Let $A$ be an $n\times n$ positive integer-valued matrix, that is every entry of $A$ is a a positive integer. Let $\lambda$ be the Perron-Frobenius eigenvalue and $x = (x_1,…,x_n)^T$ the corresponding positive probability eigenvector: $\sum_i x_i =1, \ x_i > 0$. Denote by $H(x)$ the additive subgroup of $\mathbb R$ whose generators are the coordinates of $x, \
H(x) = < x_1,…,x_n >$.
Fix any integer $k \geq 1$ and consider the set of positive integer-valued matrices $\mathcal B_k$ formed by all $n\times n$ matrices $B$ satisfying the following conditions: $\lambda^k$ is the Perron-Frobenius eigenvalue for $B$, and if $By = \lambda^k y,\ \sum_i y_i = 1, y_i >0,$ then $H(y) = H(x)$.
My questions are as follows.
(1) Is there an algorithm describing all matrices from $\mathcal B_k$?
(2) How can one find at least one matrix $B$ in the set $\mathcal B_k$ different from $PA^kP^{-1}$ where $P$ is a permutation matrix?
Comments: (i) The case when $\lambda $ is an integer is not interesting, so that one can assume that $\lambda$ is an algebraic number. (ii) I asked a similar question before but these ones seems formulated in more precise form. (iii) Of course, (2) is simpler than (1), and I actually need a constructive answer to (2).
I'll be glad to see any comments, suggestions, references.
Best Answer
Here is an example which answers several of your questions. It uses 2 different $7 \times 7$ matrices $A_G$ and $A_H$ whose entries are $0$ and $1$. These zeros are OK for your question because $A_G^4$ and $A_H^4$ are positive matrices (they are adjacency matrices of graphs each with an odd cycle)
Above are two graphs. Their respective adjacency matrices share some but not all eigenvalues. They do have the same largest eigenvalue and share an eigenvector for that eigenvalue.
The characteristic polynomial of the adjacency matrix of $G$ is $$(x+1)^2(x^2-3)(x^3-2x^2-3x+2)$$ while that of $H$ is $$(x-1)(x+1)(x^2+2x-1)(x^3-2x^2-3x+2)$$ The Perron-Frobenius eigenvalue of each comes from the shared cubic whose largest root is $\lambda \approx 2.8136.$ An eignvector for the adjacency matrix (or any other vector in $\mathbb{R}^7$ in this case) can be viewed as a weighting of the vertices of the graph. This is illustrated above where $a=\lambda-1$ and $b=\lambda^2-2\lambda-1$. This particular vector has entries which sum to $\lambda^2+1$. Dividing through by this and simplifying makes the entries $\frac{1}{\lambda^2+1}=\frac{6+\lambda-\lambda^2}{8}$ four times, $\frac{\lambda-1}{\lambda^2+1}=\frac{\lambda-2}{4}$ twice, and $\frac{\lambda^2-2\lambda-1}{\lambda^2+1}=\frac{\lambda^2-2\lambda-2}{2}$ once. Hence $H(x)=\mathbb{Z}[\frac{1}{2},\frac{\lambda}{8},\frac{\lambda^2}{8}].$ So the $\mathcal{B}_k$ for $A_G$ also contains $A_H^k$ as well as all the matrices $A_G^jA_H^{k-j}$
So I don't see any hope of classifying $\mathcal{B}_k$. I'm sure that there are also examples which don't have the same eigenvector and examples where there is a matrix $M$ with Perron-Frobenius eigenvalue $\lambda^k$ eigenvalue but such that $M$ is not the $k$th power of any matrix with $\lambda$ as an eigenvalue. Here I wanted an example that was small, simple, and did not involve two matrices with exactly the same spectrum.