Lecture 9 - Rank of Matrices Engineering Mathematics Notes | EduRev

Algebra- Engineering Maths

Engineering Mathematics : Lecture 9 - Rank of Matrices Engineering Mathematics Notes | EduRev

 Page 1


Rank of Matrices 
Institute of Lifelong Learning, University of Delhi                                                      pg. 1 
 
 
 
 
 
 
 
Subject: Algebra-I 
Semester-I 
Lesson: Rank of Matrices 
Lesson Developer: Chaman Singh and Brijendra Kumar Yadav 
College/Department: Department of Mathematics, Acharya Narendra 
Dev College , Delhi University 
 
 
 
 
 
 
 
 
 
 
 
Page 2


Rank of Matrices 
Institute of Lifelong Learning, University of Delhi                                                      pg. 1 
 
 
 
 
 
 
 
Subject: Algebra-I 
Semester-I 
Lesson: Rank of Matrices 
Lesson Developer: Chaman Singh and Brijendra Kumar Yadav 
College/Department: Department of Mathematics, Acharya Narendra 
Dev College , Delhi University 
 
 
 
 
 
 
 
 
 
 
 
Rank of Matrices 
Institute of Lifelong Learning, University of Delhi                                                      pg. 2 
 
Table of Contents 
 Chapter: Rank of Matrices 
? 1: Learning Outcomes 
? 2: Introduction 
? 3: Subspace of R
n
 
o 3.1: Column space of a matrix 
o 3.2: Null Space of a Matrix 
? 4: Basis for a Subspace 
? 5: Coordinate System 
? 6: Dimension of a Subspace 
? 7: Rank of a matrix 
o 7.1: Methods of Finding Rank of a Matix 
o 7.2: Definition of Rank 
? Exercises 
? Summary 
? References 
 
1. Learning outcomes: 
After studying this chapter you should be able to 
? Understand the subspace of R
n
. 
? Define and determine the column space and null space of a matrix. 
? Understand the basis for a subspace. 
? Calculate the dimension of a subspace. 
? Define and determine the rank of a matrix. 
 
2. Introduction:  
The rank of the matrix is the most important concept in Matrix Algebra. The 
rank is one of the fundamental pieces of data associated to a matrix. A 
matrix always represents a linear transformation between two vector spaces. 
With the help of the rank of the matrix we come to know several things 
about this linear transformation e.g. whether it is bijective, how it maps 
elements etc. 
 
Page 3


Rank of Matrices 
Institute of Lifelong Learning, University of Delhi                                                      pg. 1 
 
 
 
 
 
 
 
Subject: Algebra-I 
Semester-I 
Lesson: Rank of Matrices 
Lesson Developer: Chaman Singh and Brijendra Kumar Yadav 
College/Department: Department of Mathematics, Acharya Narendra 
Dev College , Delhi University 
 
 
 
 
 
 
 
 
 
 
 
Rank of Matrices 
Institute of Lifelong Learning, University of Delhi                                                      pg. 2 
 
Table of Contents 
 Chapter: Rank of Matrices 
? 1: Learning Outcomes 
? 2: Introduction 
? 3: Subspace of R
n
 
o 3.1: Column space of a matrix 
o 3.2: Null Space of a Matrix 
? 4: Basis for a Subspace 
? 5: Coordinate System 
? 6: Dimension of a Subspace 
? 7: Rank of a matrix 
o 7.1: Methods of Finding Rank of a Matix 
o 7.2: Definition of Rank 
? Exercises 
? Summary 
? References 
 
1. Learning outcomes: 
After studying this chapter you should be able to 
? Understand the subspace of R
n
. 
? Define and determine the column space and null space of a matrix. 
? Understand the basis for a subspace. 
? Calculate the dimension of a subspace. 
? Define and determine the rank of a matrix. 
 
2. Introduction:  
The rank of the matrix is the most important concept in Matrix Algebra. The 
rank is one of the fundamental pieces of data associated to a matrix. A 
matrix always represents a linear transformation between two vector spaces. 
With the help of the rank of the matrix we come to know several things 
about this linear transformation e.g. whether it is bijective, how it maps 
elements etc. 
 
Rank of Matrices 
Institute of Lifelong Learning, University of Delhi                                                      pg. 3 
 
3. Subspace of R
n 
: 
Let S be any subset of R
n
. Then the set S is said to be the  subspace of R
n
 if 
and only if it satisfies the following three  properties : 
 (i) The zero vector is in S.  
 (ii) For each x, y ? S, the sum x + y ? S 
 (iii) For each x ? S and to each scalar a, the vector ax ? S. 
Value Addition: Note 
(1) A set S of R
n
 is called subspace of R
n
 if and only if it is closed under    
        addition and scalar multiplication.  
(2) Three condition in the definition of subspace of R
n
 may be replaced by  
        the equivalent one condition only i.e. 
        A set S of R
n
 is called subspace of R
n
  if any only if to each x, y ? S    
        and for any scalars a, b ? R, ax+by ?S.  
(3) R
n
  is also a subspace of itself because it satisfies the three properties 
 required for a subspace.  
(4) The set consisting of only the zero vector in R
n
 is also a subspace of 
 R
n
, called the zero subspace of R
n
. 
  
