Talk:Woodbury matrix identity

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Latest comment: 15 April 2024 by Ldm1954 in topic Why so negative
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Intro

This article should begin by explaining what the Woodbury matrix identity is in general terms (what do C, U, V represent?), and what it is useful for. -- Tarquin 12:13, 14 Jan 2004 (UTC)

Done! Is this complete? -- Martin Vermeer File:Yes check.svg Done

Minus sign

Corrected a creeping minus sign -- ofer Aug 30 2004 File:Yes check.svg Done

What if C1+VA1U=0?

What if C1+VA1U=0 (for k=1,C=I: VA1U=I)? --RainerBlome 22:07, 4 October 2005 (UTC)Reply

Fonts

The correct mathematical fonts should be used. For example: A matrix should be represented by A and not A. File:Yes check.svg Done

Alternative formulation

For discussion, is the following

(A+UCV)1=A1A1UVA1C1+VA1U,

in the first place correct and second a more readable format of the theorem. Faust o 18:57, 21 January 2006 (UTC)Reply

It is correct if the right-hand side is interpreted as
A1A1U(C1+VA1U)1VA1.
I'm pretty sure it is not correct if the right-hand side is interpreted as
A1(C1+VA1U)1A1UVA1
or as
A1A1UVA1(C1+VA1U)1.
Since it is not clear how to interpret the right-hand side in your formulation of the theorem, I would say that it is not more readable. -- Jitse Niesen (talk) 19:08, 21 January 2006 (UTC)Reply
Good point, I agree to that. The form I suggested was stated in an engineering text book.Faust o 20:09, 21 January 2006 (UTC)Reply
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The point?

Why would anyone assume that inverting the matrix C1+VA1U, as required by the formula, is any easier than inverting the original matrix?

Nevermind. C is supposed to have lower dimension than A. Got it. —Preceding unsigned comment added by 92.250.74.157 (talk) 04:14, 3 January 2010 (UTC)Reply

The identity holds even if C is higher dimension than A, but from practical point of view, the intention is to have C smaller, because then it is more efficient to invert C, and to invert A directly. Such situation occurs frequently in real life application.
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Merge with article on Binomial inverse theorem?

This one is a no brainer. — Preceding unsigned comment added by 128.100.76.127 (talk) 13:47, 14 December 2011 (UTC)Reply

I think the Binomial inverse theorem article should be merged in here, because this article has more traffic: 12,435 vs. 2,493 views in the last 90 days. DavidMCEddy (talk) 17:51, 16 January 2016 (UTC)Reply
File:Yes check.svg Done Klbrain (talk) 12:23, 27 January 2018 (UTC)Reply

Is C useful?

Is a non-identity C matrix actually useful? It seems like C could easily be absorbed into the U or V matrices. This has the benefit of trivializing the computation of C1=I --Mborg (talk) 23:35, 9 May 2014 (UTC)Reply

History?

This looks like not too complex and natural extension of very well known Sherman-Morrison formula. Did Sherman or Morrison were aware of this? I did find a prior for this result: W.J. Duncan, “Some devices for the solution of large sets of simultaneous equations (with an appendix on the reciprocation of partitioned matrices)”, The London, Edinburgh and Dublin Philosophical Magazine and Journal of Science, Seventh Series, 35, p. 660, (1944).[1] I did found it as https://doi.org/10.1080/14786444408520897 . This was cited in https://arxiv.org/abs/1803.10405v1 / https://arxiv.org/pdf/1803.10405.pdf , which is reprint/reedit on arxiv from "A Sherman-Morrison-Woodbury Identity for Rank Augmenting Matrices with Application to Centering", Kurt S. Riedel, which originally appeared in SIAM J. MAT.ANAL. Vol. 12, No. 1, pp. 80–95, January 1991; and in SIAM Journal on Numerical Analysis Vol. 31, No. 4 (Aug., 1994), pp. 1219-1225.[2] I believe this is https://doi.org/10.1137/0613040 , This is the same journal where in 1989, the other paper appeared: W.W. Hager, “Updating the inverse of a matrix”, SIAM Review, 31, p.221, (1989); that focused on the Woodbury identity.[3] https://doi.org/10.1137/1031049 Also there is Bartlett's identity (~1951?), which does have similarities. Professor Duncan even in the abstract of his paper from 1944, states that the method is not new in principle, but he simply made a new clean and general formulation. Considering that Sherman-Morrison published their papers on their results in 1949 and 1950, it would be worth exploring the history more, as their results were apparently not novel. Looking at the citations in their paper, there is no mention or citations of even closely similar methods, indicating possibly they were not aware of any previous results in this area. Some other earlier papers to check - https://www.jstor.org/stable/2235999 [4] , as they do have many useful methods and additional references.

2A02:168:F609:0:93EE:57FD:B2FA:71FB (talk) 13:13, 11 April 2019 (UTC)Reply

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Dubious claim regarding numerical accuracy

This claim in the introduction does not seem to be sufficiently well cited (the citation points to a Stack Overflow question):

"The Woodbury matrix identity allows cheap computation of inverses and solutions to linear equations. However, little is known about the numerical stability of the formula. There are no published results concerning its error bounds.[citation needed] Anecdotal evidence suggests that it may diverge even for seemingly benign examples (when both the original and modified matrices are well-conditioned)."

I think it should be removed. The-erinaceous-one (talk) 09:28, 4 December 2023 (UTC)Reply

Why so negative

I think this article is inappropriately negative. The formula (and the Sherman-Morrison-Woodbury) are heavily used. I agree with @The-erinaceous-one that it seems very dubious. Ldm1954 (talk) 06:19, 15 April 2024 (UTC)Reply

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