N-dimensional vector space

Suppose the \(s \times n\) matrix on field \(K\) \[ A = {\begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & & \vdots\\ a_{s1} & a_{s2} & \cdots & a_{sn}\\ \end{pmatrix}} \] satisfies \(s \le n\), and \(2|a_{ii}|>\sum_{j=1}^{n}{|a_{ij}|},\ \ (i = 1, 2, \cdots, s)\), prove: rank of \(A\)'s row vector group, \(\gamma_{1}, \gamma_{2}, \cdots, \gamma_{s}\), equals to \(s\).

\(Proof\) Let \[ A' = {\begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1s}\\ a_{21} & a_{22} & \cdots & a_{2s}\\ \vdots & \vdots & & \vdots\\ a_{s1} & a_{s2} & \cdots & a_{ss}\\ \end{pmatrix}} \] We can know \(\det{A'} \ne 0\) because \[|a_{ii}|>\sum_{j=1 \\j\ne i}^{s}{|a_{ij}|},\ \ (i = 1, 2, \cdots, s)\], so rank of \(A'\)'s row vector group equals to \(s\), appending the row vectors, we can show that rank of \(A'\)s row vector group equals to \(s\). \(\blacksquare\)