1)
For the matrix A3, the null space denoted usually by “ker(A3)” from word
“kernel”, is defined as the set of solutions of equation:
A3  X   0
where (0) is the null vector, that is the vector (0, 0, 0) (unlike number 0, which is
a simple scalar)
Trying to solve the above equation, we get the algebraic system:
 xy z 0
1 1 1 x   0 

   

1 1 1 y    0  
 xy z 0
1 1 1 z   0 
 xy z 0


   
It can be easily observed that the system is not a Cramer one, because the rank
of main matrix is 1, the same as the extended matrix. That means the system is
consistent, but an infinite number of solutions will be available.
In order to write these solutions (we need them), we will use one of those 3
equations (equivalent) as follows:
x  y  z  0 Let y   and z   , two parameters

x    
Thus, the solution of system will be:
x, y, z     , ,   - 1, 1, 0   1, 0, 1
Since the solution is a linear combination of two vectors linearly independent,
these vectors can be considered as a basis of the subspace of null space, so that
we can write:
Ker ( A3 )  Sp 1, 1, 0 , - 1, 0, 1
The number of basis’ vectors is two, that means the dimension of null space is
also 2:
dim(Ker ( A3 ))  2
2)
The range of an operator is the set of vectors W = (u, v, w) which can be
obtained by applying A3 to all vectors of the domain. It is usually denoted by
Im(A3), from the word “image”:
Im( A3 )  u,v,w  so that A3  X   W  ,  X   R3

1 1 1 x   u 

   
1 1 1 y    v 
1 1 1 z   w 

   

 xy z u

 xy z v
 xy z w

In general, such a system is not consistent. The only case when the system is
consistent is when u = v = w, so that the ranks of main matrix and the extended
matrix are equal (to 1). In this case, one yields:
Im A3   u, u, u   u1, 1, 1  Sp1, 1, 1
and
dim(Im( A3 ))  1 (the subspace is spanned by only one vector)
Remark: This is the general rule to find the range (or image) of an linear operator
from consistency condition of the attached algebraic system
3)
The eigenvalues of the operator (A3) will be obtained from the equation:
det( A3  I3 )  0
1 
1
1

1 1 
1 0
1
1 1 
By adding, for example, all lines to the first one, we will get:
3 3 3
1 1
1
1
1 
1  3   1 1  
1 0
1
1
1 
1 1 1 
By subtracting line 1 from lines 2 and 3, one yields:
1 1
1
3    0
4)
is:
  0  2 3     0 
1   2  0 , 3  3
0 0 
For a general matrix (An), the above results can be easily extended, that
dim(Ker ( An ))  n  1
dim(Im( An ))  1
1   2  ...   n 1  0 ,  n  n
5)
A geometric transformation must be an one-to-one and on-to
transformation. Since the null space of A3 is not (only) the null vector, that
means the transformation is not one-to-one and, as a result, it cannot be a
geometric transformation. Also, the image is not full R3, that means is not an onto transformation as well.