Chapter 3: Euclidean Space
41
SECTION D Linear Independence
By the end of this section you will be able to
 understand what is meant by linear combination and linear independence
 show that given vectors are linearly independent or dependent
 prove properties about linear independence
D1 Standard Unit Vectors in n
What does the term standard unit vectors mean?
From the last section we know that unit vectors are of length 1 and the standard unit vectors
in n are column vectors with 1 in the kth position of the vector ek and zeros everywhere
else.
y
e2 =
z
0 
 
1 
e1 =
e3
e1
1 
 
0 
x
x
e2
kth position
y
0
 
 
ek   1 
 
0
 
 
Fig 35
Why are these standard unit vectors important?
Well we can write any vector of n in terms of the standard unit vectors as we showed in
Exercise 3a question 14. We proved the following Proposition:
 x1 
 
 
Proposition (3.20). Let u be any vector in n given by u   xk  . Then we have
 
 
x 
 n
u  x1e1  x2e2 
 xk ek 
 xnen
Proof. See solution to question 14 of Exercise 3a.
 xk ek 
 xnen is called a
This representation u  x1e1  x2e2 
and en . We can write this
linear combination of the standard unit vectors e1 , e2 ,
u  x1e1  x2e2  x3e3 
 xnen in matrix form as
u   e1
The matrix  e1
e2
e2
e3
e3
en 
 x1 
 
en   
x 
 n
has its first column entries e1 , its second column
is e2 , , and the nth column is en .
Moreover we can say this representation u  x1e1  x2e2 
 xnen of a vector is unique.
Chapter 3: Euclidean Space
42
Proposition (3.21). Let u be any vector in n then the linear combination
u  x1e1  x2e2 
 xk ek 
 xnen
is unique.
What does this proposition mean?
Means that there is only one way of writing any vector u of n in terms of the standard
unit vectors e1 , e2 ,
and en .
Proof.
Let the vector u be written as another linear combination:
u  y1e1  y2e2 
 yk ek 
 ynen
What do we need to show?
Required to show that the scalars are equal: y1  x1 , y2  x2 , y3  x3 ,
and yn  xn .
Equate the two linear combinations because both are equal to u:
x1e1  x2e2 
 xnen  y1e1  y2e2 
 yn e n
x1e1  x2e2
 x1  y1  e1


 xnen
 x2  y2  e2
Writing out e1 , e2 ,

y1e1 

y2e2



 xn  yn  en
yn e n  O
 O
 Factorising 
, en and O as column vectors in the last line we have

1 
 0
 0  0
 0 

 
 
   
 
0
1
0  0




Remember ek    
 x1  y1      x2  y2       xn  yn      

 1 
0

 
 
   
 
0
 
 1  0
 0  

We can write this in matrix form as
0   x1  y1   0 
 x1  y1   0 
1 0
I is the 

  

 x  y   
x

y
0
1
0
0
0
2
2
2
2
identity 
    implies I 
 










 
 
0
 matrix 
  

  


1   xn  yn   0 
0
 xn  yn   0 
and xn  yn  0 which gives our
Hence we have x1  y1  0, x2  y2  0, x3  y3  0,
and yn  xn .
result: y1  x1 , y2  x2 , y3  x3 ,
Therefore the given linear combination, u  x1e1  x2e2   xk ek   xnen , is unique. ■
D2 Linear Independence
Example 12
Determine the linear combination of the standard unit vectors in n which gives the zero
and xn in the following:
vector. This means find the values of the scalars x1 , x2 , x3 ,
x1e1  x2e2 
 xk ek 
 xnen  O
Solution
and en we have
Substituting e1 , e2 ,
1
0
 
 
0
1
x1    x2   
 
0
 
 
0
 
0 0
   
0
0
 xn     
   
   
1 0
Chapter 3: Euclidean Space
43
We can write this in matrix form as
1
0   x1   0 

   
[This is Ix  O ]

    

