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$\underline{Thm}$ (By the axiom of choice) Let $S$ be a linear independent subset of a vector space $V$. Then there exists a maximal linear independent subset $\beta$ of $V$ with $S \subset \beta$. $\underline{Proof}$ Let $\mathcal{F}$ be the set of all linear independent subsets of $V$ containing $S$. Let $\mathcal{C} \subset \mathcal{F}$ be any chain. Define $\displaystyle L_{\mathcal{C}}$ = $..

$\underline{Def}$ 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 ..

$\underline{Def}$ (극대 부분 집합) Let $\mathcal{F}$ be a set of sets. A set $M \in \mathcal{F}$ is called maximal if $ \nexists K \in \mathcal{F}$ such that $M \subsetneq K$; that is, $A \in \mathcal{F}$ & $M \subset A \Rightarrow A=M$. $\underline{Def}$ A set $\mathcal{C}$ of sets is called chain if $\forall A, B \in \mathcal{C},\; A \subset B$ or $B \subset A$. $\underline{Maximal\;Principle}$ (Zor..

$\underline{Thm}$ Let $W$ be a subspace of a finite-dimensional vector space $V$. Then $W$ is finite-dimensional, and $dim(W) \leq dim(V)$. Furthermore, if $dim(W)=dim(V)$, then $W=V$. $\underline{Proof}$ Write $dim\,V=n \in \mathbb{N}_{0}$. If $W=\{ 0 \}$, then $W$ is finite-dimensional & $dim\,W=0 \leq$. Now, assume $W \neq \{0 \}$. Then $V \neq \{ 0 \}$ & $n \geq 1$. We can choose non-zero $x..

$\underline{Def}$ 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 $dim\,V=\infty$. $\underline{Ex}$ $(1)$ $\mathb..

$\underline{Thm}$ (Replacement Theorem) Let $G, L$ be subset of a vector space $V$ such that $\vert G \vert =n \in \mathbb{N}_{0}, span(G)=V, \vert L \vert =m \in \mathbb{N}_{0}$ and $L$ is linear independent, where $\mathbb{N}_{0}=\mathbb{N} \cup \{ 0 \}$. Then $n \geq m$, and $\exists H \subset G$ with $\vert H \vert=n-m$ such that $span(H \cup L) =V$. $\underline{Proof}$ We prove by induction..

$\underline{Def}$ 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_..

$\underline{Thm}$ 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 $S_{1}$. $\underline{Rmk}$ 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' \subse..

$\underline{Def}$ $(Linear\;Dependence \; \& \;Independence)$ A subset $S$ of a vector space $V$ is linear dependent if there are finitely many distince $u_{1}, \cdots, u_{n} \in S$ and scalars $a_{1}, \cdots, a_{n}$, not all zero, such that $a_{1}u_{1}+ \cdots + a_{n}u_{n}=0$. A subset $S$ of a vector space $V$ is linear independent if it is not linear dependent. $\underline{Facts}$ $(1)$ $\phi..

$\underline{Def}$ $(Span)$ 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 $span(\phi)={0}$ $\underline{Thm}$ 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 ..