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Let $A$ be a $2n\times2n$ matrix with integer entries. Assume that $A$ has no real eigenvalue. Let the eigenvalues of $A$ be $\lambda_1,\overline{\lambda_1},\dotsc,\lambda_n,\overline{\lambda_n}$. Is it possible to construct an $n\times n$ matrix $B$ with integer entries again, with eigenvalues $\lambda_1+\overline{\lambda_1},\dotsc,\lambda_n+\overline{\lambda_n}$?

(It would be good if $B$ is somehow naturally related to $A$)

For $n=1$ it's true, because I can take $B=(\text{tr}(A))$, but I couldn't work out the general case.

EDIT: I am adding the assumption that $A$ is diagonalizbale (because Robert Israel's answer still applies with this assumption).

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No. For example, suppose $A$ is the companion matrix of the polynomial $\lambda^4-2 \lambda^3+4 \lambda^2-\lambda+4$, which is irreducible over the rationals and has no real roots. $\lambda_i + \overline{\lambda_i}$ are not quadratic irrationals (in fact they have degree $6$), so they are not eigenvalues of $2 \times 2$ matrices with rational entries.

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