Generator of Vector Space Contains Basis

Theorem

Let $E$ be a vector space of $n$ dimensions.

Let $G$ be a generator for $E$.

Then:

$G$ contains a basis for $E$.

Generator of Vector Space Contains Basis/Infinite Dimensional Case

Let $K$ be a field.

Let $X$ be a vector space over $K$.

Let $G$ be a generator of $X$.

Then:

$G$ contains a basis for $X$.

Proof

From:

Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set

Bases of Finitely Generated Vector Space have Equal Cardinality

and

Sufficient Conditions for Basis of Finite Dimensional Vector Space

all we need to do is show that every infinite generator $S$ for $E$ contains a finite generator.

Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be an ordered basis of $E$.

For each $k \in \closedint 1 n$ there is a finite subset $S_k$ of $S$ such that $a_k$ is a linear combination of $S_k$.

Hence $\ds \bigcup_{k \mathop = 1}^n S_k$ is a finite subset of $S$ generating $E$, for the subspace it generates contains $\set {a_1, \ldots, a_n}$ and hence is $E$.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.14$