# [R] Computing the rank of a matrix.

Martin Maechler maechler at stat.math.ethz.ch
Sat Apr 7 16:57:27 CEST 2007

```>>>>> "Ravi" == Ravi Varadhan <rvaradhan at jhmi.edu>
>>>>>     on Fri, 6 Apr 2007 12:44:33 -0400 writes:

Ravi> Hi, qr(A)\$rank will work, but just be wary of the
Ravi> tolerance parameter (default is 1.e-07), since the
Ravi> rank computation could be sensitive to the tolerance
Ravi> chosen.

Yes, indeed.

The point is that   rank(<Matrix>)
is well defined in pure math (linear algebra), as well as
a "singular matrix" is.

The same typically no longer makes sense as soon as you enter
the real world: A matrix "close to singular" may have to be
treated "as if singular" depending on its "singularity
closeness" {{ learn about the condition number of a matrix }}
and the same issues arise with rank(<matrix>).

Of course, the matlab programmers know all this (and much more),
and indeed, matlab's  rank(A) really is
rank(A, tol = tol.default(A))

and is based on the SVD instead of QR decomposition since the
former is said to be more reliable (even though slightly slower).

R's equivalent (with quite a bit of fool-proofing) would be the
following function (assuming correct online documentation of matlab):

rankMat <- function(A, tol = NULL, singValA = svd(A, 0,0)\$d)
{
## Purpose: rank of a matrix ``as Matlab''
## ----------------------------------------------------------------------
## Arguments: A: a numerical matrix, maybe non-square
##		tol: numerical tolerance (compared to singular values)
##	   singValA: vector of non-increasing singular values of A
##		     (pass as argument if already known)
## ----------------------------------------------------------------------
## Author: Martin Maechler, Date:  7 Apr 2007, 16:16
d <- dim(A)
stopifnot(length(d) == 2, length(singValA) == min(d),
diff(singValA) < 0)       # must be sorted decreasingly
if(is.null(tol))
tol <- max(d) * .Machine\$double.eps * abs(singValA)
else stopifnot(is.numeric(tol), tol >= 0)
sum(singValA >= tol)
}

A small scale simulation with random matrices,
i.e., things like

## ranks of random matrices; here will have 5 all the time:
table(replicate(1000, rankMat(matrix(rnorm(5*12),5, 12) )))# < 1 sec.

indicates that qr(.)\$rank  gives the same typically,
where I assume one should really use

qr(., tol = .Machine\$double.eps, LAPACK = TRUE)\$rank

to be closer to Matlab's default tolerance.

Ok, who has time to investigate further?
Research exercise:

1>>  Is there a fixed number, say  t0 <- 1e-15
1>>    for which  qr(A, tol = t0, LAPACK=TRUE)\$rank is
1>>   ``optimally close'' to rankMat(A) ?

2>> how easily do you get cases showing svd(.) to more reliable
2>> than qr(., LAPACK=TRUE)?

To solve this in an interesting way, you should probably
investigate classes of "almost rank-deficient" matrices,
and I'd also be interested if you "randomly" ever get matrices A
with  rank(A) <  min(dim(A)) - 1
(unless you construct some columns/rows exactly from earlier
ones or use all-0 ones)

Martin Maechler, ETH Zurich

Ravi> Ravi.

Ravi> -----------------------------------------------------------------
Ravi> Assistant Professor, The Center on Aging and Health
.......

Ravi> --------------------------------------------------------------------

>>       qr(A)\$rank

>> or perhaps

>>       qr(A, LAPACK=TRUE)\$rank

>> Cheers,

>> Andy

>>     Hi! Maybe this is a silly question, but I need the
>> column rank (http://en.wikipedia.org/wiki/Rank_matrix)
>> of a matrix and R function 'rank()' only gives me the
>> ordering of the elements of my matrix.  How can I
>> compute the column rank of a matrix? Is there not an R
>> equivalent to Matlab's 'rank()'?  I've been browsing
>> for a time now and I can't find anything, so any help
>> will be greatly appreciated. Best regards!

>> -- -- Jose Luis Aznarte M.
>> http://decsai.ugr.es/~jlaznarte Department of Computer
>> Science and Artificial Intelligence Universidad de
>> Granada Tel. +34 - 958 - 24 04 67 GRANADA (Spain) Fax:
>> +34 - 958 - 24 00 79

```