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1
Property:
[
u
+
v
]
B
= [
u
]
B
+ [
v
]
B
and [
c
u
]
B
=
c
[
u
]
B
for all
u
,
v
∈
V
and scalar
c
.
Definition
Let
V
be a finitedimensional vector space and
B
be a
basis for
V
.
For any vector
v
in
V
, the vector
Φ
B
(
v
) is called the
coordinate vector of
v
relative to
B
and is denoted as [
v
]
B
.
Let
T
:
V
→
V
be a linear operator on an
n
dimensional vector space
V
with a basis
B
.
Define the linear operator
Φ
B
T
(
Φ
B
)
−
1
:
R
n
→
R
n
,
and consider its standard matrix
A
, called the
matrix representation of
T
with respect to
B
and denoted as [
T
]
B
.
With the notations, [
T
]
B
=
A
and
T
A
=
Φ
B
T
(
Φ
B
)
−
1
.
V
V
R
n
R
n
Φ
B
(
Φ
B
)
−
1
T
Φ
B
T
(
Φ
B
)
−
1
Example: Let
T
:
P
2
→
P
2
be defined by
T
(
p
(
x
)) =
p
(0) + 3
p
(1)
x
+
p
(2)
x
2
for all
p
(
x
) in
P
2
. Then
T
is linear.
For
B
= {1,
x
,
x
2
},
[
T
]
B
=
A
= [
a
1
a
2
a
3
] and
22
12
31
2
3
10
[(
1
)
] [
13
]
3,
[()
3 2 ]
01
0
0
[ (
)]
[3
4
]
3 , so [ ]
[
]
3
3
3 .
41
2
4
Tx
x
T
x
x
x
x
x
T
⎡⎤
⎢⎥
==
+
+
=
=
=
+
=
⎣⎦
⎡
⎤⎡
⎤
⎢
⎥⎢
⎥
+
=
=
=
⎢
⎥
⎢
⎥
⎣
⎦⎣
⎦
aa
a
a
BB
B
B
B
Property:
If
B
= {
v
1
,
v
2
,
…
,
v
n
} , then [
T
]
B
= [ [
T
(
v
1
)]
B
[
T
(
v
2
)]
B
"
[
T
(
v
n
)]
B
].
Proof
[
T
]
B
=
A
⇒
A
e
j
=
T
A
(
e
j
) =
Φ
B
T
(
Φ
B
)
−
1
(
e
j
) =
Φ
B
T
(
v
j
) = [
T
(
v
j
)]
B
.
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View Full Document 2
Proof
[
T
(
v
)]
B
=
Φ
B
T
(
v
) =
Φ
B
T
(
Φ
B
)
−
1
Φ
B
(
v
) =
T
A
([
v
]
B
) = [
T
]
B
[
v
]
B
,
where
A
= [
T
]
B
.
10
Example:
Let
V
= Span
B
, where
B
= {
e
t
cos
t
,
e
t
sin
t
} is L.I. and thus
a basis of
V
, and the linear operator
D
:
V
→
V
be defined by
D
(
f
) =
f
′
for all
f
∈
V
.
Then
⎥
⎦
⎤
⎢
⎣
⎡
−
=
⇒
+
=
−
+
=
1
1
1
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This note was uploaded on 10/16/2010 for the course EE 155 taught by Professor Fong during the Spring '09 term at National Taiwan University.
 Spring '09
 Fong

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