Download Chapter 4: General Vector Spaces

Chapter 4: General Vector Spaces
1
Chapter 4: General Vector Spaces
SECTION A Introduction to Vector Spaces
By the end of this section you will be able to
 understand what is meant by a vector space
 give examples of vector spaces
 prove properties of vector spaces
In this chapter we examine other kinds of vector spaces such as matrices, functions and
polynomials. You will find this chapter more abstract than the previous chapter because for
Euclidean spaces 2 and 3 we can visualize the vectors.
First we state what is meant by a general vector space in terms of a set of axioms. What
does the term axiom mean?
An axiom is a statement which is assumed to be true without proof.
We define a vector space in terms of 10 axioms given below.
A1 Vector Space
Let V be a non-empty set of elements called vectors. We define two operations on the set V
– vector addition and scalar multiplication. Scalars are real numbers.
Let u, v and w be vectors in the set V. The set V is called a vector space if it satisfies the
following 10 axioms.
1. The vector addition u  v is also in the vector space V.
Generally in mathematics we say we have closure under vector addition if this property
holds.
2. Commutative law: u  v  v  u .
3. Associative law:  u  v   w  u   v  w  .
4. Neutral Element. There is a vector called the zero vector denoted by O in V which
satisfies
u  O  u for every vector u in V
5. Additive Inverse. For every vector u there is a vector u (minus u) which satisfies
the following:
u   u   O
6. Let k be a real scalar then k u is also in V.
We say we have closure under scalar multiplication if axiom 6 is satisfied.
7. Associative Law for scalar multiplication. Let k and c be real scalars then
k  cu    kc  u
8. Distributive Law for vectors. Let k be a real scalar then
k u  v   k u  k v
9. Distributive Law for scalars. Let k and c be real scalars then
 k  c  u  k u  cu
10. Identity Element. For every vector u in V we have
1u  u
We say that if the elements of the set V satisfy the above 10 axioms then V is called a
vector space and the elements are known as vectors.
Chapter 4: General Vector Spaces
2
A2 Examples of Vector Spaces
Can you think of any examples of vector spaces?
The Euclidean spaces of the last chapter - V  2 , 3 , , n are all examples of vector
spaces. Are there any other examples of a vector space?
The set of matrices M 22 which are all matrices of size 2 by 2 where matrix addition and
scalar multiplication is defined as in Chapter 1.
a b 
e f 
i
Let u  
, v  
 and w  
c d 
g h
k
above 10 axioms hold for this set M 22 .
What is the zero vector in this case?
j
 be matrices in M 22 . Check that the
l
0 0
The zero vector is the zero matrix of size 2 by 2 which is given by 
O.
0 0
The rules of matrix algebra established in Chapter 1 ensure that all 10 axioms are satisfied.
You may have to look back at your work of Chapter 1 to be satisfied that the set M 22
satisfies all 10 axioms which means it is a vector space. You are asked to check this in
Exercise 4a.
We can also show that the set M 23 which is the set of matrices of size 2 by 3 is also a
vector space. You are asked to do this in Exercise 4a.
We can also have vector spaces which are not the Euclidean space or the set of matrices.
For example the set of polynomials denoted P  t  whose elements take the form:
p  t   c0
 c1t  c2t 2

