Does the magnitude of a nonzero determinant give any information about how close to singular the matrix is?

Question

As title says, I am wondering i the magnitude of a nonzero determinant give me information about the singularity of the matrix.

Can someone prove (or help me to understand how i could) prove this to be true or false?

Answer 1

No, it does not. The determinant of a matrix is not invariant under scaling since for an $n$ by $n$ matrix,

$\det (\alpha A)=\alpha^{n} \det(A)$

Thus for example, if $n=1000$, and $\alpha=0.1$,

$\det(I)=1$

$\det(\alpha I)=1.0 \times 10^{-1000}$

A more appropriate measure of the singularity of a matrix is the condition number

$\kappa(A)=\| A \| \| A^{-1} \|$

Different norms can be used here, but when the euclidean norm is used, it can be shown that

$\kappa(A)=\sigma_{1}/\sigma_{n}$

where $\sigma_{1}$ and $\sigma_{n}$ are the largest and smallest singular values of $A$.