Check $S = \{ A \in \mathbb M_n(\mathbb R) : \mathrm{tr}(A) = 0 \}$ is nowhere dense or Dense in $\mathbb M_n(\mathbb R)$

Question

I know that $\mathbb M_n(\mathbb R) $ is homeomorphic to $\mathbb R^{n^2}$ and also I know that all the norms on $\mathbb R^{n^2}$ are equivalent. I am using this norm $$\|A\| = \max\{|a_{ij}| : 1 \leq i , j \leq n \}. $$

Any matrix $B \in \mathbb M_n(\mathbb R)$ any open ball $S_r (B)$ centered at $B$ whose radius is $r > 0$. Let $b_{ij} = \|B\|$, then define the matrix A such that $a_{11} = b_{ij}$ and $a_{nn} = -b_{ij}$ and other element are zero. Thus $A \in S_r(B)$

Thus $S$ is dense in $\mathbb M_n(\mathbb R)$

But answer is $S$ is nowhere dense.

Think of the similar case ${(x,y,z) \in \mathbb{R}^3 \mid x+y+z = 0}$. This is a plane. Does is make sense for this to be dense in $\mathbb{R}^3$? I also don’t understand how $B$ is supposed to be defined. — Jendrik Stelzner, Jan 05 '16 at 17:23
@ Jendrik : $B$ is any arbitrary matrix , i have dined a matrix $B$. — Struggler, Jan 05 '16 at 23:56
Now, i got my mistake and lot of thanks for the prompt reply. — Struggler, Jan 06 '16 at 00:05

score 4 · Accepted Answer · edited Dec 12 '19 at 18:28

4

Let $A$ such that $\operatorname*{tr}{(A)}=0$ for every $c>0$, consider $A_c=(b_{ij}): b_{11}=a_{11}+c/2$ and $b_{ij}=a_{ij}$ otherwise, $\operatorname*{tr}{(B)}\neq 0$ and $\lVert A-B\rVert<c$

edited Dec 12 '19 at 18:28

davyjones

641

answered Jan 05 '16 at 17:21

Tsemo Aristide

87,475

score 2 · Answer 2 · edited Apr 13 '17 at 12:20

2

Note that $S$ is a proper, linear subspace of the vector space $M_n(\mathbb{R})$ and as such must be have empty interior. Since the trace is continuous (as every linear functional on a finite-dimensional vector space), $S$ must be closed being the preimage of $\{0\}$.

edited Apr 13 '17 at 12:20

Community

1

answered Jan 05 '16 at 17:58

Tomasz Kania

16,361

Check $S = \{ A \in \mathbb M_n(\mathbb R) : \mathrm{tr}(A) = 0 \}$ is nowhere dense or Dense in $\mathbb M_n(\mathbb R)$

2 Answers2