matrix-inverse

Is there a package for parallel matrix inversion in R

十年热恋 提交于 2019-12-07 23:51:07
问题 This question was migrated from Cross Validated because it can be answered on Stack Overflow. Migrated 6 years ago . is there a package for matrix inversion in R using parallel computation? Thanks! Hello. I am having trouble installing the HiPLARb package, here is what I did: Download auto-installer script: http://www.hiplar.org/downloads/HiPLARb.Installer ./HiPLARb.Installer --with-openblas --no-gpu --prefix=/home/kaiyin/mylib everything went ok, all prerequesite libraries installed

Most efficient matrix inversion in MATLAB

本秂侑毒 提交于 2019-12-07 02:35:14
问题 When computing the inverse for some square matrix A in MATLAB, using Ai = inv(A) % should be the same as: Ai = A^-1 MATLAB usually notifies me that this is not the most efficient way of inverting. So what's more efficient? If I have an equation system, using the /,\ operators probably is. But sometimes I need the inverse for other computations. What's the most efficient way to invert? 回答1: I would recommend to use svd (unless you are really absolute sure that your matrix is not ill

Inverse of matrix and multiplication

耗尽温柔 提交于 2019-12-06 16:46:08
I am new to world of matrix, sorry for this basic question I could not figure out: I have four matrix (one unknown). Matrix X x <- c(44.412, 0.238, -0.027, 93.128, 0.238, 0.427, -0.193, 0.673, 0.027, -0.193, 0.094, -0.428, 93.128, 0.673, -0.428, 224.099) X <- matrix(x, ncol = 4 ) Matrix B : need to be solved , 1 X 4 (column x nrows), with b1, b2, b3, b4 values Matrix G g <- c(33.575, 0.080, -0.006, 68.123, 0.080, 0.238, -0.033, 0.468, -0.006, -0.033, 0.084, -0.764, 68.123, 0.468, -0.764, 205.144) G <- matrix(g, ncol = 4) Matrix A a <- c(1, 1, 1, 1) # one this case but can be any value A <-

matrix inversion R

血红的双手。 提交于 2019-12-06 13:17:13
问题 I want to inverse a square symmetric positive definite matrix. I know there are two functions solve() and chol2inv() in R but their results is different. I need to know why this happen? Thank you. 回答1: For solve you need to give your original matrix, but for chol2inv you use precomputed cholesky decomposition: set.seed(1) a<-crossprod(matrix(rnorm(9),3,3)) a_chol<-chol(a) solve(a) [,1] [,2] [,3] [1,] 1.34638151 -0.02957435 0.8010735 [2,] -0.02957435 0.32780020 -0.1786295 [3,] 0.80107345 -0

Symmetric Matrix Inversion in C using CBLAS/LAPACK

徘徊边缘 提交于 2019-12-06 11:18:44
问题 I am writing an algorithm in C that requires Matrix and Vector multiplications. I have a matrix Q (W x W) which is created by multiplying the transpose of a vector J (1 x W) with itself and adding Identity matrix I , scaled using scalar a . Q = [(J^T) * J + aI]. I then have to multiply the inverse of Q with vector G to get vector M . M = (Q^(-1)) * G. I am using cblas and clapack to develop my algorithm. When matrix Q is populated using random numbers (type float) and inverted using the

Most efficient matrix inversion in MATLAB

拜拜、爱过 提交于 2019-12-05 06:30:46
When computing the inverse for some square matrix A in MATLAB, using Ai = inv(A) % should be the same as: Ai = A^-1 MATLAB usually notifies me that this is not the most efficient way of inverting. So what's more efficient? If I have an equation system, using the /,\ operators probably is. But sometimes I need the inverse for other computations. What's the most efficient way to invert? I would recommend to use svd (unless you are really absolute sure that your matrix is not ill-conditioned). Then, based on singular values you make your decisions on further actions to take. This may sound like a

pseudo inverse of sparse matrix in python

时光毁灭记忆、已成空白 提交于 2019-12-04 21:10:41
问题 I am working with data from neuroimaging and because of the large amount of data, I would like to use sparse matrices for my code (scipy.sparse.lil_matrix or csr_matrix). In particular, I will need to compute the pseudo-inverse of my matrix to solve a least-square problem. I have found the method sparse.lsqr, but it is not very efficient. Is there a method to compute the pseudo-inverse of Moore-Penrose (correspondent to pinv for normal matrices). The size of my matrix A is about 600'000x2000

Symmetric Matrix Inversion in C using CBLAS/LAPACK

余生长醉 提交于 2019-12-04 16:17:29
I am writing an algorithm in C that requires Matrix and Vector multiplications. I have a matrix Q (W x W) which is created by multiplying the transpose of a vector J (1 x W) with itself and adding Identity matrix I , scaled using scalar a . Q = [(J^T) * J + aI]. I then have to multiply the inverse of Q with vector G to get vector M . M = (Q^(-1)) * G. I am using cblas and clapack to develop my algorithm. When matrix Q is populated using random numbers (type float) and inverted using the routines sgetrf_ and sgetri_ , the calculated inverse is correct . But when matrix Q is symmetrical , which

matrix inversion R

微笑、不失礼 提交于 2019-12-04 15:49:39
I want to inverse a square symmetric positive definite matrix. I know there are two functions solve() and chol2inv() in R but their results is different. I need to know why this happen? Thank you. For solve you need to give your original matrix, but for chol2inv you use precomputed cholesky decomposition: set.seed(1) a<-crossprod(matrix(rnorm(9),3,3)) a_chol<-chol(a) solve(a) [,1] [,2] [,3] [1,] 1.34638151 -0.02957435 0.8010735 [2,] -0.02957435 0.32780020 -0.1786295 [3,] 0.80107345 -0.17862950 1.4533671 chol2inv(a_chol) [,1] [,2] [,3] [1,] 1.34638151 -0.02957435 0.8010735 [2,] -0.02957435 0

numerically stable inverse of a 2x2 matrix

[亡魂溺海] 提交于 2019-12-04 10:13:30
In a numerical solver I am working on in C, I need to invert a 2x2 matrix and it then gets multiplied on the right side by another matrix: C = B . inv(A) I have been using the following definition of an inverted 2x2 matrix: a = A[0][0]; b = A[0][1]; c = A[1][0]; d = A[1][1]; invA[0][0] = d/(a*d-b*c); invA[0][1] = -b/(a*d-b*c); invA[1][0] = -c/(a*d-b*c); invA[1][1] = a/(a*d-b*c); In the first few iterations of my solver this seems to give the correct answers, however, after a few steps things start to grow and eventually explode. Now, comparing to an implementation using SciPy, I found that the