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:
| \(\text {(1)}: \quad\) | \(\ds \forall u \in U, \lambda \in K: \, \) | \(\ds \lambda u\) | \(\in\) | \(\ds U\) | ||||||||||
| \(\text {(2)}: \quad\) | \(\ds \forall u, v \in U: \, \) | \(\ds u + v\) | \(\in\) | \(\ds U\) |
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 $V$.
$\blacksquare$