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Thread: Positive definite symmetric matrix has maximal elements on diagonal

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    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.
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    Senior Member vincisonfire's Avatar
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    Re: Positive definite symmetric matrix has maximal elements on diagonal

    WLOG () $\displaystyle z^T A z = \sum\limits_{kl}z_k A_{kl} z_l = z_j A_{jj} z_j+z_i A_{ii} z_i+2 z_i A_{ij} z_j>0$
    So I guess you got up to there and you set $\displaystyle z_j = 1$ and $\displaystyle z_i = -1$. You need to set $\displaystyle z_i$ and $\displaystyle z_j$ equal to values that are functions of $\displaystyle A_{ij}$. You should get cancellations and the desired result.
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  3. #3
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    Exclamation Counterexample

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

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

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

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