# minimal Gershgorin set

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

 $G_{\mathbf{x}}(A)=\bigcup_{i=1}^{n}\left\{z\in\mathbf{C}:\left|z-a_{ii}\right|% \leq\frac{1}{x_{i}}\sum_{j\neq i}\left|a_{ij}\right|x_{j}\right\}.$

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:

$\sigma(A)=\sigma(B_{X})\subseteq\bigcup_{i=1}^{n}\left\{z\in\mathbf{C}:\left|z% -a_{ii}\right|\leq\frac{1}{x_{i}}\sum_{j\neq i}\left|a_{ij}\right|x_{j}\right\}$

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

Title minimal Gershgorin set MinimalGershgorinSet 2013-03-22 15:35:57 2013-03-22 15:35:57 Andrea Ambrosio (7332) Andrea Ambrosio (7332) 11 Andrea Ambrosio (7332) Definition msc 15A42