From: Berwin A Turlach <berwin_at_maths.uwa.edu.au>

Date: Thu 10 Aug 2006 - 11:56:50 EST

But simple enough. :)

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https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Thu Aug 10 12:01:26 2006

Date: Thu 10 Aug 2006 - 11:56:50 EST

>>>>> "YX" == Yingfu Xie <Yingfu.Xie@sekon.slu.se> writes:

YX> Thanks for reply! But I think that solution is right without YX> the constrain b'b=1. With this constrain, the solution is not YX> so simple. :(

But simple enough. :)

Write down the Lagrange function for the problem. Say, 'lam' is the Lagrange parameter for enforcing the constraint b'b=1. Then, using Rolf's notation:

RT> [...] Write M as

RT> | M_11 c |

RT> | c' m |

Then the system of equations that b and the Lagrange parameter have to fulfill is:

b = (M_11 + lam*I)^{-1} c (with I being the identity matrix) and lam = b' M_11 b - b'c

You can either use the first equation and do a (grid) search for the value of 'lam' that gives you b'b=1 (could be negative!), or start with lam=0 and then alternate between the two equations until convergence.

At least I think that this will solve your problem. :) Thinking a bit about the geometry of the problem, I actually believe that if c=0, you might have an identifiability problem, i.e. there are at least two solutions, or, depending on M_11, infinitely many.

Hope this helps.

Cheers,

Berwin

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