 cnt n
where c0 , c1 , c2 ,
and cn are real numbers called the coefficients. The following are
examples of polynomials
p  t   t 2  1 , q  t   1  2t  7t 2  12t 3  3t 4 and r  t   7
The degree of a polynomial is the highest index (power) which has a non-zero coefficient,
that is the maximum n for which cn  0 . What is the degree of p  t  , q  t  and r  t  ?
p  t   t 2  1 is of degree 2, q  t   1  2t  7t 2  12t 3  3t 4 is of degree 4 and r  t   7 is of
degree 0. Note that the last polynomial r  t   7  7t 0 is of degree 0 because 0 is the
highest index with a non-zero coefficient  c0  .
How do we define the zero polynomial?
The zero polynomial denoted by O is given by
O  0  0t  0t 2 
 0t n  0
All the coefficient c’s are zero. We say that the zero polynomial has no degree.
Let P  t  be the set of polynomials of degree less than or equal to n where n is a positive or
zero integer. The zero polynomial O is also in the set P  t  .
We define the two operations of vector addition and scalar multiplication as:
1. Vector addition p  t   q  t  is the normal addition of polynomials.
2. Scalar multiplication k p  t  is the normal multiplication of a constant, k, with the
polynomial, p  t  .
In the next example we show that this P  t  is also a vector space.
Chapter 4: General Vector Spaces
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Example 1
Show that P  t  the set of polynomials of degree n or less is indeed a vector space.
Solution
We need to check all 10 axioms with the elements or vectors (polynomials) of this space
which have the general polynomial form:
p  t   c0  c1t  c2t 2   cnt n , q  t   d0  d1t  d2t 2   dnt n and
1.
r  t   e0  e1t  e2t 2   ent n
Adding two vectors (two polynomials):
p  t   q  t   c0  c1t  c2t 2   cnt n  d 0  d1t  d 2t 2 
p t 
 d nt n
q t 
  c0  d0    c1  d1  t   c2  d 2  t 
  cn  d n  t n
2
Hence p  t   q  t  is in P  t  which means that we have closure under addition.
2.
Commutative Law:
Required to show that p  t   q  t   q  t   p  t  :
p  t   q  t   c0  c1t  c2t 2 
 cnt n  d 0  d1t  d 2t 2 
 d nt n
 d 0  d1t  d 2t 2 
 d nt n  c0  c1t  c2t 2 
 cnt n
q t 
p  t 
 q t   p t 
3.
Associative Law:
We need to show that p  t   q  t    r  t   p  t   q  t   r  t   . Substituting the above
polynomials into this yields:

p  t   q  t    r  t   c0  c1t  c2t 2 

p  t 


 cnt  d 0  d1t  d 2t   d nt 

q  t 

 e0  e1t  e2t 2   ent n
n
2
n
r  t 
 c0  c1t  c2t 
2
 cnt  d 0  d1t  d 2t 2 
n
 e0  e1t  e2t 2 
 c0  c1t  c2t 2 
 d nt n
 ent n
 cnt n
p t 

  d 0  d1t  d 2t 2 

q t 

 d nt  e0  e1t  e2t 
n
2
r  t 

 ent 


n
 p  t   q  t   r  t  
4.
What is the zero vector in this case?
This is the zero polynomial O which is the real number 0 and is in the set P  t  . Clearly
p  t   O  c0  c1t  c2t 2 
 cnt n  0
p t 
 c0  c1t  c2t 2 
 cnt n  p  t 
Chapter 4: General Vector Spaces
5.
and
4
Additive Inverse. For the polynomial p  t  we have the additive inverse as p  t 
 cnt n   c0  c1t  c2t 2 
p  t   p  t   c0  c1t  c2t 2 
 c0  c1t  c2t 2 
 cnt n  c0  c1t  c2t 2 
  c0  c0    c1  c1  t   c2  c2  t 2 
6.
 000 0  0  O
Let k be a real scalar then
k p  t   k  c0  c1t  c2t 2 
 kc0
 cnt n 
 kc1t
 kc2t 2
 cn t n
  cn  c n  t n
cnt n 



kcnt n
Hence k p  t  is also in the space P  t  because it is a polynomial of degree less than or
equal to n.
7.
Associative Law for scalar multiplication. Let k1 and k2 be real scalars:

 c1t
 k1  k2 c0
 k2 c1t
 k1 k2 c0
 k1k2 c1t
k1  k2 p  t    k1 k2 c0
 k1k2  c0
 c1t
 c2t 2

 k2c2t 2

 c2t 2
k2cnt n 

 k1k2c2t 2



k1k2cnt n
cnt n 


cnt n 

 Factorising 
p t 
8.
  k1k2  p  t 
Distributive Law. Let k be a real scalar then we have:


