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Let $A\in\operatorname{SL}(d,\mathbb{Z})$ be an irreducible positive matrix, i. e. $A=(a_{i,j})_{1\leq i,j\leq d}$ with $a_{i,j}\in\mathbb{Z}_{>0}$. From the Perron-Frobenius theorem, we know that $A$ has one real eigenvalue $\lambda_1>1$ whose absolute value is strictly larger than any other eigenvalues. Denote all eigenvalues of $A$ by $$ \lambda_1>|\lambda_2|\geq|\lambda_3|\geq\cdots\cdots\geq|\lambda_d|. $$ What can we say about $|\lambda_2|$, does it have $|\lambda_2|<1$?

If the answer is no, is that possible we have other conditions guarantee $|\lambda_2|<1$?

My feeling here is that $A$ has determinant 1, if $\lambda_1$ is large enough, it may possible that all other eigenvalues have absolutely values smaller than 1. So maybe $A$ has some type of symmetry, like some control between $\min\{a_{i,j}\}$ and $\max\{a_{i,j}\}$ which $A^n$ also satisfies for every $n\geq1$.

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  • $\begingroup$ Part of the Perron-Frobenius theorem says that the largest eigenvalue is unique... $\endgroup$
    – Asaf
    Commented Nov 10 at 14:59
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    $\begingroup$ If you start from a single positive matrix $A$ of determinant $1$, every power of $A$ will have the same property. If $A$ satisfies $|\lambda_2|\geq 1$ then so will every power of $A$. However, $\lambda_1$ will grow arbitrarily large as the power grows, as will the size of the largest entry. So none of your examples with extra conditions can be true as long as there is a single counterexample without the extra conditions. $\endgroup$
    – Will Sawin
    Commented Nov 10 at 15:11
  • $\begingroup$ @GeoffRobinson The question assumed irreducibility of the matrix (i.e., of the char. polynomial). Actually I guess ($\star$) that, as usual in this context, the $\lambda_i$ are counted with multiplicities, so if the matrix is not irreducible then $|\lambda_2|>1$ readily follows whatsoever. ($\star$) indeed OP wrote them as $\lambda_1,\dots,\lambda_d$ with $d$ the matrix size. $\endgroup$
    – YCor
    Commented Nov 20 at 16:40
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    $\begingroup$ @GeoffRobinson I guess not, because this is precisely the way to avoid the trivial counterexamples. In any case the OP hasn't much participated in the comments so is likely to not clarify this (inessential) point. $\endgroup$
    – YCor
    Commented Nov 20 at 17:03

3 Answers 3

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Did you look at any examples before asking the question?

I sat down to generate a positive integer matrix with determinant 1, and found this 4-by-4 example quickly:

$$\begin{pmatrix}1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 5\\ 9&10&11&13\\14&15&17&18\end{pmatrix}$$

Its eigenvalues are (approximately)

$$35.9829,\; -2.42941,\; -0.57343,\; 0.019949,$$

of which two have absolute value greater than 1.

Change the last matrix entry from $18$ to anything else and the determinant will still be 1. It appears that in most cases 3 of those eigenvalues will be larger than $1$ in absolute value.

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  • $\begingroup$ Just if any one needs: the characteristic polynomial is $x^4 - 33x^3 - 106x^2 - 48x + 1$. $\endgroup$
    – YCor
    Commented Nov 20 at 12:50
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There is also in size 3 (found by checking all $3\times 3$ matrices with entries in $\{1,2,3\}$): the matrix $\begin{pmatrix}1&1&1\\2&1&2\\2&3&1\end{pmatrix}$, with characteristic polynomial $x^3 - 3x^2 - 7x - 1$ has the eigenvalues $$4.577...,\; -1.424...,\; -0.153...$$

Another one has only positive eigenvalues: $\begin{pmatrix}3&1&1\\1&2&1\\2&1&1\end{pmatrix}$, with characteristic polynomial $x^3 - 6x^2 + 7x - 1$ with the eigenvalues $$4.491...,\; 1.343...,\; 0.166...$$

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  • $\begingroup$ "Half" of $3\times3$ matrices are counterexamples: either $A$ or $A^{-1}$. $\endgroup$
    – Oleg Eroshkin
    Commented Nov 20 at 17:06
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    $\begingroup$ @OlegEroshkin no... you forget the assumption that $A$ has only positive entries. $\endgroup$
    – YCor
    Commented Nov 20 at 17:07
  • $\begingroup$ Oops, yes I missed that. $\endgroup$
    – Oleg Eroshkin
    Commented Nov 20 at 17:39
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You are looking for Pisot matrices. You may take a look at Avila and Delecroix - Some monoids of Pisot matrices.

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