Gaussian elimination is stable for the inverse of a diagonally dominant matrix
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- by Alan George and Khakim D. Ikramov PDF
- Math. Comp. 73 (2004), 653-657 Request permission
Abstract:
Let $B\in M_n({\mathbf {C}})$ be a row diagonally dominant matrix, i.e., \[ \sigma _i |b_{ii}| = \sum _{\substack {j=1 j\ne i }}^n |b_{ij}|, \quad i = 1,\ldots ,n, \] 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.References
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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
- 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
- © Copyright 2003 American Mathematical Society
- Journal: Math. Comp. 73 (2004), 653-657
- MSC (2000): Primary 65F10
- DOI: https://doi.org/10.1090/S0025-5718-03-01591-6
- MathSciNet review: 2031399