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?

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    $\begingroup$ Nice question. For unique largest eigenvalue and unique positive eigenvector there is also the hypothesis of irreducibility (equivalent to the directed graph of the matrix being strongly connected), although that doesn't affect your question. As a start, if some polynomial having $\lambda$ as a root has only a single sign change that happens after the top degree, then the companion matrix fits the bill. $\endgroup$
    – Tracy Hall
    Aug 12, 2010 at 6:58

3 Answers 3


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 $\lambda$.

  • $\begingroup$ Thanks. It's cool that this works for algebraic integers and integer matrices. Also, I'm pretty sure the irreducible non-negative version will work for my purposes; positive matrices were just convenient. Some tangentially related math I came across while trying to answer this question: Apparently some special Perron numbers, Pisot and Salem numbers, come up in Diophantine analysis. en.wikipedia.org/wiki/Pisot-Vijayaraghavan_number books.google.com/books?id=yxeRXLG19cwC&printsec=frontcover $\endgroup$ Aug 13, 2010 at 5:20
  • $\begingroup$ You are welcome! PV numbers (algebraic integers $\lambda$ s.t. all Galois conjugates except $\lambda$ itself are in the open unit circle) are very interesting and for them it becomes easier to solve this and a few related problems, such as construction of self-similar tilings with expansion constant $\lambda.$ $\endgroup$ Aug 13, 2010 at 5:40
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    $\begingroup$ The strong version doesn't seem to be true. The PF eigenvalue is bounded from below by the smallest row sum. So if $\lambda$ has degree $g$, the strong version would imply $\lambda \ge g$. But for each $g$ there is a Perron number $\lambda_g$ such that $\log \lambda_g \approx 1/g$. $\endgroup$ Mar 16, 2016 at 13:51

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.


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)

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    $\begingroup$ This fails because we are requiring the entries to be rational, and lambda might not be rational. $\endgroup$ Nov 23, 2011 at 7:23

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