Equivalence of Definitions of Basis of Vector Space

Theorem
Let $K$ be a division ring.

Let $\struct {G, +_G, \circ}_K$ be an vector space over $K$.

Definition 1 implies Definition 2
Let $\BB$ be a linearly independent subset of $G$ which is a generator for $G$.

Suppose $\BB \subseteq \BB'$ is a linearly independent subset of $G$.

We aim to show that $\BB = \BB'$, proving maximality.

Suppose that $\BB \ne \BB'$.

Let $x \in \BB' \setminus \BB$.

Since $G$ is a generator for $G$, there exists $x_1, \ldots, x_n \in \BB \subseteq \BB'$ and $\alpha_1, \ldots, \alpha_n \in K$ such that:


 * $\ds x = \sum_{i \mathop = 1}^n \alpha_i x_i$

so that:


 * $\ds \sum_{i \mathop = 1}^n \alpha_i x_i - x = 0$

Since $K$ is a field, we have that $-1 \ne 0$.

This shows that $\set {x_1, \ldots, x_n, x}$ and hence $\BB'$ is linearly dependent.

This is a contradiction, so we must have $\BB = \BB'$.

Definition 2 implies Definition 1
Suppose $\BB$ is a maximal linearly independent subset of $G$.

Let:


 * $G = \map \span \BB \subseteq E$

Suppose that $E \ne F$.

Suppose that $G \ne E$.

Let $x \in E \setminus G$.

Then from Vector not contained in Linear Span of Linearly Independent Set is Linearly Independent of Set, $\BB \cup \set x$ is linearly independent, contradicting the maximality of $\BB$.

So there exists no such $x \in E \setminus G$, and we have $G = E$.

So $\BB$ is a linearly independent subset of $G$ which is a generator for $G$.