•
Explain what it means for two subspaces to be orthogonal (
V
⊥
W
) and for one subspace to
be the orthogonal complement of another (
V
=
W
⊥
).
•
State which of the fundamental subspaces are orthogonal to each other and explain why,
verify the orthogonality relations in examples, and use the orthogonality relation for
R
(
A
) to
test whether the equation
A
x
=
b
has a solution.
68
II.2.1. Nullspace
N
(
A
)
and Range
R
(
A
)
There are two important subspaces associated to any matrix. Let
A
be an
n
×
m
matrix. If
x
is
m
dimensional, then
A
x
makes sense and is a vector in
n
dimensional space.
The first subspace associated to
A
is the
nullspace
(or
kernel
) of
A
denoted
N
(
A
) (or Ker(
A
)).
It is defined as all vectors
x
solving the homogeneous equation for
A
, that is
N
(
A
) =
{
x
:
A
x
=
0
}
This is a subspace because if
A
x
1
=
0
and
A
x
2
=
0
then
A
(
c
1
x
1
+
c
2
x
2
) =
c
1
A
x
1
+
c
2
A
x
2
=
0
+
0
=
0
.
The nullspace is a subspace of
m
dimensional space
R
m
.
The second subspace is the range (or column space) of
A
denoted
R
(
A
) (or
C
(
A
)). It is defined
as all vectors of the form
A
x
for some
x
. From our discussion above, we see that
R
(
A
) is the the
span (or set of all possible linear combinations) of its columns. This explains the name “column
space”. The range is a subspace of
n
dimensional space
R
n
.
The four fundamental subspaces for a matrix are the nullspace
N
(
A
) and range
R
(
A
) for
A
together with the nullspace
N
(
A
T
) and range
R
(
A
T
) for the transpose
A
T
.
II.2.2. Finding basis and dimension of
N
(
A
)
Example: Let
A
=
1
3
3
10
2
6

1

1
1
3
1
4
.
To calculate a basis for the nullspace
N
(
A
) and determine its dimension we need to find the solutions
to
A
x
=
0
. To do this we first reduce
A
to reduced row echelon form
U
and solve
U
x
=
0
instead,
since this has the same solutions as the original equation.
>A=[1 3 3 10;2 6 1 1;1 3 1 4];
>rref(A)
ans =
1
3
0
1
0
0
1
3
0
0
0
0
69
This means that
x
=
x
1
x
2
x
3
x
4
is in
N
(
A
) if
1
3
0
1
0
0
1
3
0
0
0
0
x
1
x
2
x
3
x
4
=
0
0
0
We now divide the variables into basic variables, corresponding to pivot columns, and free variables,
corresponding to nonpivot columns. In this example the basic variables are
x
1
and
x
3
while the
free variables are
x
2
and
x
4
. The free variables are the parameters in the solution. We can solve
for the basic variables in terms of the free ones, giving
x
3
=

3
x
4
and
x
1
=

3
x
2

x
4
. This leads
to
x
1
x
2
x
3
x
4
=

3
x
2

x
4
x
2

3
x
4
x
4
=
x
2

3
1
0
0
+
x
4

1
0

3
1
The vectors

3
1
0
0
and

1
0

3
1
span the nullspace since every element of
N
(
A
) is a linear combination
of them. They are also linearly independent because if the linear combination on the right of the
equation above is zero, then by looking at the second entry of the vector (corresponding to the
first free variable) we find
x
2
= 0 and looking at the last entry (corresponding to the second free
variable) we find
x
4