MAS 213: Linear Algebra II.
Problem list for Week #6.
Tutorial on 12th October.
Problem 1:
(Problems 2.3.3 in [FIS].)
Consider two linear transformations:
•
T
:
P
2
(
R
)
→
P
2
(
R
) defined by
T
(
p
(
x
)) = (3 +
x
)
p
′
(
x
) + 2
p
(
x
).
•
U
:
P
2
(
R
)
→
R
3
defined by
U
(
a
+
bx
+
cx
2
) = (
a
+
b, c, a
−
b
).
Let
β
and
γ
be the standard ordered bases of
P
2
(
R
) and
R
3
.
(1) Check that the expected equation [
U
◦
T
]
γ
β
= [
U
]
γ
β
[
T
]
β
holds.
(2) Let
h
(
x
) = 3
−
2
x
+
x
2
∈
P
2
(
R
). Check that the expected equation
[
U
(
h
(
x
))]
γ
= [
U
]
γ
β
[
h
(
x
)]
β
holds.
Solution.
We need to calculate various ingredients for this problem. First, to determine
[
T
]
β
:
•
T
(1) = 2 = (2)
∗
1 + (0)
∗
x
+ (0)
∗
x
2
.
•
T
(
x
) = 3 + 3
x
= (3)
∗
1 + (3)
∗
x.
•
T
(
x
2
) = 4
x
2
+ 6
x
= (0)
∗
1 + (6)
∗
x
+ (4)
∗
x
2
.
[
T
]
β
= [
T
]
β
β
=
2
3
0
0
3
6
0
0
4
.
Next, to determine [
U
]
γ
β
:
•
U
(1) = (1
,
0
,
1) = (1)
∗
(1
,
0
,
0) + (0)
∗
(0
,
1
,
0) + (1)
∗
(0
,
0
,
1)
.
•
U
(
x
) = (1
,
0
,
−
1) = (1)
∗
(1
,
0
,
0) + (0)
∗
(0
,
1
,
0) + (
−
1)
∗
(0
,
0
,
1)
.
•
U
(
x
2
) = (0
,
1
,
0) = (0)
∗
(1
,
0
,
0) + (1)
∗
(0
,
1
,
0) + (0)
∗
(0
,
0
,
1)
.
Thus:
[
U
]
γ
β
=
1
1
0
0
0
1
1
−
1
0
.
Finally, to calculate [
U
◦
T
]
γ
β
:
•
(
U
◦
T
)(1) =
U
(2) = (2
,
0
,
2) = (2)
∗
(1
,
0
,
0) + (0)
∗
(0
,
1
,
0) + (2)
∗
(0
,
0
,
1)
.
•
(
U
◦
T
)(
x
) =
U
(3 + 3
x
) = (6
,
0
,
0) = (6)
∗
(1
,
0
,
0) + (0)
∗
(0
,
1
,
0) +
(0)
∗
(0
,
0
,
1)
.
•
(
U
◦
T
)(
x
2
) =
U
(6
x
+4
x
2
) = (6
,
4
,
−
6) = (6)
∗
(1
,
0
,
0)+(4)
∗
(0
,
1
,
0)+
(
−
6)
∗
(0
,
0
,
1)
.
Thus:
[
U
◦
T
]
γ
β
=
2
6
6
0
0
4
2
0
−
6
.
1