n-Dimentional Space

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

2

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}

3

4

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

5

       

(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

6

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

7

(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

8

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.