Exercise 1968Topology › Topology of Mn(ℝ)L1Added 4 Nov 2026

$GL_n(\mathbb{R})$ is open in $M_n(\mathbb{R})$

#GL_n#open set#determinant continuity

Statement

Show that $GL_n(\mathbb{R})=\{A:\det A\neq0\}$ is open in $M_n(\mathbb{R})$ .

Hint

$\det$ is continuous and $GL_n=\det^{-1}(\mathbb{R}\setminus\{0\})$ .

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Solution

Argument. $\det:M_n(\mathbb{R})\to\mathbb{R}$ is continuous, and $\mathbb{R}\setminus\{0\}$ is open. Therefore $GL_n=\det^{-1}(\mathbb{R}\setminus\{0\})$ is open.

Conclusion. $GL_n(\mathbb{R})$ is open, and its complement $SL_n$ (as well as $\{\det=0\}$ ) is closed.

$GL_n(\mathbb{R})$ is open in $M_n(\mathbb{R})$

Statement

Solution

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