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What are possible eigenvalues of a 0-1 (boolean) matrix? More specifically, if it is a symmetric matrix with 1s in it diagonal.

We know that if the matrix is symmetric then they are all real values. And if the matrix has 1s in its diagonal, then $n=\operatorname{tr}(A)=\sum_{i=1}^{n} \lambda_i$. Can you say more about its eigenvalues, like are the positive or integers?

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  • Well, the all 1's matrix is symmetric with 1's on the main diagonal, and it has at least one 0 eigenvalue. – pjs36 Apr 08 '16 at 14:27
  • The characteristic polynomial will be monic with integer coefficients, so you at least know that the eigenvalues will be algebraic integers, which is less general than general algebraic numbers. – M. Winter Jan 26 '24 at 14:20

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For the two specific questions, are the eigenvalues positive or integers, consider the matrix $$ \left( \begin{matrix} 1 & 1 & 0 \\ 1& 1 & 1 \\ 0 & 1 & 1 \end{matrix}\right).$$ The eigenvalues of this matrix are $1 \pm \sqrt{2}$ and $1$. So in particular we have $1-\sqrt{2} < 0$ and not an integer.

However, we can say something about the largest eigenvalue. Namely, let $\lambda >0$ be the greatest eigenvalue with eigenvector $x$. Then we can pick an $i \le n$ such that $|x_i| \ge |x_j|$ for all $j \le n$. We may also assume $x_i >0$ as $-x$ is also an eigenvector with eigenvalue $\lambda$. So in particular we find $$ \lambda x_i = (Ax)_i = \sum_{j=1}^n a_{ij} x_j \le \sum_{j=1}^n x_j \le n x_i.$$ Thus the greatest eigenvalue is less than or equal to $n$, where the matrix with only $1$'s gives an eigenvalue of $n$.

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