Home > Algebra, Higher Algebra > SEEMOUS 2011 Problem 2

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<j\leq n}\lambda_i\lambda_j=-\frac{1}{2}\left(\sum_{i=1}^n \lambda_i^2 \right).

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.

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