   
1   xn   0 
0
In compact form we have Ix  O where I is the identity matrix. Ix  O gives x  O .
From the zero vector x  O we have x1  0, x2  0, x3  0,
and xn  0 .
We say that the standard unit vectors e1 , e2 ,
and en are linearly independent which
means that any one of the vectors ek cannot be made by a linear combination of the others.
These standard unit vectors are not the only vectors in
Definition (3.22). Generally we say vectors v1 , v 2 , v3 ,
n
which are linearly independent.
and v n in n are linearly
independent  the only real scalars k1 , k2 , k3 ,
and kn which satisfy:
k1v1  k2 v 2  k3 v3   kn v n  O are k1  k2  k3   kn  0
What does this mean?
The only solution to the linear combination k1v1  k2 v 2 
 kn v n  O is
when all the scalars k1 , k2 , k3 ,
and kn are equal to zero. In other words you cannot
make any one of the vectors v j say, by a linear combination of the others.
We can write the linear combination k1v1  k2 v 2  k3 v3 
v
1
v2
The first column of the matrix  v1
v3
v2
v3
 kn v n  O in matrix form as
 k1   0 
   
k
0
vn   2    
   
   
 kn   0 
v n  is given by the entries in v1 ,
the second column is given by the entries in v 2 and the nth column is given by the entries
in v n .
Recall from chapters 1 and 2 that a linear system Ax  O has the unique solution x  0
 matrix A is invertible or det  A   0 [Not zero].
How is this related to linear independence?
v n  then these vectors v1 , v 2 , … and v n are
Well if the matrix A   v1 v 2
linearly independent  Ax  O has the unique solution x  O . What does this mean?
v n  is invertible or
Means we only need to check that the matrix A   v1 v 2
det  A   0 [not zero] for the vectors v1 , v 2 , … and v n to be linearly independent.
Example 13
 1
 2
Show that u    and v    are linearly independent in 2 and plot them.
 1
 3
Solution
Using the above definition (3.22) and writing out the linear combination we have
k u  c v  O [Using k and c as scalars]
Substituting the given vectors u and v into this k u  c v  O gives
Chapter 3: Euclidean Space
44
k  2c  0
 1  2   0 
k    c      or
k  3c  0
 1  3   0 
 1 2 
k 
0
Let A   u
v  
 and x    . We need to solve Ax  O where O    .
 1 3
c
0
 1 2 
Since det  A   det 
   1 3  1 2   5 is non-zero so the only solution is
 1 3
x  O [Because Ax  O and det  A   0 then x  O ] which means the values of the
scalars are k  0 and c  0 .
Hence the linear combination k u  c v  O yields k  0 and c  0 therefore the
given vectors u and v are linearly independent because all the scalars, k and c, are equal
to zero. Plotting the given vectors we have:
 2
v 
3
 1
u 
 1
Fig 36
This means that vector u is not a scalar multiple of vector v and vice-versa.
In 2 the vectors u and v are linearly independent means that they are not scalar multiples
of each other. Plotting linear independent vectors:
v
Linearly independent vectors u
and v
u
Fig 37
D3 Linear Dependence
What does linear dependence mean?
Definition (3.23). Conversely we have the vectors v1 , v 2 , v3 ,
linearly dependent  the scalars k1 , k2 , k3 ,
k1v1  k2 v 2  k3 v3 
and v n in
n
are
and kn are not all zero and satisfy
 kn v n  O
From chapters 1 and 2 we know that the linear system
Ax  O
has an infinite number of solutions if det  A   0 . This means that if
 k1 
 
A   v1 v2
vn  , x   
k 
 n
and v n are linearly dependent. Why?
and det  A   0 then these vectors v1 , v 2 , v3 ,
Because det  A   0 means the linear system Ax  O has an infinite number of solutions
so there must be non-zero solutions, that is x  O . Hence there are non-zero k’s which
satisfy
Chapter 3: Euclidean Space
45
k1v1  k2 v 2  k3 v3 
 kn v n  O
Example 14
 3 
 1 
2
Show that u    and v  
and plot them.
 are linearly dependent in
 1
 1/ 3 
Solution
Using the above definition (3.23) and writing the vectors u and v as a linear combination
we have
k u  c v  O [k and c are scalars]
Substituting the given vectors
 3   1   0 
ku  cv  k    c 
 
 1   1/ 3   0 
You can check the determinant of A   u
v  is equal to zero. However this does not
give us the relationship between the two vectors u and v. We can write out the argumented
O  and carry out row operations to determine the relationship between u and
matrix  A
v. The augmented matrix for this is given by
R1  3 1
0
 Because  u v
O  


R 2  1 1/ 3 0 
where R 1 and R 2 represent rows. Carrying out the row operation 3R 2  R1 gives
k c
 3 1 0 


3R 2  R1  0 0 0 
From the top row we have 3k  c  0 which implies c  3k . Let k  1 then
c  3k  31  3
Substituting our values k  1 and c  3 into k u  c v  O gives
u  3v  O or u  3v
We have found non-zero scalars, k  1 and c  3 , which satisfy k u  c v  O
therefore the given vectors u and v are linearly dependent and u  3v .
Plotting the given vectors u and v we have:
R1
 3 
u 
 1
 1 
v

