[선형대수학] 극대 선형 독립 (Maximal Linearly Independent)

$\underline{Def}$

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

A subset $B$ of $S$ is a maximal linearly independent subset of $S$ if

$(a)$ $B$ is linear independent.

$(b)$ if $A$ is a linear independent subset of $S% with$B \subset A$, then$A=B$.

 

$\underline{Rmk}$

We can re-write this definition as follows:

Let $\mathcal{F}$ be the set of all linear independent subsets of $S$.

A maximal linear independent subset of $S$ is a maximal set $B$ in $\mathcal{F}$.

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$\underline{Rmk}$

Let $\beta$ be a basis for a vector space $V$.

Then $\beta$ is a maximal linear independent subset of $V$.

 

$\underline{Proof}$

By definition, $\beta$ is a linear independent subset of $V$.

Let $\alpha$ e any linear independent subset of $V$ with $\beta \subset \alpha$.

If $\exists v \in \alpha \backslash \beta$, then since $v \in V = span(\beta)$, $\beta \cup \{v \}$ is linear dependent by thm, a contradiction to the fact that $\alpha$ is linear independent.

Thus, $\alpha = \beta$.

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$\underline{Thm}$

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

If $\beta$ is a maximal linear independent subset of $S$, then $\beta$ is a basis for $V$.

 

$\underline{Proof}$

It is sufficient to show that $S \subset span(\beta)$.

Suppose $\exists v \in S \backslash span(\beta)$. Then by thm, $\beta \cup \{ v \}$ is a linear independent subset of $S$ that properly contains $\beta$, a contradiction to maximality of $\beta$.

 

$\underline{Rmk}$

By the above, for a subet $\beta$ of a vector space $V$,

$\beta$ is a basis for a $V$ $\Leftrightarrow$ it is a maximal linear independent subset of $V$.