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

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.

• 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. Sep 23 '18 at 7:01
• @ChinnapparajR,,thanks u Sep 27 '18 at 9:36

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.

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!

• beautiful answer......simple and easy to understand , thanks u @Chinnapparaj Sep 23 '18 at 7:12
• You are welcome! Sep 23 '18 at 7:13
• @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. Sep 23 '18 at 7:26
• @ChinnapparajR Your proof is perfectly alright. There is nothing wrong with it, Arthur is incorrect.
– user279515
Sep 26 '18 at 16:52
• 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 $\mathbb{R}$.
– user279515
Sep 27 '18 at 5:43

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.