From: Yingfu Xie <Yingfu.Xie_at_sekon.slu.se>

Date: Fri 11 Aug 2006 - 00:08:15 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 Fri Aug 11 00:12:11 2006

Date: Fri 11 Aug 2006 - 00:08:15 EST

Hello, Berwin,

Thanks a lot for answering! I prefer the first method, which seems
easier to carry out. So, what I need now is some method to solve the
non-linear equation

c'(M_11-lam*I)^(-1) (M_11-lam*I)^(-1)c=1
for lam, where c is a vector, M_11 is a symmetric matrix and I is the
identity matrix.

I searched the R archive somehow, but didn't find anything valuable. Is
there any function in R available for this problem? I am grateful of any
hints.

When c=0, my problem reduces to the typical minimization of b'M_11b w.r.t b'b=1. The solution is the normalized eigenvector associated with the minimum eigenvalue of M_11, right?

Thanks,

Yingfu

PS: There is a type error in the first condition for b: the '+' should write to '-'.

-----Original Message-----

From: Berwin A Turlach [mailto:berwin@bossiaea.maths.uwa.edu.au] On
Behalf Of Berwin A Turlach

Sent: den 10 augusti 2006 03:57

To: Yingfu Xie

Cc: Rolf Turner; r-help@stat.math.ethz.ch
Subject: Re: [R] minimization a quadratic form with some coef fixed and
someconstrained

>>>>> "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

- Full address ============================ Berwin A Turlach Tel.: +61 (8) 6488 3338 (secr) School of Mathematics and Statistics +61 (8) 6488 3383 (self)

The University of Western Australia FAX : +61 (8) 6488 1028

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