Solutions to Homework 3.
Math 110, Fall 2006.
Prob 2.1.10.
By the linearity of
T
, we use the fact

(1
,
0) + 3(1
,
1) = (2
,
3) to obtain
T
(2
,
3) =

T
(1
,
0) + 3
T
(1
,
1) =

(1
,
4) + 3(2
,
5) = (5
,
11)
.
The map
T
is 1

1 as we see that
T
(
a
(1
,
0)+
b
(1
,
1)) =
a
(1
,
4)+
b
(2
,
5) implies
a
=
b
= 0. Thus
N
(
T
) =
{
0
}
,
so
T
is 1

1.
Prob 2.1.15.
The map
T
is linear, since indeﬁnite integration is linear:
Z
x
0
(
αf
(
t
) +
βg
(
t
))
dt
=
α
Z
x
0
f
(
t
)
dt
+
β
Z
x
0
g
(
t
)
dt.
To see that
T
is onto, suppose
Tf
is the identically zero polynomial. By diﬀerentiating
Tf
, we obtain
f
= 0.
Hence the kernel of
T
is trivial. Finally, notice that all polynomials of the form
Tf
are zero at
x
= 0. So we
cannot get any polynomial with nonzero constant term by applying
T
, i.e.,
T
is not onto.
Prob 2.1.28.
T
maps 0 to 0 by linearity and maps
V
to
V
by deﬁnition, these two subspaces are trivially
T
invariant. If
x
∈
N
(
T
), then
Tx
= 0
∈
N
(
T
), so
N
(
T
) is
T
invariant. If
x