Showing squared sum of eigenvalues is an integer

Question

I've been trying to improve my Linear algebra skills lately and I've run into this problem without not knowing where to start. Any suggestions/answers?

Suppose a $4\times 4$ matrix of integers has four distinct real eigenvalues, $\lambda_1 > \lambda_2 > \lambda_3 > \lambda_4.$ Prove that $\lambda_1^2 +\lambda_2^2 +\lambda_3^2 +\lambda_4^2 \in \mathbb Z$

Answer 1

Call the matrix $A$. Over $\mathbb{R}$, the matrix $A$ is diagonalizable and so is $A^2$, whose eigenvalues are $\lambda_i^2$. We have

$$ \operatorname{tr}(A^2) = \lambda_1^2 + \lambda_2^2 + \lambda_3^2 + \lambda_4^2 $$

but $\operatorname{tr}(A^2) = \sum_{i=1}^4 (A^2)_{ii} = \sum_{i=1}^4 \left( \sum_{j = 1}^4 a_{ij} a_{ji} \right)$ is a sum of product of integers, hence an integer.