Talk:Lanczos algorithm

[[1]] - an unfocused variety of Lanczos algorithm —Preceding unsigned comment added by 134.219.166.104 (talk • contribs) 21:23, 1 September 2005

This doesn't have much but it does have a reference to a book mathworld on Lanczos Algorithm]—Preceding unsigned comment added by RJFJR (talk • contribs) 23:36, 25 September 2005

I don't believe this is the Lanczos algorithm at all. It is the power method. —Preceding unsigned comment added by 130.126.55.123 (talk • contribs) 01:04, 5 August 2006

I don't know if the algorithm is correct, but it's certainly different than the power method, and presented pretty clearly. I think it's gotten me on the right track at least... Thanks. --Jjdonald (talk) 22:22, 17 December 2007 (UTC)[reply]

It is not easy to say it's wrong or correct, since quite some information is missing ~~in order to apply it: (a) how to choose v[1], (b) how to choose m, (c) how to recognize the eigenvalues of A among those of T_mm~~. Unfortunately, this vagueness is by no means eliminated by the Numerical stability section. — MFH:Talk 21:57, 12 September 2008 (UTC)[reply]

It is certainly not completely correct: there's at least something faulty with the indices. — MFH:Talk 19:59, 8 December 2011 (UTC)[reply]

It should state that "it applies to Hermitian matrices" at the start of the article and not somewhere in the middle. limweizhong (talk) 09:54, 11 November 2008 (UTC)[reply]

There is a paper about Non-Symmetric Lanczos' algorithm (compared to Arnoldi) by Jane Cullum. — MFH:Talk 20:07, 8 December 2011 (UTC)[reply]

In Latent Semantic Indexing, for...

I really think that this sentense has nothing to do in the first paragraph! Please someone who understand anything about it should create a separate section and explain what this is about! Alain Michaud (talk) 16:52, 19 February 2010 (UTC)[reply]

Block Lanczos algorithm

I suppose that Peter Montgomery`s 1995 paper was very good, but I do not see the need to inform everyone about its existence. This topic is much too advanced to be discussed at the top of the page. Please move this (second paragraph) towards the end of the page.

Alain Michaud (talk) 16:50, 19 February 2010 (UTC)[reply]

Extracting information from tridiagonal matrix

So Lanczos gives you a tridiagonal matrix. I think a link would be helpful which explains how to extract low eigenvalues/eigenvectors from this matrix. —Preceding unsigned comment added by 209.6.144.249 (talk) 06:30, 2 March 2008 (UTC)[reply]

Agree - or largest eigenvalues: anyway, the article starts by saying that it's for calculating eigenvalues, but then stops with the tridiag. matrix.

B.t.w., the algorithm calculates up to v[m+1], I think this could be avoided. (also, "unrolling" the 1st part of the m=1 case as initialization should allow to avoid using v[0].) — MFH:Talk 03:09, 11 September 2008 (UTC)[reply]

PS: also, it should be said what is 'm'...

Define variables

It would be nice if variables are defined before (or just after) being used. For example, at the begining, $U$ and $\sigma _{i}$ are not defined and its confusing for non-expert public.

Felipebm (talk) 13:34, 17 May 2011 (UTC)[reply]

problematic matrix decomposition

In the section "Power method for finding eigenvalues", the matrix A is represented as $A=U{\text{diag}}(\sigma _{i})U'$ , which is true only for normal matrices. For the general case, SVD decomposition should be used, i.e. $A=U{\text{diag}}(\sigma _{i})V'$ where U and V are some orthogonal matrices. — Preceding unsigned comment added by 89.139.52.157 (talk) 12:14, 24 April 2016 (UTC)[reply]

It's not stated explicitly at that point, but presumably

A

is already taken to be Hermitian (as it needs to be for the Lanczos algorithm to work), which means it has an eigendecomposition of the form stated. Instead using the SVD decomposition in this argument won't work, because the entire point is that

U'U=I

so that the product telescopes! Possibly it would be clearer to just use

A=U\operatorname {diag} (\sigma _{i})U^{-1}

, i.e., hold off on requiring orthogonality — the reason being that the paragraph in question is about the plain power method, which applies in a greater generality. 130.243.68.202 (talk) 13:01, 2 May 2017 (UTC)[reply]