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From: Mark Hoemmen (mark.hoemmen_at_[hidden])
Date: 2005-01-13 12:01:13

On Thu, 13 Jan 2005 17:38:03 +0100 (CET), Peter Schmitteckert
<peter_at_[hidden]> wrote:
> Allthough it's slightly off-topic, 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 well-conditioned. 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
2-norm condition number of the matrix is one.