Gershgorin’s circle theorem

Let $A$ be a square complex matrix. Around every element $a_{ii}$ on the diagonal of the matrix, we draw a circle with radius the sum of the norms of the other elements on the same row ${\sum }_{j\ne i}|a_{ij}|$. Such circles are called Gershgorin discs.

Theorem: Every eigenvalue of A lies in one of these Gershgorin discs.

Proof: Let $\lambda$ be an eigenvalue of $A$ and $x$ its corresponding eigenvector. Choose $i$ such that $|x_i|={\max }_j|x_j|$. Since $x$ can’t be $0$, $|x_i|>0$. Now $Ax=\lambda x$, or looking at the $i$-th component

$$(\lambda -a_{ii})x_i=\sum _{j\ne i}{a}_{ij}{x}_j.$$

Taking the norm on both sides gives

$$|\lambda -a_{ii}|=|\sum _{j\ne i}\frac{a_{ij}{x}_j}{x_i}|\le \sum _{j\ne i}|a_{ij}|.$$

Title	Gershgorin’s circle theorem
Canonical name	GershgorinsCircleTheorem
Date of creation	2013-03-22 13:14:15
Last modified on	2013-03-22 13:14:15
Owner	lieven (1075)
Last modified by	lieven (1075)
Numerical id	7
Author	lieven (1075)
Entry type	Theorem
Classification	msc 15A42
Synonym	Gershgorin’s disc theorem
Synonym	Gerschgorin’s circle theorem
Synonym	Gerschgorin’s disc theorem
Related topic	BrauersOvalsTheorem
Defines	Gershgorin disc
Defines	Gerschgorin disc