# How to prove that $GL(n,\mathbb R)$ is not connected subset and open subset of$M_n (\mathbb{R})$

let $n\gt1$ be a fixed natural number, $S:=${$A$: $M_n (\mathbb{R})$ be all real matrix,define this meter for all $A=[a_{ij}]$ $B=[b_{ij}]$

$d(A,B)$:=max{|$a_{ij}-b_{ij}$|:i,j=1,2,2...,n} and $GL(n,\mathbb R)$ is set of all $n\times n$ non singular matrix how to prove that $GL(n,\mathbb R)$ is not connected and open subset of $S$ ?

thanks in advance

• The language and math notation could be cleaned up greatly. For example, you most probably mean $M_n (\mathbb{R})$ as opposed to $M_( R)$ and $A=[a_{ij}]$ as opposed to $[a_i j]$? – Calvin Lin Jan 20 '13 at 19:37

## 2 Answers

The set $\text{GL}(n,\mathbb{R})$ consists of all non-singular $n \times n$ matrices with real entries. It is an $n^2$-dimensional real manifold. Given a matrix $X \in \text{GL}(n,\mathbb{R})$ use $x_{ij} \in \mathbb{R}$ to denote the entry in the $i^{\text{th}}$ row and $j^{\text{th}}$ column. In other words:

$$X = \left( \begin{array}{ccc} x_{11} & \cdots & x_{1n} \\ \vdots & \ddots & \vdots \\ x_{n1} & \cdots & x_{nn}\end{array}\right)$$

It follows that the determinant $\det(X)$ is a polynomial in the $n^2$-variables:

$$\det(X) \in \mathbb{R}[x_{11},\ldots,x_{1n},\ldots,x_{n1},\ldots,x_{nn}] \, .$$

The key point here is that polynomials are continuous functions.

The function $\det : \text{Mat}(n,\mathbb{R}) \to \mathbb{R}$ is continuous. By the definition of continuity, the inverse image of the open set $\{ x \in \mathbb{R} : x<0\}$ is open in $\text{Mat}(n,\mathbb{R})$. Thus, the non-singular matrices with negative determinant form an open subset of $\text{Mat}(n,\mathbb{R}).$ Similarity: the inverse image of the open set $\{ x \in \mathbb{R} : x>0\}$ is open in $\text{Mat}(n,\mathbb{R})$. Thus, the non-singular matrices with positive determinant form an open subset of $\text{Mat}(n,\mathbb{R}).$

Recall that if $X$ is connected and $f:X \to Y$ is a continuous function then $Y$ is connected. The contrapositive tells us that if $Y$ is not connected and $f : X \to Y$ is a continuous function the $X$ is not connected. To show that $X = \text{GL}^+(n,\mathbb{R}) \cup \text{GL}^-(n,\mathbb{R})$ is not connected, we need to show that $\mathbb{R}^+ \cup \mathbb{R}^-$ is not connected. This can be done because $\det : \text{Mat}(n,\mathbb{R}) \to \mathbb{R}$ is continuous.

To do this, we need to show that:

$$\begin{array}{ccc} \overline{\mathbb{R}^+} \cap \mathbb{R}^- &=& \emptyset \\ \mathbb{R}^+ \cap \overline{\mathbb{R}^-} &=& \emptyset \end{array}$$

This is clearly true since: $$\begin{array}{ccccc} \overline{\mathbb{R}^+} \cap \mathbb{R}^- &=& (-\infty,0) \cap [0,\infty) &=& \emptyset \\ \mathbb{R}^+ \cap \overline{\mathbb{R}^-} &=& (-\infty,0] \cap (0,\infty) &=& \emptyset \end{array}$$

• Note that the statement "if $X$ is connected and $f:X \to Y$ is a continuous function then $Y$ is connected" is false. (For example, $f$ could be a constant function.) What is true is "if $X$ is connected and $f:X \to Y$ is a continuous surjective function then $Y$ is connected." – Jack Lee Sep 27 '15 at 16:00
• @JackLee You're right, I should have written $f: X \to f(X)$. Fortunately, this detail does not change the validity of my answer. – Fly by Night Sep 29 '15 at 17:40
• @FlybyNight, Can you clarify the notation $\text{GL}^{+}$ and $\text{GL}^{-}$. – user225477 Oct 16 '17 at 18:49
• @Zermelo's_Choice $\mathrm{GL}^+$ is all matrices with positive determinant and $\mathrm{GL}^-$ is all matrices with negative determinant. Together they give all matrices with non-zero determinant: $\mathrm{GL}$. – Fly by Night Oct 17 '17 at 15:58

Observe that $\det$ is continuous and is either positive or negative, but never zero.