[선형대수학] 기저 (Basis)

$\underline{Def}$

A subset $\beta$ of a vector space $V$ is a basis for $V$ if $\beta$ is linear independent & $span(\beta)=V$.

 

$\underline{Ex.1}$

Since $\phi$ is linear independent & $span(\phi)= \{ 0 \}$, $\phi$ is a basis for $\{ 0 \}$.

 

$\underline{Ex.2}$

For $i=1, \cdots, n$, let $e_{i} = (0, \cdots, 0, 1, 0, \cdots, 0,) \in \mathbb{F}^{n}$ where 1 is i-th component.

If $a=(a_{1}, \cdots, a_{n}) \in \mathbb{F}^{n}$, then $a=a_{1}e_{1}+ \cdots + a_{n}e_{n}$; thus, $span(\{e_{1}, \cdots, e_{n} \})= \mathbb{F}^{n}$.

Clearly $\{ e_{1}, \cdots, e_{n} \}$ is linear independent.

Thus, $\{ e_{1}, \cdots, e_{n} \}$ is a basis for $\mathbb{F}^{n}$, called the standard basis for $\mathbb{F}^{n}$.

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$\underline{Thm}$ (Unique linear combination by a basis)

Let $\beta = \{ u_{1}, \cdots, u_{n} \}$ be a finite subset of a vector space $V$.

Then $\beta$ is a basis for $V$ iff $\forall v \in V,\; \exists (a_{1}, \cdots, a_{n}) \in \mathbb{F}^{n}$ such that $v=a_{1}v_{1}+ \cdots +a_{n}v_{n}$.

 

$\underline{Proof}$

$(\Rightarrow)$

(Existience)

Since $\beta$ is a basis for $V$, we have $v \in V=span(\beta)$, so that $\exists a_{1}, \cdots, a_{n} \in \mathbb{F}$ such that $v=a_{1}v_{1}+ \cdots +a_{n}v_{n}$.

(Uniqueness) Suppose $(a_{1}^{'}, \cdots, a_{n}^{'}) \in \mathbb{F}^{n}$ and $v=a_{1}^{'}+ \cdots +a_{n}^{'}v_{n}^{'}$.

Then $(a_{1}-a_{1}^{'})v_{1}+ \cdots +(a_{n}-a_{n}^{'})v_{n}=0$.

Since $\beta$ is linear dependent, we get $a_{i}=a_{i}^{'}$ for $i=1, \cdots, n$.

$(\Leftarrow)$

By assuption, $span(\beta)=V$.

Suppose $a_{1}, \cdots, a_{n} \in \mathbb{F}\; \& \; a_{1}v_{1}+ \cdots +a_{n}v_{n}=0$.

Since $0=0v_{1}+ \cdots +0v_{n}$, the uniqueness says $a_{i}=0$ for $i=1, \cdots, n$.

Thus $\beta$ is linear independent.

 

$\underline{Rmk}$

Let $\beta=\{ u_{1}, \cdots, u_{n} \}$ be a basis for vector space $V$.

$V=span(\beta)$ is linear isomorphic to $\mathbb{F}^{n}$, denoted by $span(\beta) \simeq \mathbb{F}^{n}$.

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$\underline{Thm}$

Suppose a vector space $V$ has a finite set $S$ with $span(S)=V$.

Then $\exists \beta \subset S$ such that $\beta$ is a basis for a $V$; hence $V$ has a finite basis $\beta$.

 

$\underline{Proof}$

If $S=\phi$ or $S=\{ 0\}$, then $V=\{ 0\}$ and $\beta=\phi \subset S$ is a basis for $V$.

So, suppose $S$ has a non-zero vector $u_{1}$. Then $\{ u_{1} \}$ is linear independent.

If $\{ u_{1}, v \}$ is linear dependent, $\forall v \in S \backslash \{ u_{1} \}$, let $\beta=\{ u_{1} \}$.

Otherwise, we choose a $u_{2} \in S \backslash \{u_{1} \}$ so that $\{ u_{1}, u_{2} \}$ is linear independent.

If $\{ u_{1}, u_{2}, v \}$ is linear dependent, $\forall v \in S \backslash \{ u_{1}, u_{2} \}$, let $\beta = \{ u_{1}, u_{2} \}$.

Otherwise, we choose a $u_{3} \in S \backslash \{u_{1}, u_{2} \}$ so that $\{ u_{1}, u_{2}, u_{3} \}$ is linear independent.

Continuing this process, since $S$ is finite, we obtain a linear independent set $\beta = \{ u_{1}, \cdots, u_{k} \} \subset S$ such that $\beta \cup \{ v \}$ is linear dependent $\forall v \in S \backslash \beta$.

We claim that $S \subset span(\beta)$.

Suppose not; i.e. $\exists v \in S \backslash span(\beta)$. Then $\beta \cup \{ v \}$ is linear independent by thm, a contradinction.

So, $S \subset span(\beta)$. Since $span(S)$ is the smallest subspace of $V$ containing $S$, we have $span(S) \subset span(\beta) \subset span(S)$; so $span(\beta)=span(S)=V$.