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 | The contents of the Binomial inverse theorem page were merged into Woodbury matrix identity on 27 January 2018 and it now redirects there. For the contribution history and old versions of the merged article please see its history. |
Woodbury
editThis seems to be the same as the Woodbury matrix identity. I guess these pages should be merged? -- Jitse Niesen (talk) 14:19, 28 December 2006 (UTC)
- Absolutely. Also, this page has no citations as to where the title is from. Jmath666 06:54, 30 September 2007 (UTC)
- Actually, it is slightly more general. But it should be merged just the same. Jmath666 07:04, 30 September 2007 (UTC)
- How is it "slightly more general"? Just giving more details? DavidMCEddy (talk) 17:26, 16 January 2016 (UTC)
- I think this should be merged into the Woodbury matrix identity page, not the other way around: The Woodbury article received 12,435 views in the last 90 days, while this only received 2,493 over the same period. Of course, we need a redirect, because I know the name "binomial inverse theorem" and not "Woodbury", and I suspect that's true for others. (On the other hand, I can't volunteer to do the merge.) DavidMCEddy (talk) 17:33, 16 January 2016 (UTC)
- How is it "slightly more general"? Just giving more details? DavidMCEddy (talk) 17:26, 16 January 2016 (UTC)
Done Klbrain (talk) 12:23, 27 January 2018 (UTC)
Incorrect passage?
editI feel that this phrase must be incorrect: "If B = Iq is the identity matrix and q = 1". Should B not be an p×p matrix in order to be added to A? 128.100.76.56 (talk) 13:38, 7 December 2010 (UTC)
- This is the same person (forgot to log in last time). Nevermind I figured it out. That special case paragraph is not an extension of the previous special case (I had thought that we had already set and V to Ip).Kiyo.masui (talk) 13:42, 7 December 2010 (UTC)
Invertibility of B
editThe first formula also requires B to be invertible. See Henderson and Searle (1981) SIAM Review 23:53-60. They provide a more general formula when B is singular. 128.148.160.246 (talk) 13:50, 8 March 2012 (UTC)
- Right--the first formula requires invertibility of [B+BVA—1UB], which equals [B(I+VA—1UB)]. The rank of the latter cannot exceed the rank of B, so it is not invertible unless B has full rank.
- I'll put this in along with the citation given in the original post here. Loraof (talk) 17:14, 10 May 2015 (UTC)