Closed or open? Three classical subspaces

Idea. Show each set is a *proper* linear subspace; then the previous problem says it is closed with empty interior, in particular not open.

Step 1: they are subspaces.
- $\mathrm{Sym}_n=\{A:A^T=A\}$ is the kernel of the linear map $A\mapsto A^T-A$.
- $\{\operatorname{tr}A=0\}$ is the kernel of the linear form $\operatorname{tr}$.
- Upper triangular matrices are those with $a_{ij}=0$ for $i>j$: the common kernel of the coordinate functionals $A\mapsto a_{ij}$ with $i>j$.

Step 2: they are proper (here $n\ge2$ is used).
- $E_{12}=\begin{pmatrix}0&1\\0&0\end{pmatrix}\oplus0$ is not symmetric (its transpose is $E_{21}\ne E_{12}$).
- $I_n$ has trace $n\neq0$.
- $E_{21}$ is not upper triangular.

Step 3: conclusion. By the previous problem each of the three sets is closed, has empty interior, and is therefore not open (a non-empty open set has non-empty interior; each set contains $0$, so it is non-empty).

Remark.
- For $n=1$ the situation is different: $M_1(\mathbb{R})=\mathbb{R}$, symmetric and upper triangular matrices are *everything* (hence open and closed), while $\{\operatorname{tr}=0\}=\{0\}$ is closed and not open.
- Dimensions: $\dim\mathrm{Sym}_n=\frac{n(n+1)}2$, $\dim\{\operatorname{tr}=0\}=n^2-1$, $\dim\{\text{upper triangular}\}=\frac{n(n+1)}2$, all $<n^2$.