# Two-Step Vector Subspace Test

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## Theorem

Let $V$ be a vector space over a division ring $K$.

Let $U \subseteq V$ be a non-empty subset of $V$ such that:

$(1): \qquad \forall u \in U, \lambda \in K: \lambda u \in U$
$(2): \qquad \forall u, v \in U: u + v \in U$

Then $U$ is a subspace of $V$.

## Proof

Suppose that $(1)$ and $(2)$ hold.

From $(1)$, we obtain for every $\lambda \in K$ and $u \in U$ that $\lambda u \in U$.

An application of $(2)$ yields the condition of the One-Step Vector Subspace Test.

Hence $U$ is a vector subspace of $V$.

$\blacksquare$