VECTOR SPACE - · PDF fileVector Space •Group and Field •Vector space and Subspace...

VECTOR SPACE

05/03/2013 budi murtiyasa ums surakarta 1

Vector Space

• Group and Field

• Vector space and Subspace

• Linear Combination

• Row Space and Column Space


Group and Field


G r o u p a set G with * operation, denote (G, *), called group if for all a, b, c G satisfy:

1. closed; a * b G

2. assosiative; (a * b) * c = a * (b * c)

3. Having identity z G, such that a * z = a

4. For all a G, a have an inverse a-1 G such that a * a-1 = z


example: Given H = { 0, 1, 2 }, under addition modulo 3, is it group ?

Solution : 1.closed

+mod3 0 1 2

0 0 1 2

1 1 2 0

2 2 0 1

2. Assosiative (please verify !)

3. Identity z = 0

4. Inverse : 0-1 = 0, 1-1 = 2, 2-1 = 1

Thus (H, +mod3) is group.


If a group satisfy (5) commutative, then called abelian group or Commutative group.

Thus Commutative Group must be satisfies : 1. closed 2. assosiative 3. having identity 4. all elements have an inverse 5. commutative

It can be shown that H = { 0, 1, 2 } under addition modulo 3 is group commutative.


Evaluate, which ones a group ? Commutative Group?

1. H = { 1, -1, i, -i } where i = √-1,

under multipliction operation.

2. Set of Natural Number under addition operation

3. Set of Integer Number under addition operation


X 1 -1 i -i

1 1 -1 i -i

-1 -1 1 -i i

i i -i -1 1

-i -i i 1 -1

Identity : Z = 1

Inverse : 1-1 = 1

-1-1 = -1

i-1 = -i

-i-1 = i Group commutative

closed?

assosiative?

commutative?

Yes

Yes

Yes

thus, Group.


F i e l d A set F with the first operation (+) and the Second operation (x), denote (F, +, x), called field if for all a, b, c F satisfies :

Under the first operation (+) satisfies : 1. closed; a + b F 2. assosiative; (a + b) + c = a + (b + c) 3. having identity element z F; such that a + z = a 4. every a there is exist a-1 F a + a-1 = z 5. commutative; a + b = b + a


Field (Cont..)

Under the second operation (x) satisfies : 6. closed; a x b F 7. assosiative; (a x b) x c = a x (b x c) 8. having identity element u F a x u = a 9. except z, every a there is exist a-1 F such that a x a-1 = u 10. commutative; a x b = b x a

The second operation (x) to the first operation (+) satisfies : 11. distributive; a x (b + c) = (a x b) + (a x c) or (a + b) x c = (a x c) + (b x c)


Based on the definition, it can be simplified : (F,+, X) is field if: 1. under the first operation (+) is group commutative 2. under the second operation (x) is group commutative 3. the operation (x) to the operation (+) satisfies distributive law.



Suppose F = {0, 1, 2, 3, 4}. The first operation on F is addition modulo 5, and the second operation is multiplication modulo 5. Is (F, +mod5, xmod5) field? Explain your answer!

Suppose H = {0, 1, 2, 3, 4, 5}. The first operation on H is addition modulo 6, and the second operation is multiplication modulo 6. Is (H, +mod6, xmod6) field? Explain your answer!

Evaluate, which ones a field under the operation + and x ?

1. Set of Integers

2. Set of Rational numbers

3. Set of Real numbers

note: Members of field called scalar.

Not field

field

field


Because set of real number under addition (+) and multiplication (x) operations are field, members of real number can be

called as scalar.


Vector Space and Subspace

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Vector Space

Suppose set of V with u, v, w V, and addition (+) operation among members of V. Given Field F with a, b F; between members of F and members of V behave multiplication (x) operation. A set of V called vector space over field F If it is satisfy :


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Vector space (cont..)

