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?



Perron-Frobenius "inverse eigenvalue problem"
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](http://www.ams.org/mathscinet-getitem?mr=766106))


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 $\lambda$.


