Gershgorin’s circle theorem result

Since the eigenvalues of $A$ and $A$ transpose are the same, you can get an additional set of discs which has the same centers, $a_{ii}$, but a radius calculated by the column ${\sum }_{j\ne i}|a_{ji}|$ (instead of the rows). If a disc is isolated it must contain an eigenvalue. The eigenvalues must lie in the intersection of these circles. Hence, by comparing the row and column discs, the eigenvalues may be located efficiently.

Title	Gershgorin’s circle theorem result
Canonical name	GershgorinsCircleTheoremResult
Date of creation	2013-03-22 13:48:47
Last modified on	2013-03-22 13:48:47
Owner	saki (2816)
Last modified by	saki (2816)
Numerical id	11
Author	saki (2816)
Entry type	Result
Classification	msc 15A42