# Is the determinant of a matrix Lipschitz continuous?

I want to know if the determinant of a matrix is Lipschitz continuous or not.

To be precise, does there exist a constant $K$ such that

$|\det(A)-\det(B)|\leq K||A-B||_F$,

for all matrices $A,B\in \mathcal{C}^{n\times n}$?

Does $|\det(A)-\det(B)|\leq K||A-B||_F^\alpha$ hold for some constant $K$ and $\alpha$?

Can anyone help me on this? Thank you in advance!

-
How does $\det A$ change when you multiply $A$ with a constant $c\neq 0$? – Daniel Fischer Feb 28 '14 at 13:37
Then it becomes $c^N det(A)$. So do you mean that when $n$ is large The value on LHS grows much faster than that on the RHS? – Victor Cheng Feb 28 '14 at 14:09
As soon as the dimension is $> 1$. – Daniel Fischer Feb 28 '14 at 14:10
Thank you, so the answer is probably no. Then what about Hölder continuous? I have now edited the question. – Victor Cheng Feb 28 '14 at 14:14
Why is this question tagged complex-analysis? – mrf Feb 28 '14 at 15:51

In the case $n = 1$, the determinant is the identity, and hence globally Lipschitz continuous.

For $n > 1$, the determinant is not globally $\alpha$-Hölder continuous for any $\alpha \in (0,1]$, since

$$\lvert \det (r\cdot I) - \det (0\cdot I)\rvert = \lvert r^n\rvert = \lVert I\rVert_F^{-1}\cdot\lvert r\rvert^{n-\alpha}\cdot \lVert r\cdot I - 0\rVert_F^\alpha,$$

and $\lvert r\rvert^{n-\alpha}$ is unbounded.

The determinant is however a polynomial in the entries of the matrix, and hence continuously differentiable everywhere, and that implies that it is locally $\alpha$-Hölder continuous for all $\alpha\in (0,1]$, in particular locally Lipschitz continuous.

-

A connection between determinant and Frobenius norm results from the fact that $|\det A|$ is the volume of the spat $A[0,1]^n$ and that this is smaller than the volume of a rectangular box with the same side lengths.

If $A$ has columns $A=(a_1,a_2,..., a_n)$ then $$|\det(A)| \le \|a_1\|\,\|a_2\|\,....\,\|a_n\| \le \left(\frac{\|a_1\|^2+\|a_2\|^2+....+\|a_n\|^2}{n}\right)^{\frac{n}{2}} = \left(\frac{\|A\|_F^2}{n}\right)^{\frac{n}{2}} =\frac{\|A\|_F^n}{n^{n/2}}$$

-