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Is the set $\mathcal{A}$ of all matrices whose trace is $0$ nowhere dense in $\mathbb{M}_n(\mathbb{R}),n \ge 2$ ?

My attempt : my answer is False

I take $A = \begin{bmatrix} 1& n \\0&-1 \end{bmatrix}$

I know that set of all invertible matrix is dense. You know that my given matrix $A$ in invertible, so I think statement must be false .

Am I right or wrong?

Thank you.

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jasmine
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    But your matrix doesn't represent the set of all invertible matrices. So that doesn't help. Integers are rationals and rationals are dense in the real numbers. That doesn't mean the integers are dense in the reals. – Arthur Sep 23 '18 at 07:01
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    @jasmine: There are lot of comments under my answer and the user Arthur saying my answer is incorrect. Actually Arthur was misunderstanding my answer. Also many other users are misunderstanding my answer. Thats why they downvote my answer. I also ask my answer is whether right or wrong in this post and peoples are responding 'my answer is correct'. So don't confuse yourself. Kindly read again my answer. It was correct! – Chinnapparaj R Sep 27 '18 at 03:00
  • @ChinnapparajR,,thanks u – jasmine Sep 27 '18 at 09:36

3 Answers3

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Your set is closed (as it is the inverse image of the closed set $\{0\}\subseteq \Bbb R$ under the continuous trace function), and it has empty interior (as any $\epsilon$-ball around a matrix with trace $0$ will contain a matrix with non-zero trace). So it is nowhere dense.

Arthur
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Result: If $A$ is a subset of a metric space $(X,d)$, then $A$ is nowhere dense in $X$ if and only if $\overline{A}^c$ is dense in $X$

[For a proof, see this]

Note that $\mathcal{A}$ is closed

So, In order to prove $\mathcal{A}$ is nowhere dense in $M_n(\Bbb{R})$, we prove $\overline{\mathcal{A}}^c=\mathcal{A}^c=M_n(\Bbb{R}) \setminus \mathcal{A}$ is dense in $M_n(\Bbb{R})$

To prove $M_n(\Bbb{R}) \setminus \mathcal{A}$ is dense in $M_n(\Bbb{R})$, we prove every point of $M_n(\Bbb{R})$ is either a point of $M_n(\Bbb{R}) \setminus \mathcal{A}$ or a limit point of $M_n(\Bbb{R}) \setminus \mathcal{A}$.

Suppose $B \in M_n(\Bbb{R}) \setminus \mathcal{A}$ ,then we are done! So asume $B \notin M_n(\Bbb{R}) \setminus \mathcal{A}$. That is $B \in \mathcal{A}$. In this case we prove $B$ is a limit point of $M_n(\Bbb{R}) \setminus \mathcal{A}$

take $B=\begin{pmatrix}a_{11} & a_{12} &\dots&a_{1n} \\ a_{21} & a_{22} \ &\dots&a_{2n} \\ \vdots \\a_{n1} & a_{n2} &\dots&a_{nn}\end{pmatrix}$ with $a_{11}+\dots+a_{nn}=0$

Then consider the sequence of elements of $M_n(\Bbb{R})\setminus \mathcal{A}$ $$A_k=\begin{pmatrix}a_{11}+1/k & a_{12} &\dots&a_{1n} \\ a_{21} & a_{22}+1/k \ &\dots&a_{2n} \\ \vdots \\a_{n1} & a_{n2} &\dots&a_{nn}+1/k\end{pmatrix}$$ Then $A_k \rightarrow B$ and so $B$ is a limit point!

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Chinnapparaj R
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  • beautiful answer......simple and easy to understand , thanks u @Chinnapparaj – jasmine Sep 23 '18 at 07:12
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    You are welcome! – Chinnapparaj R Sep 23 '18 at 07:13
  • The variable $n$ is fixed by the question. You should really add $1/k$ to the diagonal. – Theo Bendit Sep 23 '18 at 07:14
  • This just means it's a perfect set. That's not the same as "not nowhere dense". @stupid This doesn't answer your question at all. – Arthur Sep 23 '18 at 07:17
  • @Arthur Chinnapparaj is showing that the complement is dense. That is, every trace $0$ matrix is the limit of matrices with non-zero trace. This is not to say the set of $0$-trace matrices is perfect, but it isn't enough to show nowhere-density either. For example, $\mathbb{Q} \subseteq \mathbb{R}$. – Theo Bendit Sep 23 '18 at 07:20
  • Yes my proof base on this result: $A$ isnowhere dense in $X$ iff $\overline{A}^c$ is dense in $X$ – Chinnapparaj R Sep 23 '18 at 07:23
  • @TheoBandit You're right. I was too quick and thought he added $1/k$ to one entry and subtracted it from another, this making sure the entries in the sequence has trace $0$. – Arthur Sep 23 '18 at 07:24
  • Why downvote?What I'm proving wrong? – Chinnapparaj R Sep 23 '18 at 07:24
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    @CinnapparajR Your approach would also prove that $\Bbb Q$ is nowhere dense in $\Bbb R$ (take any rational number and add $\sqrt2/k$ to it). It therefore cannot be right. – Arthur Sep 23 '18 at 07:26
  • No! What I proved is $\Bbb{Q}$ is dense in R iff $\overline{\Bbb{Q}}^c=\phi$ is nowhere dense in R – Chinnapparaj R Sep 23 '18 at 07:29
  • @TheoBendit: I prove the closure of the complement is dense not the complement . In this case the set $A$ is closed and so closure of $A$ is $A$ itself – Chinnapparaj R Sep 23 '18 at 07:33
  • @Arthur: Since $\mathcal{A}$ is cloesd in $M_n(R)$, so it make sense but $\Bbb{Q}$ is not closed in $\Bbb{R}$, – Chinnapparaj R Sep 23 '18 at 07:52
  • @Arthur: I have used this result – Chinnapparaj R Sep 23 '18 at 07:56
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    @ChinnapparajR Your proof is perfectly alright. There is nothing wrong with it, Arthur is incorrect. –  Sep 26 '18 at 16:52
  • @Brahadeesh This answer looked quite different initially. I am pretty certain that I was correct then (and two people seem to have agreed). – Arthur Sep 27 '18 at 05:13
  • @Arthur oh is it? I only looked at the latest version and replied. If there was a mistake earlier that you had pointed out, then I apologise for misconstruing your comments. –  Sep 27 '18 at 05:33
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    However, it is true that the answer is correct now, as it stands. There are no errors in the logic; in particular, the proof does not show that $\mathbb{Q}$ is nowhere dense in $\mathbb{R}$. –  Sep 27 '18 at 05:43
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Your set is a finite dimensional (and hence closed) proper subspace of $\mathbb{M}_n(\mathbb{R})$ so it is nowhere dense because every proper subspace of a normed space has empty interior.

mechanodroid
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