Homework 5 Solutions
Josh Hernandez
November 4, 2009
2.4  Invertibility and Isomorphisms
4.
Let
A
and
B
be
n
×
n
invertible matrices. Prove that
AB
is invertible and (
AB
)
1
=
B
1
A
1
.
Solution:
Using the associativity of matrix multiplication,
(
AB
)(
B
1
A
1
) =
A
(
BB
1
)
A
1
=
AIA
1
=
AA
1
=
I
and
(
B
1
A
1
)(
AB
) =
B
(
AA
1
)
B
1
=
BIB
1
=
BB
1
=
I.
Thus
AB
is invertible and
B
1
A
1
is its inverse
5.
Let
A
be invertible. Prove that
A
t
is invertible and (
A
t
)
1
= (
A
1
)
t
.
Lemma:
Given
A
∈
M
m
×
n
(
F
)
and
B
∈
M
n
×
p
(
F
)
, we have the identity
(
AB
)
t
=
B
t
A
t
.
Proof:
((
AB
)
t
)
ij
= (
AB
)
ji
=
n
X
k
=1
A
jk
B
ki
=
n
X
k
=1
(
A
t
)
kj
(
B
t
)
ik
=
n
X
k
=1
(
B
t
)
ik
(
A
t
)
kj
= (
B
t
A
t
)
ij
.
The entries of (
AB
)
t
and
B
t
A
t
all agree, so the matrices are identical.
Solution:
Applying the Lemma above,
(
A
t
)(
A
1
)
t
= (
A
1
A
)
t
=
I
t
=
I
and
(
A
1
)
t
(
A
t
) = (
AA
1
)
t
=
I
t
=
I.
Thus (
A
1
)
t
= (
A
t
)
1
9.
Let
A
and
B
be
n
×
n
matrices such that
AB
are invertible. Prove that
A
and
B
are invertible. Give an
example to show that arbitrary matrices
A
and
B
need not be invertible if
AB
is invertible.
Lemma:
Given vector spaces
V
,
W
,
Z
and linear transformations
T
:
V
→
W
and
U
:
W
→
Z
,
N
(
T
)
⊆
N
(
U
◦
T
)
and
R
(
U
◦
T
)
⊆
R
(
U
)
.
Proof: Given
v
∈
N
(
T
),
(
U
◦
T
)(
v
) =
U
(
T
(
v
)) =
U
(0) = 0
,
so
v
∈
N
(
U
◦
T
). Thus
N
(
T
)
⊆
N
(
U
◦
T
). Given
z
∈
R
(
U
◦
T
), there is some
v
in
V
such that
z
= (
U
◦
T
)(
v
) =
U
(
T
(
v
)) =
U
(
w
) for
w
=
T
(
v
)
∈
W
,
1
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z
∈
R
(
U
). Thus
R
(
U
◦
T
)
⊆
R
(
U
).
Corollary:
rank(
U
◦
T
)
≤
min
{
rank(
U
)
,
rank(
T
)
}
Proof: By the ranknullity theorem,
rank(
U
◦
T
) = dim(
V
)

nullity(
U
◦
T
)
≤
dim(
V
)

nullity(
T
) = rank(
T
)
.
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 Spring '10
 SHALOM
 Linear Algebra, Algebra, Multiplication, Matrices, Vector Space, Rank

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