# Does a tiny determinant imply ill-conditioning of a matrix?

If I have a square invertible matrix and I take its determinant, and I find that $\det(A) \approx 0$, does this imply that the matrix is poorly conditioned?

Is the converse also true? Does an ill-conditioned matrix have a nearly zero determinant?

Here is something I tried in Octave:

a = rand(4,4);
det(a) %0.008
cond(a)%125
a(:,4) = 1*a(:,1) + 2*a(:,2) = 0.000000001*ones(4,1);
det(a)%1.8E-11
cond(a)%3.46E10

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The determinant shows whether a matrix is regular or singular. It does not show whether it is well- or ill-conditioned. – Allan P. Engsig-Karup Feb 16 '12 at 10:21
The magnitude of determinant cannot reflect the ill-conditioning: $\kappa(A)=\kappa(A^{-1})$ but $\det (A^{-1})=(\det A)^{-1}$. – faleichik Feb 16 '12 at 14:39
Should there be an $\approx$ or $\ne$ somewhere? – Inquest Feb 16 '12 at 15:20
If you're interested in learning more about the effects of floating-point math on matrix spectra, you should check out Nick Trefethen's book: Spectra and Pseudospectra: The Behavior of Nonnormal Matrices and Operators and the Pseudospectra Gateway. – Aron Ahmadia Feb 16 '12 at 21:35

It's the largeness of the condition number $\kappa(\mathbf A)$ that measures the nearness to singularity, not the tininess of the determinant.

For instance, the diagonal matrix $10^{-50} \mathbf I$ has tiny determinant, but is well-conditioned.

On the flip side, consider the following family of square upper triangular matrices, due to Alexander Ostrowski (and also studied by Jim Wilkinson):

$$\mathbf U=\begin{pmatrix}1&2&\cdots&2\\&1&\ddots&\vdots\\&&\ddots&2\\&&&1\end{pmatrix}$$

The determinant of the $n\times n$ matrix $\mathbf U$ is always $1$, but the ratio of the largest to the smallest singular value (i.e. the 2-norm condition number $\kappa_2(\mathbf U)=\dfrac{\sigma_1}{\sigma_n}$) was shown by Ostrowski to be equal to $\cot^2\dfrac{\pi}{4n}$, which can be seen to increase for increasing $n$.

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@Nunoxic: most certainly not; before I launch into details, are you already familiar with the singular value decomposition? – J. M. Feb 16 '12 at 13:40
Very good. That's all you need to know. The idea is that very important information on the conditioning is concentrated in $\mathbf \Sigma$. In particular, you will want to look for the largest and smallest values (remember that the decomposition is defined such that the diagonal entries of $\mathbf \Sigma$ are nonnegative) in that matrix's diagonal. The ratio of the largest to the smallest diagonal entry is the condition number $\kappa$. What size of condition number you should be concerned with depends on the machine your working on... – J. M. Feb 16 '12 at 13:59
...but in general, when solving linear equations with that matrix, you stand to lose $\approx \log_b \kappa$ base-$b$ digits in your solution. That's a rough rule of thumb for the condition number; so if you're working with only 16 digits, a $\kappa$ of $10^13$ should be cause for concern. – J. M. Feb 16 '12 at 14:02
Yes, but that is not the recommended method for determining the condition number (an explanation of which is for another question). I presume you know how to invert a diagonal matrix, no? – J. M. Feb 16 '12 at 14:08
"Regd. the loss of digits, could you give me a reference for this?" - I could, but this really is one of those things that you should be experimenting on your own in a computing environment for reinforcement. – J. M. Feb 16 '12 at 14:10

As $\det(kA)=k^n\det A$, the determinant can be made arbitrarily large or small by simple rescaling (which doesn't change the condition number). Especially in high dimensions, even scaling by an innocent factor of 2 changes the determinant by a huge amount.

Thus never use the determinant to assess condition or closeness to singularity.

On the other hand, for almost all well-posed numerical problems, the condition is closely related to the distance to singularity, in the sense of the smallest relative perturbation needed to make the problem ill-posed. In particular, this holds for linear systems.

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