Zero Subspace is Subspace

Theorem
Let $V$ be a vector space over $K$ with zero vector $\mathbf 0$.

The zero subspace $\left\{{\mathbf 0}\right\}$ is a subspace of $V$.

Proof
We use the Two-Step Vector Subspace Test.

$\left\{{\mathbf 0}\right\}$ is not empty, because it contains $\mathbf{0}$.

$\left\{{\mathbf 0}\right\}$ is closed under $+$ because:


 * $\forall \mathbf{x},\mathbf{y} \in \left\{{\mathbf 0}\right\}, \mathbf{x} + \mathbf{y} = \mathbf{0} + \mathbf{0} = \mathbf{0} \in \left\{{\mathbf 0}\right\}$

$\left\{{\mathbf 0}\right\}$ is closed under multiplication because:


 * $\forall \lambda \in K, \mathbf{x}\in \left\{{\mathbf 0}\right\}: \lambda\mathbf{x} = \lambda\mathbf{0} = \mathbf{0}\in \left\{{\mathbf 0}\right\}$

Hence the result, from the Two-Step Vector Subspace Test.