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Vector Space
• Group and Field
• Vector space and Subspace
• Linear Combination
• Row Space and Column Space
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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
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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.
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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.
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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
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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.
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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
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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)
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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.
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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
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Because set of real number under addition (+) and multiplication (x) operations are field, members of real number can be
called as scalar.
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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
by Rn. 05/03/2013 26 budi murtiyasa ums
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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:
Suppose V = R3. W is subset of V;
W = { | a – b = 0; a, b, c R }.
Is W subspace of V ?.
c
b
a
Problem 2:
Suppose V = R3. W is subset of V;
W = { | a + b = 1; a, b, c R }.
Is W subspace of V ?.
c
b
a
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
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
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
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