Under addition (+) operation on V satisfies : 1. closed; u + v V 2. assosiative; (u + v) + w = u + (v + w) 3. having identity 0 V; such that u + 0 = u 4. all of u V; there is (-u) V such that u + (-u) = 0 5. commutative; u + v = v + u


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Vector space (cont..)

between members of F and members of V satisfies: 6. closed; a u V 7. distributive; a ( u + v) = a u + a v 8. distributive; (a + b) u = a u + b u 9. assosiative; a (b u) = (a b) u 10. having identity; there is exist 1 F, such that 1 u = u


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Notes :

1.If V is vector space, members of V called vector.

2. addition operation on V called vector addition

3. multiplication operation between members of F and

members of V called scalar multiplication

4. Vector 0 V which is identity on vector addition,

called zero vector.

5. vector (-u) which is inverse of vector u called

opposite or negative of vector u.


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Based on the defition of vector space above, all of set that satisfy the 10

properties called vector space; and it members called vector.


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長庚大學電機系

Examples

• with standard vector

addition and scalar multiplication is a vector space.

• The set of all matrices with the usual matrix

addition and scalar multiplication is a vector space,

denoted by The zero vector is the zero matrix.

• Let

Define + and scalar multiplication by

for all and

Then is a vector space.

1{[ ... ] }n T

n iR x x x R

m n

m nR m n

1 2

1 2 1 0( ) : { ( ) ... }n n

n n n iP R p x a x a x a x a a R

( )( ) ( ) ( ), ( )( ) ( )p q x p x q x p x p x R

, ( )np q P R

( )nP R

example: M = { all of 3x2 matrices}. Addition operation on M is matrices addition. Multiplication operation is scalar multiplication between members of F and members of M. Is M a vector space ?


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Given P = {ao + a1t + a2 t2 } with ai Real number, that is set of polynomial order 2. Addition operation on P is polynomial addition, and multiplication operation on P is scalar multiplication. Is P a vector space ?

Is M a vector space ?, if : M = { ordered pair of two real numbers }

2

1

a

a+

2

1

b

b=

22

11

ba

ba

and multiplication operation define as:

2

1

a

ak =

2

1

ka

ka

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where addition operation define as:

Is M a vector space ?, if : M = {ordered pair of two real numbers }

2

1

a

a+

2

1

b

b=

22

11

ba

ba

and multiplication operation define as:

2

1

a

ak =

2

1

ka

ka

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where addition operation define as:

A set of ordered pair of real numbers (n – tuple) :

na

a

a

a...

2

1

with addition operation define as :

na

a

a

...

2

1

+

nb

b

b

...

2

1

=

nn ba

ba

ba

...

22

11

and multiplication define as:

na

a

a

k...

2

1

=

nka

ka

ka

...

2

1

It can be shown that a set n-tuple

satisfied the 10 properties. Thus, set of

n-tuple is vector space. Generally denote

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Theorem 1:

suppose V is vector space

with u V dan k F, then

(i) For 0 F, behave 0 u = O

(ii) For O V, behave k O = O

(iii) For -1 F, behave (-1) u = - u

(iv) If k u = 0, then k = 0 or u = O

(v) –(k u) = (-k) u = k (- u)


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0 F, behave 0 u = O

proof:

0 + 0 = 0 properties of field

(0 + 0) u = 0 u

0u + 0u = 0u propt. 8 of vector space

0u + 0u + (- 0u) = 0u + (-0u)

0u + 0 = 0 propt. 4 of vrctor space

0u = 0 propt. 3 of vector space

(proved).


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v Vector space

W M

W V

M V

If W satisfies the 10 properties of vector space, then W called Subspace of V

Subspace

Subspace

Theorem :

W is subspace of V if and only if

• W is not empty

• Closed under addition; u, v W;

u + v W

• closed under multiplication; u W,

au W; where a is scalar.

consequence: W subspace of V if and only if : O W For u, v W, then ku + lv W; where k, l are

scalar.

example:

Suppose V = R3. W is subset of V;

W = { | b = 2c; a, b, c R }.

