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).


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.