2
n
2
n

k p  t   q  t    k c0  c1t  c2t   cnt  d 0  d1t  d 2t   d nt


p t 
q t 


2
n
2
 kc0  kc1t  kc2t   kcnt  k d 0  k d1t  k d 2t   k d nt n
k  c0  c1t  c2t 2 

Factorising
 cnt n   k  d 0  d1t  d 2t 2 
p  t 
 d nt n 
q  t 
 k p t   k q t 
9.
Distributive Law for scalars. Let k1 and k2 be real scalars:
 k1  k2  p  t    k1  k2   c0
 k1  c0
10.
 c1t  c2t 2
 c1t 

cnt n 

 cnt n   k2  c0
 cnt n 
 c1t 
 k1p  t   k2p  t 
The identity element is the real number 1. We have
1p  t   1 c0  c1t  c2t 2 
 cnt n 
 1 c0  
1 c1t 

 c0
 c2t 2

 c1t
1 c t 
2
2



cnt n  p  t 
1c t 
n
n
All 10 axioms are satisfied therefore the set P  t  of polynomials of degree n or less is a
vector space.
Chapter 4: General Vector Spaces
Example 2
Let p  t   c0
 c1t  c2t 2
5
 cnt n and P  t  be the set of polynomials

p  t  of degree equal to n where n is a positive integer. Show that this set P  t  is not a
vector space.
Remark: Note the difference between Example 1 above and this example. Here the
polynomial is of degree n exactly. In Example 1 the degree of the polynomial is less than
or equal to n.
Solution
If we check the first axiom that is if we add any two vectors (polynomials) in the set P  t 
then the addition should also be in the set. Let
p  t   2  2t  2t 2   2t n1  2t n and q  t   1  t  t 2   t n  2t n
We have
p  t   q  t   2  2t  2t 2   2t n 1  2t n   1  t  t 2   t n  2t n 
p t 
  2  1   2  1 t   2  1 t 2 
q  t 
  2  1 t n 1   2  2  t n
0
 1 t  t 
2
This result p  t   q  t   1  t  t 2 
t
n 1
 t n1 is not inside the set P  t  , why not?
Because P  t  is the set of polynomials of degree n, but p  t   q  t  , is of degree n 1
polynomial which means that it cannot be a member of this set P  t  .
Hence P  t  is not a vector space because axiom 1 fails, that is we do not have closure
under vector addition.
Example 3
Let V be the set of integers . Let vector addition be defined as the normal addition of
integers, and scalar multiplication by the usual multiplication of integers by a real scalar,
which is any real number.
Show that this set is not a vector space with respect to this definition of vector addition and
scalar multiplication.
Solution
The set of integers
is not a vector space because when we multiply an integer by a
scalar, which is any real number, then the result may not be an integer.
1
For example if we consider the integer 2 and multiply this by the scalar then the result is
3
1
2
 2   which is not an integer. Our result after scalar multiplication is not in the set of
3
3
integers .
We do not have closure under scalar multiplication therefore the set of integers
is not a
vector space.
Other examples of vector spaces are real-valued functions. You should be familiar with
functions.
Let a and b be real numbers where a  b then the notation  a, b is called the interval
and is graphically displayed as:
Chapter 4: General Vector Spaces
6
a
b
Fig 1
The functions defined on this interval  a, b are normally denoted by f  a, b .
Fig 2
For example the functions in Fig 2 are defined for the interval  2, 2 .
Examples of functions are f  x   x, f  x   x2  1, f  x   sin  x  and f  x   e x .
Let F  a, b be the set of functions defined on the interval  a, b . Let f and g be
functions in F  a, b . Normally f and g being vectors are in bold but since we are used to
functions without bold symbols therefore we will write these as f and g respectively. We
define:
i. Vector Addition. The sum f  g is also in F  a, b and
 f  g  x   f  x   g  x 
ii. Scalar Multiplication. Let k be a real scalar then for all x in  a, b
The zero vector in
 k f  x   k f  x 
F  a, b is the zero function 0  x  which is equal to zero for all
the interval  a, b , that is
0 x  0
x in
for all x   a, b
The notation x   a, b means that x is a member of the interval  a, b or a  x  b which
is illustrated in figure 1 above.
For any function f there is a function  f (minus f ) for all x in the interval  a, b which
satisfies
  f  x    f  x 
You are asked to prove that F  a, b is a vector space with respect to these definitions in
Exercise 4a.
There are many other examples of vector spaces which you are also asked to show in
Exercise 4a.
Next we go on to prove some basic properties of vector spaces.
Chapter 4: General Vector Spaces
7
A3 Basic Properties of Vector Spaces
Proposition (4.1). Let V be a vector space and k be a real scalar. Then we have the
following:
(a)
For any real scalar k we have k O  O where O is the zero vector.
(b)
For the real number 0 and any vector u in V we have 0u  O .
(c)
For the real number 1 and any vector u in V we have  1 u  u .
(d)
If k u  O , where k is a real scalar and u in V , then k  0 or u  O .
We use the above 10 axioms given in subsection A1 , to prove these properties.
(a)
Proof. Need to prove k O  O . By applying Axiom 4 above u  O  u , we have
k O  k O  O
 Applying Axiom 8 which is k  u  v   k u  k v 
Adding  k O to both sides of this, k O  k O  k O , gives
 kO  kO
k O   k O   k O  k O   k O 
O
 By Axiom 5 u   u   O 
O
O  kO
Hence we have our required result, k O  O .
(b)
Proof. Required to prove 0u  O . For real numbers 0  0  0 we have
0u   0  0  u
■
 By Axiom 9  k  c  u  k u  cu 
Adding 0u to both sides of this, 0u  0u  0u , gives
0u   0u   0u  0u   0u 
 0u  0u
O
O
O  0u
We have result (b), that is 0u  O .
(c) Proof. Required to prove  1 u  u . Adding u to  1 u gives
■
u   1 u  1u   1 u
 1  1 u  0u

