Size of Linearly Independent Subset is at Most Size of Finite Generator
Theorem
Let $R$ be a division ring.
Let $V$ be an $R$-vector space.
Let $F \subseteq V$ be a finite generator of $V$ over $R$.
Let $L \subseteq V$ be linearly independent over $R$.
Then:
- $\size L \le \size F$
Proof 1
We first consider the case where $L$ is finite.
Let $S \subseteq \N$ be the set of all $n \in \N$ such that:
where:
- $L \setminus F$ denotes the set difference between $L$ and $F$
- $\card L$ and $\card F$ denote the cardinality of $L$ and $F$ respectively.
We use the Principle of Finite Induction to prove that $S = \N$.
Basis of the Induction
Let $\card {L \setminus F} \le 0$.
Then from Cardinality of Empty Set:
- $L \setminus F = \O$
By Set Difference with Superset is Empty Set:
- $L \subseteq F$
By Cardinality of Subset of Finite Set:
- $\card L \le \card F$
Hence:
- $0 \in S$
This is the basis for the induction.
Induction Hypothesis
It is to be shown that if $k \in S$ where $k \ge 1$, then it follows that $k + 1 \in S$.
This is the induction hypothesis:
It is to be demonstrated that it follows that:
- For every finite generator $F$ of $V$, if $\card {L \setminus F} \le k + 1$, then $\card L \le \card F$
Induction Step
This is the induction step:
Assume the induction hypothesis that $n \in S$.
Let $F$ be a finite generator of $V$ such that:
- $\card {L \setminus F} = n + 1$
Let $v \in L \setminus F$.
Let $L' = L \cap \paren {F \cup \set v}$.
- $L' \subseteq L$
By Subset of Linearly Independent Set is Linearly Independent, it follows that $L'$ is linearly independent over $R$.
Also by Intersection is Subset:
- $L' \subseteq F \cup \set v$
Therefore, by Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set:
- there exists a basis $B$ of $V$ such that:
- $L' \subseteq B \subseteq F \cup \set v$
Since $v \notin F$ is a linear combination of $F$, it follows that $F \cup \set v$ is linearly dependent over $R$.
Therefore:
- $B \subsetneq F \cup \set v$
By Cardinality of Subset of Finite Set:
- $\card B < \card {F \cup \set v} = \card F + 1$
Hence:
- $\card B \le \card F$
We have that:
\(\ds \card {L \setminus B}\) | \(\le\) | \(\ds \card {L \setminus L'}\) | Relative Complement inverts Subsets and Cardinality of Subset of Finite Set | |||||||||||
\(\ds \) | \(=\) | \(\ds \card {L \setminus \paren {F \cup \set v} }\) | Set Difference with Intersection is Difference | |||||||||||
\(\ds \) | \(=\) | \(\ds \card {\paren {L \setminus F} \setminus \set v}\) | Set Difference with Union | |||||||||||
\(\ds \) | \(=\) | \(\ds n + 1 - 1\) | Cardinality of Set Difference with Subset | |||||||||||
\(\ds \) | \(=\) | \(\ds n\) |
Since $n \in S$:
- $\card L \le \card B \le \card F$
Hence:
- $n + 1 \in S$
and so the induction step has been completed.
- $L \setminus F \subseteq L$
From Subset of Finite Set is Finite:
- $L \setminus F$ is finite.
Therefore, we can apply the fact that $S = \N$ to conclude that:
- $\card L \le \card F$
Let $L$ be infinite.
Then by Set is Infinite iff exist Subsets of all Finite Cardinalities, there exists a finite subset $L' \subseteq L$ such that:
- $\card {L'} = \card F + 1$
By Subset of Linearly Independent Set is Linearly Independent, it follows that $L'$ is linearly independent over $R$.
It is proven above that this is impossible.
Hence the result.
$\blacksquare$
Proof 2
Let $S \subseteq \N$ be the set of all natural numbers $n \in \N$ such that:
- For any finite generator $F$ of $V$ over $R$, if $\card {F \cap L} \ge n$, then $\card L \le \card F$.
It is to be demonstrated that $S = \N$.
That is, that $\card {F \cap L} \ge n \implies \card L \le \card F$ for all $n \in \N$.
By Intersection is Subset and Cardinality of Subset of Finite Set, we have $\card {F \cap L} \le \card F$.
Hence, it is vacuously true that $\card F + 1 \in S$.
Therefore, $S$ is non-empty.
From the well-ordering principle, $S$ has a smallest element $N$.
