TITLE: Perron-Frobenius "inverse eigenvalue problem"
QUESTION [29 upvotes]: 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?

REPLY [4 votes]: Perhaps you're thinking of the strong form given in the Roy Adler conference
Symbolic Dynamics and Its Applications (Contemporary Mathematics) [1992].
One of the papers there by me was answering a question of Doug Lind, to show that if $\lambda $ exceeds the absolute value of its algebraic conjugates, then it appears as the Perron eigenvalue of a primitive integer matrix. In general, you can arrange it so that the nonzero spectrum consists of $\lambda$ and its conjugates (multiplicity one each of course) together with a usually large number of $1$s. The presence of many more eigenvalues (hence there is no control on the size of the realizing matrix) is necessary, since we must also have $\text{tr} A^n \geq 0$ for all $n$ (and other conditions) if $A$ is a primitive integer matrix.