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

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Let $K$ be a division ring.

Let $G$ be a finitely generated $K$-vector space.

Let $H$ be a linearly independent subset of $G$.

Let $F$ be a finite generator for $G$ such that $H \subseteq F$.

Then there is a basis $B$ for $G$ such that $H \subseteq B \subseteq F$.

Outline of Proof

Of all the sets between $H$ and $F$ that generate $G$, we choose one with the fewest elements, using the Well-Ordering Principle.

Using the minimality, we show that this generator is linearly independent.


Let $\mathbb S$ be the set of all $S \subseteq G$ such that $S$ is a generator for $G$ and that $H \subseteq S \subseteq F$.

Because $F \in \mathbb S$, it follows that $\mathbb S \ne \O$.

Because $F$ is finite, then so is every element of $\mathbb S$.

Let $R = \set {r \in \Z: r = \card S \in \mathbb S}$.

That is, $R$ is the set of all the integers which are the number of elements in generators for $G$ that are subsets of $F$.

Let $n$ be the smallest element of $R$.

Let $B$ be an element of $\mathbb S$ such that $\card B = n$.

We note that as $H$ is a linearly independent set, it does not contain $0$ by Subset of Module Containing Identity is Linearly Dependent.

Then $0 \notin B$, or $B \setminus \set 0$ would be a generator for $G$ with $n - 1$ elements.

This would contradict the definition of $n$.

Let $m = \card H$.

Let $\sequence {a_n}$ be a sequence of distinct vectors such that $H = \set {a_1, \ldots, a_m}$ and $B = \set {a_1, \ldots, a_n}$.

Suppose $B$ were linearly dependent.

By Linearly Dependent Sequence of Vector Space, there would exist $p \in \closedint 2 n$ and scalars $\mu_1, \ldots, \mu_{p - 1}$ such that $\ds a_p = \sum_{k \mathop = 1}^{p - 1} \mu_k a_k$.

As $H$ is linearly independent, $p > m$ and therefore $B' = B \setminus \set {a_p}$ would contain $H$.

Now if $\ds x = \sum_{k \mathop = 1}^n \lambda_k a_k$, then:

$\ds x = \sum_{k \mathop = 1}^{p - 1} \paren {\lambda_k + \lambda_p \mu_k} a_k + \sum_{k \mathop = p + 1}^n \lambda_k a_k$

Hence $B'$ would be a generator for $G$ containing $n - 1$ elements, which contradicts the definition of $n$.

Thus $B$ must be linearly independent and hence is a basis.


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