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.
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.
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?