## SEEMOUS 2011 Problem 2

Let be a real matrix such that and . Prove that there exist two non-real numbers among the eigenvalues of .

*SEEMOUS 2011 Problem 2*

**Solution: **This is quite an easy problem. First note that and therefore all diagonal elements are , implying that . Consider the characteristic polynomial . We know that , by the hypothesis. On the other hand .

Therefore, the sum of squares of the eigenvalues is less or equal to . Since 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 , thus proving that has at least two non-real eigenvalues.

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Categories: Algebra, Higher Algebra
Algebra

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