Algebra over Field Embeds into Unitization as Vector Subspace
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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 $A$.
$\blacksquare$