Homogeneous System
Objective:
In this section, we will characterize the solutions of a homogeneous
system
Amn xn1  0 .
For a homogeneous linear system
A
Amn xn1  0 ,
- Jordan reduction
| 0 By
Gauss


 B | 0 .
Then, there are two possible situations:
(1):
1
0


B | 0  0
0


0

 There is only one solution
0  0 0
1  0 0
   

0  1 0 , n  m
.
0  0 0

   
0  0 0
xn1  0 .
(2):
Bmn
1
0



| 0   0
0


0

0  0
1

0
0

0






1
b11
 b1nr
0 b21


1 br1
0 0


0 0
 b2 nr


 brnr

0



0
0
0


0
0


0
 x1  b11 x r 1  b12 x r  2    b1n r x n
x 2  b21 x r 1  b22 x r  2    b2 n r x n

x r  br1 x r 1  br 2 x r  2    brnr x n , s1 , s 2 ,, s n r  R,
x r 1  s1

xn  s nr
 x1    b11s1  b12 s2    b1nr snr 
 x   b s  b s    b s 
2 nr nr 
 2   21 1 22 2
   


  

 x   xr     br1s1  br 2 s2    brnr snr 
 xr 1  

s1
  



  

x  

sn  r
 n 

  b11 
  b12 
  b1nr 
 b 
 b 
 b 
21
22




 2 nr 
  
  
  







b

b

b
 s1  r1   s2  r 2     snr  rnr 
 1 
 0 
 0 





,
0
1
0






  
  
  






0
0
1






 s1v1  s2v2    snr vnr
 (a) Any solution of
Amn xn1  0
v1 , v2 ,, vn r
2
is the linear combination of
(b)

v1 , v2 ,, vn r
v1 , v2 ,, vn r
are linearly independent.
are a basis of the solution space and the
dimension of the solution space is
nr .
Definition of null space and nullity:
The solution space of
Amn xn1  0 is called the null space of A.
The dimension of the null space is referred to as the nullity of A.
Note:
In the above homogeneous system, the null space of A is the vector space
with a basis
v1 , v2 ,, vn r
and the nullity of A is
nr .
Example:
Find a basis and the dimension of the solution space of the homogeneous system
1 0 1  x1 
Ax  2  1 1  x2   0
3 1 4  x3 
[solutions:]
By Gauss-Jordan reduction,
1 0 1 0
A 0  0 1 1 0  x1   x3 , x2   x3 , x3  s, s  R
0 0 0 0
 x1   s   1
  x 2    s   s  1
 x3   s   1 
3
 1
 The basis for the solution space is  1 and the dimension of the solution space
 
 1 
is 1.
Example:
Find the null space and the nullity of the matrix
1
0

A  0

1
2
1
1
0
1
1
4
2
0
0
6
1
1
1
0
0
2
1 
2 .

2
1 
[solutions:]
By Gauss-Jordan reduction,
 x1
1

0
row echelon form
A 0 inreduced

   
0
0

 0
x2
x3
x4
0
2
0
1
0
2
0
0
1
0
0
0
0
0
0
 x1
1

0
exchange 2'nd and 3'rd columns
      
 
0
0

 0

 x1 
1 0
x 
2
 1 0   
, x   x 4 
2 0
 
 x3 
0 0
 x5 

0 0
x2
x4
x3
0
0
2
1
0
0
1
2
0
0
0
0
0
0
0
Thus,
4
x5

1 0
 1 0

2 0
0 0

0 0
x5
x1  2 x3  x5
x2  2 x3  x5
x4 
 2 x5 , s, t  R.
x3  s
x5  t
 x1   2s  t 
 x1   2s  t   2   1
 x   2 s  t 
 x    2 s  t    2  1 
2
  
 2 

    

 x   x4     2t   x   x3    s   s  1   t  0 
  
  

    
x
x
s

2
t
 3 
 4 

  0    2
 x5   t 
 x5   t   0   1 
   2   1  
    
   2  1  


 The null space is span  1 ,  0   and the nullity of A is 2.
  0    2 
    
 0   1  
5