1 REAL COORDINATE SPACES

1 Real Coordinate Spaces

1 1 n 1.1 Vector Spaces R 1 REAL COORDINATE SPACES

1.1 Vector Spaces Rn

• 벡터(위치)와 벡터의 일차 함수(변화)를 연구하는 분야 = 선형대수학

• R: set of real numbers • Operation: +, · (−는 덧셈의 역, ÷는 곱셈의 역)

• Axioms of (R, +, ·): for a, b, c ∈ R 1. Addition is closed: a + b ∈ R 2. Addition is associative: a + (b + c) = (a + b) + c

3. There is an additive identity: 0 ∈ R, a + 0 = 0 + a = a 4. Every element has an additive inverse: −a ∈ R, a + (−a) = 0 5. Addition is commutative: a + b = b + a

6. Multiplication is closed: a · b ∈ R 7. Multiplication is associative: a · (b · c) = (a · b) · c

2 2 n 1.1 Vector Spaces R 1 REAL COORDINATE SPACES

8. There is a multiplicative identity: 1 ∈ R, a · 1 = a and 1 · a = a 9. Every nonzero element has a multiplicative inverse: a · b = 1 and b · a = 1 10. Multiplication is commutative: a · b = b · a 11. (Connection of +, ·) 곱셈과 덧셈을 연결시켜주는 중요한 공리 a · (b + c) = a · b + a · c

• Vector Spaces over R n R = {(u1, u2, ··· , un) | ui ∈ R},(n: 자연수)

Definition 1.1.1.

For u = (u1, u2, ··· , un), v = (v1, v2, ··· , vn),

define equality u = v if u1 = v1, ··· , un = vn

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Definition 1.1.2.

define an addition: u + v = (u1 + v1, u2 + v2, ··· , un + vn)

define a scalar product: a · u = (a · u1, a · u2, ··· , a · un) for a ∈ R

Then (Rn, +, ·) has 10 properties; 1. Addition is closed, u + v ∈ Rn idea of proof. u + v = (u1, u2, ··· , un) + (v1, v2, ··· , vn) (by replacement)

= (u1 + v1, u2 + v2, ··· , un + vn) (by definition of +)

Since ui, vi ∈ R, ui + vi ∈ R (by axiom 1) n (u1 + v1, u2 + v2, ··· , un + vn) ∈ R ⇒ u + v ∈ Rn (by definition of Rn)

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2. Addition is associative: u + (v + w) = (u + v) + w idea of proof. Let w = (w1, w2, ··· , wn), (wi ∈ R) u + (v + w)

= (u1, u2, ··· , un) + ((v1, v2, ··· , vn) + (w1, w2, ··· , wn))(by replacement)

= (u1, u2, ··· , un) + (v1 + w1, v2 + w2, ··· , vn + wn) (by definition of +)

= (u1 + (v1 + w1), u2 + (v2 + w2), ··· , un + (vn + wn)) (by definition of +)

= ((u1 + v1) + w1, (u2 + v2) + w2, ··· , (un + vn) + wn) (by axiom 2)

= (u1 + v1, u2 + v2, ··· , un + vn) + (w1, w2, ··· , wn) (by definition of +)

= ((u1, ··· , un) + (v1, ··· , vn)) + (w1, ··· , wn) (by definition of +) = (u + v) + w (by replacement)

3. Identity element 0: u + 0 = 0 + u = u

5 5 n 1.1 Vector Spaces R 1 REAL COORDINATE SPACES idea of proof. (i). Let 0 := (0, ··· , 0) (0 ∈ R) then 0 ∈ Rn by definition of Rn

(ii) u + 0 = (u1, ··· , un) + (0, ··· , 0) (by replacement)

= (u1 + 0, ··· , un + 0) (by definition of +)

Since ui + 0 = ui by Axiom 3 of R

= (u1, ··· , un) = u 같은 방식으로, 0 + u = u

Call 0: additive identity(덧셈의 항등원)

4. If v ∈ Rn, there is w ∈ Rn such that v + w = w + v = 0

6 6 n 1.1 Vector Spaces R 1 REAL COORDINATE SPACES

n idea of proof. Let v = (v1, ··· , vn) ∈ R

define w := (−v1, ··· , −vn) n Since vi ∈ R, −vi ∈ R, hence w ∈ R

v + w = (v1, ··· , vn) + (−v1, ··· , −vn)

= (v1 + (−v1), ··· , vn + (−vn)) - by definition of + = (0, ··· , 0) - by Axiom 4 = 0

∴ v + w = 0 같은 방법으로 w + v = 0 if u + w = w + u = 0, write w := −u i.e.