Example 1: Consider a set S = span {v
1
, v
2
} where v
1
, v
2 
? R
n
. Show that S 
is a subspace of R
n
. 
Solution: To show that S is  subspace of R
n
. We will show that  
 (i) Zero vector is in S. 
 (ii) for all x,y ? S ? x+y ? S 
 (iii) for all x ? S & for any scalar a, ax ? S 
(i) To show zero vector is in S.  
 we can write  
  
12
0 0 0 vv ?? 
 Thus 0 vector is a linear combination of v
1
 & v
2
 therefore 0 S ? . 
(ii) Let x, y ? S 
 Then x and y are the linear combination of v
1
 & v
2
 such that 
Page 4


Rank of Matrices 
Institute of Lifelong Learning, University of Delhi                                                      pg. 1 
 
 
 
 
 
 
 
Subject: Algebra-I 
Semester-I 
Lesson: Rank of Matrices 
Lesson Developer: Chaman Singh and Brijendra Kumar Yadav 
College/Department: Department of Mathematics, Acharya Narendra 
Dev College , Delhi University 
 
 
 
 
 
 
 
 
 
 
 
Rank of Matrices 
Institute of Lifelong Learning, University of Delhi                                                      pg. 2 
 
Table of Contents 
 Chapter: Rank of Matrices 
? 1: Learning Outcomes 
? 2: Introduction 
? 3: Subspace of R
n
 
o 3.1: Column space of a matrix 
o 3.2: Null Space of a Matrix 
? 4: Basis for a Subspace 
? 5: Coordinate System 
? 6: Dimension of a Subspace 
? 7: Rank of a matrix 
o 7.1: Methods of Finding Rank of a Matix 
o 7.2: Definition of Rank 
? Exercises 
? Summary 
? References 
 
1. Learning outcomes: 
After studying this chapter you should be able to 
? Understand the subspace of R
n
. 
? Define and determine the column space and null space of a matrix. 
? Understand the basis for a subspace. 
? Calculate the dimension of a subspace. 
? Define and determine the rank of a matrix. 
 
2. Introduction:  
The rank of the matrix is the most important concept in Matrix Algebra. The 
rank is one of the fundamental pieces of data associated to a matrix. A 
matrix always represents a linear transformation between two vector spaces. 
With the help of the rank of the matrix we come to know several things 
about this linear transformation e.g. whether it is bijective, how it maps 
elements etc. 
 
Rank of Matrices 
Institute of Lifelong Learning, University of Delhi                                                      pg. 3 
 
3. Subspace of R
n 
: 
Let S be any subset of R
n
. Then the set S is said to be the  subspace of R
n
 if 
and only if it satisfies the following three  properties : 
 (i) The zero vector is in S.  
 (ii) For each x, y ? S, the sum x + y ? S 
 (iii) For each x ? S and to each scalar a, the vector ax ? S. 
Value Addition: Note 
(1) A set S of R
n
 is called subspace of R
n
 if and only if it is closed under    
        addition and scalar multiplication.  
(2) Three condition in the definition of subspace of R
n
 may be replaced by  
        the equivalent one condition only i.e. 
        A set S of R
n
 is called subspace of R
n
  if any only if to each x, y ? S    
        and for any scalars a, b ? R, ax+by ?S.  
(3) R
n
  is also a subspace of itself because it satisfies the three properties 
 required for a subspace.  
(4) The set consisting of only the zero vector in R
n
 is also a subspace of 
 R
n
, called the zero subspace of R
n
. 
  
Example 1: Consider a set S = span {v
1
, v
2
} where v
1
, v
2 
? R
n
. Show that S 
is a subspace of R
n
. 
Solution: To show that S is  subspace of R
n
. We will show that  
 (i) Zero vector is in S. 
 (ii) for all x,y ? S ? x+y ? S 
 (iii) for all x ? S & for any scalar a, ax ? S 
(i) To show zero vector is in S.  
 we can write  
  
12
0 0 0 vv ?? 
 Thus 0 vector is a linear combination of v
1
 & v
2
 therefore 0 S ? . 
(ii) Let x, y ? S 
 Then x and y are the linear combination of v
1
 & v
2
 such that 
Rank of Matrices 
Institute of Lifelong Learning, University of Delhi                                                      pg. 4 
 
