Home > Algebra, Olympiad > IMC 2013 Problem 1

IMC 2013 Problem 1

Problem 1. Let {A} and {B} be real symmetric matrices with all eigenvalues strictly greater than {1}. Let {\lambda} be a real eigenvalue of matrix {AB}. Prove that {|\lambda| >1}.

Solution: Every symmetric real matrix is orthogonally diagonalizable, i.e. there exists an orthogonal basis formed of eigenvectors. This proves that for every vector {v} we have {\|Av\| \geq a\|v\|} and {\|Bv\| \geq b\|v\|} where {a,b>1} (we can take {a} the minimum of the eigenvalues of the matrix {A} and the same for {B}).

Therefore {\|ABv\|\geq a\|Bv\| \geq ab\|v\|}. If {\lambda} is an eivengalue of {AB} with corresponding eigenvector {v} then

\displaystyle |\lambda| \|v\| = \|ABv\| \geq ab\|v\|

and so {|\lambda| \geq ab>1}.

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