MATH 115 SOLUTION SET 5
ANSWERS TO SELECTED PROBLEMS
§
2.2:
9. The fact that
T
is linear follows from standard facts about complex numbers:
If
z
1
,z
2
∈
C
then
z
1
+
z
2
=
z
1
+
z
2
. Also, if
c
∈
R
then
cz
1
=
c
z
1
. The matrix
[
T
]
β
is
±
1

1
¶
.
10. This is the matrix with 1s along the main diagonal, 1s along the diagonal
above that, and 0s everywhere else.
15. a. For if
T
1
,T
2
∈
S
0
, and
c
is a scalar, then
cT
1
+
T
2
also lies in
S
0
because
whenever
x
∈
S
we have (
cT
1
+
T
2
)(
x
) =
cT
1
(
x
) +
T
2
(
x
) =
c
0 + 0 = 0. b. Suppose
f
∈
S
0
2
, so that
f
kills everything in
S
2
. But
S
1
⊂
S
2
, so of course
f
kills everything
in
S
1
as well. Thus
f
∈
S
0
1
. c. Suppose
f
∈
(
V
1
+
V
2
)
0
. Then
f
kills everything
in
V
1
+
V
2
. In particular it kills both
V
1
, so
f
∈
V
0
1
. Similarly
f
∈
V
0
2
. Thus
f
∈
V
0
1
∩
V
0
2
. Conversely, suppose
f
∈
V
0
1
∩
V
0
2
. Then
f
∈
V
0
1
and
f
∈
V
0
2
. Thus
f
kills both
V
1
and
V
2
. So if
v
1
∈
V
1
and
v
2
∈
V
2
, then
f
(
v
1
+
v
2
) = 0. Thus
f
kills
V
1
+
V
2
and therefore
f
∈
(
V
1
+
V
2
)
0
.
/
16. Let
{
v
1
,...,v
k
}
be a basis for
N
(
T
). Extend this to a basis
{
v
1
,...,v
n
}
for
V
. We claim that
{
T
(
v
k
+1
)
,...,T
(
v
n
)
}
is linearly independent. Indeed, if
c
k
+1
T
(
v
k
+1
) +
···
+
c
n
T
(
v
n
) = 0 for scalars
c
k
+1
,...,c
n
, then
T
(
c
k
+1
v
k
+1
+
···
+
c
n
v
n
) = 0, so that
c
k
+1
v
k
+1
+
···
+
c
n
v
n
∈
N
(
T
). But then this element must be
a linear combination of the vectors
v
1
,...,v
k
:
c
k
+1
v
k
+1
+
···
+
c
n
v
n
=
c
1
v
1