I found task written below, but I cannot prove it.
Given is system of $s$ vectors ($a_i = (a_{i1}, a_{i2}, \dots a_{in})$ for $i = 1, \dots, s$), where $s \leq n$. Prove that, if for all $1 \leq k \leq s$ is satisfied inequality: $$ |a_{kk}| > \sum_{i=1, i \neq k}^{s} |a_{ik}| $$ Then system is linear independent.
So I made (square) matrix: $$ \left\|\begin{split} a_{11}~&~a_{12}~&~\cdots~&~a_{1s}\\ a_{21}~&~a_{22}~&~\cdots~&~a_{2s}\\ \cdots~&~\cdots~&~\cdots~&~\cdots\\ a_{s1}~&~a_{s2}~&~\cdots~&~a_{ss}\\ \end{split} \right\| $$
And I'm trying to prove that determinant is not equal to zero. (I don't see easiest way, to prove it.) But without positive effect. I tried to use GM inequality. $$ |a_{kk}| > (s-1) \cdot \sum_{i=1, i \neq k}^{s} \frac{|a_{ik}|}{s-1}>(s-1)\cdot \sqrt[s-1]{\prod_{i=1, i\neq k}^{s} |a_{ik}} $$ Or induction proof. But I don't know how I can end the proof. Could you give me some hints?
It is easier to solve the linear system. Let $(x_1,..,x_n)$ be a non zero vector in the kernel, and conisder an index $k$ such that $\vert x_k\vert$ is maximal. As $\sum _{j=1}^n a_{jk}x_j=0$, we have $a_{kk} x_k=-\sum _{j\not =j}a_{jk}x_j$. Therefore $\vert a_{kk}\vert \vert x_k\vert \leq \sum _{j\not =k}\vert a_{jk}\vert \vert x_j\vert \leq (\sum _{j\not =k}\vert a_{jk}\vert ) \vert x_k\vert$. Contradiction.