search for: b'b

Dear R developers, The trace of the hat matrix H~(n,n) is computed as follows: tr(H) = tr(BS^-1B') = tr(S^-1B'B) := tr(X) = sum(diag(X)) with B~(n,p), S~(p,p). Since p is of the order 10^3 but S is sparse I would like to employ Taucs linear solver ( http://www.tau.ac.il/~stoledo/taucs/ ) on SX = B'B. (Further improvement by implying a looping over i=1,...,p, calling tauc...

Hello, all, I had problems with an extension to a classic optimization problem. The target is to minimize a quadratic form a'Ma with respect to vector b, where vector a=(b',-1)', i.e., a is the expand of b, and M is a symmetric matrix (positive definite if needed). One more constrain on b is b'b=1. I wan...

Dear R-users, I would like to know if there is any way to minimize || y - Xb||^2 under the constraint b'b=1. Thank you Giancarlo

(renaming subject as I am partially getting off-topic) On Sunday 10 August 2014 16:26:07 Richard W.M. Jones wrote: > > The next issue I see now is about the value_value function. This is > > briefly documented as: "return data length, data type and data of a > > value". > > &...

Dear Peter, thank you very much for your answer. My problem is that I need to calculate the following quantity: solve(chol(A)%*%Y) Y is a 3*3 diagonal matrix and A is a 3*3 matrix. Unfortunately one eigenvalue of A is negative. I can anyway take the square root of A but when I multiply it by Y, the imaginary part of the square root of A is dropped, a...

I want to compute B=A^{1/2} such that B*B=A. For example a=matrix(c(1,.2,.2,.2,1,.2,.2,.2,1),ncol=3) so > a [,1] [,2] [,3] [1,] 1.0 0.2 0.2 [2,] 0.2 1.0 0.2 [3,] 0.2 0.2 1.0 > a%*%a [,1] [,2] [,3] [1,] 1.08 0.44 0.44 [2,] 0.44 1.08 0.44 [3,] 0.44 0.44 1.08 > b=a%*%a i have tried to use...

Dear All, My question is simple but I need someone to help me out. Suppose I have a positive definite matrix A. The funtion chol() gives matrix L, such that A = L'L. The inverse of A, say A.inv, is also positive definite and can be factorized as A.inv = M'M. Then A = inverse of (A.inv) = inverse of (M'M) = (in...

It works! But Once I have the square root of this matrix, how do I convert it to a real (not imaginary) matrix which has the same property? Is that possible? Best, Simon >----Messaggio originale---- >Da: p.dalgaard at biostat.ku.dk >Data: 21-nov-2009 18.56 >A: "Charles C. Berry"<cb...

Thanks for your message! Actually it works quite well for me too. If I then take the trace of the final result below, I end up with a number made up of both a real and an imaginary part. This does not probably mean much if the trace of the matrix below givens me info about the degrees of freedom of a model... Simona >----Messaggio originale---- >Da: RVaradhan at jhmi.edu >Data: 25-nov-2009 18.5...

Hi, I have been working on a Python application that uses hivex. Meanwhile I have encountered some Python bindings issues which could be fixed. The next issue I see now is about the value_value function. This is briefly documented as: "return data length, data type and data of a value". For Perl,...

DeaR list, I happily use eigen() to compute the eigenvalues and eigenvectors of a fairly large matrix (200x200, say), but it seems over-killed as its rank is limited to typically 2 or 3. I sort of remember being taught that numerical techniques can find iteratively decreasing eigenvalues and corresponding orthogonal eigenvectors, which would provide a nice alternative (once I have the first 3, say, I stop th...

Hi the manpage says that crossprod(x,y) is formally equivalent to, but faster than, the call 't(x) %*% y'. I have a vector 'a' and a matrix 'A', and need to evaluate 't(a) %*% A %*% a' many many times, and performance is becoming crucial. With f1 <- function(a,X){ ignore <- t(a) %*% X %*% a } f2 <- function(a...