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 $K$.
The following definitions of the concept of Basis of Vector Space are equivalent:
Definition 1
A basis of $G$ is a linearly independent subset of $G$ which is a generator for $G$.
Definition 2
A basis is a maximal linearly independent subset of $G$.
Proof
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 $\BB = \BB'$.
$\Box$
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 $G$.
$\blacksquare$
Sources
- 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next): Appendix $\text{A}$ Preliminaries: $\S 1.$ Linear Algebra: Proposition $1.1$