 1/ 3 
Fig 38
Note that u  3v which means that the vector u is a scalar multiple  3 of the vector v.
If vectors u and v in
2
are linearly dependent then we have
k u  c v  O where k  0 or c  0
That is at least one of scalars is not zero. Suppose k  0 then
Chapter 3: Euclidean Space
46
k u  c v
c
u   v  Dividing by k 
k
This means that the vector u is a scalar multiple of the other vector v which suggests that u
is in the same (or opposite) direction as vector v. Plotting these we have:
u
Fig 39
v
Linearly dependent vectors u and v
Of course in the above Example 14 we could have let k  2, 3,  , 666 etc. Any non-zero
number will do! Generally it makes the arithmetic easier if we use k  1 .
Example 15
 2
 3 
 0




Determine whether the following vectors u   1  , v   1  and w   0  in 3 are
 0
 1 
0
 
 
 
linearly dependent or independent.
Solution
Consider the following linear combination
k1u  k2 v  k3w  O
where k’s are scalars.
Substituting the given vectors u, v and w into this, k1u  k2 v  k3w  O , yields
Let A   u
Expanding
along this
row
 2  0
 3 
 0
   




k1u  k2 v  k3w  k1  1   k2  1   k3  0    0 
 0
 1 
 0  0
 
 
   
 3 0 2 
 k1 

 and
 
v
w   1 1 0
x   k2  . We need to solve Ax  O :
 0 1 0 
k 


 3
 3 0 2 
 3 2 


det  A   det  1 1 0     1 det 
  0  2  2
1
0


 0 1 0 


Because det  A   2 is non-zero therefore Ax  O only has the solution x  O .
Solution x  O means k1  k2  k3  0 , that is all the scalars are zero:
k1u  k2 v  k3w  O implies k1  k2  k3  0
Hence the given vectors u, v and w are linearly independent.
Example 16
 2
1
 3 
 0
 
 




Determine whether the following vectors u   1  , v   1  , w   0  and x   2  in
 0
 1 
0
 3
 
 
 
 
3
are linearly dependent or independent.
Solution
Chapter 3: Euclidean Space
47
Consider the linear combination
k1u  k2 v  k3w  k4 x  O
What do we need to find?
Need to determine whether all the scalars k’s - k1 , k2 , k3 and k4 are zero or not.
Substituting the given vectors into this linear combination:
 2
1   0 
 3 
 0
 
   




k1u  k2 v  k3w  k4 x  k1  1   k2  1   k3  0   k4  2    0 
 0
 1 
0
3  0
 
 
 
   
The augmented matrix of this is given by
R1  3 0 2 1
0


 Using  u v w x
R2  1 1 0 2
0
O  
R 3  0 1 0 3
0 
Carrying out the row operation R 2  R 3 gives:
k1 k2 k3 k4
 3 0

R2  R3  1
0

R3
 0 1
From the bottom row we have k2  3k4 
R1
0

0 5
0
0 3
0 
0 which gives k2
2 1
 3k4 . Let k4  1 :
 3k4  3 1  3
k2
From the middle row we have k1  5k4  0 implies that k1  5k4  5 1  5 .
The top row gives 3k1  2k3  k4  0 . Substituting k1  5 and k4  1 into this:
3  5  2k3  1  0
implies
k3  8
Our scalars are k1  5 , k2  3 , k3  8 and k4  1 . Substituting these into the above linear
combination k1u  k2 v  k3w  k4 x  O gives the relationship between the
vectors:
5u  3v  8w  x  O or x  5u  3v  8w
Since we have non-zero scalars (k’s) so the given vectors are linearly dependent.
We can write the
vector:
x  5u  3v  8w
1
 
x   2
 3
 
 3 
 
u   1
 0
 
 0
 
v   1
 1 
 
 2
 
w  0
0
 
Fig 40
In general if vectors u , v and w in
3
are linearly dependent then we have
Chapter 3: Euclidean Space
48
k1u  k2 v  k3w  O gives k1  0 or k2  0 or k3  0
That is at least one of scalars is not zero. Suppose k1  0 then
k1u   k2 v  k3w
k 
k 
u    2 v   3 w
 Dividing by k1 
 k1 
 k1 
This means that the vector u in 3 is a scalar multiply of the other two vectors v and w.
Hence we can make vector u out of vectors v and w.
In the above Example 16 we have x  5u  3v  8w which means we can make the
vector x out of the vectors u, v and w.
D4 Properties of Linear Dependence
Proposition (3.24). Let v1 , v 2 , v3 ,
and v n be vectors in n . If one (or more) of these
and v n are linearly
vectors, v j say, is the zero vector then the vectors v1 , v 2 , v3 ,
dependent.
Proof.
Consider the linear combination
k1 v1  k2 v 2  k3 v 3 
 kjv j 
 kn v n  O
(*)
In (*) take k j  0 [Non-zero number] and all the other scalars equal zero, that is
k1
 k2
 k3

