[선형대수학] 스팬 (Span)

$\underline{Def}$ $(Span)$

Let $S$ be a non-empty subset of a vector space $V$. The span of $S$, denoted by $span(S)$, is the set of all linear combination of vectors in $S$. We also define $span(\phi)={0}$

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$\underline{Thm}$

Let $S$ be any subset of a vector space $V$. Then $span(S)$ is a subspace of $V$. Moreover, $span(S)$ is the smallest subspace of $V$ containing $S$. That is, $span(S)=\cap W$ where $W$ is a subspace of $V$ with $S \subset W$.

 

$\underline{Proof}$

If $S=\phi$, then by definition, $span(S)={0}$ is the zero vector space.

Assume $S \neq \phi$. Choose any $z \in S$. Then $0=0 \times z \in span(S)$. Let $x,y \in span(S),\; c \in \mathbb{F}$.

Then $x=a_{1}u_{1}+ \cdots +a_{m}u_{m},\; y=b_{1}v_{1}+ \cdots + b_{n}v_{n}$ for some $u_{1}, \cdots, u_{m}, v_{1}, \cdots ,v_{n} \in S$ and some $a_{1}, \cdots , a_{m}, b_{1}, \cdots ,b_{n} \in \mathbb{F}$.

So, $cx+y=(ca_{1})y_{1}+ \cdots + (ca_{m}u_{m}+b_{1}v_{1}+ \cdots b_{n}v_{n} \in span(S)$. Thus, $span(S)$ is a subspace of $V$.

Next, let $W$ be any subspace of $V$ containing $S$. Then, $a_{1}u_{1}+ \cdots + a_{n}u_{n} \in W,$ $\forall n \in \mathbb{N},$ $\forall a_{1}, \cdots, a_{n} \in \mathbb{F},$ $\forall u_{1}, \cdots, u_{n} \in S$.

Hence, $span(S) \subset W$.

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$\underline{Def}$

A subset $S$ of a vector space $V$ spans $V$ if $span(S)=V$. In this case, we also say that the vectors in $S$ span $V$.