Thus (i) and (ii) are satisfied and the transformation IS linear.
Matrix multiplication gives a linear transformation
IMPORTANT REMARK:
Given any
m
×
n
matrix
A
, the transformation from
R
n
to
R
m
given by
L
(
x
) =
A
x
IS LINEAR.
In section 4.2 we will see that all
linear transformations can be written in the form
L
(
x
) =
A
x
for some
matrix
A
.
Transformations on vector spaces of matrices
EXAMPLES
Determine which of the following transformations are linear.
1.
T
:
R
6
×
4
→
R
4
×
6
defined by
T
(
A
) =
A
T
.
Solution:
(i)
T
(
A
+
B
) = (
A
+
B
)
T
=
A
T
+
B
T
=
T
(
A
) +
T
(
B
).
(ii)
T
(
αA
) = (
αA
)
T
=
αA
T
=
αT
(
A
).
Thus (i) and (ii) are satisfied and the transformation IS linear.
2.
T
:
R
3
×
3
→
R
3
×
3
define by
T
(
A
) =
A
2
.
Solution:
(i)
T
(
A
+
B
) = (
A
+
B
)
2
= (
A
+
B
)(
A
+
B
) =
A
2
+
AB
+
BA
+
B
2
.
On the other hand:
T
(
A
) +
T
(
B
) =
A
2
+
B
2
.
Thus (i) is NOT satisfied in general and the transformation IS NOT linear.
Note that (ii) fails as well.
In general,
if the output involves squares, or products or lengths,
x
2
1
or
x
1
x
2
or

x

,
then
L
is not linear.
4
4.1
3. Let
S
=
bracketleftbigg
2
4
−
1
0
bracketrightbigg
. We define
T
:
R
2
×
2
→
R
2
×
2
as
T
(
A
) =
SAS
−
1
.
Solution:
(i)
T
(
A
+
B
) =
S
(
A
+
B
)
S
−
1
=
S
(
A
)
S
−
1
+
S
(
B
)
S
−
1
=
T
(
A
) +
T
(
B
) .
(ii)
T
(
αA
) =
S
(
αA
)
S
−
1
=
αS
(
A
)
S
−
1
=
αT
(
A
) .
Thus (i) and (ii) are satisfied and the operator IS linear.
Transformations on vector spaces of functions
EXAMPLES
Determine which of the following transformations are linear. Assume the transformations are defined
from
C
∞
to
C
∞
where
C
∞
denotes the set of all functions whose derivatives of any order are defined
and continuous on the real line .
1.
L
(
f
(
t
)) =
f
′
(
t
) + 3
f
(
t
)
Solution:
(i) Let
f
and
g
be two functions in
C
∞
L
(
f
+
g
) = (
f
+
g
)
′
+ 3(
f
+
g
) = (
f
′
+ 3
f
) + (
g
′
+ 3
g
) =
L
(
f
) +
L
(
g
).
(ii)
L
(
αf
) = (
αf
)
′
+ 3(
αf
) =
αf
′
+
α
3
f
=
α
(
f
′
+ 3
f
) =
αL
(
f
)
Thus (i) and (ii) are satisfied and the operator IS linear.
2.
L
(
f
(
t
)) = 3(
f
(
t
))
2
+ 9
f
(
t
)
Solution:
(i)
L
(
f
+
g
) = 3(
f
+
g
)
2
+ 9(
f
+
g
) = 3(
f
2
+ 2
fg
+
g
2
) + 9(
f
+
g
). On the other hand:
L
(
f
) +
L
(
g
) = (3
f
2
+ 9
f
) + (3
g
2
+ 9
g
).
Thus (i) is obviously not satisfied and the operator IS NOT linear.
Note that (ii) is also not satisfied.
3.
L
(
f
(
t
)) =
f
(5)
Solution:
(i)
L
(
f
+
g
) = (
f
+
g
)(5) =
f
(5) +
g
(5) =
L
(
f
) +
L
(
g
)
(ii)
L
(
αf
) = (
αf
)(5) =
αf
(5) =
αL
(
f
)
.
Thus (i) and (ii) are satisfied and the operator IS linear.
The Image and Kernel
Definition
Let
L
:
V
→
W
be a linear transformation, then
•
The KERNEL
of
L
, denoted
ker
(
L
), is defined by
ker
(
L
) =
{
v
∈
V

L
(
v
) = 0
w
}
=
{
set of input vectors that are mapped into 0
}
•
The RANGE
of
L
(or IMAGE
of
L
) is defined by
{
w
∈
W

w
=
L
(
v
)
for some
v
∈
V
}
= set of all output vectors
REMARK:
•
ker
(
L
) is a subspace of
V
.
4.1
5
•
The Range of
L
is a subspace of
W
.
EXAMPLE:
Let
L
be the linear operator in
R
2
defined by
L
(
x
) = (2
x
1
+ 4
x
2
,
−
6
x
1
−
12
x
2
)
T
.
Find a basis for the Kernel and Image of
L
.
Solution.
Ker
(
L
) =
{
(
x
1
, x
2
)
T

L
(
x
1
, x
2
) = 0
}
=
braceleftbigg
(
x
1
, x
2
)
T
vextendsingle
vextendsingle
vextendsingle
vextendsingle
2
x
1
+ 4
x
2
=
0
−
6
x
1
−
12
x
2
=
0
bracerightbigg
.