# 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 $\sum_{j\neq i}|a_{ij}|$. 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

 $(\lambda-a_{ii})x_{i}=\sum_{j\neq i}a_{ij}x_{j}.$

Taking the norm on both sides gives

 $|\lambda-a_{ii}|=|\sum_{j\neq i}\frac{a_{ij}x_{j}}{x_{i}}|\leq\sum_{j\neq i}|a% _{ij}|.$
 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