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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:

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...

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    $\begingroup$ Yes. that's correct! For other part, see this post $\endgroup$ – Chinnapparaj R Oct 27 '18 at 13:57
  • $\begingroup$ 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? $\endgroup$ – John Hughes Oct 27 '18 at 14:02
  • $\begingroup$ Sorry, I really made some mistakes. I've corrected (1), (2) and (3). And yes I know the leibniz formula $\endgroup$ – user605984 Oct 27 '18 at 14:21
  • $\begingroup$ 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. $\endgroup$ – Joppy Oct 27 '18 at 14:37
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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 $\mathbb{R^{n^2}}$.

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