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

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...

• Yes. that's correct! For other part, see this post – Chinnapparaj R Oct 27 '18 at 13:57
• There are several problems with your question: (1) I don't know what the first $\Bbb R$ is doing in your title or question. (2) "The domain of $A$..." doesn't make any sense, as $A$ is not a function. Perhaps you meant "the domain of det". (3) the next statement about $GL_n$ being the inverse image also makes sense only when $A$ is replaced by "det". And "since $A$ is continuous" should be "since det is continuous", but the conclusion that the inverse image is open requires that you note that $\Bbb R \setminus \{0\}$ is open. As for the last line, do you know a formula for det? – John Hughes Oct 27 '18 at 14:02
• Sorry, I really made some mistakes. I've corrected (1), (2) and (3). And yes I know the leibniz formula – user605984 Oct 27 '18 at 14:21
• If you know the formula for the determinant as an alternating sum over permutations of the matrix entries, then you know it just looks like a (large) polynomial in the matrix entries, and polynomials are continuous. – Joppy Oct 27 '18 at 14:37

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 $$\mathbb{R^{n^2}}$$.