# Growth of eigenvalues for certain sequences of matrices

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!

• 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 $log(\Lambda(B))$. – user35593 Nov 30 '20 at 13:09
• 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 $t \to\infty$. – Zestylemonzi Nov 30 '20 at 13:39
• yes it only works for $t\rightarrow -\infty$ sorry – 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 $$t<0$$.