Purdue University
MA 353: Linear Algebra II with Applications
Homework 8, due Mar. 12, Some Solutions
(I)
Sec. 5.1#8
(a) Prove that a linear operator
T
on a ﬁnitedimensional vector space is invertible
if and only if zero is not an eigenvalue of
T
.
Proof.
Let us call this ﬁnitedimensional vector space
V
.
(=
⇒
) Assume that
T
is invertible,
i.e.
T

1
exists. Let
λ
be an eigenvalue of
T
. Then
T
(
v
) =
λv
for some nonzero
v
∈
V
. By applying
T

1
to both sides, we
get
T

1
T
(
v
) =
T

1
(
λv
), which gives
v
=
λT

1
(
v
). Now if
λ
= 0, then this would
give
v
= 0, but
v
is not zero, which is a contradiction. Hence
λ
6
= 0.
(
⇐
=) Assume zero is not an eigenvalue of
T
. Then
N
(
T
) =
{
0
}
because if
N
(
T
)
6
=
{
0
}
, then there is nonzero
v
∈
V
such that
T
(
v
) = 0,
i.e.
T
(
v
) = 0
F
v
,
which would imply that 0
F
is an eigenvalue. Hence
T
is 11. But
V
is ﬁnite
dimensional, this implies that
T
is invertible.
±
(b) Let
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Staff
 Linear Algebra, Algebra, Vector Space

Click to edit the document details