalues for certain sequences of matrices
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 $t<0$.


