Growth rates of ‘sub-traces’ of matrices

Question

Consider an aperiodic, non-negative and non-zero $m\times m$ matrix $A$. Here aperiodic means that there exists an integer $n$ such that every entry of $A^n$ is strictly positive. By the Perron–Frobenius theorem $A$ has a strictly positive, real, simple maximal eigenvalue $\lambda >0$. In particular the trace of $A^n$ grows exponentially like $\lambda^n$, i.e. $$ \operatorname{Tr}{(A^n)} = \lambda^n(1+O(\theta^n)) $$ for some $0 < \theta <1$. Let us write $P_n$ for the set of periodic $n$-long strings that are ‘allowed by $A$’: $$P_n = \{I = i_1,\dotsc,i_n : i_j \in \{1,\dotsc, m\} \text{ with } i_1= i_n \text{ and } A_I >0\} $$ where each $i_j \in \{1,\dotsc,m\}$ denotes a row/column of $A$ and for each $I = i_1,\dotsc,i_n$ we write $A_I = A_{i_1, i_2}A_{i_2,i_3}\dotsm A_{i_{n-1},i_n}$ ($A_{i,j}$ is the $i,j$th entry of $A$). In this new notation, $$ \operatorname{Tr}(A^n) = \sum_{I \in P_n} A_I. $$ Note that the growth rate of the cardinality of $P_n$ is purely exponential (indeed it is the growth rate if the trace of the matrix $\widetilde{A}$ where $\widetilde{A}_{i,j} = 1$ if $A_{i,j} >0$ and $\widetilde{A}_{i,j}=0$ otherwise).

My question is the following: if for each $n \ge 1$ we have a subset $Q_n \subset P_n$ such that $$ \limsup_{n\to\infty} \frac{1}{n} \log \#Q_n < \limsup_{n\to\infty} \frac{1}{n} \log \#P_n $$ is it true that $$ \limsup_{n\to\infty} \frac{1}{n} \log \sum_{I \in Q_n} A_I < \limsup_{n\to\infty} \frac{1}{n} \log \sum_{I \in P_n} A_I = \lambda\ \ ? $$ That is, if we looks at the contribution to the trace of $A^n$ coming from an ‘exponentially smaller’ subset of strings, do we see a drop in the exponential growth rate of $\operatorname{Tr}(A^n)$?

Answer 1

If I understand your question correctly the answer is negatif: Take the matrix $\left(\begin{array}{ccc}100&0&0\\0&1&1\\0&1&1\end{array}\right)$. Consider $Q_n=\{1^n\}$ and $P_n$ is the union of $Q_n$ with $\{2,3\}^n$. Only $Q_n$ contributes asymptotically.

Added: In order to get your result you need a 'connectivity-condition': Every word in $Q_n$ can be realized as a subword of $u$ in $P_m\setminus Q_m$ for some $m>n$. This can be rephrased in terms of the directed graph $\Gamma$ with oriented edges $(i,j)$ if $A_{i,j}>0$. If for every $i,j$ there is an oriented path in $\Gamma$ which starts at $i$ and ends at $j$ then your result holds. Otherwise you get an invariant subspace generated by a subset of basis vectors ($\mathbb R(1,0,0)$ in the above example) which can contain the Frobenius-vector. This gives a counterexample.