Is W subspace of V ?.

c

b

a

(i)0 = belong on W, because 0 = 2.0

(ii) Take u = where b1 = 2c1

v = where b2 = 2c2

0

0

0

1

1

1

c

b

a

2

2

2

c

b

a

ku + lv = + =

1

1

1

kc

kb

ka

2

2

2

lc

lb

la

21

21

21

lckc

lbkb

laka

kb1 + lb2 = 2kc1 + 2lc2 = 2(kc1 + lc2)

So, ku + lv is belong on W.

Thus W subspace of V

Problem 1:


W = { | a – b = 0; a, b, c R }.


c

b

a

Problem 2:


W = { | a + b = 1; a, b, c R }.


c

b

a

Theorem:

Suppose U and W are subspace of V, then U W is subspace of V

proof:

• Because of U and W are subspace of V, then O U and O W O (U W).

• Take u, v (U W), its mean :

u, v U au + bv U

u, v W au + bv W

au + bv (U W).

Thus U W is subspace of V.

example:

suppose V = R3. If U and W are subspace of V, where :

U = { | a + b = 0; a, b, c R }, and

W = { | a = 2c; a, b, c R }, then:

UW = { | a + b = 0; a = 2c; where a,b, c R }.

Show that U W is subspace of V.

c

b

a

c

b

a

c

b

a

Linear Combination

Linear Combination

Suppose a vector space V over field F, with vectors u1, u2, …, un V. Any vector on V (instance: v V) which is can be denote in form :

v = a1 u1 + a2 u2 + … + an un; with ai F

called linear combination of vectors u1, u2, ..., un.

example:

Suppose s, u, v, w V; where

u = , v = , w = , and s = .

If maybe, express v as linear combination of vectors

u, s, and w !

2

1

1

1

0

1

1

1

2

6

3

1

solution:

v = xu + ys + zw

6

3

1

= x + y + z

2

1

1

1

0

1

1

1

2

Syst.of Linear Equation:

x – y + 2z = -1

-x – 3y + z = 0

2x + 6y – z = 1

So,

x = -2, y = 1, dan z = 1

Thus, v is linear combination of u, s, and w ; v = -2u + s + w

Practice: Suppose s, u, v, w V; where u = , v = , w = , and s = . If maybe, express s as linear combination of vectors u, v, and w !

3

1

1

1

2

1

1

1

1

10

3

2

Generating System

A set of vectors { u1, u2, …, um} called generating system of vector space V; denoted V = L{u1, u2, …, um}; if every vectors v V can be expressed as linear combination of vectors {u1, u2, …, um}.

example:

Suppose V = R2, vectors u1 = , u2 = , u3 =

It can be shown that u1, u2, and u3 are generating system for

R2; because for all v V can be expressed as linear

combination of vectors u1, u2 and u3.

Instantly, v = v = 2u1 – u2 – 3u3

Instantly, v = v = -3u1 + u2 + 2u3 ; etc.

0

1

3

2

1

0

0

4

1

5

example: Suppose V = R3, vectors u1 = , u2 = , u3 = It can be shown that u1, u2, and u3 are generating system for R3; because every v V can be expressed as linear combination of vectors u1, u2 and u3.

0

0

1

0

1

1

1

1

1

Instantly, v = v = u1 – u2 + 2u3

2

1

2

Instantly, v = v = 3u1 + 2u2 + u3 ; etc.

1

3

4

Row Space & Column Space

Row Space & Column Space

A =

mnmm

n

n

aaa

aaa

aaa

...

............

...

...

21

22221

11211

Row Space = Rn = { , , …, }

na

a

a

1

12

11

...

na

a

a

2

22

21

...

mn

m

m

a

a

a

...

2

1

Coloumn Space = Rm = { , , …, }

1

21

11

...

ma

a

a

2

22

12

...

ma

a

a

mn

n

n

a

a

a

...

2

1

Suppose A =

• Row Space of A is { , , }

• Column Space of A is { , , , }


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4

3

1

2

0

2

5

5

6

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8

8

5

2

9

5

1

7

2

3

6

0

4

VECTOR SPACE - · PDF fileVector Space •Group and Field •Vector space and Subspace...

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Transcript of VECTOR SPACE - · PDF fileVector Space •Group and Field •Vector space and Subspace...