vector spaces 2003 (cd burn)
TRANSCRIPT
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TMA 2023 / SMA 3013
Semester 2010/ 2011
Assignment
Vector Spaces
No. Name Student ID
1 NAJWA HANIM BINTI YUSOFF D20091034560
2 NUR AMANINA NAJIHAH BINTI AB
LATIB
D20091034561
3 NOR ATIQAH BINTI FAHRUL RADZI D20091034569
4 NUR FATIHAH BINTI MOHD KARIM D20091034574
5 JENIFER JOSLI D20091034575
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VECTOR SPACES
Definition:
A set V is called a vector space over the real numbers provided that there are two operations-
addition, denoted by + , and scalar multiplication, denoted by - that satisfy all the following
axioms. The axioms must hold for all vectors u, v, w in V and all scalars c and d R.
Vector addition: This assigns to any u, v V a sum u + v in V Scalar multiplication: This assign to any u V, k K, a product ku V.
These are the 10 axioms in Vector Space:
1. u + vV( Vis closed under addition)2. u + v = v + u (commutative law)3. (u + v ) + w = u (v + w) (associative law)4. There exists a vector Ov, V where u + Ov = u = Ov + u, all elements in u ( the
existing of a zero vector)
5. For each vector u V, there exists a vector u V where u + (-u) = Ov. (existinginverse of addition or a negative vector).
6. c uV(Vis closed under multiplication)7. c (u +v) = cu + cv (left distributive law)8. (c+ d) u= cu + du9. c(du) = (cd)u10.I u = u
In this section, we use the special symbols + and of the previous definition to distinguish
vector addition and scalar multiplication from ordinary addition and multiplication of real
numbers (R).
Euclidean Vector Spaces
The set V = R with the standard operations of addition and scalar multiplication is a vector
space.
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Vector Spaces of Matrices
Show that the set V = M mxn of all m x n matrices is a vector space over the scalar field R,
with + and defined componentwise.
Since addition of matrices is componentwise, the sum of two m x n matrices is another m
x n matrix as is a scalar times an m x n matrix. Thus, the closure axioms (axioms 1 and 6) are
satisfied.
Exercises
1. Determine whether these matrices are vector spaces or not.A = , B =
i. A + BV( Vis closed under addition)A + B =
=
ii. A + B = B + A (commutative law)LHS A + B =
=
RHS B + A =
=
=
iii. (A + B ) + C = A + (B + C) (associative law) which is C =LHS (A + B) + C =
=
=
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RHS A + (B + C) =
=
=
iv. A + 0= A = 0 + A
v. A + (-A) = 0
=
vi. cAV(Vis closed under multiplication) which is c = x, x is a scalar.
vii. c(A + B) = cA + cB (left distributive law)LHS c(A + B) =
=
=
=
RHS cA + cB =
=
=
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viii. (c + d) A= cA + dA, which is d = y, y is a scalar.LHS (c + d) A =
=
=
RHS cA + dA =
=
=
ix. c(dA) = (cd)ALHS c(dA) =
=
=
RHS (cd)A =
=
x. IA = A, which is I =
=
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=
Therefore, the 10 axioms are satisfied. Thus, these matrices are vector spaces.
2. Let V = R. Show that this addition A + B = 3A + 3B is commutative but notassociative.
i. To show whether + is commutative.A + B = B + A
A + B = 3A + 3B
= 3 (A + B)
= 3 (B + A) commutative
= B + A
This operation is commutative.
ii. To show whether + is associative.(A + B) + C = A + (B + C)
LHS (A + B) + C = (3A + 3B) + C
= 3(3A + 3B) + 3C
= 9A + 9B + 3C
RHS A + (B + C) = A + (3B + 3C)
= 3A + 3(3B + 3C)
= 3A + 9B + 9C
This operation is not associative.
Thus, V is not a vector spaces.
Counter example:
Let A = 1, B = 2 and C = 3.
LHS (1 + 2) + 3 = [3(1) + 3(2)] + 3
= (3 + 6) + 3
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= 9 + 3
= 3(9) + 3(3)
= 27 + 9
= 36
RHS 1 + (2 + 3) = 1 + [3(2) + 3(3)]
= 1 + (6 + 9)
= 1 + 15
= 3(1) + 3(15)
= 3 + 45
= 48
LHS RHS.
Thus, it is not a vector spaces.
3. Show that the operation is a vector space or not.i) To show whether + is a commutative
A + B = B + A
A + B = 5A + 5B
= 5(A + B)
= 5(B + A) commutative
= B + A
Therefore, the operation is commutative
ii) To show whether + is associative(A + B) + C = A + (B + C)
LHS : (A + B) + C = (5A + 5B) + C
= 5 (5A + 5B) + 5C
= 25A + 25B + 5C
RHS : A + (B + C) = A + (5B + 5C)
= 5A + 5(5B + 5C)
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= 5A + 25B + 25C
LHS RHS Thus, V is not a vector space.
Counter example:
Let A = 3, B = 6, C = 9
LHS : (3 + 6) + 9 = [5(3) + 5(6)] + 9
= [15 + 30] + 9
= 45 + 9
= 5(45) + 5(9)
= 270
RHS : 3 + (6 + 9) = 3 + [5(6) + 5(9)]
= 3 + [30 + 45]
= 3 + 75
= 5(3) + 5(75)
= 390
LHS RHS Thus, it is not a vector space.
4. Given A = , B = , C =Show that this operation from two matrices is a vector space or not.
i) To show whether + is commutative: A + B = B + ALHS: A + B = 3 + 3
= +
=
RHS: B + A = 3 + 3
= +
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=
Thus, the operation is commutative.
ii) To show whether + is associative: (A + B) + C = A + (B + C)LHS: (A + B) + C = ( + ) +
= (3 + 3 ) +
= ( + ) +
= 3 + 3
= +
=
RHS: A + (B + C) = + ( + )
= + (3 + 3 )
= + ( + )
= 3 + 3
= +
=
The operation is not associative. Thus, this is not a vector space.Counter example: Given A = , B = , C =
LHS: (A + B) + C = ( + ) +
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= (3 + 3 +
= ( + ) +
= 3 + 3
= +
=
RHS: A + (B + C) = + ( +
= + (3 + 3 )
= + ( + )
= +
= 3 + 3
= +
=
LHS RHS Thus, this is not a vector space.
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REFERENCES
Bronson, R., Costa, G, B. (2007). Linear Algebra: an Introduction. Elsevier Inc. pp: 8597.
Cullen, C. (1997).Linear algebra with application. Addison-Wesley. pp: 136 - 139.
Lipschutz, S., Lipson, M, L. (2001). Theory and Problems of Linear Algebra(3). McGRAW-
HILL International Edition. pp:
Lipschutz, S., Lipson, M, L. (2009). Linear Algebra(4). McGRAW-HILL International
Edition.
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Group discussion / Meeting report
No Date Activities Task Assign
1 5 March 2011 Group discussion at library.
Finding reference books
From this discussion, we
have found out the
definition and also the 10
axioms in Vector Spaces.
2 9 March 2011 Preparing for the Vector
Spaces module at library.
From this discussion, we
have prepared the module of
Vector Spaces.
3 12 March 2011 Prepare the powerpoint and
ready for the presentation.
From this group discussion,
we have prepared for the
presentation slides in VectorSpaces.