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HOMEWORK 13 SOLUTIONS
Problem
(
§
7.1 # 1)
.
Label the following statements as true or false.
Proof.
(a) Eigenvectors of a linear operator
T
are also generalized eigenvectors of
T
.
True.
By Theorem 7.1.a,
K
λ
⊇
E
λ
.
(b) It is possible for a generalized eigenvector of a linear operator
T
to correspond to
a scalar that is not an eigenvalue of
T
.
False.
Let
x
be a generalized eigenvector, for the scalar
λ
. Let
p
be the
smallest number so that (
T

λI
)
p
(
x
) = 0. Setting
y
= (
T

λI
)
p

1
(
x
)
6
= 0 we
have (
T

λI
)(
y
) = (
T

λI
)
p
(
x
) = 0, so
y
is an eigenvector with eigenvalue
λ
.
Another way to see this is that if
λ
is not an eigenvalue, we can think of it as
an eigenvalue but with multiplicity 0. Then by Theorem 7.4, dim(
K
λ
) = 0.
(c) Any linear operator on a ﬁnitedimensional vector space has a Jordan canonical
form.
False.
This is true if and only if the characteristic polynomial splits. For a
speciﬁc counterexample, consider
A
=
±
0
1

1 0
²
, over
R
. This matrix has no
eigenvalues, and so no eigenvectors. But every matrix made up of Jordan blocks
has at least one eigenvector. Therefore,
A
does not have a Jordan canonical form
over
R
.
(d) A cycle of generalized eigenvectors is linearly independent.
True.
This is just the Corollary on page 490.
(e) There is exactly one cycle of generalized eigenvectors corresponding to each eigen
value of a linear operator on a ﬁnitedimensional vector space.
False.
Let
T
be the identity operator on
R
2
. Then
T
has two cycles corre
sponding to its single eigenvalue
λ
= 1.
The zero operator will also work to give a counterexample.
(f) Let
T
be a linear operator on a ﬁnitedimensional vector space whose character
istic polynomial splits, and let
λ
1
,λ
2
,...,λ
k
be the distinct eigenvalues of
T
. If,
for each
i
,
β
i
is a basis for
K
λ
i
then
β
1
∪
β
2
∪ ··· ∪
β
k
is a Jordan canonical basis
for
T
.
False.
We need each
β
i
to actually be a basis consisting of cycles. To give
a concrete counterexample, consider the operator
T
=
L
A
on
R
2
, where
A
=
±
0 1
0 0
²
. The only eigenvalue is
λ
= 0, and
K
λ
=
R
2
. Let
β
=
{
(1
,
1)
t
,
(0
,
1)
t
}
.
Then
β
is a basis for
K
0
. However, [
T
]
β
=
±

1

1
1
1
²
is not a Jordan canonical
form. Therefore,
β
is not a Jordan canonical basis for
T
.
(g) For any Jordan block
J
, the operator
L
J
has Jordan canonical form
J
.
True.
Let
β
be the standard basis. Then [
L
J
]
β
=
J
is a Jordan block, and
hence the Jordan canonical form for
L
J
.
1
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HOMEWORK 13 SOLUTIONS
(h) Let
T
be a linear operator on an
n
dimensional vector space whose characteristic
polynomial splits. Then, for any eigenvalue
λ
of
T
,
K
λ
=
N
((
T

λI
)
n
).
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 Fall '08
 GUREVITCH
 Eigenvectors, Vectors

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