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$\underline{Thm}$ 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 ..

$\underline{Def}\;(Transpose)$ $\mathbb{F}$를 체(field)라 하고, $A \in \mathbb{M}_{n \times m} (\mathbb{F})$라 하자. $A$의 전치행렬(transpose matrix) $A^{t}$는 $A^{t} \in \mathbb{M}_{m \times n} (\mathbb{F})$이고, $(A^{t})_{ij}=A_{ji} \; \forall i,j$를 만족하는 행렬이다. $\underline{Prop}$ $\mathbb{F}$를 체라 하면, 다음이 성립한다. $(aA+bB)^{t}=aA^{t}+bB^{t} \; \forall A, B \in \mathbb{M}_{m \times n}, \; \forall a,b \in \mathbb{F}..

$\underline{Def} (Subspace)$ 벡터공간 $V$의 부분 집합 $W$가 벡터공간이 될 조건을 만족시키면 이를 $V$의 부분공간이라 한다. $\underline{Rmk}$ $V$가 벡터공간이면, ${0}$은 항상 $V$의 부분공간이 되며, 이를 zero subspace라 부른다. $\underline{Thm}$ $W$를 벡터공간 $V$의 부분집합이라 하자. $W$가 $V$의 부분공간이 되는 것은 다음 세 조건이 성립하는 것과 동치다. $(a)\; 0 \in W$ $(b)\; x+y \in W \; \forall x,y \in W$ $(c)\; cx \in W \; \forall x \in W, \forall c \in F$ $\underline{Proof}$ $(\Rightarrow)$ ..

$\underline{Thm}\;(Cancellation\;law)$ $V$를 벡터공간이라 하자. 만약, $x, y, z \in V$이고 $x+z=y+z$라면, $x=y$이다. $\underline{Proof}$ 벡터공간 정의에 의해 벡터 공간 내의 모든 원소는 덧셈에 대학 역원을 갖는다. $\exists -z \in V\;s.t.\;z+(-z)=0.$ 그러므로, $x=x+0=x+(z+(-z))=(x+z)-z=(y+z)-z=y$ $\underline{Coro.1}\;(Uniqueness\;of\;0)$ $V$를 벡터공간이라 하자. 그러면 임의의 $x \in V$에 대해 $x+0=x$를 만족하는 $0$은 오직 하나 존재한다. $\underline{Proof}$ 벡터공간의 정의에 의해 $0 \in V$. 만..