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Gaussian elimination is stable for the inverse of a diagonally dominant matrix


Authors: Alan George and Khakim D. Ikramov
Journal: Math. Comp. 73 (2004), 653-657
MSC (2000): Primary 65F10
DOI: https://doi.org/10.1090/S0025-5718-03-01591-6
Published electronically: October 17, 2003
MathSciNet review: 2031399
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Abstract | References | Similar Articles | Additional Information

Abstract: Let $B\in M_n({\mathbf{C}})$ be a row diagonally dominant matrix, i.e.,

\begin{displaymath}\sigma_i \vert b_{ii}\vert = \sum_{\substack{j=1 \\ j\ne i }}^n \vert b_{ij}\vert, \quad i = 1,\ldots,n, \end{displaymath}

where $0 \le \sigma_i < 1, i= 1,\ldots,n,$ with $ \sigma = \max_{1\le i \le n} \sigma_i.$We show that no pivoting is necessary when Gaussian elimination is applied to $A = B^{-1}.$ Moreover, the growth factor for $A$ does not exceed $1 + \sigma.$ The same results are true with row diagonal dominance being replaced by column diagonal dominance.


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Additional Information

Alan George
Affiliation: School of Computer Science, University of Waterloo, Waterloo, Ontario, Canada

Khakim D. Ikramov
Affiliation: Faculty of Computational Mathematics and Cybernetics, Moscow State University, 119992 Moscow, Russia
Email: ikramov@cs.msu.su

DOI: https://doi.org/10.1090/S0025-5718-03-01591-6
Keywords: Gaussian elimination, growth factor, diagonally dominant matrices, Schur complement
Received by editor(s): February 13, 2002
Received by editor(s) in revised form: August 1, 2002
Published electronically: October 17, 2003
Additional Notes: This work was supported by Natural Sciences and Engineering Research Council of Canada grant OGP0008111
Article copyright: © Copyright 2003 American Mathematical Society

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