# One-Step Vector Subspace Test

## 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:

- $\forall u, v \in U: \forall \lambda \in K: u + \lambda v \in U$

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

## Proof

We need to verify the vector space axioms for $U$.

We start with observing that the properties for a unitary module are true for all elements of $V$.

Hence, since $U \subseteq V$, they hold for all elements of $U$ as well.

The same holds for the axioms $(G1)$ and $(C)$.

From Vector Inverse is Negative Vector, we have for all $u \in U$:

- $u + \paren {- 1_K} u = 0_V$

which by assumption is an element of $U$.

Since $U$ is non-empty, this means $0_V \in U$.

Hence it is seen that axioms $(G2)$ and $(G3)$ are satisfied.

The last axiom that remains is $(G0)$.

To this end we employ the knowledge that for all $v \in U$, we have:

- $1_K v = v$

and hence:

- $u + 1_K v = u + v \in U$

Having verified all the vector space axioms, we conclude that $U$ is a subspace of $V$.

$\blacksquare$