The subspace of traceless matrices

Linearity. The trace is linear: $\operatorname{tr}(A+B)=\operatorname{tr}(A)+\operatorname{tr}(B)$ and $\operatorname{tr}(\lambda A)=\lambda\operatorname{tr}(A)$. So $F$ is the kernel of a linear form, hence a subspace.

Dimension. The linear form $\operatorname{tr}:\mathcal{M}_3(\mathbb{R})\to\mathbb{R}$ is nonzero (e.g. $\operatorname{tr}(E_{11})=1$), so it is surjective. By rank-nullity, $\dim F=9-1=8$.

Conclusion. $F$ is a hyperplane of $\mathcal{M}_3(\mathbb{R})$, of dimension $8$.