1. 벡터공간


1.2 벡터공간

정의:
$F$에 대한 벡터공간(선형공간) $\mathsf{V}$
($F$는 덧셈과 스칼라곱, 두 연산에 대해 닫힌 집합)

s.t. $^{\forall} x, y \in \mathsf{V}, x + y \in \mathsf{V}$,
$^{\forall} a \in \mathsf{F}, ^{\forall} x \in \mathsf{V}, ax \in \mathsf{V}$ (닫혀있다)

  • VS1) $^{\forall} x, y \in \mathsf{V}, \quad x + y = y + x$ (덧셈의 교환법칙)
  • VS2) $^{\forall} x, y, z \in \mathsf{V}, \quad (x + y) + z = x + (y + z)$ (덧셈의 결합법칙)
  • VS3) $\exists 0 \in \mathsf{V} \quad s.t. \quad x + 0 = x, ^{\forall} x \in \mathsf{V}$ (덧셈의 항등원)
  • VS4) $^{\forall} x \in \mathsf{V}, \quad \exists y \in \mathsf{V} \quad s.t. \quad x + y = 0$ (덧셈의 역원)
  • VS5) $^{\forall} x \in \mathsf{V}, \quad 1 \cdot x = x \quad (\rm{where}\ 1 \in \mathsf{F})$ (곱셈의 항등원)
  • VS6) $^{\forall} a, b \in F, \quad ^{^{\forall}}x \in \mathsf{V},\ (ab)x = a(bx)$ (곱셈의 결합법칙)
  • VS7) $^{\forall} a \in \mathsf{F}, \quad ^{^{\forall}}x, y \in \mathsf{V},\ a(x + y) = ax + by$
  • VS8) $^{\forall} a, b \in \mathsf{F}, \quad ^{^{\forall}} x \in \mathsf{V},\ (a + b)x = ax + bx$

이상의 성질을 만족하는 공간이 벡터공간(vector space), 벡터공간에 속한 원소들이 벡터
수학에서 일반화된 벡터공간에서는 물리학에서 벡터를 다룰 때와 달리 방향 개념이 전제되지 않는다.

정리 1.1 (벡터 덧셈의 소거법칙)

$F$에 대한 벡터공간 $\mathsf{V} \ni x, y, z$ s.t. $x + z = y + z \\ \implies x = y$

증명:
$z \in \mathsf{V} \implies \exists v \in \mathsf{V} \quad s.t. \quad z + y = 0$
$\therefore\ x$$= x + 0 \qquad \rm{(VS3)} \\ = x + (z + v) \\ = (x + z) + v \qquad \rm{(VS2)} \\ = (y + z) + v \\ = y + (z + v) \qquad \rm{(VS2)} \\ = y + 0 \\ = y$

정리 1.2

벡터공간 $\mathsf{V}$

a) $0\ x = \vec{0} \quad ^{\forall} x \in \mathsf{V}$
$\because\ 0 \cdot x + 0 \cdot x$ $= (0 + 0) x \qquad \rm{(VS8)} \\ = 0\ x \\ = 0\ x + 0 \qquad \rm{(VS3)} \\ = 0 + 0\ x \qquad \rm{(VS1)}$
$\implies 0 \cdot x = 0 \qquad \rm{(by\ 정리\ 1.1)}$

b) $^{\forall} a \in \mathsf{F}, x \in \mathsf{V} \\ (-a)x = -(ax) = a(-x)$
$\rm{consider}$ $-(ax)\ \rm{is\ the\ unique\ element\ of}\ \mathsf{F} \\ s.t.\ ax + [-(ax)] = 0$
$\begin{matrix} \rm{But},\ &ax + (-ax)& &=& &\left\{a + (-a)\right\} x& \qquad \rm{(VS8)} \\ && &=& &0 \cdot x& \\ && &=& &0& \qquad \rm{(Thm 1.2)} \end{matrix}$
$\therefore (-a)x$ 역시 $ax$의 역원
덧셈의 역원은 유일하므로 $(-a)x = -(ax)$

