Chapter 3: Euclidean Space
53
SECTION E Basis and Spanning Set
By the end of this section you will be able to
 show that given vectors span n
 determine whether given vectors are a basis for
 prove properties of a basis
n
The first two subsections E1 and E2 are not difficult but you should be comfortable in
scalar multiplication of matrices and finding the inverse of a 2 by 2 matrix.
However the last subsection E3 is more challenging because you need to prove results
which means that you should be confident in applying definitions of linear algebra.
E1 Spanning Sets
Example 18
Consider the standard unit vectors in 3 :
1 
0
0
 
 
 
e1   0  , e2  1  and e3   0 
0
0
1 
 
 
 
Show that we can write any vector as a linear combination of these standard unit vectors
e1 , e2 and e3 . (These vectors are illustrated in Fig 35 on page 41.)
Solution
What does linear combination of e1 , e2 and e3 mean?
Linear combination is an expression of the form
k1e1  k2e2  k3e3 where the k’s are real scalars
How do we write any vector in 3 (3-space)?
a
 
Let v   b  be any vector in 3 because a, b and c are any real numbers. What are we
c 
 
trying to show?
We need to find the values of scalars k’s such that k1e1  k2e2  k3e3  v :
1
0
 0
 
 
 
v  k1e1  k2e 2  k3e3  k1  0   k2 1   k3  0 
0
0
1 
 
 
 
e1
e2
 e3

 k1   0   0   k1   a 
 a 

         
 
  0    k2    0    k2    b 
 Because v   b  
 0  0  k   k  c 
 c 

     3  3  
 
We have k1  a, k2  b and k3  c .
Hence we can write the vector v  k1e1  k2e2  k3e3 as
v  ae1  be2  ce3
That is we can write the vector v as a linear combination of vectors e1 , e2 and e3 .
Chapter 3: Euclidean Space
54
3
We say the vectors e1 , e2 and e3 span or generate
k1e1
 k2 e 2
 k3e3 , produces any vector in
3
because their linear combination,
. These are not the only vectors which
span 3 , there are other vectors which also span 3 . We define the term span as follows:
Definition (3.27). Consider the n vectors v1 , v 2 , v3 ,
and v n in the n space, n . If
every vector in n can be produced by a linear combination of these vectors
v1 , v 2 , v3 ,
and v n then we say these vectors span or generate the n space,
Sometimes the vectors v1 , v 2 , v3 ,
S  v1 , v 2 , v3 ,
n
.
and v n are written in a set such as
, v n  and we call this the spanning set. We also say that the set S
spans the n space or S spans
n
.
1 
0
For example the standard unit vectors e1    and e 2    span 2 because the linear
0
1 
combination, k e1  ce2 , of these vectors e1 and e2 produce any vector in 2 .
To check that given vectors v1 , v2 , v3 ,
, v n  span
n
, we carry out the following:
 u1 
1. Let u    be an arbitrary vector in n .
u 
 n
2. Show that this vector u is a linear combination of vectors v1 , v2 , v3 ,
u  k1v1  k2 v 2 
 kn v n where k’s are scalars.
If we cannot write an arbitrary vector u as a linear combination of vectors
v1, v2 , v3 , , vn  then these vectors do not span n .
Example 19
Consider the following vectors in
(i)
(ii)
2
:
1 
 1
u    and v   
 2
 1
2
Show that the vectors u and v span
.
 3
Write the vector   in terms of the given vectors u and v.
 2
Solution
a
(i) Step 1. Let w    be an arbitrary vector in 2 .
b 
Step 2. Consider the linear combination
ku  cv  w
Substituting the given vectors u, v and w yields
1   1  a 
k    c    
 2  1 b 
u
v
w
Representing this in matrix form we have
 1 1 k   a 

    
 2 1  c   b 
, vn  :
Chapter 3: Euclidean Space
55
 1 1
1  1 1
Since det 
  1   2   3 [Not zero] so the inverse of this is 
 . Hence
3  2 1
 2 1
1
 k   1 1  a 
 c    2 1  b 
  
  
1  1 1  a  1  a  b   k 
 


3  2 1  b  3  2a  b   c 
ab
b  2a
and c 
therefore we can write an arbitrary
3
3
vector w  k u  c v which means that these vectors u and v span or generate 2 .
We can illustrate these vectors in 2 as:
w
a
Any vector w    in 2
b
can be written as:
 1
v 
ab
 b  2a 
 1
w
u  
v
 3 
 3 
1 
u 
 2
We have found the scalars k 
Fig 41
 3
1 
 1
(ii) How do we write the vector   in terms of u    and v    ?
 2
 1
 2
We can use part (i) with a  3 and b  2 because we have shown above that any vector in
2
can be generated by the vectors u and v. Substituting these, a  3 and b  2 , into the
ab
b  2a
above derived scalars, k 
and c 
, gives
3
3
a b 3 2 5
k


