Real Linear Subspace Contains Zero Vector

Theorem
The two definitions for a vector subspace of $\R^n$ are equivalent.

Specifically, the two versions of the first criterion:


 * $\mathbf{0} \in \mathbb{W}$


 * $\varnothing \ne \mathbb{W}$

imply one another.

Proof
Suppose $\mathbf{0} \in \mathbb{W}$.

Then $\mathbb{W}$ contains an element and cannot be the empty set.

Suppose $\varnothing \ne \mathbb{W}$. Then there exists some element in $\mathbb{W}$. Call it $\mathbb{v}$.

By the third criterion, that $\mathbb{W}$ is closed under scalar multiplication,


 * $0\mathbf{v} \in \mathbb{W}$

where $0$ is the zero scalar.

But $0\mathbf{v} = \mathbf{0}$, and so $\mathbf{0} \in \mathbb{W}$.