A difficult problem for me.

let ($\displaystyle a_{ij}$) = A so that $\displaystyle a_{ij} <= 0$ and $\displaystyle \sum\limits_{j=1}^{n} {a_{ij}}$ $\displaystyle > 0$. Prove that det A > 0

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- May 18th 2008, 02:38 PMkezmanDeterminant
A difficult problem for me.

let ($\displaystyle a_{ij}$) = A so that $\displaystyle a_{ij} <= 0$ and $\displaystyle \sum\limits_{j=1}^{n} {a_{ij}}$ $\displaystyle > 0$. Prove that det A > 0 - May 18th 2008, 03:55 PMKGene
- May 18th 2008, 04:08 PMkezman
Sorry I forgot:

the first condition $\displaystyle a_{ij} \leq{0} $is for $\displaystyle i \neq j$

the second condition is for all $\displaystyle a_{ij}$ - May 19th 2008, 12:21 AMKGene
So did you mean $\displaystyle a_{ij}\leq 0$ if $\displaystyle i\neq j$ and $\displaystyle \sum_{1\leq i,j\leq n}a_{ij}>0$? But this can't be true, a counterexample,

$\displaystyle \left [\begin{array}{cc} 0&-1\\ -1&3\end{array}\right ]$ where the sum of all entries is 1>0 and $\displaystyle a_{12}=a_{21}=-1\leq 0$. Please double check again (Itwasntme) - May 19th 2008, 02:20 AMOpalg
Such a matrix is called diagonally dominant. The result that you want follows from the Gershgorin circle theorem.