On Gersgorin Theorem (Solved)

Importance: Medium ✭✭
Subject: Algebra
Recomm. for undergrads: no
Posted by: Miwa Lin
on: October 11th, 2008
Solved by: two commenters to the problem

Gersgorin theorem states that: all the eigenvalues of $ A=[a_{ij}]\in M_n $ are located in the union of $ n $ discs $ \bigcup\limits_{i=1}^n\{z\in C:|z-a_{ii}|\leq \sum\limits_{j=1,j\neq i}^n|a_{ij}|\} $. For some special matrices, the region can be confined to $ \bigcup\limits_{i=1}^n\{z\in C:|z-a_{ii}|\leq \sum\limits_{j=1,j\neq i}^n|a_{ij}|\}\backslash\{z\in C:|z-a_{kk}|<\sum\limits_{j=1,j\neq k}^n|a_{kj}|\} $ for some $ k $. I wonder if the new region above is valid in general?


* indicates original appearance(s) of problem.

This is false

Take any diagonal matrix. Then the eigenvalues are the diagonal entries but the the Gerschgorin discs are points centred at the diagonal entries. So we can't remove one of the discs from the set.


Doesn't the matrix $ \begin{pmatrix} 1 & 2 <br> 2 & 2 \end{pmatrix} $ provide a counterexample?

It remains open.

It remains open.

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