3
3
3
b  2a 2  2  3 
4
c


3
3
3
3
5
4
u 
v    . This is illustrated below:
We have k u  c v 
3
3
 2
5
u
3
Fig 42
 3
 
 2
4
 v
3
You may like to check this by arithmetic means:
3
5
4
u v  
3
3
 2
3
The vector   is
 2
made up of adding
5
4
u and  v .
3
3
Chapter 3: Euclidean Space
56
Example 20
 1
0
Show that the vectors u    and v    do not span 2 .
 2 
0
Solution
a
Step 1. Let w    be an arbitrary vector in 2 .
b 
Step 2. Consider the linear combination k u  c v  w where k and c are real
 1
0
a
scalars. Substituting u    , v    and w    into this k u  c v  w :
 2 
0
b 
 1
 0
k u  cv  k    c  
 2 
 0
u
v
 k   0  k   a 

  
 
  2 k   0   2 k   b 
We have the simultaneous equations
k a
2k  b or b  2a

 a 
 Because w    
 b 

 Because
k  a
 a
 1
This case only works if b  2a that is for the vector w  
  a   . Vectors u and v
 2a 
 2 
 1
only span vectors which are scalar multiplies of   :
 2 
The given vectors
u and v only span
vectors along this
line (1 – space).
Fig 43
Since we want the vectors u and v to span all of 2 therefore it has to work for any real
a
values of a and b , that is w    , not just when b  2a . We conclude that the vectors
b 
u and v do not span the whole of 2 . Any vector away from the line in Fig 43 cannot be
made by a linear combination of vectors u and v.
E2 Basis
In this subsection we discuss the nature of a basis for the n space, n . A basis is a set of
vectors which describes the n space.
and v n in the n space, n .
Definition (3.28). Consider the n vectors v1 , v 2 , v3 ,
These vectors form a basis of n 
v1 , v 2 , v3 ,
and v n span n and
(i)
v1 , v 2 , v3 ,
and v n are linearly independent
(ii)
We can write the vectors v1 , v 2 , v3 ,
and v n as a set B  v1 , v2 , v3 ,
, vn  .
Chapter 3: Euclidean Space
57
These are called the basis vectors.
Example 21
 1
 1
Show that the vectors u    and v    forms a basis for 2 .
 1
 1
Solution
How do we show the given vectors u and v forms a basis for 2 ?
Required to show two things, that u and v - (i) span 2 and (ii) are linearly independent.
(i) How do we verify that u and v span 2 ?
a
Apply the above stated procedure. Let w    be an arbitrary vector in 2 and consider
b 
the linear combination k u  c v  w . Substituting the given vectors u, v and w
yields
 1  1   a 
1 1   k   a 
k    c      or in matrix form 
    
1  1  b 
 1 1  c   b 
u
v
w
The determinant of the 2 by 2 matrix is:
1 1 
det 
   1  1  2 [Non Zero]
 1 1
1
1 1 
1  1 1 1 1 1 
The inverse of the 2 by 2 matrix is 
  
 
 . Hence
2  1 1  2 1 1
1 1
1
 k  1 1 
 c    1 1
  

1
1
Thus k   a  b  and c   a  b 
2
2
 a  1 1 1   a  1  a  b 
 b   2  1 1  b   2  a  b 
 

 


a 1
1
which means w      a  b  u   a  b  v
2
b 2
because w  k u  c v . Therefore we conclude that the vectors u and v span
a
w 
b
2
:
1
u 
1
 1
v 
 1
Fig 44
(ii) What else do we need to show for vectors u and v to be a basis?
We need to verify that they are linearly independent. How?
By showing k u  c v  O gives k  c  0 .
Let w be the zero vector, O, and substitute into part (i) with a  0 and b  0 :
0 1
1
1
1


