Thread: Positive definite symmetric matrix has maximal elements on diagonal

1. Positive 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.

2. Re: Positive definite symmetric matrix has maximal elements on diagonal

WLOG () $z^T A z = \sum\limits_{kl}z_k A_{kl} z_l = z_j A_{jj} z_j+z_i A_{ii} z_i+2 z_i A_{ij} z_j>0$
So I guess you got up to there and you set $z_j = 1$ and $z_i = -1$. You need to set $z_i$ and $z_j$ equal to values that are functions of $A_{ij}$. You should get cancellations and the desired result.

3. Counterexample

I wrestled with this problem for a while, and have found a counterexample.

Let $A= \left[ \begin{array}{ccc} 1 & 0 & 0.4 \\ 0 & 1 & 0 \\ 0 & 0 & 0.4 \end{array} \right]$.

It has full rank so $x \ne 0 \Longrightarrow Ax \ne 0 \Longrightarrow x^T A^T Ax = (Ax)^T Ax = \left\Vert{Ax}\right\Vert _2 ^2 > 0 \Longrightarrow B=A^T A$ is symmetric positive-definite.

$B$ is $\left[ \begin{array}{ccc} 1 & 0 & 0.4 \\ 0 & 1 & 0 \\ 0.4 & 0 & 0.32 \end{array} \right]$.

This matrix is not diagonally dominant $\left( \forall i, \vert b_{ii} \vert \ge \sum_{j \ne i}{\vert b_{ij} \vert} \right)$. We don't even have the weaker $\forall i, \vert b_{ii} \vert \ge \left\vert \sum_{j \ne i}{ b_{ij}} \right\vert$, 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 $B$'s maximal entries lie on its diagonal?