 
1 1 2 2 1 2
,, x av a v a a R ? ? ? and  
1 1 2 2 1 2
,, y bv bv b b R ? ? ? . 
 Then ? ? ? ?
1 1 2 2 1 1 2 2
x y av a v bv bv ? ? ? ? ? 
   ? ? ? ?
1 1 1 2 2 2 1 1 2 2
, ( ),( ) a b v a b v a b a b R ? ? ? ? ? ? ? 
 This show that x + y is a linear combination of v
1
 and v
2
. Hence         
 x+y ? H. 
(iii) Let x ? S and let a be any scalar, then 
    
1 1 2 2 1 2
,, x av a v a a R ? ? ? 
 Now ? ?
1 1 2 2
ax a av a v ?? 
  ? ? ? ?
1 1 2 2 1 2
,, ax aa v aa v aa aa R ? ? ? 
This shows that ax is a linear combination of v
1
 and v
2
. Hence ax ? s. Since 
S satisfies all the three condition. Therefore S is a subspace of R
n
. 
Value Addition: Note 
Defining a set by S = span{v
1
, v
2
 ,. . ., v
n
} means that every vector of the 
set S can be written as the linear combination of the vectors v
1
, v
2
,. . ., v
n
. 
i.e. let S = span { v
1
, v
2
,. . ., v
n
} and let u ?S, then there exists the scalars 
a
1
, a
2
, . . ., a
n
 such that  u = a
1
v
1
 + a
2
v
2
 + . . . + a
n
v
n
. 
 
3.1. Column space of a matrix: 
 The column space of a matrix A is denoted by Col A and defined as the 
set of all linear combinations of the columns of A.  
Value Addition: Note 
1. Consider a matrix A = [a
1
  a
2
 . . . a
n
] where a
1
, a
2
,. . ., a
n
 are the 
 columns in R
m
. Then Col A = span [a
1
, a
2
,. . ., a
n
] 
2. The column space of an m ? n matrix is a subspace of R
m
. 
3. To check whether a vector b is in the column space of a given matrix 
 A, or not, we will find the solution of the system Ax = b. 
 If the system is consistent i.e. the system has a solution then the 
 vector b is in the column space of A. 
  
Page 5


Rank of Matrices 
Institute of Lifelong Learning, University of Delhi                                                      pg. 1 
 
 
 
 
 
 
 
Subject: Algebra-I 
Semester-I 
Lesson: Rank of Matrices 
Lesson Developer: Chaman Singh and Brijendra Kumar Yadav 
College/Department: Department of Mathematics, Acharya Narendra 
Dev College , Delhi University 
 
 
 
 
 
 
 
 
 
 
 
Rank of Matrices 
Institute of Lifelong Learning, University of Delhi                                                      pg. 2 
 
Table of Contents 
 Chapter: Rank of Matrices 
? 1: Learning Outcomes 
? 2: Introduction 
? 3: Subspace of R
n
 
o 3.1: Column space of a matrix 
o 3.2: Null Space of a Matrix 
? 4: Basis for a Subspace 
? 5: Coordinate System 
? 6: Dimension of a Subspace 
? 7: Rank of a matrix 
o 7.1: Methods of Finding Rank of a Matix 
o 7.2: Definition of Rank 
? Exercises 
? Summary 
? References 
 
1. Learning outcomes: 
After studying this chapter you should be able to 
? Understand the subspace of R
n
. 
? Define and determine the column space and null space of a matrix. 
? Understand the basis for a subspace. 
? Calculate the dimension of a subspace. 
? Define and determine the rank of a matrix. 
 
2. Introduction:  
The rank of the matrix is the most important concept in Matrix Algebra. The 
rank is one of the fundamental pieces of data associated to a matrix. A 
matrix always represents a linear transformation between two vector spaces. 
With the help of the rank of the matrix we come to know several things 
about this linear transformation e.g. whether it is bijective, how it maps 
elements etc. 
 
Rank of Matrices 
Institute of Lifelong Learning, University of Delhi                                                      pg. 3 
 
3. Subspace of R
n 
: 
Let S be any subset of R
n
. Then the set S is said to be the  subspace of R
n
 if 
and only if it satisfies the following three  properties : 
 (i) The zero vector is in S.  
 (ii) For each x, y ? S, the sum x + y ? S 
 (iii) For each x ? S and to each scalar a, the vector ax ? S. 
Value Addition: Note 
(1) A set S of R
n
 is called subspace of R
n
 if and only if it is closed under    
        addition and scalar multiplication.  
(2) Three condition in the definition of subspace of R
n
 may be replaced by  
        the equivalent one condition only i.e. 
        A set S of R
n
 is called subspace of R
n
  if any only if to each x, y ? S    
        and for any scalars a, b ? R, ax+by ?S.  
(3) R
n
  is also a subspace of itself because it satisfies the three properties 
 required for a subspace.  
(4) The set consisting of only the zero vector in R
n
 is also a subspace of 
 R
n
, called the zero subspace of R
n
. 
  