w  O      0  0 u   0  0 v
Because w   a  b  u   a  b  v 

2
2
2


0 2
 0u  0 v
Chapter 3: Euclidean Space
58
This means that k u  c v  O implies that k  0 and c  0 . Since both scalars (k
and c) are zero so vectors u and v are linearly independent.
These vectors (u and v) both span 2 and are linearly independent therefore we conclude
that they are a basis for 2 .
2
The vectors e1 and e2 (illustrated in Fig 35 on page 41) are another basis for
generally called the natural basis for
and e3 are a natural basis for
natural basis for
n
3
2
. This is
. Similarly we can show that the vectors e1 , e2
. In general the vectors e1 , e2 , e3 ,
and en form a
. Sometimes the natural basis is called the standard basis.
E3 Properties of Basis
Proposition (3.29). Any n linearly independent vectors in n form a basis for n .
Proof. Exercise 3(e).
Note that this is an important result because it means that given n vectors in the n – space,
n
, it is enough to show that they are linearly independent to form a basis. We don’t need
to show that they span n as well.
Proposition (3.30). Any n vectors which span n form a basis for n .
Proof. Miscellaneous Exercise.
Again only need to show that n vectors span n to prove that they are a basis for n .
For the next example you will need to revise your work from the last section on linear
independence. Remember Ax  O has the only solution x  O provided
det  A   0 [Not Zero]
From the last section we know that if A   v1
v1 , v 2 ,
v2
vn  then these vectors
, v n are linearly independent  det  A   0 .
Example 22
1 
 0
 2 
Show that the vectors u   0  , v   1  and w   3  forms a basis for
 
 
 
1 
 1 
 0
 
 
 
Solution
1
u   0 
1
The given vectors
 
u, v and w are
illustrated in this
 2 
figure.
w   3 
 0
 
Fig 44
 0
v   1 
 1 
 
3
.
Chapter 3: Euclidean Space
59
What do we need to show?
Required to prove that the given vectors are linearly independent in 3 . Why don’t we have
to also prove that these vectors span 3 ?
Because by the above Proposition (3.29) it is enough to show that 3 vectors are linearly
independent for them to be a basis for the 3 – space, 3 .
We show linear independence by proving that the linear combination k1u  k2 v  k3w  O
has the only solution k1  0, k2  0 and k3  0 . How do we prove this?
By showing that the determinant of A   u v w  is non-zero. Let A   u v w  :
 1 0 2 
 k1 

 and
 
A   0 1 3
x   k2 
 1 1 0 
k 


 3
Evaluating the determinant of this gives
 1 0 2 
 1 3
 0 1


det  A   det  0 1
3   det 
  2 det 

 1 0 
 1 1
 1 1 0 


  0  3  2  0  1  5
We have det  A   5  0 [Not Zero] therefore Ax  O has the only solution x  O which
means that all the scalars are zero, k1  k2  k3  0 .
We conclude that the given vectors u, v and w are linearly independent.
By Proposition (3.29) the given vectors u, v and w form a basis for 3 . Hence any vector x
in 3 can be written as x  k1 u  k2 v  k3 w . The given vectors are illustrated:
If determinant of A   v1
vectors do not form a basis for
v1,
v2 ,
Example 23
vn  is zero then we conclude that the given set of
v2
n
. Since det  A   0 means the given vectors
, v n  are linearly dependent.
 1
 9
5
0
 
 
 
 
4
3
2
0
Determine whether the vectors v1    , v 2    , v 3    and v 4    forms a
 3 
 1 
 11 
0
 
 
 
 
 6
 6 
 1
0
4
basis for
.
Solution
Since v 4  O is the zero vector therefore by:
Proposition (3.24) of the last section which says that if one (or more) is the zero vector(s)
then the vectors are linearly dependent.
Therefore the vectors v1 , v 2 , v3 and v 4 form a linearly dependent set which means that
these vectors cannot form a basis for 4 . Remember for a basis we need the vectors to be
both linearly independent and span 4 .
The remaining work in this section is more difficult than the above. You will need to know
the definitions of spanning set, linear independence and basis of n .
Chapter 3: Euclidean Space
60
Proposition (3.31). Let the vectors v1 , v 2 , v3 ,
and v n be a basis for
n
. Every vector
n
in
can be written uniquely as a linear combination of the vectors in the basis
v1 , v 2 , v3 ,
and v n .
What does this proposition mean?
There is only one way of writing any vector as a linear combination of the basis vectors.
Proof.
Let u be an arbitrary vector in n . We are given that the vectors v1 , v 2 , v3 ,
and v n
forms a basis therefore they span n which means that we can write any vector in n as a
linear combination of v1 , v 2 , v3 ,
and kn
and v n . There exists scalars k1 , k2 , k3 ,
which satisfy
u  k1v1  k2 v 2  k3v3 
 kn v n
Suppose we can write this vector u as another linear combination of the basis vectors
u  c1v1  c2 v 2  c3v3 
 cn v n
where the c’s are real scalars. What do we need to prove?
Required to prove that the two sets of scalars are equal: k1  c1 , k2  c2 ,
and kn  cn .
Equating the two linear combinations because both are equal to u gives
k1v1  k2 v 2 
 kn v n  c1v1  c2 v 2 
 cn v n
k1v1  k2 v 2
 k1  c1  v1


