The Tits alternative for $\operatorname{Out}(F_n)$ Not sure if this is the right place to ask this, but the paper I am reading seems to be too specialised for mathstack (if you do not agree, pleas let me know and I will take down this question)
I am reading the paper 'The Tits alternative for $\operatorname{Out}(F_n)$ I: dynamics of exponentially growing automorphisms' by Bestvina, Feighn and Handel. In it, the following lemma occurs:

There is a homomorphism $PF_{\Lambda^+}: \operatorname{Stab}(\Lambda^+) \to \mathbb{Z}^k$ such that $\psi \in \ker PF_{\Lambda^+}$ if and only if $\Lambda^+ \notin \mathcal{L}(\psi)$ and $\Lambda^+ \notin \mathcal{L}(\psi^{-1})$ ($\psi \in \operatorname{Out}(F_n)$).

This lemma can be found on page 546 of the paper. The proof is the following:

and proposition 3.3.3. is

question I do not understand why each $\mu(\Psi)$ other than $1$ occurs as the Perron-Frobenius eigenvalue of some irreducible matrix and why the set of Perron-Frobenius eigenvalues is discrete. Any chance someone knows this paper/has read this paper or could point me to some extra lecture which could clarify this proof for me?
My reasoning My confusion came from the following reasoning, which dr. Bestvina (through mail) told me was false (I can not figure out why): I think I could show from the definition of $\mu$ that $\mu(\operatorname{Id}) = 1$. Let $\psi \in \operatorname{Stab}(\Lambda^+)$ be an outer automorphism with $\mu(\psi) > 1$, then $\mu(\psi)$ is indeed a Perron-Frobenius eigenvalue (proposition 3.3.3.(4)). However, this would imply that $\mu(\psi^{-1})$ is not, by proposition 3.3.3.(2): $mu(\psi^{-1}) < 1$ and Perron-Frobenius eigenvalues $\lambda$ of non-negative, integer irreducible matrices satisfy $\lambda \geq 1$. I can not see where my reasoning is flawed, nor can I see why then every $\mu$ has to be a Perron-Frobenius eigenvalue
 A: I think the first sentence in the proof of 3.3.1, with "$...PF_\Lambda^*(\psi^{-1}) = -PF_\Lambda^*(\psi)...$" was meant to tell you to just think about when that is positive because, as you mention, $PF^*_\Lambda(\psi)$ could not have come from a Perron-Frobenius matrix if it was negative, but up to taking inverse it will, unless $\mu(\psi)=1$. This also makes sense since the theorem is about whether or not $\Lambda^+ \in \mathcal{L}(\phi^\pm)$, and proposition 3.3.3 implies that $\mu(\psi^{-1})<1$ then $\Lambda^+ \in \mathcal{L}(\psi).$
As for discreteness of the eigenvalues, this takes a bit of an argument and it is in [BH1]=Train tracks and Automorphisms of Free Groups, and is in the proof of theorem 1.7, along with the comment about the more general setting on page 37. The outline of the proof that irreducible integer Perron-Frobenius matrices, of bounded dimensions will have discrete set of Perron-Frobenius eigenvalues goes like this:


*

*Bound the eigenvalue $\lambda$, of $M=(m_{ij})$, from below by the minimum row sum of the matrix.

*Show that there is a uniform $k$ so that $M^k$ will have row sums larger than the largest entry of the original matrix $M$. I would use the graph theoretic definition of irreducible(which is in the paper) and there is a uniform bound on the dimensions to get the "mixing of the largest term"

*The above shows $\lambda^k \geq $ min row sum of $M^k$ $\geq \max{m_{ij}}$, which bounds the eigenvalues with the terms of a matrix, so only finitely many eigenvalues below any $K$, so the set of the eigenvalues is discrete. 

