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Let $\{(a_{ij},...,a_{in}) \in T^n \mid i = 1,...,s \le n\}$ be a set of vectors. Prove that if $$|a_{jj}| \gt \sum_{i=1, i\neq j}^s|a_{ij}|, \quad 1 \leq j \leq s$$ then the set is linearly independent.

My attempt: I was trying an example with $n=s=3$,

$|a_{11}| \gt $$\sum_{i=1, i\neq j}^{n=3}|a_{i1}|= |a_{21}|+|a_{31}|$

$|a_{22}| \gt $$\sum_{i=1, i\neq j}^{n=3}|a_{i1}|= |a_{12}|+|a_{32}|$

$|a_{33}| \gt $$\sum_{i=1, i\neq j}^{n=3}|a_{i1}|= |a_{13}|+|a_{23}|$

For the linear independence we would have to satisfy, that $x=y=z=0$

$$ \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ \end{pmatrix} $$

But I just don't know what's next. Like the $|a_{jj}|$ are on the places where are pivots and if these elements are after Gauss elimination nonzero, then our given set in linearly independent. But I don't know how to prove it. Thanks for help.

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    What is $T$? See [Geršgorin circle theorem](https://en.wikipedia.org/wiki/Gershgorin_circle_theorem). – 2017-02-09
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    Not sure but I think it is just division ring or the number field. I don't know how to translate if from my language but we are working with it as with the real numbers. – 2017-02-09
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    It's weird because nothing about the problem claims you only have have three vectors in the particular case of $n=s=3$. If $T=\mathbb R$, then the set is infinite. – 2017-02-09
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    I just tried to visualizate it, I don't understand that example very much. – 2017-02-09
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    My point was that if the set is infinite, given the context, it most likely is linearly dependent. – 2017-02-09
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    @GitGud Probably my edit inadvertently caused a confusion, OP probably meant $$\{(a_{11},...,a_{1n}),\ldots,(a_{s1},...,a_{sn})\}$$ – 2017-02-09
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    @ZoranLoncarevic Ohhh, this makes much more sense now. **Edit:** And your edit wasn't the issue. – 2017-02-09
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    But it is not in my textbook like that. – 2017-02-09

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You are talking about Diagonal-dominant matrices, which are non-singular. You can show this, as pointed out in the commets, by using Gersgorin-circles.

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    I have never heard about it, but I will try, thank you. My example is from the basic course of linear algebra. – 2017-02-09
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    These parts of linear algebra are more advanced then I have ever encountered yet, so I think I will leave it. – 2017-02-09