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$\underline{Def}$ Let $\beta=\{u_{1},\cdots,u_{n}\}$ be an ordered basis for an n-dimensional vector space $V$. $\forall x \in V$, let $a_{1},\cdots,a_{n} \in \mathbb{F}$ be the unique scalars such that $\displaystyle x=\sum_{i=1}^{n} a_{i}u_{i}$; then we define the coordinate vector of $x$ by $\beta$ (relative to $\beta$), denoted by $[x]_{\beta}$, as $$\displaystyle [x]_{\beta}=\begin{bmatrix}..

$\underline{Def}$ Let $V$ be a finite dimensional vector space. An ordered basis for $V$ is a finite sequence of linear independent vectors in $V$ that spans $V$. $\underline{Ex}$ In $\mathbb{F}^{3},\, \{ e_{1},e_{2},e_{3} \}$ is an ordered basis for $\mathbb{F}^{3}$. Also, $\{e_{2},e_{1},e_{3} \}$ is another ordered basis for $\mathbb{F}^{3}$. $\underline{Rmk}$ For a given basis $\beta=\{v_{1},..

$\underline{Thm}$ Let $V,W$ be finite-dimensional vector space of equal dimension. Let $T \colon V \rightarrow W$ be linear. Then TFAE: $(a)$ $T$ is 1-1. $(b)$ $T$ is onto. $(c)$ $rank(T)=dim(V)$ $\underline{Proof}$ Write $n=dim(V)=dim(W)$. $(a) \Rightarrow (b)$ Since $T$ is 1-1, by thm, $nullity(T)=0$. By Dimension Theorem, $n=dim(V)=nullity(T)+rank(T)=0+rank(T)$,and so $dim(R(T))=n$. Thus, $R(..

$\underline{Thm}$Let $T \colon V \rightarrow W$ be linear. Then $T$ is 1-1 $\Leftrightarrow N(T)=\{ 0 \}$. $\underline{Proof}$$(\Rightarrow)$Since $T(0)=0$ & $T$ is 1-1, $N(T)=\{0\}$.$(\Leftarrow)$Let $u,v \in V$, and suppose $N(T)=\{ 0 \}$.Then $T(u-v)=0, u-v \in N(T)=\{ 0\}$.Thus, $u=v$, Hence $T$ is 1-1.

$\underline{Thm}$ (Dimension Theorem) Let $T \colon V \rightarrow W$ be linear. If $dim(V) < \infty$, then $dim(V)=nullity(T)+rank(T)$. $\underline{Proof}$ Since $N(T)$ is a subspace of $V$, $N(T)$ is finite-dimensional & $dim(N(T))=nullity(T)=k \leq n$. If $k=n$, then $N(T)=V$, and so $R(T)=\{0\}$; thus $dim(V)=n=n+0=nullity(T)+rank(T)$. Suppose $k

$\underline{Def}$ Let $V,W$ be vector spaces, and let $T \colon V \rightarrow W$ be linear. We define the null space $N(T)$, the range $R(T)$ of $T$. $$ N(T)=\{ x \in V \colon Tx=0 \} $$ $$ R(T)=\{ Tx \in W \colon x \in V \} $$ Null space and range are called kernel and image respectively. $\underline{Thm}$ Let $V, W$ & $T$ be an in the previous definition. Then $N(T), R(T)$ are subspaces of $V,..

$\underline{Def}$ (Linear Transformation) Let $V,W$ be vector spave over $\mathcal{F}$. A function $T \colon V \rightarrow W$ is called a linear map (transformation) if $\forall x,y \in V,\, \forall c \in \mathcal{F}$, $(a) T(x+y)=T(x)+T(y)$, $(b) T(cx)=cT(x)$. $\underline{Rmk}$ $(1)$ If $T \colon V \rightarrow W$ is linear, then $T(0)=0$. $\because T(0)=T(0+0)=T(0)+T(0) \Rightarrow T(0)=0$ $(2)..

$\underline{Thm}$ (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..