# On positive matrices and their eigenvectors

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).

• To avoid a trivial answer to (2), you must ask for a $B$ different from $PA^kP^{-1}$ where $P$ is any permutation matrix. – Denis Serre Apr 28 '11 at 8:50

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.

• Thank you for your answer. All above, I don't understand why $A^k$ is not in $\mathcal B_k$. Clearly, $A^k x = \lambda^k x$, so that $H(x)$ is the same here. I know examples of matrices $B$ whose size is different from that of $A$. Unfortunately, I don't have examples with matrices of the same size. – SIB Apr 30 '11 at 8:00
• You are right, I was making a silly mistake. Here $x=y$ so of course $H(x)=H(y).$ I'll fix it. – Aaron Meyerowitz Apr 30 '11 at 17:58

This question is very closely related to problems in dimension groups (partially ordered abelian groups with various other properties), specifically stationary ones, for which there is a history going back to the early 80s [e.g., by me, Positive matrices and dimension groups affiliated to C*-algebras and topological Markov chains, J of Operator Theory 6 (1981) 55--74].

For example, if $A$ has determinant $\pm 1$, then $H(x)$ is an invariant of shift equivalence (related to conjugacy, but somewhat coarser). More generally (without the determinant condition), one considers $\cup_{n=0}^{\infty} H(x)\lambda^{-n}$, the direct limit group, an invariant of shift equivalence. In particular, examples size two examples (with irreducible characteristic polynomial) exist with the same $H$ values, but aren't shift equivalent. Of course, we can also arrange non-shift equivalence by making sure the spectra are different.

However, it is also true that if $\det A =\pm 1$ and $A$ and $B$ are primitive, and the characteristic polynomials of $A$ and $B$ are equal to each other and irreducible over the integers, then $H_A = H_B$ entails that $A$ is conjugate to $B$. This however, is rather special. Drop the irreducibility of the characteristic polynomial and the result fails, as the example above shows. Drop the determinant condition (keeping irreducibility of the char poly), and the union determines the shift equivalence class, almost the same as conjugacy.

Likely what you are looking for is the connection between ideal class structure of orders in number fields and classification of the matrices.