# Show that for a $n×n$-matrix $A$, $\operatorname{rank}(A ) = n$

Could you give me please some tips or direction on how should I deal with the following problem?

$A = \left( a_{i,j} \right)$ is a Matrix from $M(n\times n, \mathbb{R})$ so that for all $i\in \left \{ 1,\ldots,n \right \}$ exists $j$ such that:

$$\left|a_{i,j}\right|>\left|a_{i,1}\right|+\left|a_{i,2}\right|+\ldots+\left| a_{i,i-1}\right|+\left|a_{i,i+1}\right|+ \cdots +\left | a_{i,n} \right|.$$

I have to prove, that $\operatorname{rank}\left ( A \right ) = n$.

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What is rg(A) ? – Babak Miraftab Nov 16 '11 at 12:44
@babgen It is the German version of rank. I'll fix and hope that it is what OP meant :) – user13838 Nov 16 '11 at 15:22

You wrote

$$\left | a_{i,j} \right | > \left | a_{i,1} \right | + \left | a_{i,2} \right | + ... + \left | a_{i,i-1} \right | + \left | a_{i,i+1} \right | + ... +\left | a_{i,n} \right |$$

This makes only sense for $j=i$ because otherwise you will get an immediate contradiction.

Matrices that satisfy this property are called strictly diagonally dominant and the Levy-Desplanques theorem is what you want, it shows that those matrices are non-singular (even over $\mathbb{C}$).

Also it seems you have quite a few old answers you did not accept yet, maybe you should recheck them and see if they were answered in the meantime.

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Hint If the rows of the matrix are linearly dependent, then

$$c_1r_1+...c_nr_n =0 \,. (*)$$

For some $c_1,..c_n$, whicha re not all zero... Pick the $i$ so that $|c_i|$ takes the largest value, and use $(*)$ to calculate $r_i$ for that $i$...

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You can right multiply it by some permutation matrices to change it into a strictly row diagonally dominant matrix. Then I think is should be clear.

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