Overview of Next 2 Weeks
Some of the main results of linear algebra will be presented:
For any vector space (covered last week), we will study the relationship between the
dimension of the solution of a linear system (the nullity) and the ‘actual’ number of
constraints provided by the linear system (linear independence of the equations, rank of
the corresponding augmented matrix).
Linear Dependence
Definition : Linear Dependence of a Vector
Linear dependence of a vector w: A vector w is linearly dependent on a set of
vectors{v1,v2,…,vk} if w is equal to a linear combination of {v1,v2,…,vk}. I.e., if
w=a1v1+a2v2+…+akvk for some scalars a1, …, an.
Example: Write down 2 vectors that are linearly dependent on {[0,2], [1,1]}.
The vector [?,?] is linearly dependent on {[0,2], [1,1]}.
The vector [??,??] is linearly dependent on {[0,2], [1,1]}.
1 0 1 0 
Example: Write down 2 vectors that are linearly dependent on { 
, 
 .
1 0 0 1 
? ?
?? ??
The vector 
and 
.

?? ??
? ?
Property of a set: If a vector in a set can be expressed as a linear combination of the other
vectors, then the other vectors can also be expressed as linear combination of other
vectors in the set.
Example:
Suppose w  2u  3v (thus w is linearly dependent on u and v)
Then we also have:
u  1 w 3 v
(thus u is linearly dependent on v and w)
2
2
v 2 u 1 w
(thus v is linearly dependent on u and w)
3
3
Definition : Linear Dependence of a Set
Linear dependence of a set: A set that contains a linearly dependent vector is a linearly
dependent set.
Example: {[0,2],[1,1],[?,?]} is a linearly dependent set.
Three Equivalent Understandings of a Linearly Dependent Set
1. A set of vectors is linearly dependent if one of vectors can be expressed as a linear
combination of the other vectors. (definitional)

1
 4
11

1
 2
7





S   u  ,v 
,w  




9
1
3





 

1
1
5






Example:
This set is linearly dependent because
w=3u+2v.
2. A set of vectors is linearly dependent if the matrix generated with each row being one
of the vectors has a row of zeros in its row reduced echelon form. (Recall, that
forming a row of zeros means that you ‘knocked it out’ by subtracting a multiple of
another row so this makes sense.).
 1 1 1 1
M   4 2 3 1 ,
11 7 9 5
1 0

RREF M   0 1

0 0
1 
2
3 
1
2
2 
0
0 

1
2
Rank = 2
3. Test for Linear Dependence: A set of vectors v1 , v 2 ,..., v n is linearly dependent if
the following equation has a non-trivial solution:
a1 v1  a 2 v 2  ...a n v n  0
a. First, observe that setting all coefficients equal to 0 is a solution. This is always the
case for homogeneous equations.
b. Second, observe that if there is a non-trivial solution, then you can express any of the
vectors as a linear combination of the others.
c. Observe in this example how solving for the non-trivial solution forms a link between
linear dependence and zero rows in either the row matrix or its transpose; and linear
dependence with the existence of infinite solutions:
1
 4
11 0
1
 2
 7  0 
a1    a 2    a 3     
1
3
 9  0 

 
   
1
1
 5  0 
yields
augmented
matrix
1

1
1

1
4 11
2
3
1
7
9
5
0

0
0

0
which row
reduces to
1

0
0

0
0 3
1 2
0 0
0 0
0

0
0

0
Rank = 2 (not a coincidence), and we have a non-pivot column that represents a
free parameter
Two important results that are coming up:


The rank of a matrix M is equal to the rank of the transpose MT and corresponds to
the number of pivots which corresponds to the number of linearly independent
vectors.
Linear independence corresponds to fewer solutions of the corresponding
homogeneous system. We have the rank nullity theorem:
The rank of a matrix plus the dimension of the solution set (nullity) of a homeogeneous
system is equal to the number of variables.
In other words, the rank and the nullity are inversely proportional.
Example Book 4.21
Show f (t )  sin t , g t   cos t , ht   t are independent.
Not clear how to represent these as vectors, so use the definition of linear dependence
or the test for linear dependence.
We want to find out if af (t )  bg (t )  ch(t )  0 has solutions besides the trivial
solution.
The equation must be true for all t. We will use this…
Must be true for t=0:
a sin t  b cos t  ct  0
so b=0
0b0  0
Must be true for t=п/2:
a sin t  b cos t  ct  0
a  0  c  0
2
Must be true for t=3п/2:
a sin t  b cos t  ct  0
a  0  c 3  0
the only way a  0  c   a  0  c 3 is if c=0.
2
2
2
And then we get that a=0.