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EE221A Section 3
9/9/11
1
Functions, linear maps
1.1
Solutions to linear equations
Theorem.
(range and nullspace of linear operators) [LN3 p. 4]
Consider
A
:
U
→
V
with
(
U,F
)
,
(
V,F
)
linear spaces. Let
b
∈
V
. Then:
a)
A
(
u
) =
b
has at least one solution
⇐⇒
b
∈ R
(
A
)
b) If
b
∈ R
(
A
)
then
i)
A
(
u
) =
b
has a unique solution
⇐⇒ N
(
A
) =
{
θ
U
}
ii) Let
x
0
be such that
A
(
x
0
) =
b
. Then
A
(
x
) =
b
⇐⇒
x

x
0
∈ N
(
A
)
Proof.
a)
=
⇒
:
Let
x
be one solution. Then
b
=
A
(
x
)
for some
x
∈
U
, thus
b
∈ R
(
A
)
by the deﬁnition of
R
(
A
)
.
⇐
=:
There exists an
x
∈
U
such that
A
(
x
) =
b
. Therefore there is at least one
solution
x
.
b)
i)
=
⇒
:
Call the unique solution
x
0
. Now suppose that there exists a nonzero vector
x
n
in
N
(
A
)
, and let
z
=
x
0
+
x
n
6
=
x
0
. Then by the linearity of
A
,
A
(
z
) =
A
(
x
0
+
x
n
) =
A
(
x
0
) +
A
(
x
n
) =
b
+ 0 =
b
. But this says that there is another solution
z
distinct from
x
0
which is unique.
→←
(thus there are no nonzero vectors in
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 Fall '10
 ClaireTomlin

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