O
by part (b)
By axiom 5 we have u   u   O . Equating this with the above, u   1 u  O , we have
u   1 u  u   u 
Adding u to both sides of this gives:
u  u   1 u  u  u   u 
O
O
O   1 u  O   u 
 1 u  u
Hence we have our result,  1 u  u .
(d) Proof. Need to prove that k u  O implies k  0 or u  O . We can write the Right
Hand Side O  0u :
■
Chapter 4: General Vector Spaces
8
k u  0u which gives k  0
Suppose k  0 (non-zero) then
ku  O
1
1
ku  O
k
k
1

Multplying
both
sides
by

k 
O
1 
 k  u  O gives 1u  O
k 
Since 1u  u therefore u  O which is our required result.
■
Proposition (4.2). Let V be a vector space. The zero vector O (neutral element) of V is
unique.
What does this proposition mean?
There is only one zero vector in the vector space V. Remember by Axiom 4 there is a
vector called the zero vector denoted by O which satisfies
u  O  u for every vector u
Proof.
Suppose there is another vector, say w, which also satisfies
uw u
What do we need to prove?
Required to prove that w  O . Equating these two, u  O  u and u  w  u , gives
uO  uw
Adding u to both sides of this yields:
u   u  O   u   u  w 
 u  u   O   u  u   w
O
O
O  O  O  w gives O  w
Hence w  O which means that the zero vector is unique.
■
Proposition (4.3). Let V be a vector space and k be a real scalar. The vector u which
satisfies Axiom 5
u   u   O
is unique.
Proof.
Suppose the vector w also satisfies this, that is
uw O
What do we need to prove?
Required to prove that w  u . Equating these two, u   u   O and u  w  O , gives
u   u   u  w
Adding u to both sides yields:
Chapter 4: General Vector Spaces
9
u   u    u     u   u  w 
  u  u    u     u  u   w
O
O
O   u   O  w
u  w
Hence the vector u is unique.
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SUMMARY
A vector space is a non-empty set within which the operation of vector addition and scalar
multiplication satisfies the above 10 axioms.
2
, 3 , , n are examples of vector spaces.
Other examples of vector spaces include the set of matrices, polynomials and functions.
Some properties of vector spaces are:
(a) k O  O (b) 0u  O
(c)  1 u  u
(d) If k u  O then k  0 or u  O
The zero vector O and the vector u have the following properties
u   u   O and u  O  u
and are unique.