Generator of Vector Space Contains Basis
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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
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 $27.14$