Homework 1 Solutions
Josh Hernandez
October 27, 2009
1
2.1  Linear Transformations, Null Spaces, and Ranges
For 3 and 6, prove that
T
is a linear transformation, find bases for both
N
(
T
) and
R
(
T
), compute the nullity
and rank of
T
, and verify the dimension theorem. Finally, determine whether
T
is onetoone or onto:
3
T
:
R
2
→
R
3
defined by
T
(
a
1
, a
2
) = (
a
1
+
a
2
,
0
,
2
a
1

a
2
).
Solution:
Linearity:
T
(
(
a
1
, a
2
) +
c
(
b
1
, b
2
)
)
=
T
(
(
a
1
+
cb
1
, a
2
+
cb
2
)
)
=
(
(
a
1
+
cb
1
) + (
a
2
+
cb
2
)
,
0
,
2(
a
1
+
cb
1
)

(
a
2
+
cb
2
)
)
=
(
(
a
1
+
a
2
) +
c
(
b
1
+
b
2
)
,
0
,
(2
a
1

a
2
) +
c
(2
b
1

b
2
)
)
= (
a
1
+
a
2
,
0
,
2
a
1

a
2
) +
c
(
b
1
+
b
2
,
0
,
2
b
1

b
2
)
=
T
(
a
1
, a
2
) +
c
T
(
b
1
, b
2
)
.
As for the nullspace,
N
(
T
) =
{
(
a
1
, a
2
) :
a
1
+
a
2
= 2
a
1

a
2
= 0
}
The linear system
a
1
+
a
2
=
0
2
a
1
+

a
2
=
0
has only the trivial solution. Therefore, the nullspace of
T
is
{
0
}
. The range of
T
is
span
{
(1
,
0
,
2)
,
(1
,
0
,
1)
}
= span
{
(1
,
0
,
0)
,
(0
,
0
,
1)
}
;
each of these sets is also a basis of
R
(
T
). Thus rank(
T
) = 2. This verifies the ranknullity theorem:
2 = dim(
V
) = nullity(
T
) + rank(
T
) = 0 + 2
.
This function is onetoone, since its nullity is zero. It is not onto, since the second coordinate of
any vector in the range must be zero.
6
T
:
M
n
×
n
(
F
)
→
F
defined by
T
(
A
) = tr(
A
).
Solution:
Linearity:
T
(
A
+
cB
) =
n
X
i
=1
(
A
+
cB
)
ii
=
n
X
i
=1
(
A
ii
+
cB
ii
)
=
n
X
i
=1
A
ii
+
c
n
X
i
=1
B
ii
=
T
(
A
) +
c
T
(
B
)
.