minimal Gershgorin set

related to Gershgorin’s theorem is the so called ”minimal Gershgorin set”.For every $A\in {𝐂}^{n,n}$, $𝐱>0$ meaning $x_i>0\mathit{\hspace{1em}}\forall i$, let’s define its minimal Gershgorin set $G(A)$ as:

$$G(A)=\bigcap _{𝐱>0}{G}_{𝐱}(A),$$

where

$$G_{𝐱}(A)=\bigcup _{i=1}^n\{z\in 𝐂:\left|z-a_{ii}\right|\le \frac{1}{x_i}\sum _{j\ne i}\left|a_{ij}\right|x_j\}.$$

Theorem: Let $A\in {𝐂}^{n,n}$, 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 $𝐱>0$, let $X=diag\{x_1,x_2,\mathrm{\dots },x_n\}$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 $X^{-1}=diag\{x_1^{-1},x_2^{-1},\mathrm{\dots },x_n^{-1}\}$, we have $b_{ij}^{(X)}=a_{ij}\frac{x_j}{x_i}$and, applying Gershgorin theorem to $B_X$, we get:

$\sigma (A)=\sigma (B_X)\subseteq {\bigcup }_{i=1}^n\{z\in 𝐂:\left|z-a_{ii}\right|\le \frac{1}{x_i}{\sum }_{j\ne i}\left|a_{ij}\right|x_j\}$

and, since this is true for any $𝐱>0$, we finally get the thesis.∎