Home > Algebra, Higher Algebra, Olympiad, Problem Solving > IMC 2014 Day 2 Problem 2

IMC 2014 Day 2 Problem 2


Let {A=(a_{ij})_{i,j=1}^n} be a symmetric {n \times n} matrix with real entries, and let {\lambda_1,...,\lambda_n} denote its eigenvalues. Show that

\displaystyle \sum_{1 \leq i<j\leq n} a_{ii}a_{jj} \geq \sum_{1 \leq i<j\leq n}\lambda_i \lambda_j

and determine all matrices for which equality holds.

IMC 2014 Day 2 Problem 2

Hint: The inequality is equivalent to tr(A)^2-\sum_{i=1}^n a_{ii}^2 \geq tr(A)^2-\sum_{i=1}^n \lambda_i^2. This is true since if A is symmetric then \sum_{i=1}^n \lambda_i^2 = tr(A^2)=tr(AA^T) which is the sum of the squares of the elements of A.

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  1. IMC2010
    August 4, 2014 at 6:19 pm

    I proved the opposite inequality. Seems like you got the inequality sign reversed, or was it like that at the competition?

  2. IMC2010
    August 4, 2014 at 6:40 pm

    Never mind, I got the sign reversed in one place. Very simple and easy solution using the fact that sum of the squares of the eigenvalues equal the sum of the squares of the entries of A, given A is real symmetric.

    • Artur Kirkoryan
      August 4, 2014 at 11:53 pm

      Is this a famous fact, I’ve never heard it before? Without it, the problem can be solved using diagonalization and the fact that similar matrices have the same trace.

      • August 5, 2014 at 1:36 pm

        You can prove it using diagonalization as you say. A real symmetric matrix can be diagonalized using an orthonormal matrix. The sum of the squares of the entries of A is the trace of AA^T. Using the diagonal form you can prove that the trace of this matrix is the sum of the squares of the eigenvalues.

      • IMC2010
        August 5, 2014 at 6:33 pm

        Or you can just use the fact that squares of the eigenvalues of $A$ are precisely the eigenvalues of $A^2$ and compute $tr(A^2)$ in two different ways.

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