Example 1: Consider a set S = span {v
1
, v
2
} where v
1
, v
2 
? R
n
. Show that S 
is a subspace of R
n
. 
Solution: To show that S is  subspace of R
n
. We will show that  
 (i) Zero vector is in S. 
 (ii) for all x,y ? S ? x+y ? S 
 (iii) for all x ? S & for any scalar a, ax ? S 
(i) To show zero vector is in S.  
 we can write  
  
12
0 0 0 vv ?? 
 Thus 0 vector is a linear combination of v
1
 & v
2
 therefore 0 S ? . 
(ii) Let x, y ? S 
 Then x and y are the linear combination of v
1
 & v
2
 such that 
Rank of Matrices 
Institute of Lifelong Learning, University of Delhi                                                      pg. 4 
 
 
1 1 2 2 1 2
,, x av a v a a R ? ? ? and  
1 1 2 2 1 2
,, y bv bv b b R ? ? ? . 
 Then ? ? ? ?
1 1 2 2 1 1 2 2
x y av a v bv bv ? ? ? ? ? 
   ? ? ? ?
1 1 1 2 2 2 1 1 2 2
, ( ),( ) a b v a b v a b a b R ? ? ? ? ? ? ? 
 This show that x + y is a linear combination of v
1
 and v
2
. Hence         
 x+y ? H. 
(iii) Let x ? S and let a be any scalar, then 
    
1 1 2 2 1 2
,, x av a v a a R ? ? ? 
 Now ? ?
1 1 2 2
ax a av a v ?? 
  ? ? ? ?
1 1 2 2 1 2
,, ax aa v aa v aa aa R ? ? ? 
This shows that ax is a linear combination of v
1
 and v
2
. Hence ax ? s. Since 
S satisfies all the three condition. Therefore S is a subspace of R
n
. 
Value Addition: Note 
Defining a set by S = span{v
1
, v
2
 ,. . ., v
n
} means that every vector of the 
set S can be written as the linear combination of the vectors v
1
, v
2
,. . ., v
n
. 
i.e. let S = span { v
1
, v
2
,. . ., v
n
} and let u ?S, then there exists the scalars 
a
1
, a
2
, . . ., a
n
 such that  u = a
1
v
1
 + a
2
v
2
 + . . . + a
n
v
n
. 
 
3.1. Column space of a matrix: 
 The column space of a matrix A is denoted by Col A and defined as the 
set of all linear combinations of the columns of A.  
Value Addition: Note 
1. Consider a matrix A = [a
1
  a
2
 . . . a
n
] where a
1
, a
2
,. . ., a
n
 are the 
 columns in R
m
. Then Col A = span [a
1
, a
2
,. . ., a
n
] 
2. The column space of an m ? n matrix is a subspace of R
m
. 
3. To check whether a vector b is in the column space of a given matrix 
 A, or not, we will find the solution of the system Ax = b. 
 If the system is consistent i.e. the system has a solution then the 
 vector b is in the column space of A. 
  
Rank of Matrices 
Institute of Lifelong Learning, University of Delhi                                                      pg. 5 
 
Example 2 :  Let 
2 3 4
8 8 6
6 7 7
A
?? ??
??
??
??
?? ??
??
   and   
6
10
11
b
??
??
??
??
??
??
. Determine whether 
b is in the column space of A.  
Solution : For the solution of the equation Ax = b, we have the augmented 
matrix  
  ? ?
2 3 4 6
8 8 6 10
6 7 7 11
Ab
?? ??
??
? ? ?
??
?? ??
??
 
    
2 3 4 6
0 4 10 14
0 2 5 7
?? ??
??
??
??
?? ?
??
 
2 2 1
3 3 1
4
3
R R R
R R R
??
??
 
           
2 3 4 6
0 4 10 14
0 0 0 0
?? ??
??
??
??
??
??
 
3 3 2
1
2
R R R ? . 
There is a free variable therefore, the system Ax = b has a non-zero solution 
i.e. the system is consistent, thus b is in the column space of A.  
Example 3: Let 
1 2 3
3 2 0
0 , 2 , 6
6 3 3
v v v
?? ? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
, 
6
10
11
p
??
??
??
??
??
??
 and ? ?
1 2 3
A v v v ? , then 
 (i) How many vectors are in {v
1
, v
2
, v
3
}  
 (ii) How many vectors are in Col A ?  
  Is p in Col A ? Why or why not ? 
Solution: (i) There are three vectors in {v
1
, v
2
, v
3
}. 
(iii) We have   
 ? ?
1 2 3
A v v v ?
3 2 0
0 2 6
6 3 3
? ??
??
??
??
??
??
 
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