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I want to prove that a strictly (row) diagonally dominant matrix $A$ is invertible. Im using the Gershgorin circle theorem. This is my approach:

Gershgorin circle theorem says that every eigenvalue of $A$ satisfies :

$|\lambda - a_{ii}|\leq \sum_{i \neq j} |a_{ij}|$ for some $i $.

Strict dominance implies :

$\sum_{i \neq j} |a_{ij}|< |a_{ii}| $

Putting both together we get the following:

$|\lambda - a_{ii}| < |a_{ii}|$, which implies that $\lambda \neq 0 $. This shows that matrix $A$ is invertible because any matrix $A$ is invertible if and only if $\lambda_{i} \neq 0$.

Is this proof valid?

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    $\begingroup$ Valid, but you should be more precise about "if and only if $\lambda \ne 0$": what are the quantifiers on $i$? Also, it's overkill to use Gershgorin's theorem when this can be proved by a more direct elementary argument, but that doesn't make the proof invalid. $\endgroup$ – Erick Wong Sep 26 '17 at 23:05
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It is valid but as noted the Gershgorin theorem is quite an overkill. A more elementary approach follows.

Let $A=D-B$, where $D$ is the diagonal part of $A$. Let $A$ be row diagonally dominant, that is, $$\tag{1} |a_{ii}|>\sum_{j\neq i}|a_{ij}|\;\iff\; 1>\frac{1}{|a_{ii}|}\sum_{j\neq i}|a_{ij}|, \quad i=1,\ldots,n. $$ The latter inequality is equivalent to $$\tag{2}1>\|D^{-1}B\|_\infty.$$

If $A$ was singular, $Ax=0$ and hence $x=D^{-1}Bx$ for some nonzero $x$. We would have $$ \|x\|_\infty=\|D^{-1}Bx\|_\infty\leq\|D^{-1}B\|_\infty\|x\|_\infty $$ and, dividing by $\|x\|_\infty$, $$ 1\leq\|D^{-1}B\|_\infty, $$ which contradicts (2).

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