−u = (−u1, ··· , −un), call an additive inverse

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5,6,7,8 - 숙제

9. distributive law: (a + b)v = av + bv(a, b ∈ R) idea of proof.

(a + b)v = (a + b)(v1, ··· , vn)

= ((a + b)v1, ··· , (a + b)vn) - by definition of ·

= (av1 + bv1, ··· , avn + bvn) - by Axiom of distribute law

= a(v1, ··· , vn) + b(v1, ··· , vn) - by definition of · = av + bv - by replacement ⇒ (a + b)v = av + bv

Exercise 1.1.3. (a) The vector“0” is unique ; u + 0 = 0 + u = u

8 8 n 1.1 Vector Spaces R 1 REAL COORDINATE SPACES idea of proof. Suppose that there is another 0¯ ∈ R such that(s.t.) u + 0¯ = 0¯ + u = u Then u + 0 = u = u + 0¯ - by assumption ⇒ u + 0 = u + 0¯ - by definition of ”=” ⇒ u + 0 + (−u) = u + 0¯ + (−u) - by proposition 4 ⇒ (−u) + (u + 0) = (−u) + (u + 0)¯ by proposition 5 ⇒ (−u + u) + 0 = (−u + u) + 0¯ by proposition 2 ⇒ 0 + 0 = 0 + 0¯ by proposition 4 ⇒ 0 = 0¯ by proposition 3 Hence “0” is unique

(b)c · u = 0 ⇐⇒ either c = 0 or u = 0 (u · c : no such form!)

9 9 n 1.1 Vector Spaces R 1 REAL COORDINATE SPACES idea of proof. (⇒) If c = 0, then 0 · u = (0 + 0) · u = 0 · u + 0 · u - by proposition 10 ⇒ 0 · u + (−0 · u) = (0 · u + 0 · u) + (−0 · u) - by proposition ⇒ 0 = 0 · u + (0 · u + (−0 · u)) = 0 · u - by proposition ⇒ 0 · u = 0

If c 6= 0, then c · u = 0

⇒ c−1 · (c · u) = c−1 · 0 (for c 6= 0, c−1 ∈ R) ⇒ (c−1 · c) · u = c−1 · 0 ⇒ 1 · u = (c−1 · 0, ··· , c−1 · 0) ⇒ u = 0 - by proposition 10

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(⇐) Assume either c = 0 or u = 0 If c = 0, then 0 · u = 0 If u = 0, then c · 0 = c(0, ··· , 0) = (c · 0, ··· , c · 0) = (0, ··· , 0) = 0

∴ c · u = 0 Another Proof: c · u = c(u1, ··· , un) = (cu1, ··· , cun) = (0, ··· , 0)

⇔ c · ui = 0 for all i

⇔ c = 0 or ui = 0 for all i ⇔ c = 0 or u = 0

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(c) The vector −u ∈ R which u + (−u) = 0 is unique idea of proof. Assume w ∈ R such that u + w = 0 u + (−u) = 0 = u + w ⇒ u + (−u) = u + w ⇒ −u + (u + (−u)) = −u + (u + w) ⇒ (−u + u) + (−u) = (−u + u) + w ⇒ 0 + (−u) = 0 + w ⇒ −u = w

∴ such −u is unique

(d) −u = (−1) · u − u : additive inverse(u + (−u) = 0)

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(−1) · u : scalar product (−1) ∈ R, u ∈ Rn Show (−1) · u is an additive inverse idea of proof. u + (−1) · u = 1 · u + (−1) · u - by proposition 10 = (1 + (−1)) · u = 0 · u - by Axiom 3 = 0 - by Exercise(b)

∴ (−1) · u = −u ( by Exercise(c))

(e) The vector u − v is by definition the vector w s.t. v + w = u −v; v + (−v) = 0 (−1) · v; −v = (−1) · v u − v? Prove that u − v = u + (−v)!

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idea of proof. by definition w := u − v v + w = u ⇒ v + w + (−1)v = u + (−1)v ⇒ v + (−1) · v + w = u + (−1)v ⇒ v + (−v) + w = u + (−1)v ⇒ w = u + (−v) ⇒ u − v = u + (−1) · v

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