TITLE: Showing that every rational eigenvalue of a graph is integral
QUESTION [5 upvotes]: (This is taken from the exercises in Bondy and Murty's Graph Theory.) Let the adjacency matrix of a graph $G$ be denoted by $\mathbf{A}$. The eigenvalues $\lambda$ of $G$ are defined as the roots of the characteristic polynomial of $\mathbf{A}$, $\lvert \mathbf A-\lambda\mathbf I\rvert=0$. I am asked to show that, if $\lambda\in\mathbb Q$, then $\lambda\in\mathbb Z$.
I have already managed to show that $\lambda$ is real. This is because $\mathbf A$ is symmetric by definition, then it becomes a simple linear algebra fact that its eigenvalues are real. Then, if $G$ is simple, all diagonal entries of $\mathbf A$ are $0$, and since the eigenvalues of a matrix have the same signs as the diagonal entries (I hope!) then all the eigenvalues are $0$. However, I'm not given that $G$ is simple and so this line of reasoning doesn't work, and I have no idea where to use the fact that $\lambda$ is rational in the proof anyway. Any help is appreciated!

REPLY [5 votes]: More generally, if $A$ is a square matrix whose entries are integers, the characteristic polynomial of $A$ is a monic polynomial with integer coefficients, so the eigenvalues of $A$ are algebraic integers.  Every algebraic integer that is a rational number is an ordinary integer (see e.g. here).