# 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 $E$.

## Proof

From:

- Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set
- Bases of Finitely Generated Vector Space have Equal Cardinality

and

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 $\displaystyle \bigcup_{k \mathop = 1}^n S_k$ 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): $\S 27$: Theorem $27.14$