Is it true that all the eigenvalues of $D\pm A$ are also integers?

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.

• How can it be that eigenvalue of $A$ is not an integer? – enedil Feb 10 at 17:55
• @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 $1\pm\sqrt{3}$. – Servaes Feb 10 at 17:57
• @Servaes And what are the associated eigenvectors? They are not from $\mathbb Z^n$, right? – enedil Feb 10 at 17:59
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 $$3\pm\sqrt{5}$$.