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Strictly related to Gershgorin's theorem is the so called "minimal Gershgorin set". For every $A\in\mathbf{C}^{n,n}$ , $\mathbf{x}>0$ meaning $x_i>0\quad \forall i$ , let's define its minimal Gershgorin set $G(A)$ as: $$ G(A)=\bigcap_{\mathbf{x}>0} G_{\mathbf{x}}(A), $$ where $$ G_{\mathbf{x}}(A)=\bigcup_{i=1}^n \left\{z\in\mathbf{C}:\left|z-a_{ii}\right|\leq\frac{1}{x_i}\sum_{j\ne i}\left|a_{ij}\right|x_j\right\}. $$
Theorem: Let $A\in\mathbf{C}^{n,n}$ , let $\sigma(A)$ be the spectrum of $A$ and let $G(A)$ be its minimal Gershgorin set defined as above. Then $$ \sigma(A)\subseteq G(A). $$
Proof. Given $\mathbf{x}>0$ , let $X=diag\left\{x_1, x_2, \ldots, x_n\right\}$ and let $B_X=X^{-1}AX$ . Then $A$ and $B_X$ share the same spectrum, being similar. Due to definition, and keeping in mind that $X^{-1}=diag\left\{x_1^{-1}, x_2^{-1}, \ldots, x_n^{-1}\right\}$ , we have $b_{ij}^{(X)}=a_{ij}\frac{x_j}{x_i}$ and, applying Gershgorin theorem to $B_X$ , we get:
$\sigma(A)=\sigma(B_X)\subseteq\bigcup_{i=1}^n \left\{z\in\mathbf{C}:\left|z-a_{ii}\right|\leq\frac{1}{x_i}\sum_{j\ne i}\left|a_{ij}\right|x_j\right\}$
and, since this is true for any $\mathbf{x}>0$ , we finally get the thesis. 
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