Question

The Perron-Frobenius theorem says that the largest eigenvalue of a positive real matrix (all entries positive) is real. Moreover, that eigenvalue has a positive eigenvector, and it is the only eigenvalue having a positive eigenvector.

Now suppose we want to construct a positive rational matrix with a particular Perron-Frobenius eigenvalue. Specifically, consider a positive real algebraic number $\lambda$ which is greater in absolute value than all of its Galois conjugates. Does there exist a positive rational matrix $A$ with $\lambda$ as its Perron-Frobenius eigenvalue?

Answer 1

The answer to a sharper question involving integers, rather than rationals, is affirmative.

Let $\lambda$ be a positive real algebraic integer that is greater in absolute value than all its Galois conjugates ("Perron number" or "PF number"). Then $\lambda$ is the Perron–Frobenius eigenvalue of a positive integer matrix.

(The converse statement is an integer version of the Perron–Frobenius theorem, and is easy to prove.)

In a slightly weaker form (aperiodic non-negative matrix), this is theorem of Douglas Lind, from

The entropies of topological Markov shifts and a related class of algebraic integers. Ergodic Theory Dynam. Systems 4 (1984), no. 2, 283--300 (MR)

I don't have a good reference for the strong form, but it was discussed at Thurston seminar in 2008-2009. One interesting thing to note is that, while the proof can be made constructive, it is non-uniform: the size of the matrix can be arbitrarily large compared to the degree of $\lambda$.

Answer 2

For the problem of constructing a strictly positive real matrix $A$ of size $N \times N~$ with Perron-Frobenius eigenvalue $\lambda$, I offer the following solution without proof:

Choose any strictly positive real column vector $\pi$ of length $N~$ s.t. $||\pi||_1 = \lambda$.

Then $A = \left|\begin{array}{ccc} \pi & \pi & ... & \pi \end{array}\right|$, i.e. the $N \times N$ matrix with every column equal to $\pi$.

$\lambda$ is the Perron-Frobenius eigenvalue of the matrix $A$ and the corresponding eigenvector is $\pi$.

(note: this implies a family of solutions for any $\lambda$, however I make no claim that these are the only solutions)