$\begin{matrix} \rm{In\ particular}, &(-1)x& &=& &-(1 \cdot x)& \\ && &=& &-x& \qquad \rm{(VS5)} \end{matrix}$
$\begin{matrix} \rm{And}, &a(-x)& &=& &a \left\{ (-1) x \right\}& \\ && &=& &\left\{ a\ (-1) \right\}\ x& \qquad \rm{(VS6)} \\ && &=& &(-a)\ x& \end{matrix}$

c) $a \cdot 0 = 0 \quad ꟻ^{\forall} a \in \mathsf{F}$
$\begin{matrix} \because &a\ 0 + a\ 0& &=& &a\ (0 + 0)& \qquad \rm{(VS7)} \\ && &=& &a\ (0)& \qquad \rm{(VS3)} \\ && &=& &a\ (0) + 0& \qquad \rm{(VS3)} \\ && &=& &0 + a\ (0)& \qquad \rm{(VS1)} \\ \therefore &a \cdot 0 = 0& && && \rm{(Thm\ 1.1)} \end{matrix}$

1.3 부분공간

정의:
$\mathsf{W}가\ \mathsf{V}의\ 부분공간 \\ \iff \mathsf{W} \subset \mathsf{V}\ 그리고\ \mathsf{W}가\ 벡터공간$

정리 1.3

$\begin{matrix} \mathsf{V}가\ 벡터공간 \\ \mathsf{W} \subset \mathsf{V}\ 가\ 부분공간 \end{matrix} \iff \begin{cases} a)\ &0 \in \mathsf{W}& \\ b) &x + y \in \mathsf{W},& \quad ꟻ^{\forall}\ x, y \in \mathsf{W} \\ c) &c \cdot x \in \mathsf{W},& \quad ꟻ^{\forall}\ c \in \mathsf{F}, x \in \mathsf{W} \end{cases}$

i) if $W$ a subspace of $V$, $W$ a vector space with addition and scalar pultiplation on $V$
$\exists \rm{zero\ vector}\ \vec{0} \in V$
$\vec{0} + x = x \quad ꟻ^{\forall} x \in W\ \rm{also}\ W \subset V\ \implies ꟻ^{\forall} x \in W,\ x \in V$
$0 + x = x, \vec{0} + x = 0 + x. \\ \therefore \vec{0} = 0$

ii) Check! $^{\forall} x \in W,\ -x \in W \\ \rm{by\ c)} \quad -1 \in F\ \implies (-1) \cdot x \in W \\ \rm{But}\ (-1) \cdot x = -x \\ \therefore -x \in W$

예제 3:
$M_{n \times n}(F) \\ D_{n \times n}(F) \subset M_{n \times n}(F) =$ the set of all diagonal matrix
a) $0_{n \times n} \in D_{n \times n}(F)$
b) $A, B \in D_{n \times n}(F) \\ (A + B)_{ij} = 0, \implies A + B \in D_{n \times n}(F)$
c) $ꟻ^{\forall} c \in F, (c\ A)_{ij} = 0 \\ \therefore c\ A \in D_{n \times n}(F) \\ \therefore D_{n \times n}(F)$ is subspace of $M_{n \times n}(F)$

정리 1.4

벡터공간 $V$의 부분공간의 교집합도 $V$의 부분공간

증명:
suppose $C = \left\{ W \right\} ^{n}_{n+1},$ $C$$V$의 부분공간들의 집합
Let $W = \bigcap^{n}_{i=1}\ W_i,$ claim that $W$ is subspace of $V$
 $ꟻ^{\forall}\ i, 0 \in W_{i}, \quad 0 \in \bigcap^{n}_{i=1}\ W_i = W$
 $ꟻ^{\forall}\ x, y \in W = \bigcap\ W_i$, $x, y \in W_{i} \quad ꟻ^{\forall}\ i = 1, \cdots, n \\ \implies x, y \in W_{i} \quad ꟻ^{\forall}\ i = 1, \cdots, n \\ \implies x, y \in \bigcap\ W_i = W$
 $ꟻ^{\forall}\ x \in W, c \in F$, $^{\forall}i \quad c\ x \in W_i \\ \implies c\ x \in \bigcap^{n}_{i=1}\ W_i = W$
 $\therefore\ W$ is subspace of $V$

1.4 선형결합

$V$: 벡터공간
$S(+ \phi) \subset V$
for $v \in V$, if $\exists\ u_1, \cdots, u_n$ and $a_1, \cdots, a_n \in F$
$v = a_1 u_1 + a_2 u_2 + \cdots + a_n u_n \in V$
$v$$S$의 벡터들의 선형결합(linear combination)이라고 한다

