[선형대수학] 선형 종속, 독립의 성질 (Property of Linear Dependence, Independence)

$\underline{Thm}$

Let $V$ be a vector space and let $S_{1} \subset S_{2} \subset V$.

If $S_{1}$ is linear dependent, then so is $S_{2}$.

If $S_{2}$ is linear independent, then so is $S_{1}$.

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$\underline{Rmk}$

Let $S \subset V$, where $V$ is a vector space. Consider the space $span(S)$.

When is there a proper subset $S' \subsetneq S$ such than $span(S')=span(S)$.

 

First, suppose $\exists S' \subsetneq$ with $span(S')=span(S)$.

Then $\exists u_{0} \in S \backslash S' \subset S \subset span(S) = span(S')$.

So, $\exists$ finitely many distinct $u_{1}, \cdots, u_{n} \in S'$ and $\exists a_{1}, \cdots, a_{n} \in \mathbb{F}$ such that $u_{0}=a_{1}u_{1}+ \cdots + a_{n}u_{n}$ or $(-1)u_{0}+a_{1}u_{1}+ \cdots +a_{n}u_{n}=0$ where $u_{0}, u_{1}, \cdots, u_{n}$ are distinct, since $u_{0} \notin S'$.

Thus, $S$ is linear dependent.

 

Conversely, suppose $S$ is linear dependent.

If $0 \in S$, note that $span(S')=span(S)$, where $S' = S \backslash \{ 0 \} \subsetneq S$.

Assume $0 \notin S$.

There are finitely many distinct $u_{1}, \cdots, u_{n} \in S$ and $a_{1}, \cdots, a_{n} \in \mathbb{F}$, not all zeros, such that $a_{1}u_{1}+ \cdots +a_{n}u_{n}=0$.

If$n=1$, then $a_{1}u_{1}=0$ and $a_{1} \neq 0$, and so $u_{1}=0 \in S$, a contradictaion.

So, $u_{1}=(-a_{2}a_{1}^{-1})u_{2}+ \cdots + (-a_{n}a_{1}^{-1})u_{n}$

Put, $S' = S \backslash \{ u_{1} \}$. Then $span(S')=span(S)$

 

So, we have proved the following

 

$\underline{Thm}$

Let $S$ be a subset of a vector space $V$. Then $S$ is linear dependent iff $\exists S' \subsetneq S$ such that $span(S')=span(S)$.

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$\underline{Thm}$

Let $S$ be a linear independent subset of a vector space $V$. Let $v \in V \backslash S$.

Then $S \cup \{ v \}$ is linear dependent iff $v \in span(S)$.

 

$\underline{Proof}$

If $S \cup \{ v \}$ is linear dependent, then by above thm, $\exists S' \subsetneq S \cup \{ v \}$ such that $span(S')=span(S \cup \{ v \})$.

If $S' \subsetneq S$, then $S$ is linear dependent by above thm, a contradiction. So, $S \subset S'$. Since $S' \subsetneq S \cup \{ v \}$, we have $S'=S$. Thus, $v \in span(S \cup \{ v \})=span(S')=span(S)$.

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$\underline{Coro}$

Let $S$ be a subset of a vector space $V$.

Then $S$ is linear independent iff $\nexists S' \subsetneq S$ such that $span(S')=span(S)$; that is, $\forall S' \subsetneq S, span(S') \subsetneq span(S)$.

 

$\underline{Rmk}$

Let $S$ be a subset of a vector space $V$ with $V=span(S)$.

Then $S$ is linear independent iff $\nexists S' \subsetneq$ such that $span(S')=V$.