Algebra over Field Embeds into Unitization as Vector Subspace

Theorem
Let $K$ be a field.

Let $A$ be an algebra over $K$ that is not unital.

Let $A_+$ be the unitization of $A$.

Let:
 * $A_0 = \set {\tuple {x, 0_K} : x \in A} \subseteq A_+$.

Then $A_0$ is a vector subspace of $A$.

Proof
Clearly $A_0 \ne \O$.

From One-Step Vector Subspace Test, it is sufficient to show that for each $u, v \in A_+$ and $\lambda \in K$, we have:
 * $u + \lambda v \in K$

Let $u, v \in A_+$ and $\lambda \in K$.

Then there exists $x, y \in A$ such that:
 * $u = \tuple {x, 0_K}$

and:
 * $v = \tuple {y, 0_K}$

Then by the definition of the unitization, we have:
 * $\tuple {x, 0_K} + \lambda \tuple {y, 0_K} = \tuple {x + \lambda y, 0_K} \in A_0$

So by the One-Step Vector Subspace Test, $A_0$ is a vector subspace of $A$.