# If all eigenvalues of $A \in \operatorname{M}_2(\mathbb{Z})$ are rational, then they are integers.

How to show that if $A$ is a $2\times 2$ matrix with all integer entries and all eigenvalues are in $\mathbb{Q}$, then all eigenvalue are integers?

I cannot find why eigenvalues in $\mathbb{Q}$, in this case, will precisely belong to $\mathbb{Z}$?

What is the proof?

• en.wikipedia.org/wiki/Rational_root_theorem Sep 1, 2018 at 3:37
• Are you familiar with the theorem that says a monic polynomial with integer coefficients cannot have rational roots that are not integers? Sep 1, 2018 at 3:38
• The characteristic polynomial of $A$ is monic of degree two, has integer entries, and rational roots... Sep 1, 2018 at 3:38
• how? please explain in answer not in comment spaceisdarkgreen Sep 1, 2018 at 4:27

This actually implies the result for an $n\times n$ matrix, not just a $2\times 2.$ So perhaps they want you to give a more direct proof of the special case. If the characteristic polynomial is $\lambda^2 + b\lambda + c,$ the roots are $$\lambda = \frac{-b\pm\sqrt{b^2-4c}}{2}.$$ If these are to be rational, we need $b^2-4c$ to be a perfect square, since it is known that the square root of an integer is either irrational or an integer. (And thus $\sqrt{b^2-4c}$ is an integer.)
Can you show that $-b\pm \sqrt{b^2-4c}$ is always even? I suggest splitting it up into the case where $b$ is even and odd.