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This question is inspired by an old question on here.

Prove:

$ {\mathbf{GL}_{n}}(\mathbb{R}) $ is dense in $ {\mathbf{M}_{n}}(\mathbb{R}) $

Proof:

$ {\mathbf{GL}_{n}}(\mathbb{R}) = \{A \in {\mathbf{M}_{n}}(\mathbb{R}) : \det A \neq 0\}$ where $ {\mathbf{M}_{n}}(\mathbb{R}) $ consists of all $n \times n$ matrices with real entries.

We want to prove that $ {\mathbf{GL}_{n}}(\mathbb{R}) $ is dense in $ {\mathbf{M}_{n}}(\mathbb{R}) $.

$ {\mathbf{GL}_{n}}(\mathbb{R}) $ is dense in $ {\mathbf{M}_{n}}(\mathbb{R}) $ if every matrix $A \in {\mathbf{M}_{n}}(\mathbb{R}) $ either belongs to $ {\mathbf{GL}_{n}}(\mathbb{R}) $ or is a limit point of $ {\mathbf{GL}_{n}}(\mathbb{R}) $.

A matrix $P$ is a limit point of $ {\mathbf{GL}_{n}}(\mathbb{R}) $ in $ {\mathbf{M}_{n}}(\mathbb{R}) $ if and only if every open set $U$ containing $P$ also contains a point of $ {\mathbf{GL}_{n}}(\mathbb{R}) $ different from $P$.


And now when we should start to think I am not sure what to do as I have never worked with $ {\mathbf{GL}_{n}}(\mathbb{R}) $ or $ {\mathbf{M}_{n}}(\mathbb{R}) $ before.

I would appreciate a hint or a few on how to continue.

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    $\begingroup$ If you think of the characteristic polynomial of $A$, can you find matrices close to $A$ whose determinant is nonzero? $\endgroup$ Mar 11, 2016 at 16:36
  • $\begingroup$ Under what topology? Spectral norms? $\endgroup$
    – Vim
    Mar 11, 2016 at 17:12
  • $\begingroup$ @Vim Yes, that is of course important. It is the usual topology of euclidean $n^2$-space, i.e. $\mathbb{R}^{n^2}$ $\endgroup$
    – JKnecht
    Mar 11, 2016 at 19:20

6 Answers 6

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Let $A\in M_n$. Choose $\epsilon >0$ such that $\epsilon$ is less than the smallest positive eigenvalue of $A$ (if any exists, otherwise any positive $\epsilon$ will do). Observe that $A-\epsilon I$ is invertible (otherwise, $\epsilon$ is an eigenvalue for $A$), therefore $A_\epsilon \in GL_n$.

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  • $\begingroup$ I am a little out of my depth here when it comes to linear algebra because i have not studied linear algebra in a long time. But these things you mention I might be able to work with. Thanks! $\endgroup$
    – JKnecht
    Mar 11, 2016 at 16:58
  • $\begingroup$ Sorry but I have a basic question. If $A$ is not invertible so there is no such $\epsilon$? If I understood you well it not just simply necessary to chose any $\epsilon >0$ that is different of any eigen value $\endgroup$
    – X0-user-0X
    Dec 15, 2023 at 18:46
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I particularly like this argument. Take any matrix $A$ and any invertible matrix $B$. Consider $$p(t) = \det((1-t)A+tB)$$ which is clearly a polynomial in one variable, and it is not zero polynomial because $p(1)\neq 0$. Thus, it has finitely many zeroes. That means that you can find arbitrarily small $t$ such that $C_t = (1-t)A+tB$ is invertible and we have $$\|A-C_t\| = |t|\|A-B\|<\varepsilon,\quad |t|<\frac{\varepsilon}{\|A-B\|}$$ Geometrically, we consider a line segment through matrices $A$ and $B$ and can find invertible matrix on it arbitrarily close to $A$.

