rank of a matrix

For the example that Duncan just gave, using the "matrix.rank" function of
Spencer Graves (which uses singular value decomposition) I obtained the
following result:

exponent <- -(7:16)
eps <- 10^exponent
sapply(eps,mat=h9times4,function(x,mat)matrix.rank(mat,x))

[1] 6 7 7 8 8 9 9 9 9 9

This tells me that the correct rank should be 9, since the rank stabilizes
for smaller tolerances.  I realize that this may not work generally, and one
could create counter-examples to defeat this strategy.

Best,
Ravi.

--------------------------------------------------------------------------
Ravi Varadhan, Ph.D.
Assistant Professor,  The Center on Aging and Health
Division of Geriatric Medicine and Gerontology
Johns Hopkins University
Ph: (410) 502-2619
Fax: (410) 614-9625
Email:  rvaradhan at jhmi.edu
--------------------------------------------------------------------------

-----Original Message-----
From: r-help-bounces at stat.math.ethz.ch [mailto:r-help-
bounces at stat.math.ethz.ch] On Behalf Of Duncan Murdoch
Sent: Wednesday, May 04, 2005 3:32 PM
To: Gabor Grothendieck
Cc: mingan; r-help at stat.math.ethz.ch; Huntsinger,Reid
Subject: Re: [R] rank of a matrix

Gabor Grothendieck wrote:

In this case, try a lower tolerance (1e-7 is the default):

qr(hilbert(9), tol = 1e-8)$rank

[1] 9

But don't trust the results.  For example, create a matrix with 4
identical copies of hilbert(9).  This still has rank 9.  It's hard to
find, though:

 > h9 <- hilbert(9)
 > temp <- cbind(h9, h9)
 > h9times4 <- rbind(temp, temp)
 >
 > qr(h9times4,tol=1e-7)$rank

[1] 7

 > qr(h9times4, tol=1e-8)$rank

[1] 10

 > qr(h9times4, tol=1e-9)$rank

[1] 11

 > qr(h9times4, tol=1e-10)$rank

[1] 12


There's a tolerance that gives the right answer (1.5e-8 works for me),
but how would I know that in a real problem where I didn't already know
the answer?

Duncan Murdoch