2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.