 kn v n
 k2  c2  v 2

 c1v1  c2 v 2

 kn  cn  v n

 O
 cn v n  O
Collecting Terms
and v n are linearly independent therefore all the
The basis vectors v1 , v 2 , v3 ,
scalars are equal to zero.
k1  c1  0, k2  c2  0, k3  c3  0,
and kn  cn  0
k1  c1 , k2  c2 , k3  c3 ,
and kn  cn
Hence the arbitrary vector u can be written uniquely as a linear combination of the basis
and v n .
vectors v1 , v 2 , v3 ,
■
The next two proofs are of the type “proof by contradiction”. The procedure to prove
statement P implies statement Q denoted P  Q for proof by contradiction is to suppose
not Q is true and then derive at a contradiction. For example in the field of Number
Theory to prove that there are an infinite number of primes we suppose there are a finite
number of primes and then derive a contradiction to this supposition. Proof by
contradiction is a common technique in mathematics and is a powerful tool to prove a
mathematical result.
Next we prove a lemma. A lemma is like a proposition or theorem which means it is a
statement which requires a proof. Normally a lemma is a statement which is used to prove a
more important result. However there are many lemmas in mathematics which have
become important results in themselves such as Zorn’s lemma, Euclid’s lemma and
Gauss’s lemma.
Lemma (3.32). If the set of vectors S  v1 , v 2 , v3 , , v n  span n and
T  w1 , w2 , w3 , , w m  are a set of m vectors which are linearly independent in
then m  n .
What does this lemma mean?
Means that there can only be n or less than n independent vectors in n .
n
Chapter 3: Euclidean Space
61
Proof.
Suppose m  n then by Proposition (3.25) of the last section which says that if the number
of vectors is greater than the value of n in the n –space, n , then the vectors are linearly
dependent. This contradicts that T is a set of vectors which are linearly independent.
Hence we have our required result m  n .
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Proposition (3.33). Every basis of n contains exactly n vectors.
Proof.
Let B  v1 , v2 , v3 ,
v m  be a basis for n . We consider two cases, where m  n and
m  n . In each case we arrive at a contradiction which means that m  n [m cannot be
greater than n] and m  n [m cannot be less than n] therefore m  n .
Case I: Suppose m  n , that is m is greater than n, then by Proposition (3.25) of the last
section which says that if the number of vectors is greater than the value of n in the n –
space then the vectors are linearly dependent. Hence the vectors in the set B cannot be a
basis because they are linearly dependent. This contradicts that B is a basis. Therefore we
cannot have m  n .
Case II: Suppose m  n , that is m is less than n. The set
B  v1 , v2 , v3 ,
vm
is a basis therefore it spans
n
. Let
T  e1 , e2 , e3 ,
, en 
be the natural basis of n . Then the vectors in the set T are linearly independent and so by
the above Lemma (3.29) we have n  m . This n  m can be written the other way round,
that is m  n . This contradicts our supposition m  n . Hence m cannot be less than n.
Cases I and II prove that we cannot have m  n or m  n therefore m  n .
Hence every basis of n has exactly n vectors.
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This result means that in a basis there are enough vectors to span (generate) the whole n –
space, n , but not too many so that the vectors become dependent. Hence we need exactly
n vectors to form a basis for n .
Proposition (3.34). Any n non-zero orthogonal vectors in n are linearly independent.
Proof. Exercise 3(e).
SUMMARY
The set of vectors S  v1 , v 2 , v3 , , v n  is said to span or generate
can be produced by a linear combination of these vectors.
The set S is a basis for n 
(i) S  v1 , v 2 , v3 , , v n  spans n and
(ii) S  v1 , v 2 , v3 ,
, v n  is linearly independent
Every vector in n can be written uniquely in terms of its basis vectors.
Every basis of n contains exactly n vectors.
The unit vectors e1 , e2 , e3 , , en  form a natural basis for n .
n
if every vector