정의:
$S (+ \varnothing) \subset V$, the span of $S = \operatorname{span}(S)$ is the set of all linear combinations of the vectors in $S$
$S = \left\{ u_1, u_2, \cdots, u_n \right\} \implies$ $\operatorname{span}(S) = \left\{ a_1u_1 + a_2u_2 + \cdots + a_nu_n \right\} \quad ^{\forall}\ a_1, \cdots, a_n \in F \\ \rm{If}\ S = \varnothing, \operatorname{span}(\varnothing) = \left\{ 0 \right\}$

정리 1.5

  1. $^{\forall}\ \mathsf{S} \subset \mathsf{V}, \operatorname{span}(\mathsf{S})$$\mathsf{V}$의 부분공간
  2. $^{\forall}\ \mathsf{V}의\ 부분공간\ \mathsf{W}$, $\mathsf{S} \subset \mathsf{W} \implies \operatorname{span}(\mathsf{S}) \subset \mathsf{W}$

증명:
$\mathrm{If}\ S = \varnothing, \operatorname{span}(\varnothing) = \left\{ 0 \right\} \subset V \\ \mathrm {if}\ S \neq \varnothing, \mathrm{consider\ on\ element}\ z \in S \\ \qquad ①\ 0 = 0\ z \in \operatorname{span}(S) \\ \qquad ②\ \mathrm{Let}\ x, y \in \operatorname{span}(S) \\ \quad \qquad \exists\ \mathrm{vectors}\ u_1, \cdots u_n \in S\ \mathrm{and}\ a_1, \cdots, a_n \in F \\ \qquad \qquad x = u_1 a_1 + \cdots u_n a_n \\ \quad \qquad \exists\ \mathrm{vectors}\ v_1, \cdots v_n \in S\ \mathrm{and} b_1, \cdots, b_n \in F \\ \qquad \qquad y = v_1 b_1 + \cdots + v_n b_n \\ \quad \qquad \implies x + y = u_1 a_1 + \cdots + u_n a_n + v_1 b_1 + \cdots + v_n b_n \quad : \mathrm{a\ linear\ combination\ of}\ S \in \operatorname{span}(S) \\ \mathrm{for\ any\ scalar}\ c \in F, \\ \quad c\ x = (c a_1)u_1 + (c a_2) u_2 + \cdots + (c a_m) u_m \in \operatorname{span}(S) \\ \operatorname{span}(S) \mathrm{is\ subspace\ of}\ V$

정의:
$\operatorname{span}(\mathsf{S}) = \mathsf{V} \implies \mathsf{S}가\ \mathsf{V}를\ 생성한다$

예제 5:
 $S = \left\{ {\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}}, {\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}}, {\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}}, {\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}} \right\}$ generates $M_{2 \times 2} (\mathbb{R})$

i) $\operatorname{span}(S) = M_{2 \times 2} (\mathbb{R})$ trivial
ii) $M_{2 \times 2} (\mathbb{R}) \subset \operatorname{span}(S) \quad$ $\because)\ ꟻ^{\forall}\ {\begin{pmatrix} a & b \\ c & d \end{pmatrix}} \subset M_{2 \times 2} (\mathbb{R}), \quad \exists\ x, y, z, k \in \mathbb{R} \\ \begin{matrix} \mathrm{prove\ that}\ {\begin{pmatrix} a & b \\ c & d \end{pmatrix}} & = & x{\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}} + y{\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}} + z{\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}} + k{\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}} \\ & = & {\begin{pmatrix} x + y + z & x + y + k \\ x + z + k & y + z + k \end{pmatrix}} \end{matrix} \\ \implies \begin{cases} a = x + y + z \\ b = x + y + k \\ c = x + z + k \\ d = y + z + k \end{cases} \implies \begin{matrix} x = {1 \over 3}(a + b + c - 2d) \\ y = {1 \over 3}(a + b + d - 2c) \\ z = {1 \over 3}(a + c + d - 2b) \\ k = {1 \over 3} (b + c + d - 2a) \end{matrix} \\ \therefore\ M_{2 \times 2} (\mathbb{R}) \subset \operatorname{span}(S) \quad \mathrm{i.e.}\ M_{2 \times 2} (\mathbb{R}) = \operatorname{span}(S)$

