Thread: Determinant

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

Originally Posted by kezman

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

How can these $\displaystyle a_{ij}\leq 0$ and $\displaystyle \sum_{j=1}^na_{ij}>0$ both true simultaneously?

Sorry I forgot:
the first condition $\displaystyle a_{ij} \leq{0} $is for $\displaystyle i \neq j$
the second condition is for all $\displaystyle a_{ij}$

Originally Posted by kezman

Sorry I forgot:
the first condition $\displaystyle a_{ij} \leq{0} $is for $\displaystyle i \neq j$
the second condition is for all $\displaystyle a_{ij}$

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

Originally Posted by kezman

A difficult problem for me.
let ($\displaystyle a_{ij}$) = A so that $\displaystyle a_{ij} \leqslant 0$ for $\displaystyle i \neq j$ and $\displaystyle \sum\limits_{j=1}^{n} {a_{ij}}$ $\displaystyle > 0$ 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.