How to show that subset $GL_{n}(R)$ of $M_{n}(R)$ consisting of all invertible matrices is dense in $M_{n}(R)$.
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3Hint: add $-\lambda I$ to any arbitrary matrix $M$. – Gribouillis Sep 23 '17 at 13:25
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2@JoséCarlosSantos: That question is about density in the Zariski topology, not in the usual Euclidean topology. – Nate Eldredge Sep 23 '17 at 13:32
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@NateEldredge When a set is dense with respect the Zariski topology, then it is also dense with respect to the usual topology. And, in this case, the proof is the same for both topologies. – José Carlos Santos Sep 23 '17 at 13:34
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3@JoséCarlosSantos It is true that the proof is the same, but it is not true that all Zariski dense sets are euclidean dense. – Sep 23 '17 at 13:36
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@G.Sassatelli You're right. Sorry about that. – José Carlos Santos Sep 23 '17 at 13:37
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@JoséCarlosSantos Can we show the Euclidean denseness just by noting that the determinant is continuous? – ViktorStein Aug 09 '19 at 14:08
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Identifying $M_n(\mathbb{R})$ with $\mathbb{R}^{n^2}$, the space $GL_n(\mathbb{R}) = \{g\in \mathbb{R}^{n^2}:\, P(g)\not = 0\}$ for some polynomial $P(g) = \det g\in \mathbb{R}[g_{11}, \dots, g_{nn}]$. If $P$ vanishes on some neighborhood of $g\in \mathbb{R}^{n^2}$, then expanding the Taylor series of $P$ around $g$ forces $P = 0$ everywhere, which is clearly false.
anomaly
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A matrix $M\in M_n(\mathbb{R})$ is singular iff $0\in\text{Spec}(M)$, and since the identity matrix $I$ belongs to the center of $M_n(\mathbb{R})$ we have $$ \text{Spec}(A+\varepsilon I) = \varepsilon + \text{Spec}(A) $$ for any $\varepsilon\in\mathbb{R}$, granting that the non-singular (i.e. invertible) matrices are a dense subset of $M_n(\mathbb{R})$.
Jack D'Aurizio
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