A Perron number is an algebraic number which is greater than one in absolute value and is greater than all of its Galois conjugates in absolute value as well. Lind's theorem states that any Perron number is spectral radius of some Perron-frobenius matrix (a matrix $A$ is Perron-Frobenius if all of its entries are non-negative integers and there is some $n>0$ such that all entries of $A^n$ are positive).

Question: Given a Perron number $\lambda$, is there any lower bound for the size of smallest Perron-Frobenius matrix $A$ with spectral radius equal to $\lambda$ (called Perron-frobenius degree)? Besides algebraic degree off course

PS: In particular I'm interested in Perron numbers which come from a topological construction, namely as stretch factors of pseudo-Anosov homeomorphisms of hyperbolic surfaces. There is an upper bound for size of such matrix , 6g-6 where g is the genus of closed surface, however I'm not aware of lower bounds.


1 Answer 1


If a Perron number $\lambda$ has negative trace, then any Perron-Frobenius matrix must have size strictly greater than the algebraic degree of $\lambda$, for example the largest root of $x^3 + 3x^2-15x-46$.

If $B$ denotes the $d\times d$ companion matrix of the minimal polynomial of $\lambda$ (which of course can have negative entries), then $\mathbb{R}^d$ splits into the dominant 1-dimensional eigenspace $D$ and the direct sum $E$ of all the other generalized eigenspace. Although I've not worked this out in detail, roughly speaking the smallest size of a Perron-Frobenius matrix for $\lambda$ should be at least as large as the smallest number of sides of a polyhedral cone lying on one side of $E$ (positive $D$-coordinate) and invariant (mapped into itself) under $B$.This is purely a geometrical condition, and there are likely further arithmetic constraints as well. For example, if $\lambda$ has all its other algebraic conjugates of roughly the same absolute value, then $B$ acts projectively as nearly a rotation, and this forces any invariant polyhedral cone to have many sides, so the geometric lower bound will be quite large.

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    $\begingroup$ Thanks @Douglas ! I guess after reading your paper more carefully, I should be able to fill in the details. It was really nice that you answered this question :) $\endgroup$ Commented Jan 20, 2016 at 2:19

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