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?

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

1 Answer 1


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.

  • $\begingroup$ what is the case when (b^2-4*a*c)^(1/2) is irrotational? spaceisdarkgreen $\endgroup$
    – Priya Dey
    Sep 1, 2018 at 4:07
  • $\begingroup$ and how to show b^2>=4*c? $\endgroup$
    – Priya Dey
    Sep 1, 2018 at 4:10
  • $\begingroup$ In those cases the solution won’t be rational... you don’t have to worry about them $\endgroup$ Sep 1, 2018 at 4:15
  • $\begingroup$ what is the exact theorem says? then we cannot say any monic polynomial with integer cofficient has roots in integer.. then could be anhthing.. then what is the specility we have here? $\endgroup$
    – Priya Dey
    Sep 1, 2018 at 4:22
  • $\begingroup$ yes I'm able to show what may be "b", numeretor always is even integer whenever the sq. root gives us a perfect squre $\endgroup$
    – Priya Dey
    Sep 1, 2018 at 4:24

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