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Math Help - Determinant

  1. #1
    Member kezman's Avatar
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    Determinant

    A difficult problem for me.
    let ( a_{ij}) = A so that a_{ij} <= 0 and \sum\limits_{j=1}^{n} {a_{ij}} > 0. Prove that det A > 0
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  2. #2
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    Quote Originally Posted by kezman View Post
    A difficult problem for me.
    let ( a_{ij}) = A so that a_{ij} <= 0 and \sum\limits_{j=1}^{n} {a_{ij}} > 0. Prove that det A > 0
    How can these a_{ij}\leq 0 and \sum_{j=1}^na_{ij}>0 both true simultaneously?
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  3. #3
    Member kezman's Avatar
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    Sorry I forgot:
    the first condition a_{ij} \leq{0} is for i \neq j
    the second condition is for all a_{ij}
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    Quote Originally Posted by kezman View Post
    Sorry I forgot:
    the first condition a_{ij} \leq{0} is for i \neq j
    the second condition is for all a_{ij}
    So did you mean a_{ij}\leq 0 if i\neq j and \sum_{1\leq i,j\leq n}a_{ij}>0? But this can't be true, a counterexample,
    \left [\begin{array}{cc} 0&-1\\ -1&3\end{array}\right ] where the sum of all entries is 1>0 and a_{12}=a_{21}=-1\leq 0. Please double check again
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  5. #5
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    Quote Originally Posted by kezman View Post
    A difficult problem for me.
    let ( a_{ij}) = A so that a_{ij} \leqslant 0 for i \neq j and \sum\limits_{j=1}^{n} {a_{ij}} > 0 for 1≤i≤n. Prove that det A > 0
    Such a matrix is called diagonally dominant. The result that you want follows from the Gershgorin circle theorem.
    Last edited by Opalg; May 19th 2008 at 03:19 AM.
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