Generator of Vector Space Contains Basis

Theorem

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

Every generator for $G$:

$(1): \quad$ has at least $n$ elements;

$(2): \quad$ contains a basis for $G$;

$(3): \quad$ is a basis for $G$ iff it contains exactly $n$ elements.

Proof

From Linearly Independent Subset of Basis of Vector Space, Bases of Finitely Generated Vector Space and Basis of Vector Space is Linearly Independent and a Generator, all we need to do is show that every infinite generator $S$ for $G$ contains a finite generator.

Let $\left \langle {a_n} \right \rangle$ be an ordered basis of $G$.

For each $k \in \left[{1 .. n}\right]$ there is a finite subset $S_k$ of $S$ such that $a_k$ is a linear combination of $S_k$.

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

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965): $\S 27$: Theorem $27.14$