This presentation is useful all the students who study in engineering because it is a part of your Math 2 subject. Subject name is vector calculus and linear algebra. Also useful for those student who learn about vector space.

1. Name: Panchal Dhrumil Indravadan
Activity: VCLA(ALA)
Branch: Computer Engineering(B.E.)
Semester: Second Sem
Year: 2017-18

2. Null space and Dimension
theorem

3.  Definition: The null space of an matrix A,
written as Nul A, is the set of all solutions of the
homogeneous equation In set notation,
.
m n
Nul {x : x is in and x 0}n
A A ¡
x 0A 

4.  Solution: The first step is to find the general solution
of in terms of free variables.
 Row reduce the augmented matrix to reduce
echelon form in order to write the basic variables in
terms of the free variables:
3 6 1 1 7
1 2 2 3 1
2 4 5 8 4
A
   
   
 
   
x 0A 
 0A

5.  =
 The general solution is ,
, with x2, x4, and x5 free.
 Next, decompose the vector giving the general
solution into a linear combination of vectors where the
weights are the free variables. That is,
1 2 0 1 3 0
0 0 1 2 2 0
0 0 0 0 0 0
  
 
 
  
1 2 4 5
3 4 5
2 3 0
2 2 0
0 0
x x x x
x x x
   
  

1 2 4 5
2 3x x x x  
3 4 5
2 2x x x  

6.  Every linear combination of u, v, and w is an element
of Null A.
1 2 4 5
2 2
3 4 5 2 4 5
4 4
5 5
2 3 2 1 3
1 0 0
2 2 0 2 2
0 1 0
0 0 1
x x x x
x x
x x x x x x
x x
x x
           
         
         
              
         
         
                 
2 4 5
u v wx x x  

7.  If A is an m×n matrix, then:
a) rank(A)=the number of leading variables in the
solution of Ax=0.
b) nullity(A)=the number of parameters in the
general solution of Ax=0.
 If A is an m×n matrix then,
rank(A) + nullity(A) = n (number of columns)
 If A is an m×n matrix then nullity (A) represents
the number of parameter in the general solution
of Ax=0

8.  Prove the dimension theorem for
 Augmented matrix is,

 Corresponding system of equations are:

9.  Which can be expressed as follow:

10.  Rank (A) = 2
 Nullity (A) = 2
 Dimension (A) = 4 = no. of columns
 rank(A) + nullity (A) = 2 + 2
= 4
= no. of columns
 So, here dimension theorem is verified.

11.  Inspiration from Prof. Bhavesh V. Suthar
 Notes of Vector Calculus Linear Algebra
 Textbook of VCLA
 Image from Google images
 Some my own knowledge