Let us now exhibit a basis for
𝒩A
𝒩
A
.
We exploit the already mentioned fact that
𝒩A=𝒩
A
red
𝒩
A
𝒩
A
red
.
Regarding the latter, we partition the elements of xx into so called
pivot variables,
x
c
j
j=
1
,
...
,
r
x
c
j
j
1
,
...
,
r
and free variables
x
k
k∉
c
j
j=
1
,
...
,
r
x
k
k
c
j
j
1
,
...
,
r
There are evidently
n−r
n
r
free variables. For convenience, let us denote these in the future by
x
c
j
j=
r
+
1
,
...
,
n
x
c
j
j
r
+
1
,
...
,
n
One solves
A
red
x=0
A
red
x
0
,
by expressing each of the pivot variables in terms of the
nonpivot, or free, variables. In the example above,
x
1
x
1
,
x
2
x
2
,
x
3
x
3
,
x
4
x
4
,
x
5
x
5
,
and
x
7
x
7
are pivot while
x
6
x
6
and
x
8
x
8
are free. Solving for the pivot in terms of the free we find
x
7
=0
x
7
0
x
5
=0
x
5
0
x
4
=
x
8
x
4
x
8
x
3
=0
x
3
0
x
2
=
x
6
x
2
x
6
x
1
=0
x
1
0
or, written as a vector,
x=
x
6
01000100+
x
8
00010001
x
x
6
0100
0100
x
8
0001
0001
(2)
where
x
6
x
6
and
x
8
x
8
are free. As
x
6
x
6
and
x
8
x
8
range over all real numbers, the
xx
above traces out a plane in
ℝ
8
ℝ
8
.
This plane is precisely the null space of
AA and
Equation 2 describes a generic element as the linear
combination of two basis vectors. Compare this to what MATLAB
returns when faced with
null(A,'r')
. Abstracting
these calculations we arrive at
Suppose that AA is
mbyn with pivot indices
c
j
j=
1
,
...
,
r
c
j
j
1
,
...
,
r
and free indices
c
j
j=
r
+
1
,
...
,
n
c
j
j
r
+
1
,
...
,
n
.
A basis for
𝒩A
𝒩
A
.
may be constructed of
n−r
n
r
vectors
z
1
z
2
...
z
n

r
z
1
z
2
...
z
n

r
where
z
k
z
k
,
and only
z
k
z
k
,
possesses a nonzero in its
c
r
+
k
c
r
+
k
component.