Proving that $GL_n$ ($ \mathbb{R}$) $: = \{A \in \mathbb{R}^{n \times n}: \det A \neq 0\}$ is open in $\mathbb{R}^{n \times n}$

Question

How can one prove that the set of invertible matrices $GL_n := \mathbb{R} : = \{A \in \mathbb{R}^{n \times n}: \det A \neq 0\}$ is open in $\mathbb{R}^{n \times n}$?

${\mathbb{R}^{n \times n}}$ has the following norm:

The domain of $A \to \det (A)$ is $\mathbb{R}^{n \times n}$ and $A \to \det A \in \mathbb{R} $ is the transformation rule, so $A$ can be a random matrix in in $\mathbb{R}^{n \times n}.$

$GL_n(\mathbb R)$ is the inverse image $\det^{-1} (\mathbb{R}$ \ {$0$}).

Since $\det$ is continuous, the inverse image is also open in $\mathbb{R}^{n \times n}$. Is that correct?

And how can I show that $\mathbb{R}^{n \times n} \ni A \to \det A \in \mathbb{R}$ is continuous? I don't know how I should use the Laplace expansion here...

Answer 1

The determinant function $$\det: \mathbb{R^{n^2} \to \mathbb{R}}\\A \mapsto \det A$$ is continuous because it is a polynomial one.

The set $GL_n(\mathbb{R})=\{A \in M_n(\mathbb{R}) | \det A \ne 0\}$ is the preimage of the open set $\mathbb{R} \setminus \{0\}$, so, for the continuity of the determinant function, it is open in $\mathbb{R^{n^2}}$.