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) %0.008cond(a)%125a(:,4) = 1*a(:,1) + 2*a(:,2) = 0.000000001*ones(4,1);det(a)%1.8E-11cond(a)%3.46E10
Does a tiny determinant imply ill-conditioning of a matrix?
linear algebra;condition number
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 $n$.