Thanks all of you for the very interesting discussion. I think I get it
now.
Feng
On 03/16/2011 09:25 PM, rex.dw...@syngenta.com wrote:
Feng,
Your matrix is *not* (practically) singular; its inverse is.
The message said that the *system* was singular, not the matrix.
Remember Cramer's Rule: xi = |Ai| / |A|
The really, really large determinant of your matrix is going to appear in the
denominator of your solutions, so, essentially, you get underflow. Try working
out the entire solution with Cramer's Rule if you still don't see the problem.
solve doesn't really use Cramer's Rule, but it will give you a feel for the
issue.
HTH
Rex
-----Original Message-----
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Berend Hasselman
Sent: Wednesday, March 16, 2011 1:33 PM
To: r-help@r-project.org
Subject: Re: [R] Singularity problem
Peter Langfelder wrote:
On Wed, Mar 16, 2011 at 8:28 AM, Feng Li<m...@feng.li> wrote:
Dear R,
If I have remembered correctly, a square matrix is singular if and only
if
its determinant is zero. I am a bit confused by the following code error.
Can someone give me a hint?
a<- matrix(c(1e20,1e2,1e3,1e3),2)
det(a)
[1] 1e+23
solve(a)
Error in solve.default(a) :
system is computationally singular: reciprocal condition number = 1e-17
You are right, a matrix is mathematically singular iff its determinant
is zero. However, this condition is useless in practice since in
practice one cares about the matrix being "computationally" singular,
i.e. so close to singular that it cannot be inverted using the
standard precision of real numbers. And that's what your matrix is
(and the error message you got says so).
You can write your matrix as
a = 1e20 * matrix (c(1, 1e-18, 1e-17, 1e-17), 2, 2)
Compared to the first element, all of the other elements are nearly
zero, so the matrix is numerically nearly singular even though the
determinant is 1e23. A better measure of how numerically unstable the
inversion of a matrix is is the condition number which IIRC is
something like the largest eigenvalue divided by the smallest
eigenvalue.
svd(a) indicates the problem.
largest singular value / smallest singular value=1e17 (condition number)
--> reciprocal condition number=1e-17
and the standard solve can't handle that.
(pivoted) QR decomposition does help. And so does SVD.
Berend
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Feng Li
Department of Statistics
Stockholm University
106 91 Stockholm, Sweden
http://feng.li/
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