Gershgorin cicle theorem §

It’s possibile to gain some information about eigenvalues locations easly using Gershgorin’s theorem.

Let $D(c,r)$ denote the closed disk centered at $c\in \mathbb{C}$ of radius $r\in \mathbb{R}$.

Theorem Let $A\in {\mathbb{C}}^{n\times n}$, then the eigenvalues of $A$ are all contained in the union of the Gershgorin disks:

$$\bigcup _{i=1}^{n}D(a_{ii},R_i)R_i=\sum _{j=1,j\ne i}|a_{ij}|$$

The disks are centered at the diagonal elements of $A$, and their radius is the sum of the modulus of the row elements, exept the diagonal one.

Remark Since the spectrum of $A^h$ is the same, one can comput the gershgorin disks of $A^h$ and take the intersection with the disks of $A$ to improve the location exstimate.

Proof Consider an eigenvalue $\lambda$ and its corresponding norm $1$ eigevector $v$, so that $Av=\lambda v$. Let $v_i$, $i=1,\dots ,n$ be the biggest (in modulus) cordinate of $v$, then

$$\sum _{j=1}^n{a}_{ij}{v}_j=\lambda v_i$$

we now separate from the sum the diagonal term $a_{ii}{v}_i$ and rearrenge:

$$\lambda v_i-a_{ii}{v}_i=\sum _{j=1j\ne i}^n{a}_{ij}{v}_j$$

we now use the fact that $v_i\ge v_j$, we divide and get:

$$\lambda -a_{ii}\le \sum _{j=1j\ne i}^n{a}_{ij}$$

now we take norms and use the triangle inequality:

$$|\lambda -a_{ii}|\le \sum _{j=1j\ne i}^n|a_{ij}|=R_i$$

this is the definition of a closed disk centerd at $a_{ii}$ of radius $R_i$. $\square$

Corollary §

From this theorem follows easly that a striclty diagonal dominant matrix is invertible, since all its eigenvalues are non-zero.

Also, an hermitian and diagonal dominant matrix is positive semidefinite.