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Let $A$ be a symmetric matrix over $\Bbb Z^{+}\cup \{0\}$. Assume that all the eigenvalues of $A$ are integers. Also all the diagonal elements of $A$ are $0$. If $D$ is a diagonal matrix with all entries in $\Bbb Z^{+}\cup \{0\}$ , is it true that all the eigenvalues of $D\pm A$ are also integers?

Prove or give a counterexample.

I tried as examples for some $3\times 3$ matrices. I found the result true in all cases taking entries to be non-negative as defined above.

I don’t understand how to prove this fact.

Any help, please.

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  • $\begingroup$ How can it be that eigenvalue of $A$ is not an integer? $\endgroup$ – enedil Feb 10 at 17:55
  • $\begingroup$ @enedil Why should the eigenvalues be integers? Consider the matrix $$\begin{pmatrix}0&1&2\\1&0&1\\2&1&0\end{pmatrix}.$$ It has eigenvalues $-2$ and $1\pm\sqrt{3}$. $\endgroup$ – Servaes Feb 10 at 17:57
  • $\begingroup$ @Servaes And what are the associated eigenvectors? They are not from $\mathbb Z^n$, right? $\endgroup$ – enedil Feb 10 at 17:59
  • $\begingroup$ @enedil Obviously they cannot be. Given that the OP states explicitly that the eigenvalues should be integers, it seems to me that we are considering the eigenvalues over an algebraic closure. $\endgroup$ – Servaes Feb 10 at 18:16
  • $\begingroup$ @enedil Either way of no immediate consequence to the question, and counterexamples exist, see my answer. $\endgroup$ – Servaes Feb 10 at 19:20
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This is not true; consider the matrices $$A=\begin{pmatrix} 0&1&3\cr 2&0&2\cr 3&1&0 \end{pmatrix} \qquad\text{ and }\qquad D=\begin{pmatrix} 1&0&0\\ 0&2&0\\ 0&0&1 \end{pmatrix}.$$ Then $A$ has integer eigenvalues $4$, $-3$ and $-1$, but $D+A$ has eigenvalues $-2$ and $3\pm\sqrt{5}$.

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