1.5 선형독립과 선형종속

정의:
$S \subset \mathsf{V} \quad \mathrm{linearly\ dependent} \\ \iff \exists\ u_1, \cdots, u_n \in S,\ \mathrm{scalars}\ a_1, \cdots, a_n\ (\mathrm{not\ all\ zero}) \\ \qquad \mathrm{s.t.} \quad u_1 a_1 + \cdots + u_n a_n = 0$

if $S \subset \mathsf{V}$ is not linearly dependent, $\implies S \quad \mathrm{linearly\ independent} \\ ( u_1 a_1 + \cdots + u_n a_n = 0 \iff a_1 = \cdots = a_n = 0 )$

정리 1.6

$\mathsf{V}$ a vector space
$S_1 \subseteq S_2 \subseteq \mathsf{V} \quad \begin{cases} S_1 \quad \mathrm{linearly\ dependent}\ & \implies & S_2 \quad \mathrm{linearly\ dependent} \\ S_2 \quad \mathrm{linearly\ independent}\ & \implies & S_1 \quad \mathrm{linearly\ independent} \end{cases}$

$\ast \mathrm{for}\ S \subset \mathsf{V} \quad$ consider $\operatorname{span}(S)$
$\quad \mathrm{If}\ S\ \mathrm{the\ smallest\ generating\ set\ for}\ \operatorname{span}(S) \\ \qquad < \mathrm{No\ proper\ subset\ of}\ S\ \mathrm{generates\ the}\ \operatorname{span}(S) >$

$\quad \mathrm{If\ not},\ \exists w(\neq 0) \in S \quad \mathrm{s.t.} \quad$ $w = a_1 u_1 + a_2 u_2 + \cdots a_n u_n, \quad u_1, \cdots, u_n, a_1, \cdots, a_n \in F \\ (\because w \in \operatorname{span}(S) ) \\ 0 = -w + a_1 u_1 + a_2 u_2 + \cdots + a_n u_n \\ S\ \mathrm{is\ linearly\ dependent}$
$\qquad \qquad \qquad \qquad \Uparrow \\ \qquad \qquad \qquad \qquad 대우 \\ \qquad \qquad \qquad \qquad \Downarrow$
$\quad S\ \mathrm{is\ linearly\ independent}\ \implies S\ \mathrm{is\ the\ smallest\ generating\ set\ for}\ \operatorname{span}(S)$

정리 1.7

Let $S$ a linearly independent subset of $V$
for $v \in V$, $v \notin S$, $S\ \cup \left\{ v \right\}$ linearly dependent $\iff\ v \in \operatorname{span}(S)$

증명:
supp $S \cup \left\{ v \right\}$ linearly dependent, $\implies\ \exists\ u_1, \cdots, u_n\ \in\ S \cup \left\{ v \right\} \\ \mathrm{s.t.} \quad a_1 u_1 + \cdots + a_n u_n = 0$ for some non-zero scalars $a_1, \cdots, a_n$
Since $S$ is linearly independent, $u_i = v$ for some $i$
i.e. $a_1 v_1 + a_2 v_2 + cdots + a_n u_n = 0 \\ v = - (a_1 + a_2) u_2 - \cdots - (a_1 + a_n) u_n \\ \therefore\ v \in \operatorname{span}(S)$
$\iff$ $\mathrm{let}\ v \in \operatorname{span}(S) \quad \mathrm{then} \quad \exists\ v_1, \cdots, v_n\ \in S\ \mathrm{and\ scalars}\ a_1, \cdots, a_n \in F \\ \mathrm{s.t.} \quad v = a_1 v_1 + \cdots + a_n v_n \\ \mathrm{i.e.} \quad a_1 v_1 + \cdots a_n v_n + (-1) v = 0 \\ \mathrm{consider}\ -1 \neq 0, \mathrm{the\ set}\ \left\{ v_1, \cdots, v_n \right\}\ \mathrm{linearly\ independent} \\ \left\{ v_1, \cdots, v_n, v \right\} \subset\ S \cup \left\{ v \right\}$
$\therefore S \cup \left\{ v \right\}$ is linearly dependent

1.6 기저와 차원

정의:
$\begin{cases} \beta\ : \mathsf{V}의\ 선형독립인\ 부분집합 \\ V = \operatorname{span}(\mathsf{S}) \end{cases} \\ \implies \beta는\ \mathsf{V}의\ 기저(\mathrm{basis})$

