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On Gersgorin Theorem (Solved)

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Subject: Algebra
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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?


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This is false

On November 18th, 2009 alext87 says:

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.

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Counterexample?

On June 23rd, 2009 Anonymous says:

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

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It remains open.

On October 12th, 2008 Miwa Lin says:

It remains open.

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