[Rd] Inverting a square matrix using solve() with LAPACK=TRUE (PR#13762)
Peter Dalgaard
p.dalgaard at biostat.ku.dk
Thu Jun 18 09:14:02 CEST 2009
rvaradhan at jhmi.edu wrote:
> Full_Name: Ravi Varadhan
> Version: 2.8.1
> OS: Windows
> Submission from: (NULL) (162.129.251.19)
>
>
> Inverting a matrix with solve(), but using LAPACK=TRUE, gives erroneous
> results:
Thanks, but there seems to be a much easier fix.
Inside coef.qr, we have
coef[qr$pivot, ] <-
.Call("qr_coef_real", qr, y, PACKAGE = "base")[seq_len(p)]
which should be [seq_len(p),]
(otherwise, in the matrix case, the RHS will recycle only the 1st p
elements, i.e., the 1st column).
>
> Here is an example:
>
> hilbert <- function(n) { i <- 1:n; 1 / outer(i - 1, i, "+") }
> h5 <- hilbert(5)
> hinv1 <- solve(qr(h5))
> hinv2 <- solve(qr(h5, LAPACK=TRUE))
> all.equal(hinv1, hinv2) # They are not equal
>
> Here is a function that I wrote to correct this problem:
>
> solve.lapack <- function(A, LAPACK=TRUE, tol=1.e-07) {
> # A function to invert a matrix using "LAPACK" or "LINPACK"
> if (nrow(A) != ncol(A)) stop("Matrix muxt be square")
> qrA <- qr(A, LAPACK=LAPACK, tol=tol)
> if (LAPACK) {
> apply(diag(1, ncol(A)), 2, function(x) solve(qrA, x))
> } else solve(qrA)
> }
>
> hinv3 <- solve.lapack(h5)
> all.equal(hinv1, hinv3) # Now, they are equal
>
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> R-devel at r-project.org mailing list
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--
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