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  • $\begingroup$ Yes, this argument i really like too! $\endgroup$
    – JKnecht
    Mar 11, 2016 at 17:50
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The function $\det g$ is a polynomial in the $g_{ij}$. In particular, it's equal to its Taylor series everywhere. It follows that the closed set $X = \{g\in M_n(\mathbb{R}):\, \det g = 0\}$ has empty interior, and thus its complement $GL_n(\mathbb{R})$ is dense.

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  • $\begingroup$ That's nice and quick. $\endgroup$
    – zhw.
    Mar 11, 2016 at 18:03
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    $\begingroup$ +1 Why does it follow that X has empty interior? $\endgroup$
    – JKnecht
    Mar 11, 2016 at 19:28
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    $\begingroup$ @JKnecht: If it contained an open ball, then the Taylor series of $\det g$ around the center of that ball would vanish. $\endgroup$
    – anomaly
    Mar 11, 2016 at 20:46
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The set $X$ of matrices with zero determinant is a submanifold of codimension $1$ (with some singularities). Then by Sard's theorem it has measure zero. It follows that its complement is dense.

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    $\begingroup$ +1 Sard's theorem was new for me, and reading about it I also learned about null sets, Lebesgue measure etc. $\endgroup$
    – JKnecht
    Mar 11, 2016 at 20:08
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    $\begingroup$ Could you provide more detail about how Sard's theorem applies here? I'm a little unclear what mapping you're applying it to, since it doesn't appear that $\det: \mathbf{M}_n(\mathbb R) \to \mathbb R$ is what you want. Thanks! $\endgroup$
    – Erick Wong
    Mar 28, 2016 at 21:58
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    $\begingroup$ Assume $N$ is an embedded submanifold of $M$ of codimension $>0$. Work in a chart $U\subseteq\mathbb{R}^{\dim M}$ of $M$ and apply Sard to the inclusion of (a subset of) $N$ in $\mathbb{R}^{\dim M}$. $\endgroup$ Mar 29, 2016 at 0:37
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Let $A \in M_n(K)$, with $K=\mathbb{R}$ or $\mathbb{C}$, then $\chi_A(\lambda)=\det\left(A-\lambda I_n \right)$ is a nonzero polynomial, so it has a finite number of roots. So it exits a $p \in \mathbb{N}$ such that $\forall k>p \text{, }$ $\chi_A(1/k)\neq 0$ and then the sequence $$\left(A-\frac{1}{k} \cdot I_n\right)_{k>p}\longrightarrow A$$

is made of invertible matrices.

QED.

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  • $\begingroup$ Great answer thank you! $\endgroup$
    – X0-user-0X
    Dec 15, 2023 at 18:50
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(1) If $A=[a_1\cdots a_n]$ with ${\rm rank}\ A=k,\ k<n $, then we can assume that $\{ a_1,\cdots,a_k\}$ is linearly independent. If $ \{a_1,\cdots,a_k, b_{k+1},\cdots, b_n\}$ is linearly independent and if $B:= [ 0\cdots 0\ b_{k+1}\ \cdots\ b_n]$, then ${\rm det}\ (A+ \epsilon B) = {\rm det} \ [a_1\cdots a_k \ \epsilon b_{k+1}\ \cdots\ \epsilon b_n]\neq 0 $

(2) $M_n({\bf R})$ has a inner product, i.e., $|X|^2={\rm trace}\ XX^T$ and we have a metric $d(X,Y)=|X-Y|$. So since $|A-(A+ \epsilon B)|=\epsilon |B|$ Hence $A$ has an invertible matrix within arbitrary small distance. That is $Gl_n({\bf R})$ is dense.

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  • $\begingroup$ +1 The last step there, how does it follow that it is dense? $\endgroup$
    – JKnecht
    Mar 11, 2016 at 19:33
  • $\begingroup$ I add more explanation $\endgroup$
    – HK Lee
    Mar 11, 2016 at 19:43

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