Use qr tmp <- cbind(matrix(rnorm(30), 5, 6), 0)[, c(1,2,3,4,7,5,6)] tmp tmp.qr <- qr(tmp) tmp.qr tmp.qr$pivot tmp.subset <- tmp[, tmp.qr$pivot[1:tmp.qr$rank]] solve(tmp.subset)
On Sun, Oct 2, 2016 at 2:19 PM, Bertrand Marc <beberk...@gmail.com> wrote: > Dear R helpers, > > I am looking for an efficient way to extract (any) one of the biggest > invertible submatrix. > > I have a rectangular matrix A (p x n), with rank k <= min(p, n). I would like > to get a submatrix (k x k) invertible, or even better, the list of rows and > columns of A which > would form the submatrix (A[rows, columns] would be invertible, with > length(rows)=length(columns)=k). > > This is the general problem, but in my particular R code, the rank of A would > be p (p<n), so I only need to select p columns to get the submatrix. But I am > not sure it is > easier. > For now, my (very bad) solution would be to try every submatrix until I find > one invertible. > > Do you think of any solution which would be more efficient ? > > Best regards, > Bertrand > > ______________________________________________ > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see > 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. ______________________________________________ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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.
