MATH321 – HOMEWORK SOLUTIONS
HOMEWORK #5
Section 2.4: Problems 1, 2, 3, 4, 5, 6, 7, 13, 14, 22
Section 2.5: Problems 1, 2(a)(b), 3(a), 4, 6, 7, 9, 10
Krzysztof Galicki
Problem 2.4.1
(See
Answers to Selected Exercises
).
Problem 2.4.2
(a)
T
:
R
2
→
R
3
cannot be invertible. However,
N
(
T
) =
{
(0
,
0)
}
is trivial so that
T
can
be inverted on
R
(
T
) which is a plane in
R
3
.
(b)
T
:
R
2
→
R
3
cannot be invertible.
N
(
T
) =
{
(0
,
0)
}
is trivial so that
T
can be inverted
on
R
(
T
) which is a plane in
R
3
.
(c) Here
N
(
T
) =
{
(0
,
0
,
0)
}
and it follows that
R
(
T
) =
R
3
by nullityrank equation.
Hence,
T
is invertible.
(d) Here
N
(
T
) is the space of constant polynomials
f
(
x
) =
c
. Hence
T
is not onetoone
and cannot be inverted.
(e) Here
N
(
T
) consists of matrices of the form
0
0
c

c
.
Hence
T
is not oontoone
and cannot be inverted.
(f) Here
N
(
T
) is trivial so it follows that
R
(
T
) =
M
2
×
2
(
R
) by nullityrank equation.
Hence,
T
is both onetoone and onto.
T
is invertible.
Problem 2.4.3
(a)
F
3
and
P
3
(
F
) are of dimension 3 and 4, respectively so thay cannot be isomorphic.
(b)
F
4
and
P
3
(
F
) are isomorphic. One could take
T
:
P
3
(
F
)
→
F
4
to be
T
(
a
+
bx
+
cx
2
+
dx
3
) = (
a, b, c, d
)
.
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Clearly,
T
is invertible.
(c) Take
T
a
b
c
d
=
a
+
bx
+
cx
2
+
dx
3
. This map is linear and invertible.
(d)
V
is 3dimensional so it cannot be isomorphic to
R
4
.
Problem 2.4.4
Let
A
,
B
∈ M
n
×
n
(
F
) be invertible with inverses
A

1
and
B

1
. Consider
the product
C
=
B

1
A

1
. We have
C
(
AB
) =
B

1
A

1
AB
=
B

1
B
=
I
n
and, similarly,
λ
(
AB
)
C
=
ABB

1
A

1
=
AA

1
=
I
n
.
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 Winter '07
 Liu
 Math, Linear Algebra, Algebra, Binary relation, Isomorphism

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