Rn AS A NORMED VECTOR SPACE PROF. SEBASTIAN VATTAMATTAM
1. Real Vector Space Definition 1.1. Real Vector Space: Let X be a non-empty set. (1) On X there is defined a binary operation (x, y) → x + y, called addition such that (a) For all x, y ∈ X x + y = y + x (Commutativity) (b) For all x, y, z ∈ X, (x + y) + z = x + (y + z), (Associativity) (c) There exists 0 ∈ X such that x + 0 = 0 + x = x, ∀x ∈ X (Existence of null element) (d) For every x ∈ X, ∃ − x ∈ X such that x + (−x) = (−x) + x = 0 (Existence of inverse element) (2) There is defined a function (α, x) → αx from R×X → X, called scalar multiplication such that (a) For α, β ∈ R, x ∈ X α(βx) = (αβ)x, (b) For α, β ∈ R, x, y ∈ X α(x + y) = αx + αy, (α + β)x = αx + βx, . (c) For all x ∈ 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. 1
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PROF. SEBASTIAN VATTAMATTAM
Note (X, +) is an Abelian Group. Problem 1.2. If X is a vector space and for x, y ∈ X, x − y := x + (−y) then show that (1.1) (1.2) (1.3) (1.4)
0x (−1)x x − (y + z) α(x − y)
= = = =
0 −x (x − y) − z αx − αy
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) − z
REAL N-SPACE
(4) α(x − y) = α(x + (−1)y) = αx + α((−1)y) = αx + (α(−1))y = αx + ((−1)α)y = αx + (−1)(αy) = αx − αy 2. Normed Vector Space Definition 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 ∈ X, α ∈ R kxk kxk kx + yk kαxk
≥ = ≤ =
0 0⇔x=0 kxk + kyk |α|kxk
Definition 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 ∈ 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 Definition 3.1. For n ∈ N Rn := {(x1 , x2 , · · · , xn )T : xk ∈ R, k = 1, 2, ..., n}
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PROF. SEBASTIAN VATTAMATTAM
x1 x2 where (x1 , x2 , · · · , xn )T = ... x is called a vector in xn Rn and xk is called its k th coordinate. Properties
x1 y1 x2 , y = y2 (1) If x = ... ... xn yn
then
x = y ⇔ xk = yk , k = 1, 2, ..., n x1 y1 x2 , y = y2 then (2) If x = ... ... xn yn
x1 + y1 x2 + y2 x+y = ... x n + yn x1 x2 (3) If α ∈ R, x = ... then xn
αx1 αx2 αx = ... αxn
REAL N-SPACE
(4) The null vector 0 =
0 0 . . . 0
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(5) Opposite vector x1 x2 If x = ... then xn
−x1 −x2 −x = ... −xn (6) Standard Basic Vectors
e1 =
1 0 . . 0
, e2 =
0 1 . . 0
, · · · , en =
0 0 . . 1
(7) Linear Combination Every x ∈ Rn is a linear combination of the basic vectors. T
x = (x1 , x2 , ..., xn ) =
n X k=1
xk ek
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PROF. SEBASTIAN VATTAMATTAM
Problem 3.2. If 0 1 1 0 , e2 = . . e1 = . . 0 0
, · · · , en =
0 0 . . 1
and x =t (x1 , x2 , · · · , xn ) then show that (1) n X x= xk ek k=1
Pn
(2) If αk ∈ R, k = 1, 2, ..., n and x = show that αk = xk , k = 1, 2, ..., n
k=1 αk ek
then
Solution (1) x1 e1 +x2 e2 +· · ·+xn en = x1 =
x1 0 . . 0
+
1 0 . . 0
0 x2 . . 0
0 1 +x2 . . 0 + ···
x1 x2 = ... = x xn
+· · · xn
0 0 . . xn
0 0 . . 1
REAL N-SPACE
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(2) x=
n X
xk ek
k=1
and x=
n X
αk ek
k=1
then
n X
xk ek =
k=1
n X
αk e k
k=1 t
⇒ (x1 , x2 , · · · , xn ) = (α1 , α2 , · · · , αn )t xk = αk , k = 1, 2, · · · , n Theorem 3.3. If α ∈ R, x ∈ R2 then (1) if |α| > 0 the vector x is stretched by |α| in αx (2) if |α| < 0 the vector x is Also (1) If α > 0, multiplication by α does not change the direction of x. (2) If α < 0, multiplication by α reverses the direction of x. Problem 3.4. If a, b ∈ R2 and Gab = {a + t(b − a)|t ∈ R} then show that x Gab = { ∈ R2 |Ax + By + C = 0 where A, B, C ∈ R} y Interpret geometrically. Solution Since a, b ∈ R2 , t ∈ R, a + t(b − a) ∈ R2 Let a1 b1 a= ,b = a2 b2
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PROF. SEBASTIAN VATTAMATTAM
Then a + t(b − a) =
a1 + t(b1 − a1 ) a2 + t(b2 − a2 )
=
x y
, say
Then x = a1 + t(b1 − a1 ), y = a2 + t(b2 − a2 ) y − a2 x − a1 = ⇒ b1 − a1 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 = b2 a1 − b1 a2 This is of the form Ax+By+C = 0where A = b2 −a2 , B = a1 −b1 , C = b2 a1 −b1 a2 ∈ R Hence the conclusion. Geometrical Interpretation We know that the linear equation Ax + By + C = 0 represents 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 ∈ Gab and when t = 1, a + t(b − a) = b ∈ Gab so, Gab represents the line passing through the points a and b. T heEnd For online classes in Mathematics, please contact vattamattam@gmail.com References [1] H. P. Petersson,Lineare Optimierung [2] J K Sharma, Operations Research Theory and Applications,4th Ed, Macmillan [3] George F. Simmons, Topology and Modern Analysis,International Student Edition, McGrawHill, 1963.