TITLE: Proof that determinant is continuous using $\epsilon-\delta $ definition
QUESTION [9 upvotes]: I need to prove that the determinant $\det: M(n, \mathbb{R}) \rightarrow \mathbb{R}$ is a continuous function given the euclidean metric on the vector space of all $n \times n$ matrices over $\mathbb{R}$, i.e. $\Vert M \Vert = \sqrt{\sum_{i,j=1}^n M_{i,j}^2}$.
So what I need to prove, I think, is that there exists a $\delta > 0$ such that if $\Vert M - M' \Vert < \delta$, for any $M' \in M(n,\mathbb{R})$ then it follows that for all $\epsilon > 0: |\det(M) - \det(M')|< \epsilon$.
Unfortunately I have no idea how to derive the correct inequalities between the given euclidean metric and a determinant, since I'm kind of struggling with the permutation part in the definition of a determinant. Can anybody help, please?

REPLY [5 votes]: For a more geometric approach, you can use the following inequality: 
\begin{equation}\tag{1}
|\det M|\le \prod_{j=1}^n \lVert M_j\rVert.
\end{equation}
Here $\lVert M_j\rVert$ denotes the Euclidean norm of the $j$-th column of $M$. This inequality is best understood if the determinant of $M$ is interpreted as the signed volume of the parallelogram spanned by the vectors $M_1\ldots M_n$. It expresses the intuitive fact that a skew parallelogram has smaller volume than the right parallelogram having the same sides. (Look here for more information and a proof). 
Observing that $\det(M)$ is a linear map in each column $M_j$, one obtains the following formula, in which the right hand side is a telescoping sum:
$$
\det M-\det M'=\sum_{j=1}^n\det\begin{bmatrix} M'_1&\ldots& M'_{j-1}& M_j-M'_j& M_{j+1}&\ldots& M_n\end{bmatrix}.$$
Taking absolute values and using inequality $(1)$ one has 
$$\tag{2}
\lvert \det M-\det M'\rvert \le \sum_{j=1}^n\lVert M'_1\rVert\ldots \lVert M_j-M'_j\rVert\ldots\lVert M_n\rVert.$$

The inequality $(2)$ shows that $\det$ is locally Lipschitz continuous.

Indeed, if $C>0$ is such that 
$$\lVert M_k\rVert\le C,\ \lVert M'_k\rVert\le C, \qquad \forall k=1\ldots n, 
$$
then using the elementary inequality 
$$
\sum_{j=1}^n  a_j\le \sqrt{n}\left(\sum_{j=1}^n  a_j^2\right)^{\frac{1}{2}}\qquad \forall a_1\ldots a_n\ge 0, $$
one sees from $(2)$ that 
$$
\begin{split}
\lvert \det M-\det M'\rvert&\le C^{n-1}\sum_{j=1}^n\lVert M_j-M'_j\rVert \\
&\le \sqrt{n}C^{n-1}\sqrt{\sum_{j=1}^n \lVert M_j-M'_j\rVert^2} \\ 
&=\sqrt{n}C^{n-1}\lVert M-M'\rVert_{\mathrm{matrix}}
\end{split}
$$
where $\lVert M\rVert_{\mathrm{matrix}}^2=\sum_{k=1}^n\lvert M_{i\,j}\rvert^2.$