$\mathcal{I} = \{X \in M_{n \times n} \mid X \text{ is invertible }\}$ is not-connected.

Question

Let $\mathcal{I}$ be the set of all invertible real matrix $n \times n$. I have to prove that $\mathcal{I}$ is open and not-connected.

My attempt: Take any $A \in \mathcal{I}$. Puting $\varepsilon = \frac{1}{\|A^{-1}\|}$ we have that $\|A-B\| < \varepsilon \Rightarrow B \in \mathcal{I}$. Therefore $\mathcal{I}$ is open.

But I really don't know how I can show that $\mathcal{I}$ is not-connected.

Real matrices? Consider the determinant function. (It's continuous and never $0$ in $\cal I$...) — anon, Aug 11 '16 at 23:51
Determinant also solves the openness question: set of open matrices is exactly the preimage of an open set $\mathbb{R} - {0}$, thus is open. — xyzzyz, Aug 11 '16 at 23:52

score 1 · Accepted Answer · answered Aug 11 '16 at 23:58

1

The image of a connected set under a continuous map is connected. Consider the image of $\mathcal{I}$ under the determinant map, as suggested in the comments.

answered Aug 11 '16 at 23:58

user66081

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score 1 · Answer 2 · answered Aug 11 '16 at 23:59

For example (with real matrices)

$$\mathcal I=\left\{X\in M_{n\times n}\;/\;\det X<0\right\}\cup\left\{X\in M_{n\times n}\;/\;\det X>0\right\}$$

and for example

$$\left\{X\in M_{n\times n}\;/\;\det X<0\right\}=\Delta^{-1}(-\infty,\,0)\;,\;\;\Delta:=\text{ the determinant function}$$

and $\;\Delta\;$ is clearly continuous.

$\mathcal{I} = \{X \in M_{n \times n} \mid X \text{ is invertible }\}$ is not-connected.

2 Answers2