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1.7 Linear Independence
A homogeneous system represented by the a
matrix equation such as
1 2  3  x1  0
3 5 9   x   0

 2   
5 9 3   x3  0
Can be viewed as a vector equation
1
 2
 3 0
x1 3  x2 5  x3  9   0
9
5
 3  0
This version emphasizes the vectors instead
of the solution.
Clearly, this has the trivial solution.
Does it have any other solutions?
Definition: A set of vectors v1 , v 2 ,, v p  in Rn
is said to be linearly independent if the vector
equation x1v1  x2 v 2    x p v p  0 has only the
trivial solution.
1
The set v1 , v 2 ,, v p  is said to be linearly
dependent if there exist weights c1 , c2 ,, c p
not all zero, such that
c1v1  c2 v 2    c p v p  0 .
Example: Let
1
 2
 3
v1  3 v 2  5 v 3   9 
 3 
5
9
Are they linearly independent?
To answer this, we ask the following question:
What are the solutions to the equation
 2
1
 3 0
x1 3  x2 5  x3  9   0
   
9
5
 3  0 ?
The corresponding augmented matrix is:
1 2  3 0 1 0 33 0
3 5 9 0 

0
1

18
0

 

5 9 3 0 ~ 0 0
0 0
2
Note: x3 is free, so  an infinite number of
solutions. The trivial solution is not the only
solution,  the vectors are not linearly
indeptndent.
Can we find a linear dependence relation
among the vectors?
Note: the solution to the homogeneous
equation is x1 = -33x3 , x2 = 18x3, and x3 is
free.
We can choose any non-zero value for x3, say
x3 = 1.
Then x1 = –33, x2 = 18, x3 = 1 is a non-trivial
solution and
2  3 0
1
 333  185  1 9   0
   
9  3  0
5
Write the matrix equation with this non-trivial
solution:
3
1 2  3   33 0
3 5 9   18   0


  
5 9 3   1  0
A
x = 0
In this linear dependence relation, each value
of x3 gives a non-trivial solution.
Fact: The columns of the matrix A are linearly
independent iff Ax = 0 has only the trivial
solution.
Special Cases:
1. A set with one vector:
Consider the set containing only one vector
{v}. The only solution to
x1v = 0 is x1 = 0.
Thus, any set w/ only one vector is linearly
independent.
So, {v} is linearly independent when v ≠ 0.
When v = 0, x1v = 0 has infinitely many
solutions.
4
2. Set of two vectors
Example:
4
 2
 2
 2
u1    u2    v1    v 2   
2 ,
1 ,
3
1  ,
“Is {u1, u2} linearly independent or linearly
dependent?”
Notice that u2 = 2u1, so
2u1 + –u2 = 0. This is not the trivial solution,
so {u1, u2} is linearly dependent.
What about {v1, v2}? Is this set linearly
independent or linearly dependent?
We need to look at the possible solutions of
cv1 + dv2 = 0.
Suppose cv1 + dv2 = 0, and c, d ≠ 0.
v1  
d
v2 ,
c
but this is impossible since v1 is not
a multiple of v2.
5
c
v 2   v1
v
Also, we could solve for 2 :
but
d
this is also impossible since v2 is not a
multiple of v1.
So, the assumption c, d ≠ 0 must be false and
c = d = 0.
So, {v1, v2} linearly independent.
Fact: A set of two vectors is linearly
dependent if at least one vector is a multiple
of the other. And a set of two vectors is
linearly independent iff neither of the vectors
is a multiple of the other.
Theorem 1.7: An indexed set S  v1 , v 2 ,..., v p 
of two or more vectors is linearly dependent iff
at least one of the vectors in S is a linear
combination of the others.
In fact, if S is linearly dependent and v1 ≠ 0,
then some vector vj (j ≥ 2) is a linear
combination of the preceding vectors.
6
3. A Set Containing the zero vector 0 is
always linearly dependent. This is Theorem
1.9.
Theorem 1.9: A set of vectors
S  v1 , v 2 ,, v p  containing the zero vector is
linearly dependent.
Proof: Let v1 , v 2 ,, v p  in Rn be a set
containing the zero vector. We can arrange
the set so that the zero vector is the first one
in the list: 0, v 2 ,, v p 
Now the equation x1v1  x2 v 2    x p v p  0
becomes x1 0  x2 v 2    x p v p  0 .
Thus, x1 can be any real number, so there are
infinitely many solutions to the equation.
4. A Set Containing Too Many Vectors
Theorem 1.8: If a set contains more vectors
than there are entries in each vector, then the
set is linearly dependent (i.e. any set
v1, v 2 ,, v p  is linearly dependent if p > n).
7


Proof: v1 v 2  v p is an n x p matrix.
If p > n, then Ax = 0 has more variables than
equations, so it has non-trivial solutions.
Thus, the columns of A are linearly
dependent.
Example: With the least amount of work
possible, decide which of the following sets is
linearly independent, and give a reason.
  3  9  


a. 2, 6 
 1   4  
    
This is linearly independent since it is a set of
two vectors, and neither is a multiple of the
other.
1
6
b. 
9

4
2 3 4 5
7 8 9 0
8 7 6 5

3 2 1 8
8
This is linearly dependent by Theorem 1.8,
since  more column vectors than entries per
vector.
3 9 0 


c. 2, 6, 0 
 1   3  0  
      
This is linearly dependent since it contains the
zero vector (Theorem 1.9) and v2 = 3v1.
 8  
  
  2 
d.  


1
 
4  This set is linearly independent
since any set containing one non-zero vector
is linearly independent.
9
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