정리 1.8

벡터공간 $V$에 대하여
$\beta = \left\{ u_1, \cdots, u_n \right\}\ \subset\ V$$V$의 기저 $\iff\ \forall\ v \in V,$ $\exists!\ 스칼라\ a_1, \cdots, a_n \\ \quad \mathrm{s.t.}\ v = a_1 u_1 + \cdots + a_n u_n$

증명:
Let $\beta$ a basis for $V$
If $v \in V$ i.e. $v \in \operatorname{span}(\beta) = V$
$\implies v = a_1 u_1 + \cdots + a_n u_n$ for $u_1, \cdots, u_n \in \beta \\ a_1, \cdots, a_n \in F$

for uniqueness, let $v = b_1 u_1 + \cdots + b_n u_n$ for $b_1, \cdots, b_n \in F$
$a_1 u_1 + a_2 u_2 + cdots + a_n u_n = b_1 u_1 + b_2 u_2 + cdots + b_n u_n \\ \implies (a_1 - b_1) u_1 + (a_2 - b_2) u_2 + cdots + (a_n - b_n) u_n = 0 \\ \implies a_1 = b_1, a_2 = b_2, \cdots, a_n = b_n$
$\iff$ $①\ \forall v \in V, \exists\ a_1, \cdots, a_n \in F\ \mathrm{s.t.}\ v = a_1 u_1 + \cdots + a_n u_n \in \operatorname{span}(\beta) \\ \therefore\ V \subset \operatorname{span}(\beta) \\ \mathrm{since}\ \operatorname{span}(\beta) \subset V, \operatorname{span}(\beta) = V \\ ②\ \mathrm{supp}\ x_1 v_1 + \cdots + x_n v_n = 0\ \mathrm{and\ the\ representation\ of}\ \vec{0}\ \mathrm{is\ also\ unique} \\ x_1 = x_2 = \cdots = x_n = 0$

  • $V \mathrm{vector\ space}, \beta = \left\{ a_1, \cdots, a_n \right\}\ \mathrm{a\ basis\ for}\ V \\ \forall v \in V \iff v = a_1 u_1 + \cdots + a_n u_n \iff \left( a_1, \cdots, a_n \right) \in F^n \\ n = \operatorname{number}(\beta)$

정리 1.9

$\begin{pmatrix} V\ \mathrm{a\ vector\ space}, & \mathrm{a\ finite\ set}\ S \subset V, \\ V = \operatorname{span}(S) \implies & \exists\ \mathrm{some\ subset}\ \beta\ \mathrm{of}\ S \\ & \mathrm{s.t.} \quad \beta \mathrm{is\ a\ basis\ for}\ V \end{pmatrix}$

증명:
If $S = \varnothing$ or $S = \left\{ 0 \right\} \implies$ $V = \left\{ 0 \right\}\ \mathrm{and}\ \varnothing \subset \left\{ 0 \right\} \\ \varnothing\ \mathrm{is\ a\ basis\ for}\ V \quad \mathrm{(linearly\ independent!)}$

If $S \neq \varnothing$ and $V = \operatorname{span}(S)$,
choose $u_1 \in S$. $\left\{ u_1 \right\}$ is linearly independent
let $\beta = \left\{ u_1, \cdots, u_n \right\}$ linearly independent
and $\beta \cup \left\{ u_{k+1} \right\}$ is linearly independent where $u \neq u_{i, \cdots, n} \\ u \in S$
claim that $\beta$ is a basis for $V$
$\beta$ linearly independent
$\operatorname{span}( \beta ) \subset \operatorname{span}( S ) = V \quad (\because \beta \subset S) \\ \mathrm{consider}\ v \in S \\ \mathrm{if}\ v \in \beta\ \mathrm{then}\ v \in \operatorname{span}( \beta ) \\ \mathrm{supp}\ v \in (S - \beta ), \beta \cup \left\{ v \right\}\ \mathrm{linearly\ independent} \\ \therefore\ v \in \operatorname{span}(\beta)\ \mathrm{i.e.}\ v = a_1 u_1 + \cdots + a_k u_k\ \mathrm{for\ some}\ a_1, \cdots, a_k \in F$

