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?
This follows from the rational root theorem, which implies that any rational root of a polynomial with integer coefficients and leading coefficient one is an integer. The characteristic polynomial of the matrix is a polynomial that meets this criterion.
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.