I think there is a simpler approach. Let $\alpha$ be any algebraic integer, and let $A$ be an $n \times n$ integer matrix with $\alpha$ as a rootan eigenvalue. Take $B$ to the the direct sum of 2 copies of $A$, ie a $2n \times 2n$ matrix, and let $J$ be the $2n \times 2n$ matrix with all entries $1.$ For large enough positive integers $m,$ $B +mJ$ is a matrix with non-negative integer entries, and still has $\alpha$ as an eigenvalue. This is because the $\alpha$ eigenspace of $B$ is (at least) $2$-dimensional, while the $0$-eigenspace of $J$ has codimension $1,$ so there is a nonzero vector $u$ with $Bu = \alpha u$ and $Ju = 0.$ Hence $(B+mJ)u = \alpha u$ for all positive integers $m.$
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I think there is a simpler approach. Let $\alpha$ be any algebraic integer, and let $A$ be an $n \times n$ matrix with $\alpha$ as a root. Take $B$ to the the direct sum of 2 copies of $A$, ie a $2n \times 2n$ matrix, and let $J$ be the $2n \times 2n$ matrix with all entries $1.$ For large enough positive integers $m,$ $B +mJ$ is a matrix with non-negative integer entries, and still has $\alpha$ as an eigenvalue. This is because the $\alpha$ eigenspace of $B$ is (at least) $2$-dimensional, while the $0$-eigenspace of $J$ has codimension $1,$ so there is a nonzero vector $u$ with $Bu = \alpha u$ and $Ju = 0.$ Hence $(B+mJ)u = \alpha u$ for all positive integers $m.$ |
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