Let $A$ be an $n \times n$ matrix such that $|a_{ii}|>\sum_{j=1,j\neq i}^n|a_{ij}|$ for each $i$. Show that $A$ is invertible. $(complex matrix)

The straight forward way is to show that the determinant is non-zero, but it seems to be labor intensive. Then I tried (not sure if this is the right approach) to show that the column rank is $n$ by the following:

Suppose one column, say, the $i^{th}$ column, is the linear combination of the other columns, then there exists complex entries $b_j$ such that $a_{ii}=\sum_{j=1,j\neq i}^nb_j a_{ij}$.

By the triangle inequality we now have

$\sum_{j=1,j\neq i}^n |a_ij| <|a_{ii}| \leq \sum_{j=1,j\neq i}^n |b_j||a_{ij}| \leq \max|b_j|(\sum_{j=1,j\neq i}^n |a_{ij}|)$

So $\max|b_j|\geq 1$. This inequality is rather weak so I'm not sure if it'd be useful, and from now I don't know how to continue, any hints, tips would be appreciated.


As you assume , $$ a_{ii} = \sum_{j \not = i} b_ja_{ij}$$ with $$max(|b_j|)>1$$

In general , we have $$ a_{ki} = \sum_{j \not = i} b_ja_{kj}$$

However, $$\exists r \ s.t \ |b_r| =max(|b_j|)>1$$

$$b_ra_{rr} = a_{ri}-\sum_{j \not = i,r}b_ja_{rj} \Rightarrow a_{rr} = \frac{1}{b_r}a_{ri}-\sum_{j \not = i,r}\frac{b_j}{b_r}a_{rj} $$

Which means that $$|a_{rr}| =| \frac{1}{b_r}a_{ri}-\sum_{j \not = i,r}\frac{b_j}{b_r}a_{rj}| \leq \sum_{j \not = r} |a_{rj}| $$ Contradict to the given condition.


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