Re: [R] Singularity problem

From: Peter Langfelder <>
Date: Wed, 16 Mar 2011 09:59:12 -0700

On Wed, Mar 16, 2011 at 8:28 AM, 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 mailing list PLEASE do read the posting guide and provide commented, minimal, self-contained, reproducible code. Received on Wed 16 Mar 2011 - 17:01:33 GMT

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