2
2.
Let
T
:
R
3
→
R
2
be defned by
T
(
x
1
,x
2
3
) = (
x
1
+
x
3
,
0).
[2] a) Prove that
T
is linear.
Solution: Let
!x
=(
x
1
2
3
),
!
y
y
1
,y
2
3
) and
k
∈
R
. Then
T
(
k!x
+
!
y
)=
T
(
k
(
x
1
2
3
)+(
y
1
2
3
))
=
T
(
kx
1
+
y
1
, kx
2
+
y
2
, kx
3
+
y
3
)
1
+
y
1
+
3
+
y
3
,
0)
=
k
(
x
1
+
x
3
,
0) + (
y
1
+
y
3
,
0)
=
kT
(
)+
T
(
!
y
)
[3] b) Find the standard matrix o±
T
.
Solution: We have
T
(1
,
0
,
0) = (1
,
0)
T
(0
,
1
,
0) = (0
,
0)
T
(0
,
0
,
1) = (1
,
0)
Hence
[
T
]=
±
T
(1
,
0
,
0)
T
(0
,
1
,
0)
T
(0
,
0
,
1)
²
=
³
101
000
´
.
[2] c) Find a spanning set ±or the nullspace o±
T
.
Solution: The nullspace o±
T
is the set o± all (
x
1
2
3
) such that
T
(
x
1
2
3
) = (0
,
0).
Hence, we must have
x
1
+
x
3
= 0 or
x
3
=

x
1
. There±or a vector in the nullspace has the
±orm
(
x
1
2
,

x
1
x
1
(1
,
0
,

1) +
x
2
(0
,
1
,
0)
.
Thus
null(
T
) = span
{
(1
,
0
,

1)
,
(0
,
1
,
0)
}
.
Moreover, since the vectors are not scalar multiples o± each other, they are linearly inde
pendent and so
{
(1
,
0
,

1)
,
(0
,
1
,
0)
}
is a basis ±or the nullspace o±
T
.