Thanks for the response. Yes, in that situation a solution of x = 1 would be
just as good as x = 1000 or any other value of x for me (but in my problem the
matrix has nonzero rank, so I can't just randomly choose a vector and have it
be a solution). If it helps, what I'm interested in is the R equivalent ofÂ
x = A\b
in MATLAB, for these particular kinds of A matrices. I looked into irlba, and
it seems to be able to calculate some of the singular values/vectors for the
large dataset without taking too much time. I'll look more into seeing how I
can solve the system with it.
On Wednesday, April 20, 2016 11:01 AM, Jeff Newmiller
<jdnew...@dcn.davis.ca.us> wrote:
This is kind of like asking for a solution to x+1=x+1. Go back to linear
algebra and look up Singular Value Decomposition, and decide if you really want
to proceed. See also ?svd and package irlba.
--
Sent from my phone. Please excuse my brevity.
On April 20, 2016 4:22:34 AM PDT, A A via R-help <r-help@r-project.org> wrote:
I have a situation in R where I would like to find any x (if one exists) that
solves the linear system of equations Ax = b, where A is square, sparse, and
singular, and b is a vector. Here is some code that mimics my issue with a
relatively simple A and b, along with three other methods of solving this
system that I found online, two of which give me an error and one of which
succeeds on the simplified problem, but fails on my data set(attached). Is
there a solver in R that I can use in order to get x without any errors given
the structure of A? Thanks for your time.
#CODE STARTS HEREA =
as(matrix(c(1.5,-1.5,0,-1.5,2.5,-1,0,-1,1),nrow=3,ncol=3),"sparseMatrix")b =
matrix(c(-30,40,-10),nrow=3,ncol=1)
#solve for x, Error in LU.dgC(a) : cs_lu(A) failed: near-singular A (or out of
memory)solve(A,b,sparse=TRUE,tol=.Machine$double.eps)
#one x that happens to solve Ax = bx = matrix(c(-10,10,0),nrow=3,ncol=1)A %*% x
#Error in
lsfit(A, b) : only 3 cases, but 4 variableslsfit(A,b)#solves the system, but
fails belowsolve(qr(A, LAPACK=TRUE),b)#Error in qr.solve(A, b) : singular
matrix 'a' in solveqr.solve(A,b)
#matrices used in my actual problem (see attached files)A = readMM("A.txt")b =
readMM("b.txt")
#Error in as(x, "matrix")[i, , drop = drop] : subscript out of
boundssolve(qr(A, LAPACK=TRUE),b)
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