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# 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

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

Defines:

Gershgorin disc, Gerschgorin disc

Related:

BrauersOvalsTheorem

Synonym:

Gershgorin's disc theorem, Gerschgorin's circle theorem, Gerschgorin's disc theorem

Type of Math Object:

Theorem

Major Section:

Reference

## Mathematics Subject Classification

15A42*no label found*

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new correction: Error in proof of Proposition 2 by alex2907

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new correction: Error in proof of Proposition 2 by alex2907

Jun 24

new question: A good question by Ron Castillo

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new question: A trascendental number. by Ron Castillo

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new question: Banach lattice valued Bochner integrals by math ias