Let A be a positive definite symmetric matrix. Prove that for any column (or row) the maximal element of that list is on the diagonal. I've only been able to prove the weaker statement a_(i,i)+a_(j,j)>2*a_(i,j) for i not equal to j.

- Feb 4th 2013, 10:31 AMblmalikovPositive definite symmetric matrix has maximal elements on diagonal
Let A be a positive definite symmetric matrix. Prove that for any column (or row) the maximal element of that list is on the diagonal. I've only been able to prove the weaker statement a_(i,i)+a_(j,j)>2*a_(i,j) for i not equal to j.

- Feb 4th 2013, 01:33 PMvincisonfireRe: Positive definite symmetric matrix has maximal elements on diagonal
WLOG ((Wondering))

So I guess you got up to there and you set and . You need to set and equal to values that are functions of . You should get cancellations and the desired result. - Oct 27th 2013, 07:20 PMtbemroseCounterexample
I wrestled with this problem for a while, and have found a counterexample.

Let .

It has full rank so is symmetric positive-definite.

is .

This matrix is not diagonally dominant . We don't even have the weaker , nor do the entries on the diagonal correspond with the 3 largest entries of the matrix. In what way can one unambiguously define "maximal entries" in which 's maximal entries lie on its diagonal?