[선형대수학] 차원 (Dimension)

$\underline{Def}$

A vector space $V$ over $\mathbb{F}$ is called finite-dimensional if it has a finite basis $\beta$. In this case, $\vert \beta \vert$ is called the dimension of $V$ and denoted by $dim\,V=dim_{\mathbb{F}}\,V=\vert \beta \vert$.

A vector space $V$ that is not finite-dimensional is called infinite-dimensional. In this casem we write $dim\,V=\infty$.

 

$\underline{Ex}$

$(1)$ $\mathbb{C}$ is a vector space over itself with a basis $\{ 1 \}$. So, $dim_{\mathbb{C}}\,\mathbb{C}=1$.

$(2)$ $\mathbb{C}$ is a vector space over $\mathbb{R}$ with a basis $\{ 1, i \}$. So, $dim_{\mathbb{R}}\,\mathbb{C}=2$.

$(3)$ $\mathbb{C}^{n}$ is a vector space over $\mathbb{C}$ with a baisis $\{ e_{1}, \cdots, e_{n} \}$; hence, $dim_{\mathbb{C}}\,\mathbb{C}^{n}=n$.

$(4)$ $\mathbb{C}^{n}$ is a vector space over $\mathbb{R}$ with a basis $\{e_{1}, \cdots, e_{n}, ie_{1}, \cdots, ie_{n} \}$; so $dim_{\mathbb{R}}\,\mathbb{C}^{n}=2n$

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$\underline{Coro}$ (기저 대체 정리의 따름 정리)

Let $V$ be a vector space with dimension $n \in \mathbb{N}_{0}$.

$(a)$ Let $\beta \subset V$. Then $\beta$ is a basis for $V$. $\cdots (1)$

  $\Leftrightarrow \; \beta$ is linear independent & $\vert \beta \vert=n \cdots (2)$

  $\Leftrightarrow \; span(\beta)=V$ & $\vert \beta \vert = n \cdots (3)$

$(b)$ Let $\gamma \subset V$ be such that $span(\gamma)=V$. Then $\vert \gamma \vert \geq n$.

$(c)$ If  $\omega \subset V$ is linear independent, then there is a basis $\beta$ for $V$ with $\gamma \subset \beta$, so that $\vert \gamma \vert \leq \vert \beta \vert = n$.

 

$\underline{Proof}$

By assumption, $\exists$ a basis $B$ for $V$ with $\vert B \vert=n$.

$(b)$

If $\vert \gamma \vert = \infty$, there is nothing to show.

So, assume $\vert \gamma \vert \in \mathbb{N}_{0}$. Since $span(\gamma)=V$ and $B$ is linear independent, by RT, $\vert \gamma \vert \geq \vert b \vert =n$.

 

$(a)$

$(1) \Rightarrow (2)$

Assume $(1)$. Then $\beta$ is linear independent. Also by Coro, $\vert \beta \vert =n$.

$(2) \Rightarrow (3)$

Assume $(2)$. Since $span(\beta)=V$, using the RT, we see that $span(\beta)=V$.

$(3) \Rightarrow (1)$

Assume $(3)$. We have to show that $\beta=\{u_{1},\cdots,u_{n} \}$ is linear independent. Suppose not; then after re-labaeling if necessary, $u_{1}=a_{2}u_{2}+\cdots+a_{n}u_{n}$ for some $a_{2}, \cdots, a_{n} \in \mathbb{F}$.

This implies $span(\beta)=span(\{ u_{2}, \cdots, u_{n} \})$. Since $B$ is linear independent with $\vert B \vert=n$, by the RT, $n=\vert B \vert \geq n-1$, a contradiction. Therefore, $\beta$ is linear independent.

 

$(c)$

Since $\omega$ is linear independent & $span(B)=V$, by the RT, $\vert \omega \vert \leq n$, and $\exists G \subset B$ with $\vert G \vert = n-\vert \omega \vert$ such that $span(\omega \cup G)=V$.

Put $\beta=\omega \cup G$. Note $\vert \beta \vert \leq \vert \omega \vert + \vert G \vert =n$.

By $(b)$, $n \leq \vert \beta \vert$; thus $\vert \beta \vert =n$. Since $\vert \beta \vert = n$ & $span(\beta)=V$, it follows from $(a)$ that $\beta$ is a basis for $V$ with $\omega \subset \beta$.