Positive definite symmetric matrix has maximal elements on diagonal

**blmalikov** · February 4th 2013, 09:31 AM

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.

**vincisonfire** · February 4th 2013, 12:33 PM

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.

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