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 • ¡0(위X)@ ¡0X |( h수(변T)| 연lX는 분| = 형대수Y • R: set of real numbers • Operation: +; · (−는 gHX 역, ÷는 ñHX 역) • Axioms of (R; +; ·): for a; b; c 2 R 1. Addition is closed: a + b 2 R 2. Addition is associative: a + (b + c) = (a + b) + c 3. There is an additive identity: 0 2 R; a + 0 = 0 + a = a 4. Every element has an additive inverse: −a 2 R; a + (−a) = 0 5. Addition is commutative: a + b = b + a 6. Multiplication is closed: a · b 2 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 2 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 +; ·) ñH과 gHD 연결시켜주는 중요\ õ¬ a · (b + c) = a · b + a · c • Vector Spaces over R n R = f(u1; u2; ··· ; un) j ui 2 Rg,(n: 자연수) Definition 1.1.1. For u = (u1; u2; ··· ; un); v = (v1; v2; ··· ; vn), define equality u = v if u1 = v1; ··· ; un = vn 3 3 n 1.1 Vector Spaces R 1 REAL COORDINATE SPACES 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 2 R Then (Rn; +; ·) has 10 properties; 1. Addition is closed, u + v 2 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 2 R; ui + vi 2 R (by axiom 1) n (u1 + v1; u2 + v2; ··· ; un + vn) 2 R ) u + v 2 Rn (by definition of Rn) 4 4 n 1.1 Vector Spaces R 1 REAL COORDINATE SPACES 2. Addition is associative: u + (v + w) = (u + v) + w idea of proof. Let w = (w1; w2; ··· ; wn); (wi 2 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 2 R) then 0 2 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 같@ )식<\, 0 + u = u Call 0: additive identity(gHX mñÐ) 4. If v 2 Rn, there is w 2 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) 2 R define w := (−v1; ··· ; −vn) n Since vi 2 R; −vi 2 R; hence w 2 R v + w = (v1; ··· ; vn) + (−v1; ··· ; −vn) = (v1 + (−v1); ··· ; vn + (−vn)) - by definition of + = (0; ··· ; 0) - by Axiom 4 = 0 ) v + w = 0 같@ 방법<\ w + v = 0 if u + w = w + u = 0, write w := −u i.e. −u = (−u1; ··· ; −un), call an additive inverse 7 7 n 1.1 Vector Spaces R 1 REAL COORDINATE SPACES 5,6,7,8 - 숙제 9. distributive law: (a + b)v = av + bv(a; b 2 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¯ 2 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 2 R) ) (c−1 · c) · u = c−1 · 0 ) 1 · u = (c−1 · 0; ··· ; c−1 · 0) ) u = 0 - by proposition 10 10 10 n 1.1 Vector Spaces R 1 REAL COORDINATE SPACES (() 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 11 11 n 1.1 Vector Spaces R 1 REAL COORDINATE SPACES (c) The vector −u 2 R which u + (−u) = 0 is unique idea of proof. Assume w 2 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) 12 12 n 1.1 Vector Spaces R 1 REAL COORDINATE SPACES (−1) · u : scalar product (−1) 2 R; u 2 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)! 13 13 n 1.1 Vector Spaces R 1 REAL COORDINATE SPACES 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 14 14.