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## SEEMOUS 2011 Problem 2

Let $A=(a_{ij})$ be a real $n \times n$ matrix such that $A^n \neq 0$ and $a_{ij}a_{ji} \leq 0,\ \forall i,j$. Prove that there exist two non-real numbers among the eigenvalues of $A$.
SEEMOUS 2011 Problem 2

Solution: This is quite an easy problem. First note that $a_{ii}^2\leq 0$ and therefore all diagonal elements are $0$, implying that $tr(A)=0$. Consider the characteristic polynomial $p(\lambda)=\lambda^n+a_{n-2}\lambda^{n-2}+...$. We know that $a_{n-2}=\sum \displaystyle \left| \begin{matrix} 0 & a_{ij} \\ a_{ji} & 0 \end{matrix} \right|\geq 0$, by the hypothesis. On the other hand $a_{n-2}=\sum_{1\leq i.

Therefore, the sum of squares of the eigenvalues is less or equal to $0$. Since $A$ is not nilpotent, it has a non-zero eigenvalue, which means that we cannot have all the eigenvalues real. Moreover, the polynomial has real coefficients, and together with one non-real eigenvalue, its complex conjugate is also an eigenvalue for $A$, thus proving that $A$ has at least two non-real eigenvalues.