2011
DOI: 10.1090/s0002-9939-2011-10959-6
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Groups of banded matrices with banded inverses

Abstract: Abstract. A product A = F 1 . . . F N of invertible block-diagonal matrices will be banded with a banded inverse: A i j = 0 and also (A −1 ) ij = 0 for |i−j| > w. We establish this factorization with the number N controlled by the bandwidths w and not by the matrix size n. When A is an orthogonal matrix, or a permutation, or banded plus finite rank, the factors F i have w = 1 and we find generators of that corresponding group. In the case of infinite matrices, the A = LP U factorization is now established but … Show more

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Cited by 10 publications
(10 citation statements)
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“…This section describes how to find an appropriate fairing operation from Strang's groups of banded matrices with banded inverses [78].…”
Section: Methodsmentioning
confidence: 99%
See 1 more Smart Citation
“…This section describes how to find an appropriate fairing operation from Strang's groups of banded matrices with banded inverses [78].…”
Section: Methodsmentioning
confidence: 99%
“…In a banded matrix, non-zero entries are in a band w along the diagonal and a ij = 0 for |i − j| > w. Strang [77] proves that a banded matrix has a banded inverse if it can be factorized into a product of block diagonal matrices with invertible blocks. Using this property he establishes a group of banded matrices with banded inverses [78].…”
Section: Chasing Gamementioning
confidence: 99%
“…This short section establishes the factorizations P c = BC and Both factorizations are particularly simple cases of a more general theorem [22,23] for banded matrices with banded inverses. For permutations, P and P −1 = P ⊤ have the same bandwidth w. Figure 4.1 below shows a typical matrix P c with w = 2.…”
Section: Factorizations Of Banded Permutationsmentioning
confidence: 99%
“…Suppose q > 0. By observation (5), k may be moved to the right side by q greedy steps, in which k is most difficult for k 0 to overtake. We may assume that this worst case happens, and k is moved to the right by q steps before it is overtaken by k 0 , so k will be at the i l þ q position.…”
mentioning
confidence: 99%
“…Assume that there are r elements on the left of k that are also larger than k 0 , and among them, r 1 elements are on the left of k 0 so that i j − r 1 ≥ u 1 . Then by observation (5), k 0 may be moved to the left by r 1 greedy steps before k 0 overtakes k. After that, all r elements may force k 0 to remain stationary for r þ 1 greedy steps, by observation (6).…”
mentioning
confidence: 99%