Use Elementary Column
Operations to Calculate the Basis
of the Null Space of a Matrix
by Chen Bihong
In linear algebra, the kernel or null space (also nullspace) of a matrix A is the set of all vectors x for which
Ax=0. The kernel of a matrix with n columns is a linear subspace of n-dimensional Euclidian space. The
dimension of null space of A is called the nullity of A. With traditional method, for calculating the basis of
null space of A, we must use row reduction to find a basis for the null space. That is, first use elementary row
operations to put A in reduced row echelon form, then interpreting the reduced row echelon form as a
homugeneous linear system, determine which of the variables in terms of the free variables, then write
equations for the dependent variables in terms of the free variables, then for each free variable xi, choose
the vector in the null space for which xi=1 and the remaining free variables are zero then resulting collection
of vectors is a basis for the null space of A. In this paper we call it elementary row operations method.
Intruduction
Now I give another method which is called elementary column operations method. It is given by theorem as
follows.
Chen Bihong Theorem:
Given a matrix Amn, rank(A)=r<n
 A mn  Elementary Column Operations  B mr O m( n  r ) 

 E    P
Q
n

r
n( n  r ) 
 n 

n-r columns of Q are basis of null space of
matrix A.
Given a m rows n columns matrix A with rank(A)=r, and r less than n, construct a partition metrix by adding
identity matrix E below A, then do elementary column operations to it to let A become (B, O) where O is a
zero matrix with n-r columns, so the matrix is changed to partition matrix B,O,P,Q , and the all n minul r
columns of Q are the basis of null space of matrix A.
Proof this theorem is easy, but I will proof it use some special method as follows to suggest some thinking
methods and teaching methods.
Chen Bihong Theorem:
Given a matrix Amn, rank(A)=r<n
 A mn  Elementary Column Operations  B mr
 E    P
 n 
 nr
O m( n  r ) 

Q n( n  r ) 
n-r columns of Q are basis of null space of
matrix A.
Pinciple
Old system:
Ax=b
New system:
PAQy=Pb
where P and Q are invertible matrix.
We call a usual system of linear equations Ax equal to b as old system, where A is a matrix with m rows and
n columns and b is a column vector of n entrays. Given any invertible square matrix P of order m and
invertible square matrix Q of order n, change the old system to PAQx equals Pb witch are called new
system.
Old system:
New system:
Ax=b
PAQx=Pb
Given a solution kesai of the old system, so
Ax=b
-1
and let h=Q x, then h is solution of new system.
because
PAQh=PAQQ-1x=PAx=Pb.
Otherwise, given a solution h of the new system,
PAQh=Pb
Left multiply P-1 on two side of the equation can get
AQh=b
so x=Qh is a solution of the old system.
Old system:
Ax=b
New system:
PAQx=Pb
If x is a solution of old system,
then h=Q-1x is a solution of new system.
If h is a solution of new system,
then x=Qh is a solution of old system.
So the two sets of solutions of old and new systems have relation of one to one and invertible linear
transformation. If we select suitable P and Q to let new system have simple form, we can find the new
method to calculate solutions of system of linear equations
Standard matrix:
O and E are standard matrix.
Add some zero rows or zero columns to a
standard matrix to get a new standard
matrix.
Use Dmn to express a standard matrix,
rank(D)=r
Standard system:
Dmnx=d
where D is standard matrix,
r(D)=r.
d=(d1,d2,…,dr,dr+1,…,dm)
Standard system if solutable, it
must be
dr+1=dr+2=…=dm=0, if they are
exist.
Then system can write as:
xr+1 to xn can be any value.
 x1  d1 ,
x  d ,
2
 2


 xr  d r ,
 0  0,


 0  0.

So the resolution x can write as
 x1   d1   d1 
0
0
 x   d   




d
0
0
2
2
2

 
  
 
 

 
  
 
 

 
  
0
0
x
d
dr 
r 
r 



x


 c1    c2   
 xr 1   c1   0 
1
0

 
  
 
 
0
1
 xr  2   c2   0 
 
 

 
  
 
 

 
  
 
 
0
0
 xn   cn  r   0 
   c1e r 1  c2e r  2   cn  r e n
0
0
 
 
0
 cn  r  
0
 
0
 
 
 
1
change the note x to y, then Dy=d, its
solution are:
y=+c1er+1+c2er+2+…+cn-ren,
where ei is i-th column of En.
and old system is
Ax=b
use invertible matrix P and Q, let D=PAQ,
this means do elementary operations to A.
then old system's solutions are
x=Qy=Q+c1Qer+1+c2Qer+2+…+cn-rQen
Old system: Ax=b
New system: PAQy=Pb or Dy=d
Old system's solution:
x=Q+c1Qer+1+c2Qer+2+…+cn-rQen
and
Qei is just the i-th column of Q!
Q is the record of doing elementary column
operation to A to change A to standard
matrix D, so add E below A, to record the
Q
construct matrix
A
E
 
do elementary operations to A's part to
change A to standard matrix D, then E
becomes Q.
 A  D
 E   Q
   
But row operations are not need to do.
Example: Let
construct
 1 1 1 1 


A  2 2 2 2


 1 1 2 2 


 1 1 1 1 
 2 2 2 2 


 1 1 2 2 
A 

E  1 0 0 0 
  0 1 0 0


0
0
1
0


0 0 0 1


 1 1
 2 2

 1 1
A 
E  1 0
  0 1

0 0
0 0

1 0
2 0

 1 0
c4  c3

 1 1

0 1

0 0
0 0

1
2
2
0
0
1
0
0
4
3
1
0
1
0
1 
1 0 0 0
 2 0 4 4 
2



2  c  c  1 0 3 3 
2
1

 1 1 1 1 
c3  c1

0  
c4  c1


0 1 0 0
0



0
0
0
1
0


0 0 0 1
1 


0  a basis of null space of matrix A:
0

1 0

0





1 0
0




,
0
0 1

1
0 1

1
   
Old method:
 1 1
A   2 2

 1 1

1
r2 ( 4)


