Real n-Space as a Normed Linear Space

Rn AS A NORMED VECTOR SPACE PROF. SEBASTIAN VATTAMATTAM 1. Real Vector Space Denition 1.1. Real Vector Space: Let X be a non-empty set. (1) On X there is dened a binary operation (x; y) ! x + y, called addition such that (a) For all x; y 2 X x + y = y + x (Commutativity) (b) For all x; y; z 2 X; (x + y) + z = x + (y + z); (Associativity) (c) There exists 0 2 X such that x + 0 = 0 + x = x; 8x 2 X (Existence of null element) (d) For every x 2 X; 9 x 2 X such that x+(x) = (x)+x = 0 (Existence of inverse element) (2) There is dened a function (; x) ! x from RX ! X, called scalar multiplication such that (a) For ; 2 R; x 2 X (x) = ()x; (b) For ; 2 R; x; y 2 X (x + y) = x + y; (+ )x = x + x; : (c) For all x 2 X 1x = x Then X with addition and scalar multiplication is called a vector space over R or a real vector space or a linear space. 12 PROF. SEBASTIAN VATTAMATTAM Note (X; +) is an Abelian Group. Problem 1.2. If X is a vector space and for x; y 2 X; xy := x + (y) then show that 0x = 0 (1.1) (1)x = x (1.2) x (y + z) = (x y) z (1.3) (x y) = x y (1.4) Solution (1) 0x + x = (0 + 1)x = 1x = x = 0 + x ) 0x = 0 (2) x + (1)x = 1x + (1)x = (1 + (1))x = 0x = 0 ) x = (1)x (3) x (y + z) = x + ((y + z)) = x + ((1)(y + z)) x + ((1)y + (1)z) = (x + (1)y) + (1)z = (x y) zREAL N-SPACE 3 (4) (x y) = (x + (1)y) = x + ((1)y) = x + ((1))y = x + ((1))y = x + (1)(y) = x y 2. Normed Vector Space Denition 2.1. Let X be a real vector space. A function k:k : X ! R is called a norm on X if for x; y 2 X; 2 R kxk 0 kxk = 0 , x = 0 kx + yk kxk + kyk kxk = jjkxk Denition 2.2. Let X be a non-empty set. A metric on X is a function d: X X ! R such that, for x; y; z 2 X; (1) d(x; y) 0 (2) d(x; y) = 0 , x = y (3) d(x; y) = d(y; x)(symmetry) (4) d(x; y) d(x; z) + d(z; y)(triangle inequality) (X; d) is called a metric space. 3. Rn as a Vector Space Denition 3.1. For n 2 N Rn := f(x1; x2; ; xn)T : xk 2 R; k = 1; 2; :::; ng4 PROF. SEBASTIAN VATTAMATTAM where (x1; x2; ; xn)T = 0BB@ x1 x2 ::: xn 1CCAx is called a vector in Rn and xk is called its kth coordinate. Properties (1) If x = 0BB@ x1 x2 ::: xn 1CCA; y = 0BB@ y1 y2 ::: yn 1CCAthen x = y , xk = yk; k = 1; 2; :::; n (2) If x = 0BB@ x1 x2 ::: xn 1CCA; y = 0BB@ y1 y2 ::: yn 1CCAthen x + y = 0BB@ x1 + y1 x2 + y2 ::: xn + yn 1CCA (3) If 2 R; x = 0BB@ x1 x2 ::: xn 1CCAthen x = 0BB@ x1 x2 ::: xn 1CCAREAL N-SPACE 5 (4) The null vector 0 = 0BBBBBB@ 00:::0 1CCCCCCA(5) Opposite vector If x = 0BB@ x1 x2 ::: xn 1CCAthen x = 0BB@ x1 x2 ::: xn 1CCA(6) Standard Basic Vectors e1 = 0BBBB@ 10::0 1CCCCA ; e2 = 0BBBB@ 01::0 1CCCCA ; ; en = 0BBBB@ 00::1 1CCCCA(7) Linear Combination Every x 2 Rn is a linear combination of the basic vectors. x = (x1; x2; :::; xn)T = n Xk=1 xkek6 PROF. SEBASTIAN VATTAMATTAM Problem 3.2. If e1 = 0BBBB@ 10::0 1CCCCA ; e2 = 0BBBB@ 01::0 1CCCCA ; ; en = 0BBBB@ 00::1 1CCCCAand x =t (x1; x2; ; xn) then show that (1) x = n Xk=1 xkek (2) If k 2 R; k = 1; 2; :::; n and x = Pnk=1 kek then show that k = xk; k = 1; 2; :::; n Solution (1) x1e1+x2e2++xnen = x10BBBB@ 10::0 1CCCCA +x20BBBB@ 01::0 1CCCCA +xn0BBBB@ 00::1 1CCCCA = 0BBBB@ x1 0::0 1CCCCA + 0BBBB@ 0 x2 ::0 1CCCCA + 0BBBB@ 00:: xn 1CCCCA = 0BB@ x1 x2 ::: xn 1CCA= xREAL N-SPACE 7 (2) x = n Xk=1 xkek and x = n Xk=1 kek then n Xk=1 xkek = n Xk=1 kek ) (x1; x2; ; xn)t = (1; 2; ; n)t xk = k; k = 1; 2; ; n Theorem 3.3. If 2 R; x 2 R2 then (1) if jj > 0 the vector x is stretched by jj in x (2) if jj < 0 the vector x is Also (1) If > 0, multiplication by does not change the direction of x. (2) If < 0, multiplication by reverses the direction of x. Problem 3.4. If a; b 2 R2 and Gab = fa + t(b a)jt 2 Rg then show that Gab = fxy 2 R2jAx + By + C = 0 where A;B;C 2 Rg Interpret geometrically. Solution Since a; b 2 R2; t 2 R; a + t(b a) 2 R2 Let a = a1 a2 ; b = b1 b2 8 PROF. SEBASTIAN VATTAMATTAM Thena + t(b a) = a1 + t(b1 a1) a2 + t(b2 a2) = xy ; say Then x = a1 + t(b1 a1); y = a2 + t(b2 a2) ) x a1 b1 a1 = y a2 b2 a2 ) (b2 a2)(x a1) = (b1 a1)(y a2) ) (b2 a2)x (b2 a2)a1 = (b1 a1)y (b1 a1)a2) ) (b2 a2)x + (a1 b1)y = (b2 a2)a1 (b1 a1)a2) ) (b2 a2)x + (a1 b1)y = b2a1 b1a2 This is of the form Ax+By+C = 0where A = b2a2;B = a1b1;C = b2a1b1a2 2 R Hence the conclusion. Geometrical Interpretation We know that the linear equation Ax + By + C = 0 rep- resents a line in the Cartesian plane. Since Gab is the set of all the points satisfying this equation,Gab represents the same line. When t = 0; a + t(b a) = a 2 Gab and when t = 1; a + t(b a) = b 2 Gab so, Gab represents the line passing through the points a and b: TheEnd For online classes in Mathematics, please contact vattamattam@gmail.com