$S \subset \operatorname{span}(\beta)$
$\implies$ $\operatorname{span}(S) \subset \operatorname{span}(\beta) \\ \therefore \operatorname{span}(\beta) = \operatorname{span}(S) = V$

정리 1.10 (Replacement Theorem)

Let $V$ a vector space, $G$ a set $\subset V$, $V = \operatorname{span}(G)$ and $\operatorname{n}(G) = n$
and let $L$ a linearly independent subset of $V$ with $\operatorname{n}(L) = m$
Then, $m \le n$ and $\exists$ a subset $H \subset G \quad \mathrm{s.t.}\ \operatorname{n}(H) = n-m$
$\operatorname{span}( L \cup H ) = V$

증명:
By induction on $m$,
i ) $m = 0, L = \varnothing \\ \mathrm{taking}\ H = G - \varnothing\ \mathrm{then}\ L \cup H = \varnothing \cup G \\ \therefore \operatorname{span}(L \cup H ) = \operatorname{span}(G) = V$

ii ) supp the Thm is true for some integer $m > 0$

iii ) $\mathrm{consider}\ m+1 \\ \mathrm{Let}\ L = \left\{ v_1, \cdots, v_{m+1} \right\}\ \mathrm{a\ linearly\ independent\ subset} \\ \implies \left\{ v_1, \cdots, v_m \right\}\ \mathrm{linearly\ independent} \quad \mathrm{(by\ corollary\ of\ Thm\ 1.6)}$

By the hypothesis of the above stage,
$m \leq n$ and $\exists$ a subset $H = \left\{ u_1, \cdots, u_{n-m} \right\} \subset G$
$\mathrm{s.t.} \quad \left\{ v_1, \cdots, v_m, u_1, \cdots, u_{n-m} \right\}$ generates $V$
$v_{m+1} =$ $a_1 v_1 + a_2 v_2 + \cdots + a_m v_m + b_1 u_1 + \cdots + b_{n-m} u_{n-m} \\ \mathrm{for\ some}\ a_1, \cdots, a_m, b_1, \cdots, b_{n-m} \in F$
$n - m > 0 \quad ( n - m \ge 1) \quad \mathrm{i.e.}\ n \ge m+1$ $\because\ \left\{ v_1, \cdots, v_{m+1} \right\}\ \mathrm{linearly\ independent}$

Moreover, $b \mathrm{some}\ \neq 0 \quad (b_1 \neq 0) $ \\ \therefore u_1 = (- {b_1}^{-1} a_1 ) v_1 - \cdots - {b_1}^{-1} b_{n-m} u_{n-m} + {b_1}^{-1} v_{m+1}$

Let $\tilde{H} = \left\{ u_2, \cdots, u_{n-m} \right\} \subset G \\ \mathrm{then}\ u_1 \in \operatorname{span}(L \cup \tilde{H} ) \\ \left\{ v_1, \cdots, v_{m+1}, \cdots, u_1, \cdots, u_{n-m} \right\} \subseteq \operatorname{span}(L \cup \tilde{H} )$
$V = \operatorname{span}( \left\{ v_1, \cdots, v_{m+1}, u_1, \cdots, u_{n-m} \right\} \subset \operatorname{span}(L \cup \tilde{H} )$
But, $\operatorname{span}(L \cup \tilde{H} ) \subset V \\ V = \operatorname{span}(L \cup \tilde{H} )$

$\operatorname{n}(\tilde{H}) = (n-m)-1 = n - (m+1)$
the theorum is true for $m+1$

예제 6:
$S = \left\{ (2, -3, 5), (8, -12, 20), (1, 0, -2), (-a, 2, -1), (7, 2, 0) \right\}$

$\operatorname{span}(S) = \mathbb{R}^3$
$\mathrm{select}\ \left\{ (2, -3, 5) \right\} \quad \mathrm{linearly\ independent} \\ \mathrm{select}\ \left\{ (2, -3, 5), (8, -3, 20), (1, 0, -2) \right\} \quad \mathrm{linearly\ independent} \\ \mathrm{select}\ \left\{ (2, -3, 5), (1, 0, 2), (0, 2, -1) \right\} \quad \mathrm{linearly\ independent} \\ \mathrm{select}\ \left\{ (2, -3, 5), (1, 0, 2), (7, 2, 0) \right\} \quad \mathrm{linearly\ dependent} \\$

