
Ublas : 
From: Mark Hoemmen (mark.hoemmen_at_[hidden])
Date: 20050113 12:01:13
On Thu, 13 Jan 2005 17:38:03 +0100 (CET), Peter Schmitteckert
<peter_at_[hidden]> wrote:
> Allthough it's slightly offtopic, isn't the determinant a precise measure
> for singularity? One just has to be careful that the
> singularity has nothing to do with convergence rate in usual iterative
> methods, where the condition number  A  A^{1}  provides
> a better measure? Or have I messed up namings?
In exact arithmetic, (determinant == 0) is equivalent to (matrix is singular).
In finite precision arithmetic there are no such guarantees. Furthermore,
a matrix can have a tiny determinant and be perfectly wellconditioned. For
example, consider the n x n diagonal matrix whose diagonal entries are all
the smallest positive number representable in double precision. Then the
product of the diagonal entries is zero in finite precision, even though the
2norm condition number of the matrix is one.
mfh