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Show that set of all invertible $n\times n$ matrices with real entries (denoted by $GL_n(R)$) is open set in $M_n(R)$.

My attempt: by open set I think it means neighborhood of every point in set is contained in set, but this definition doesn't seem to help solve this question. Any other approach?

Martin Sleziak
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singularity
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1 Answers1

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If you can show that its complement is a closed set the also you prove that $$GL_{n}(\mathbb R)$$ is open.

Now see that the complement of $GL_{n}(\mathbb R)$ is the set of all matrices that have determinant $0$. And $$det : M_{n}(\mathbb R)\rightarrow \mathbb R$$ is a continuous function.

For continuous functions, pre-image of a closed set is closed and $\{0\}$ is a closed set in $\mathbb R$.

So the pre-image of this singleton closed set under the determinant map, i.e., the set of all non-invertible matrices, i.e., the complement of the set $GL_{n}(\mathbb R)$ is a closed set.

Hence $GL_{n}(\mathbb R)$ is open in $M_{n}(\mathbb R)$.

Martin Sleziak
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user118494
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    Why did you use the complement? The preimage of an open set by a continuous function is open as well, and $\mathbb{R} \setminus {0}$ is open in $\mathbb{R}$. Exactly the same proof without needing to consider the complement. – Najib Idrissi Sep 05 '15 at 08:22