Talk:Biconjugate gradient stabilized method

Latest comment: 9 years ago by Fromain in topic Is the preconditioned algorithm correct?

Step 2

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Step 2 is confusing, there seems to be two different r0? —Preceding unsigned comment added by 130.37.28.128 (talkcontribs) 09:38, April 13, 2010 (UTC)

Could be a problem with your browser. The second r-zero is in fact "r hat"-zero. Kxx (talk | contribs) 14:07, 13 April 2010 (UTC)Reply
I was confused too, as the spanish page says:
Elige un vector arbitrario   tal que  , por ejemplo,  
while the english one says:
Choose an arbitrary vector r̂0 such that (r̂0, r0) ≠ 0, e.g., r̂0 = r0
Which one is right? Should the english page be changed? Nicolas Bigaouette (talk) 21:40, 26 April 2010 (UTC)Reply
This is how your reply looks on my machine: http://img153.imageshack.us/img153/1053/61070581.png. I don't see any problems. Kxx (talk | contribs) 08:14, 27 April 2010 (UTC)Reply

Fails if Residual is Eigenvector

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  • I'm wondering if anyone else has noticed the problem that if   is any eigenvector of   and  , step 5.8 will evaluate as  , thus wreaking havoc. Mind you, if this case does crop up, there's an exact solution of  , where  . Another instant failure case is if   and  , in which case step 5.5 divides by zero, though it might be that this only happens when either the above is true or the initial guess is orthogonal to the solution. Are there are any other instant failure cases to worry about? --Ndickson (talk) 05:36, 22 February 2012 (UTC)Reply
Even if it does not fail instantly, it can still fail after some iterations because there is no guarantee that the method will converge. So people generally do not care if the failure is instant or not. When it fails, you can choose another initial guess or move to another method such as GMRES. Kxx (talk | contribs) 07:18, 28 July 2012 (UTC)Reply

Is the preconditioned algorithm correct?

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I'm rather confused by how the preconditioners are defined. From modified system  , it looks like   and   are intended to be left and right preconditioners respectively. However as I understand it, this means that we should have either    and   is the identity matrix, or    and   is the identity matrix, depending on if a left or right preconditioner is to be used. At worst it should be that   should be well conditioned, but I can't figure out where the relation    comes from.

With that in mind, by naively solving the modified system   using the unpreconditioned algorithm, I find that the following steps should be modified:

1.  

4.  

5.  

8.  

9.  

10.  

where   and   are left and right preconditioners. Only one can be used by simply setting the other to  , or both can be used if they make   more suitable to iterative solving.

Is that correct, or am I missing something? — Preceding unsigned comment added by Fromain (talkcontribs) 21:32, 9 June 2015 (UTC)Reply