Why is the determinant $M_n(\mathbb R) \to \mathbb R$ continuous? Why is the determinant as a function from $M_n(\mathbb{R})$ to $\mathbb{R}$ continuous, please can anyone explain precisely and rigorously? So far I know the explanation which comes from the facts: polynomials are continuous, sum and product of continuous functions are continuous. Also I have the confusion regarding the metric on $M_n(\mathbb{R})$
 A: $M_n(\mathbb R)$ is just $\mathbb R^{n^2}$ with the euclidian metric.
det is countinous, because it is a polynomial in the coordinates
$$
\text{det}(x_{i,j})= \sum_\sigma \text{sgn}(\sigma) \prod_{i=1}^{n} x_{\sigma(i),i}
$$
A: It's continuous because it's computable as a function from $\mathbb{R}^{n^2}$ to $\mathbb{R}$.
A: Recall that the determinant can be computed by a sum of determinants of minors, that is "sub"-matrices of smaller dimension.
Now we can prove by induction that $\det$ is continuous:


*

*For $n=1$, $A\in M_1(\mathbb R)$ is simply a scalar we have that $\det A=A$, and surely the identity function is continuous.

*Suppose that for $n$ we have that $\det$ is continuous on $M_n(\mathbb R)$, let $A\in M_{n+1}(\mathbb R)$. We know that $\det A$ can be calculated as the alternating sum over one of first row, when calculating the $\det$ of the appropriate minor. 
So $\det A$ is written as a sum and scalar multiplication of $\det$ on a smaller dimension. From the induction hypothesis these are continuous and therefore $\det$ is continuous on $n+1\times n+1$ matrices.
A: The coefficient maps $A\longmapsto a_{i,j}$ are continuous because they are linear on the finite-dimensional vector space $M_n(\mathbb{R})$. Here you want to refer to the topology of the latter as a normed space, which does not depend on the norm since they are all equivalent in finite dimension. Then the determinant is a polynomial in the coefficients, so it is continuous by composition of continuous maps.
