Okay, so I'm having some trouble with this topic. If a question asks you to show that a set is a subspace (or not a subspace), do you show the following:

$\displaystyle \bullet$ The set $\displaystyle S$ is not empty

$\displaystyle \bullet$ The set is closed under vector addition, i.e. $\displaystyle \mathbf{v,u}\in S \Longrightarrow \mathbf{v}+\mathbf{u}\in S$

$\displaystyle \bullet$ The set is closed under scalar multiplication, i.e. $\displaystyle \mathbf{u}\in S \ \mbox{and} \ \lambda \in \mathbb{F}\Longrightarrow \lambda\mathbf{u}\in S$

1) Show that the set

$\displaystyle S=\{\mathbf{x}\in\mathbb{R}^3:2x_1+3x_2-4x_3=6\}$

is not a subspace of $\displaystyle \mathbb{R}^3.$

Would I do the following:

$\displaystyle 2(\mathbf{0})+3(\mathbf{0})-4(\mathbf{0})=6$

Therefore, it does not contain the $\displaystyle \mathbf{0}$ vector, hence not a subspace in $\displaystyle \mathbb{R}^3$