Section 7.1: 1
TFFTFFTT
Section 7.1: 2b
The characteristic polynomial is (
t

4)(
t

1), so the eigenvalues are
distinct, and the matrix is diagonalizable. The basis of eigenvectors can be chosen as (2
,
3)
t
,
and (1
,

1)
t
, leading to a Jordan Form of
±
4
0
0

1
²
Section 7.1: 2c
The characteristic polynomial is

(
t
+ 1)(
t

2)
2
. For
λ
=

1, the
eigenvector can be chosen to be (1
,
3
,
0)
t
. For
λ
= 2,
E
2
has dimension 1.
K
2
must therefore
have dimension 2, and we can pick any vector in
K
2
which is not in
E
2
to generate a cycle.
To calculate
K
2
ﬁnd the nullspace of (
T

2
I
)
2
. We can use
x
0
= (1
,

1
,
0)
t
, and then
(
T

2
I
)
x
0
= (1
,
1
,
1)
t
. Thus we can use for a basis
β
=
{
(1
,
3
,
0)
t
,
(1
,
1
,
1)
t
,
(0
,
1
,

1)
t
}
, and
with respect to
β
, the matrix of
T
is

1 0 0
0
2 1
0
0 2
Section 7.1: 3a
Using the standard basis for the polynomials of degree less than or
equal to 2, the matrix of
T
is
2

1
0
0
2

2
0
0
2
. the dimension of the nullspace of
T

2
I
is
one, of (
T

2
I
)
2
is 2, and of (
T

2
I
)
3
is 3. Thus there is a single 3 cycle. We can generate
it from
x
2