Example:
A
=
1
5
3

1

5

3
!
,
rank(
A
) = 1
range(
A
) =
±
y
:
y
=
α
a
1
=
α
²
1

1
³
for some
α
´
Observe that
z
=
1
1
!
6∈
range(
A
)
because there is no
α
such that
1
1
!
=
α
1

1
!
UCSD Center for Computational Mathematics
Slide 5/24, Wednesday, January 19, 2011
The vectors
z
=
1
1
!
and
a
1
=
1

1
!
are independent.
Moreover, they are
orthogonal
, i.e.,
z
T
a
1
=
(
1 1
)
1

1
!
= 0
But
a
1
spans all of range(
A
)
⇒
z
is orthogonal to
every
vector in range(
A
)
⇒
z
satisﬁes
(
1 1
)
1
5
3

1

5

3
!
=
z
T
A
= 0
.
UCSD Center for Computational Mathematics
Slide 6/24, Wednesday, January 19, 2011
Result
Let
z
be a
nonzero
vector that is orthogonal to every column of
A
=
µ
a
1
a
2
···
a
n
¶
i.e.,
z
T
a
j
= 0 for
j
= 1, 2, .
. . ,
n
. Then
z
6∈
range(
A
).
Proof: Let
z
∈
range(
A
) with
z
T
a
j
= 0 for
j
= 1, 2, .
. . ,
n
.
As
z
∈
range(
A
), there must be some
y
such that
z
=
Ay
, i.e.,
z
=
Ay
=
µ
a
1
a
2
···
a
n
¶
y
=
n
X
j
=1
a
j
y
j
If we form
z
T
z
, we get
z
T
z
=
z
T
Ay
=
n
X
j
=1
z
T
a
j
y
j
=
n
X
j
=1
(
z
T
a
j
)
y
j
= 0
UCSD Center for Computational Mathematics
Slide 7/24, Wednesday, January 19, 2011
⇒
z
T
z
= 0.