Is determinant uniformly continuous?

Question

The determinant map $\det$ sending an $n\times n$ real matrix to its determiant is continuous since it's a polynomial in the coefficients. Is it also uniformly continuous?

Answer 1

For $n=1$, clearly yes as $\det (a)=a$ is linear.

For $n\geq 2$, no. Consider the sequence of diagonal matrices $D_n=\mbox{diag}(\sqrt{n},\sqrt{n},1\ldots,1)$ and use, say, the operator norm inherited from the $\ell^2$ norm. Then $\|D_{n+1}-D_n\|=\frac{1}{\sqrt{n}+\sqrt{n+1}}\longrightarrow 0$ while $\det D_{n+1}-\det D_n=1$ for every $n$.

Note: since all the norms are equivalent on $M_n$, the determinant will not be uniformly continuous for any matrix norm. I assume you are asking your question with a normed space structure on $M_n$ in mind. But with an ad hoc distance, the determinant could become uniformly continuous. Trivial example: the discrete distance...

Answer 2

The determinant function on the set of $n\times n$ matrices is a non-zero polynomial which is homogeneous of degree $n$.

Since the restriction of a uniformly continuous function to a subspace is still uniformly continuous, it is enough to show that the restriction of $\det$ to the line spanned by any matrix with non-zero determinant is not uniformly continuous.

That is easy, because that restriction is (up to more or less obvious identifications) the function $t\in\mathbb R\mapsto t^n\in\mathbb R$.

A different way to say this is as follows: let $A\in M_n(\mathbb R)$ be any matrix with determinant equal to $1$. Then the map $f:t\in\mathbb R\mapsto tA\in M_n(\mathbb R)$ is linear, so uniformly continuous. If $\det:M_n(\mathbb R)\to\mathbb R$ were uniformly continuous, then the composition $\det\circ f:\mathbb R\to\mathbb R$ would also be uniformly continuous, and it isn't unless $n=1$, for $\det f(t)=t^n$ for all $t\in\mathbb R$.