 0
r3 3 r2

r1  r2
0

1 
 1 1 1 1 
r2  2 r1
 0 0 4 4 
2 2  

 r3  r1 

 0 0 3 3 
2 2 


1 0 0 

0 1 1

0 0 0 
1
Old method:
 1 1 1 1   1 1 0 0 
A   2 2 2 2    0 0 1 1

 

 1 1
 0 0 0 0 
2

2

 

go back to system of linear equations:
 x1  x2 ,

 x3  x4 .
Let x2=c1,x4=c2,
c1,c2 can be any number.
Ole method:
 x1  x2 ,
Let x2=c1,x4=c2,

c1,c2 can be any number.
 x3  x4 .
Then we get
 x1   c1 
1
0
x  c 
1
0
2
1


x

 c1    c2  
 x3   c2 
0
1
 x  c 
0
1
 
 
 4  2
Conclusion
After Chen Bihong theorem is discovered,
a lot of concepts of linear algebra, like
"free variable", "reduced row echelon",
"Gaussian elimination", etc. become stupid
and will be discarded or less use. Entire
textbook of linear algebra should rewrite,
and a lot of computer programs to calculate
the solution of system of linear equations
should be reprogrammed.
Thank you!
Example 1: Let
1
2

1
A 
 E   1
  0

0

0

Then
2 1
5 2
3 1
0 0
1 0
0 1
0 0
1 2 1 3 


A 2 5 2 1


 1 3 1 1


3

1

1

0
0

0

1
1 2 1 3 
1 0 0 0 
2 5 2 1 
 2 1 0 5 




 1 3 1 1 c  2 c  1 1 0 4 
2
1
 1 0 0 0  


c3  c1

1

2

1

3
c

3
c

 4 1 

0 1 0 0 
0 1 0 0 




0
0
1
0
0
0
1
0




0 0 0 1 
0 0 0 1 




(-1,0,1,0)T is a basis of nulspace of A.
Example 2: For a lunch, a big monk need 3
breads, 3 small monk need 1 bread, 100
monk just eat 100 breads, calculate the
number of big monk and small monk.
Example 2: For a lunch, a big monk need 3
breads, 3 small monk need 1 bread, 100
monk just eat 100 breads, calculate the
number of big monk and small monk.
Solution: Let x and y is the number of big
monk and small monk, so
 x  y  100,

1

3 x  3 y  100.

 1 1 100 


1
3
100 
3


0
0 
 1 1 100 
1
 3 1 / 3 100 
 3 8 / 3 200 
c2  c1

 


c3 100 c1
0 
1 100 
1 0
1
0 1



0 
1
0 

0
0
0
1


600
c3 
c2
3 8 / 3 0
8




c3 ( 1)
1 25 
1
0

1
75 

so, x=25, y=75.
Example 3: Determine whether the vectors
v1=(1 2 3), v2=(1 2 -1), v3=(3 -1 0), and
v4=(2 1 2) form a basis of F3. If not, choose,
if possible, a basis of F3 consisting of
vector of the given set of vectors.
Example 3: Determine whether the vectors
v1=(1 2 3), v2=(1 2 -1), v3=(3 -1 0), and
v4=(2 1 2) form a basis of F3. If not, choose,
if possible, a basis of F3 consisting of
vector of the given set of vectors.
Solution: Construct matrix A:
1 1 3 2


A  2 2 1 1


 3 1 0 2 


1 1 3
 2 2 1

 3 1 0
A 
 E   1 0 0
  
0 1 0

0
0
1

0 0 0

2
1 0 0 0 



1
2 0 7 3



2  c c
 3 4 9 4 
2
1



c3 3c1
 1 1 3 2
0  
c4  2 c1


0 1
0
0 0



0
0
0
1
0


0 0 0 1 
1 


1
2

3
1

0

0
0

0 
1 0 0
 2 0 7 0 
0 0 0


1

0 7 3 

9

4

3



7
4 9 4 


3
c4  ( ) c3
5

7

 1 1 3 
1 3 2  

7
1 0 0

0 1 0
0



0 1 0
3

0 0 1 
0 0 1  7 


1 
0 0 0
0 
0 
1 0 0
1 0 0
 2 0 7 0 
 2 0 7
0 




1
0 

 3 4 9
 3 4 9  7 


19


 1 1 3  
1
c4 
28 
28
 1 1 3  5  



7
1 
0 1
0  
0 1


0
0
28 



3
3 


0 0 1  7 
0 0 1  7 




1 
1 
0 0 0
0 0 0
so v1,v2,v3 is a basis of F3, and
19
1
3
v4  v1  v2  v3
28
28
7
Example 4: Get the general resolution of
equation x1+x2+2x3=1.
Example 4: Get the general resolution of
equation x1+x2+2x3=1.
resolution: construct matrix as follows:
1
1

0
0

1 2 1
1 0 0


c2  c1
0 0 0
1

1

2
c

2
c
3
1
 


c4  c1
1 0 0  c4 ( 1)  0 1 0


0 1 0
0 0 1
0

1

0

0
1
1

0
0

1 2 1
1 0 0


c2  c1
0 0 0
1

1

2
c

2
c
3
1
 


c4  c1
1 0 0  c4 ( 1)  0 1 0


0 1 0
0 0 1
0

1

0

0
 x1   1 
 1
 2 
 x    0   k  1   k  0  ,(k , k  F )
 2    1  2   1 2
0
1
 x  0
 
 
 3  
x1+x2+2x3=1