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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 $\lambda_1^2 +\lambda_2^2 +\lambda_3^2 +\lambda_4^2 \in \mathbb Z$

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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.

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  • $\begingroup$ The trace! Ah I always forget the trace. Thank you for the perfectly clear answer. $\endgroup$ – Merkh Apr 24 '16 at 10:59

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