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If $\mathbf a_1,...,\mathbf a_n$ are n vectors in $\mathbb C^n$ with $\mathbf a_i = (a_{i1},...,a_{in})$ and such that $$\lvert a_{jj} \rvert \gt \sum_{\overset{i=1}{i\ne j}}^{n} \lvert a_{ij} \rvert \space\space \text{for j=1,...,n}$$ how can one show that they are linearly independent?

For n=3, for example, I tried assuming that one of the vectors is a linear combination of the other two and then using the property given above and the usual inequalities ($\lVert \mathbf x + \mathbf y \rVert \leq \lVert \mathbf x \rVert + \lVert \mathbf y \rVert$ and $\lvert \space \lVert \mathbf x \rVert - \lVert \mathbf y \rVert \space \rvert \leq \lVert \mathbf x + \mathbf y \rVert$) to show that you would have both $\lvert \lambda \rvert \lt \lvert \mu \rvert$ and $\lvert \mu \rvert \lt \lvert \lambda \rvert$ if, say, $\mathbf a_1 = \lambda \mathbf a_2 + \mu \mathbf a_3$, but only got $$\begin{align} \lvert\lambda\rvert + \lvert\mu\rvert\geq 1\\ (\lvert\lambda\rvert - 1)\cdot \lvert a_{22}\rvert\lt(\lvert\mu\rvert - 1)\cdot \lvert a_{32}\rvert \\ (\lvert\mu\rvert - 1)\cdot \lvert a_{33}\rvert\lt(\lvert\lambda\rvert - 1)\cdot \lvert a_{23}\rvert\end{align}$$

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This question has already been answered here: Proof that a strictly diagonally dominant matrix is invertible

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