[R] Singularity problem
Peter Langfelder
peter.langfelder at gmail.com
Wed Mar 16 17:59:12 CET 2011
On Wed, Mar 16, 2011 at 8:28 AM, Feng Li <m at 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.
Peter
More information about the R-help
mailing list