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 ${|\lambda| \geq ab>1}$.