정리 1.11

$\begin{pmatrix} V\ \mathrm{a\ finite\ dimensional\ vector\ space} \\ W\ \mathrm{a\ subset\ of}\ V \\ \implies W\ \mathrm{finite\ dimensional}\, \operatorname{dim}(W) \le \operatorname{dim}(V)\ \\ \mathrm{If} \operatorname{dim}(V) = \operatorname{dim}(W) \implies V = W \end{pmatrix}$

증명:
Let $\operatorname{dim}(V) = n$
If $W = \left\{ 0 \right\}, \quad \operatorname{dim}(W) = 0 \le n$
Let $W \neq \left\{ 0 \right\}, \quad$ $\mathrm{choose}\ x ( \neq 0 ) \in W \\ \implies \left\{ x_1 \right\} : \mathrm{a\ linearly\ independent} \\ \mathrm{choose}\ x_2 ( \neq 0 ) \in W \\ \mathrm{s.t.}\ \left\{ x_1, x_2 \right\} \mathrm{a\ linearly\ independent} \\ \cdots \cdots \\ \mathrm{continue\ this\ process} \\ \left\{ x_1, \cdots, x_k \right\} \mathrm{a\ linearly\ independent} \\ \mathrm{and\ for\ any}\ x_{k+1} \in W, \\ \left\{ x_1, \cdots, x_k \right\} \cup \left\{ x_{k+1} \right\} \mathrm{linearly\ independent} \\ \mathrm{By\ replacement\ theorum,}\ k \le n\ \mathrm{and}\ x_{k+1} \in \operatorname{span}( \left\{ x_1, \cdots, x_k \right\} ) \\ \therefore W = \operatorname{span}( \left\{ x_1, \cdots, x_k \right\} ) \quad \mathrm{by\ Thm\ 1.7}$
$\therefore \left\{ x_1, \cdots, x_k \right\}\ \mathrm{a\ basis\ for}\ W \\ \therefore \operatorname{dim}(W) = k, \quad k \le n$
If $\operatorname{dim}(W) = n, \\ \beta = \mathrm{basis\ for}\ W \qquad \mathrm{(Try\ Proof!)}$
$\therefore V = W$

예제 19:
$M_{n \times n} (F)$
$W =$ the set of symmetric $n \times n$ matrixes
$\implies$ a basis for $W = \left\{ A^{ij} | 1 \le i \le j \le n \right\} \\ A^{ij} = \begin{cases} a_{ij} = a_{ji} = 1 \\ \mathrm{other\ elements}\ = 0 \end{cases}$
$\quad \operatorname{dim}(W) = {1 \over 2} n(n+1)$

라그랑주 내삽공식(Lagrange Interpolation Formula)
$c_0, c_1, \cdots, c_n$ distinct
Define $f_{i}(x)$ $= {{(x - c_0) \cdots (x - c_{i-1})(x - c_{i+1}) \cdots (x - c_n)} \over {(c_i - c_0) \cdots (c_i - c_{i-1})(c_i - c_{i+1}) \cdots (c_i - c_n)}} \in P_n (F) \\ = \prod_{k=0 \\ k \neq i}^n {{x - c_k} \over {c_i - c_k}}, \quad i = 1, \cdots, n$
$\implies \left\{ f_0, f_1, \cdots, f_n \right\}$ a basis for $P_n (F) \qquad \mathrm{(check!)}$

$\therefore$ $g \in P_n (F) \\ g = \sum_{i=o}^n b_i f_i \quad \mathrm{for\ some}\ b_0, \cdots, b_n$
$\ast$ $g(c_j) = \sum_{i=0}^n b_i f_i (c_j) = b_j \\ therefore g = \sum_{i=0}^n g( c_i ) f_i$

EX.
$(1, 8), (2, 5), (3, -4)$
the real polynomial $g$ of degree $\le 2$
$g(1) = 8, g(2) = 5, g(3) = -4$
$c_0 = -1, c_1 = 2, c_3 = 3$
$f_0 (x) = {{(x-2)(x-3)} \over {(1-2)(1-3)}} = \cdots = {1 \over 2} (x^2 - 5x + 6)$
$f_1 (x) =$
$f_2 (x) =$
$g(x)$ $= 8 f_0 (x) + 5 f_1 (x) + 4 f_2 (x) \\ = \cdots = -3x^2 + 6x +5$

1.7 최대선형독립부분공간