MATH 223, Linear Algebra
Winter, 2008
Solutions to Assignment 5
1. Let
V
=
P
3
(
X
), the space of real polynomials with degree at most 3.
Deﬁne
T
:
V
→
V
by
T
(
p
(
x
)) = (
x
2

2
x
)
p
00
(
x
) + (2
x
+ 2)
p
0
(
x
)

12
p
(
x
)
.
(a) Verify that
T
is a linear operator on
V
.
Solution: For any
p
1
(
x
)
,p
2
(
x
)
∈
V
,
T
[
p
1
(
x
) +
p
2
(
x
)] =
(
x
2

2
x
)(
p
1
+
p
2
)
00
(
x
) + (2
x
+ 2)(
p
1
+
p
2
)
0
(
x
)

12(
p
1
+
p
2
)(
x
) =
(
x
2

2
x
)(
p
00
1
(
x
) +
p
00
2
(
x
)) + (2
x
+ 2)(
p
0
1
(
x
) +
p
0
2
(
x
))

12(
p
1
(
x
) +
p
2
(
x
)) = [(
x
2

2
x
)
p
00
1
(
x
) + (2
x
+ 2)
p
0
1
(
x
)

12
p
1
(
x
)]+
[(
x
2

2
x
)
p
00
2
(
x
) + (2
x
+ 2)
p
0
2
(
x
)

12
p
2
(
x
)] =
T
(
p
1
(
x
)) +
T
(
p
2
(
x
))
.
Also, for any real
r
and
p
(
x
)
∈
V
,
T
(
rp
(
x
)) =
(
x
2

2
x
)(
rp
)
00
(
x
) + (2
x
+ 2)(
rp
)
0
(
x
)

12(
rp
(
x
)) =
(
x
2

2
x
)(
rp
00
(
x
)) + (2
x
+ 2)(
rp
0
(
x
))

12(
rp
(
x
)) =
r
[(
x
2

2
x
)
p
00
(
x
) + (2
x
+ 2)
p
0
(
x
)

12
p
(
x
)] =
rT
(
p
(
x
))
.
That does it.
(b) Find [
T
]
B
where
B
= (1
,x,x
2
,x
3
) is the standard ordered basis for
V
.
Solution:
T
1 = (
x
2

2
x
)0 + (2
x
+ 2)0

12(1) =

12 =

12(1) +
0
x
+ 0
x
2
+ 0
x
3
, so the ﬁrst column of [
T
]
B
is

12
0
0
0
.
Tx
= (
x
2

2
x
)0 + (2
x
+ 2)1

12
x
= 2

10
x
and the second column
is
2

10
0
0
. Tx
2
= (
x
2

2
x
)(2) + (2
x
+ 2)(2
x
)

12
x
2
=

6
x
2
and
the third column is
0
0

6
0
. Tx
3
= (
x
2

2
x
)(6
x
)+(2
x
+2)(3
x
2
)

12
x
3
=

6
x
2
so the fourth column is the same as the third. The
matrix [
T
]
B
is

12
2
0
0
0 10
0
0
0
0

6

6
0
0
0
0
.
(c) Find [
T
]
B
0
where
B
0
= (1
,x
+1
,x
3

x
2
,x
2
) is a nonstandard ordered
basis for
V
.
Solution: The ﬁrst column here is the same as above.
T
(
x
+ 1) =
2

10
x

12 = 0(1)

10(
x
+1)+0(
x
3

x
2
)+0(
x
2
) so the second column
1
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 Linear Algebra, Algebra, Polynomials, Vector Space, UK, 0 W

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