Example 2: Rank, nullity, nullspace, basis for a nullspace
Find the rank, the nullity, the nullspace and a basis for the nullspace of the matrix
A =   1  2  1  4   . 
2  4  3  5 
1  2  6  7 

Solution:

1  2  1  4  
2  4  3  5 
1  2  6  7  

~  
1  2  1  4  
0  0  5  3 
0  0  5  3  

~  
1  2  1  4  
0  0  5  3 
0  0  0  0  

Add the 1st row multiplied by 2 to the 2nd row 
Add the 1st row to the 3rd row 
 

Subtract the 2nd row from the 3rd row 

 

Multiply the 2nd row by 1/5 

 

~  
1  2  1  4  
0  0  1  3/5 
0  0  0  0  

~  
1  2  0  17/5   = A_{r} 
0  0  1  3/5 
0  0  0  0  

Add the 2nd row to the 1st row 

 


=>dim R(A) = 2 (the rank of A)
=>dim N(A) = n  dim R(A) = 4  2 = 2 (the nullity of A)
The nullspace: A =
<=>  
1  2  1  4  
2  4  3  5 
1  2  6  7  


x_{1}  
x_{2} 
x_{3} 
x_{4}  

=  
0  
0 
0  

=>  
1  2  1  4  0   ~ ... 
2  4  3  5  0 
1  2  6  7  0  

~  
1  2  0  17/5  0  
0  0  1  3/5  0 
0  0  0  0  0  

The matrix on the right is obtained from the reduced echelon form of A, which we already know.
The use of elementary row operations does not affect the zero column on the right.
<=>  
x_{1} + 2 x_{2} + 17/5 x_{4}  = 0 
x_{3}  3/5 x_{4}  = 0 


2 equations in 4 unknowns;  
Denote:  
x_{4} = t, x_{2} = s,  s, t R 

=>x_{3} = 3/5 t
=> x_{1} = 2 x_{2}  17/5 x_{4} = 2 s 17/5 t
=> = ( 2s 
17/5t, s, 3/5t, t ), s, t R
N(A) = {  = ( 2s  17/5t, s, 3/5t, t ), s, t R} ( the nullspace of A)
= ( 2s , s, 0, 0 ) + ( 17/5t, 0, 3/5t, t )
= s ( 2, 1, 0, 0 ) + t ( 17/5, 0, 3/5, 1 )
=> A basis for the nullspace: { ( 2, 1, 0, 0 ), ( 17/5, 0, 3/5, 1 ) }
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