One-Step Vector Subspace Test

Theorem
Let $V$ be a vector space over a division ring $K$ whose unity is $1_K$.

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:

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


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

which is an element of $U$.

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

Hence it is seen that axioms:

are satisfied.

The last axiom that remains is.

From and :


 * $\forall u, v \in U: u + v = u + 1_K v \in U$

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

Also see

 * Vector Subspace of Real Vector Space
 * Two-Step Vector Subspace Test