If $N = 0$, the theorem immediately follows.
Aiming for a contradiction, suppose $N \ge 1$.
Let $\card {F \cap L} \ge N - 1$.
If $L \subseteq F$, the theorem follows from Cardinality of Subset of Finite Set.
Otherwise, there exists a $v \in L$ such that $v \notin F$.
Let $F' = F \cup \set v$.
By Intersection is Subset, we have $F' \cap L \subseteq L$, so it follows by Subset of Linearly Independent Set is Linearly Independent that $F' \cap L$ is linearly independent over $R$.
Also, by Intersection is Subset:
- $F' \cap L \subseteq F'$
We have that $F'$ is a generator of $V$ over $R$.
By Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set, there exists a basis $B$ for $V$ such that:
- $F' \cap L \subseteq B \subseteq F'$
Since $v \notin F$ is a linear combination of $F$, it follows that $F'$ is linearly dependent over $R$.
By the definition of a basis:
- $B \subsetneq F'$
By Cardinality of Subset of Finite Set and Cardinality is Additive Function:
- $\card B < \card {F'} = \card F + 1$
Hence:
- $\card B \le \card F$
We have that:
\(\ds \card {B \cap L}\) | \(\ge\) | \(\ds \card {\paren {F' \cap L} \cap L}\) | Set Intersection Preserves Subsets and Cardinality of Subset of Finite Set | |||||||||||
\(\ds \) | \(=\) | \(\ds \card {F' \cap L}\) | Intersection is Associative and Set Intersection is Idempotent | |||||||||||
\(\ds \) | \(=\) | \(\ds \card {\paren {F \cap L} \cup \paren {\set v \cap L} }\) | Intersection Distributes over Union | |||||||||||
\(\ds \) | \(=\) | \(\ds \card {\paren {F \cap L} \cup \set v}\) | Intersection with Subset is Subset | |||||||||||
\(\ds \) | \(\ge\) | \(\ds N\) | Cardinality is Additive Function |
Since $N \in S$, it follows by the definition of a basis that:
- $\card L \le \card B \le \card F$
Hence:
- $N - 1 \in S$
But this contradicts the assumption that $N$ is the smallest element of $S$.
Therefore:
- $N = 0$
Hence the result.
$\blacksquare$
Proof 3
Let $\alpha_1, \alpha_2, \ldots, \alpha_n$ be a generator of $V$.
Let $\xi_1, \xi_2, \ldots, \xi_r$ be a linearly independent set of elements of $V$.
Hence the sequence $\sequence {\xi_1, \alpha_1, \alpha_2, \ldots, \alpha_n}$ is a linearly dependent sequence of elements of $V$.
One of these elements, which cannot be $\xi_1$, is a linear combination of the preceding elements.
Let this elements be $\alpha_i$.
So we can omit $\alpha_i$ from that sequence, and the remaining set is still a generator of $V$.
Therefore $\xi_2$ is a linear combination of these.
Thus $\sequence {\xi_2, \xi_1, \alpha_1, \alpha_2, \ldots, \alpha_{i - 1}, \alpha_{i + 1}, \ldots, \alpha_n}$ is a linearly dependent sequence of elements of $V$.
Again, one of them is a linear combination of the preceding elements.
This cannot be $\xi_2$, as $\xi_2$ has no preceding elements.
Neither can it be $\xi_1$, as $\xi_1$ and $\xi_2$ are linearly independent.
Thus we can omit whichever $\alpha_j$ it is, and we have a new set which is a generator of $V$.
This consists of $\xi_1$, $\xi_2$ and whichever $n - 2$ of the remaining elements of $\set {\alpha_1, \alpha_2, \ldots, \alpha_n}$.
After $p$ such steps, we have a set which is a generator of $V$ which consists of:
- $\xi_1, \xi_2, \ldots, \xi_p$ and $n - p$ of the elements of $\set {\alpha_1, \alpha_2, \ldots, \alpha_n}$.
Aiming for a contradiction, suppose suppose $n < r$.
Then when $p = n$, the remaining set which is a generator of $V$ consists of:
- $\xi_1, \xi_2, \ldots, \xi_n$
and there is at least one more element $\xi_{n + 1}$.
This is a linear combination of $\set {\xi_1, \xi_2, \ldots, \xi_n}$.
But this contradicts the supposition that $\set {\xi_1, \xi_2, \ldots, \xi_n, \xi_{n + 1} }$ is a linearly independent set.
Hence, by Proof by Contradiction, $n \ge r$.
The result follows.
$\blacksquare$