 k j 1
 k j 1

 kn
 0
Since v j  O we have k j v j  O which means that all scalars in (*) are not zero ( k j  0 ).
By definition (3.23) we have that the vector v1 , v 2 ,
and v n are linearly dependent.
■
What does Proposition (3.24) mean?
and v n , one of these is the zero vector then they
If amongst the vectors v1 , v 2 , v3 ,
are linearly dependent.
and v m be distinct vectors in n . If n  m , that is
Proposition (3.25). Let v1 , v 2 , v3 ,
value of n in the n – space is less than the number, m , of vectors, then the vectors
v1 , v 2 , v3 ,
and v m are linearly dependent.
Proof.
and v m :
Consider the linear combination of the given vectors v1 , v 2 , v3 ,
k1v1  k2 v 2 
 kn v n  kn1v n1 
 km v m  O
(*)
The number of equations is n because each vector belongs to the n – space n but the
number of unknowns k1 , k2 , k3 , , kn , kn1 , , km is m. Writing this out we have
 v11 
 v21 
 
 
v12 
v

k1
 k2  22  
 
 
 
 
 v1n 
 v2 n 
 v n 11 
 vn1 


 
v
v

n

1
2
  
 kn  n 2   kn 1 

 


 


v
v
 nn 
  n 1n 
 vm1   0 


vm 2   0 

 km


  

  
 vmn   0 



 n equations


m unknowns
By the following Proposition (1.24) case (II) of chapter 1:
In a linear system, if the number of equations is less than the number unknowns then the
system has an infinite number of solutions.
(3.23) If non-zero k’s satisfy k1v1  k2 v 2 
 kn v n  O then vectors v’s are dependent
Chapter 3: Euclidean Space
49
In our linear system the number of equations n is less than the number of unknowns m.
Therefore we have an infinite number of k’s which satisfy (*) and this means all the k’s are
not zero. By Definition (3.23) the given vectors are dependent.
Example 17
Determine whether the following vectors
 2
 2 
2
 4 
 
 




u   3  , v   19  , w   6  and x   2 
 7
 5 
 3
 5 
 
 
 
 
3
in
are linearly dependent or independent.
Solution
Since we have 4 vectors u, v, w and x in 3 and 4  3 therefore by the above Proposition
(3.25) we conclude that the given vectors u, v, w and x are linearly dependent.
Normally we write the vectors as a collection in a set. A set is denoted by { } and is a
collection of objects. The objects are called elements or members of the set.
We can write the set of vectors v1 , v 2 , v3 ,
and v n as
S  v1 , v 2 , v3 ,
, vn 
We use the symbol S for a set.
Proposition (3.26). Let S  v1 , v 2 , v3 ,
, v n  be n vectors in the n - space
be the n by n matrix whose columns are given by the vectors v1 , v 2 , v3 ,
n
. Let A
and v n :
A   v1 v2
vn 
If the reduced row echelon form (rref) of A contains a zero row then the set of vectors
S  v1 , v 2 , v3 , , v n  are linearly dependent.
Proof.
See Exercise 3(d).
SUMMARY
and v n be vectors in n and k1 , k2 , k3 ,
and kn be scalars.
Let v1 , v 2 , v3 ,
Consider the following linear combination
k1v1  k2 v 2  k3 v3 
 kn v n  O
(*)
 kn  0 (all scalars are
If the only solution to this (*) is k1  k2  k3 
and v n are linearly independent.
zero) then the vectors v1 , v 2 , v3 ,
and kn are not all zero satisfy (*) then the vectors
If the scalars k1 , k2 , k3 ,
v1 , v 2 , v3 ,
and v n are linearly dependent.
and v n is zero then they are linearly
If one or more of the vectors v1 , v 2 , v3 ,
dependent.
and v m in n where n  m then these vectors
If there are m vectors v1 , v 2 , v3 ,
v1 , v 2 , v3 ,
and v m are linearly dependent.