§
2.4, Problem 7
Let
A
be an
n
×
n
matrix.
Theorem.
If
A
2
= 0
, then
A
is not invertible
Proof.
For a contradiction assume that
A
2
= 0 and
A
is invertible. Then
A
has
an inverse
A

1
and
A
2
=
0
,
A

1
AA
=
A

1
0
,
IA
=
0
,
A
=
0
.
But then
A

1
A
=
A

1
0 = 0 =
I
– Contradiction.
Theorem.
If
AB
= 0
for some nonzero
n
×
n
matrix
B
, then
A
is not invertible.
Proof.
Again, by contradiction. Assume
AB
= 0 and
A
is invertible. Then
AB
=
0
,
A

1
AB
=
A

1
0
,
IB
=
0
,
B
=
0
.
Contradiction: we assumed
B
was nonzero.
§
2.4, Problem 15
Theorem.
Let
V
and
W
be finitedimensional vector spaces, and let
T
:
V
→
W
be a linear transformation. Suppose that
β
=
{
β
1
, . . . , β
n
}
is a basis for
V
. Then
T
is an isomorphism if and only if
T
(
β
)
is a basis for
W
.
Proof.
(
⇒
) First assume that
T
is an isomorphism. Since
T
an isomorphism it
is onetoone and onto. Problem 14(c) on page 75 applies to tell us that
T
(
β
)
is a basis.
(
⇐
) Conversely, assume that
T
(
β
) is a basis for
W
. (We want to prove that
T
is an isomorphism. Since we already know
T
is linear we need prove that
T
is onetoone and onto.)
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 Winter '07
 Liu
 Linear Algebra, Algebra, Vector Space, basis, v0, finitedimensional vector spaces

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