Return to Answer

Yes: if $\alpha$ is an algebraic integer which obeys $|\alpha| < 2$ for all archimedean norms $|\ |$ then $\alpha$ is an eigenvalue of a sum of two permutations matrices. sqrt{

Yes: if $\alpha$ is an algebraic integer which obeys $|\alpha| < 2$ for all archimedean norms $|\ |$ then $\alpha$ is an eigenvalue of a sum of two permutations matrices.

Before heading into step $2$, it is convenient to modify this argument slightly. Take $M$ large enough that any ball of radius $M$ contains at least $3$ points of $A$. In this way, we can ensure that, for any $z \in B_R \cap A$ we can write $\alpha z = z_1 + z_2$ with $z_1$, $z_2 \in B_R \cap A \setminus \{ 0 \}$. We can then take $z_1$, $z_2$, \dots, $z_N$ be the points of $B_R \cap A \setminus \{ 0 \}$. It will be convenient at the next step to make sure none of the entries of our eigenvector are $0$.

Suppose we could permute the rows and columns, so $C = \left( \begin{smallmatrix} C_{11} & 0 \\ C_{21} & C_{22} \end{smallmatrix} \right)$ and $C \vec{z} = \alpha \vec{z}$ where we can write the eigenvector $\vec{z}$ as $\left( \begin{smallmatrix} \vec{z}_1 \\ \vec{z}_2 \end{smallmatrix} \right)$. Then $C_{11} = \alpha \vec{z}_1$ We arranged above that none of the components of $\vec{z}$ is $0$, so $\vec{z}_1 \neq 0$. We see $C_{11}$ is a smaller matrix with eigenvalue $\alpha$ and row sums $2$, and we may consider it instead.

Before heading into step $2$, it is convenient to modify this argument slightly. Take $M$ large enough that any ball of radius $M$ contains at least $3$ points of $A$. In this way, we can ensure that, for any $z \in B_R \cap A$ we can write $\alpha z = z_1 + z_2$ with $z_1$, $z_2 \in B_R \cap A \setminus \{ 0 \}$. We can then take $z_1$, $z_2$, $\dots$, $z_N$ be the points of $B_R \cap A \setminus \{ 0 \}$. It will be convenient at the next step to make sure none of the entries of our eigenvector are $0$.

Suppose we could permute the rows and columns, so $C = \left( \begin{smallmatrix} C_{11} & 0 \\ C_{21} & C_{22} \end{smallmatrix} \right)$ and $C \vec{z} = \alpha \vec{z}$ where we can write the eigenvector $\vec{z}$ as $\left( \begin{smallmatrix} \vec{z}_1 \\ \vec{z}_2 \end{smallmatrix} \right)$. Then $C_{11}\vec{z}_1 = \alpha \vec{z}_1$ We arranged above that none of the components of $\vec{z}$ is $0$, so $\vec{z}_1 \neq 0$. We see $C_{11}$ is a smaller matrix with eigenvalue $\alpha$ and row sums $2$, and we may consider it instead.

Yes: if $\alpha$ is an algebraic integer which obeys $|\alpha| < 2$ for all archimedean norms $|\ |$ then $\alpha$ is an eigenvalue of a sum of two permutations matrices.

Yes: if $\alpha$ is an algebraic integer which obeys $|\alpha| < 2$ for all archimedean norms $|\ |$ then $\alpha$ is an eigenvalue of a sum of two permutations matrices. sqrt{

I remark that this is not true when $|\alpha|=2$, for example, $(1+\sqrt{-15})/2$ is an algebraic integer with absolute value $2$ which is not twice a root of unity.