MATH2040 HW 12 Solution
Tsang Chi Shing, Yu Rongfeng
Sec. 7.1
: 2, 4, 5, 8, 10;
Sec. 7.2
: 2, 3, 4, 6, 7, 11, 12, 13, 14, 15, 16, 17.
Sec. 7.1:
2)
(a) The characteristic polynomial of
A
is:
det(
A

tI
) = det
1

t
1

1
3

t
= (
t

2)
2
Hence
λ
= 2 is the only eigenvalue of
A
with multiplicity 2 (so
dim(
K
λ
) = 2). We have
A

λI
=

1
1

1
1
So dim(
E
λ
) = 2

rank
(
A

λI
) = 1. We have
(
A

λI
)
2
=
0
0
0
0
Thus we can choose
v
=
1
0
such that (
A

λI
)
v
=

1

1
6
= 0
and (
A

λI
)
2
v
= 0. We have
β
=

1

1
,
1
0
is a Jordan canonical basis for
A
. Therefore a Jordan canonical form
of
A
is
J
= [
L
A
]
β
=
2
1
0
2
(b) The characteristic polynomial of
A
is:
det(
A

tI
) = det
1

t
2
3
2

t
= (
t
+ 1)(
t

4)
Hence
λ
1
=

1 and
λ
2
= 4 are eigenvalues of
A
with the same
multiplicity 1 (so dim(
K
λ
1
) = dim(
K
λ
2
) = 1). Thus we have
E
λ
1
=
1
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N
(
A

λ
1
I
) =
K
λ
1
and
E
λ
2
=
N
(
A

λ
2
I
) =
K
λ
2
. Consider that
1

1
is an eigenvector of
A
corresponding to

1, and
2
3
is
an eigenvector of
A
corresponding to 4, therefore
β
1
=
1

1
is a basis for
K
λ
1
, and
β
2
=
2
3
is a basis for
K
λ
2
. We have
β
=
β
1
∪
β
2
=
1

1
,
2
3
is a Jordan canonical basis for
A
. Therefore a Jordan canonical form
of
A
is
J
= [
L
A
]
β
=

1
0
0
4
(c) The characteristic polynomial of
A
is:
det(
A

tI
) = det
11

t

4

5
21

8

t

11
3

1

t
=

(
t
+ 1)(
t

2)
2
Hence
λ
1
=

1 and
λ
2
= 2 are eigenvalues of
A
with multiplicities 1
and 2 respectively (so dim(
K
λ
1
) = 1 and dim(
K
λ
2
) = 2).
For
λ
1
=

1, we have
E
λ
1
=
N
(
A

λ
1
I
) =
K
λ
1
.
Consider that
1
3
0
is an eigenvector of
A
corresponding to

1, therefore
β
1
=
1
3
0
is a basis for
K
λ
1
.
For
λ
2
= 2, we have
A

λ
2
I
=
9

4

5
21

10

11
3

1

2
So dim(
E
λ
2
) = 3

rank
(
A

λ
2
I
) = 1. We have
(
A

λ
2
I
)
2
=

18
9
9

54
27
27
0
0
0
Thus we can choose
v
=
1
2
0
such that (
A

λ
2
I
)
v
=
1
1
1
6
= 0
and (
A

λ
2
I
)
2
v
= 0. We have
β
2
=
1
1
1
,
1
2
0
2
is a basis for
K
λ
2
. We have
β
=
β
1
∪
β
2
=
1
3
0
,
1
1
1
,
1
2
0
is a Jordan canonical basis for
A
. Therefore a Jordan canonical form
of
A
is
J
= [
L
A
]
β
=

1
0
0
0
2
1
0
0
2
(d) The characteristic polynomial of
A
is:
det(
A

tI
) = det
2

t
1
0
0
0
2

t
1
0
0
0
3

t
0
0
1

1
3

t
=

(
t

2)
2
(
t

3)
2
Hence
λ
1
= 2 and
λ
2
= 3 are eigenvalues of
A
with the same multi
plicity 2 (so dim(
K
λ
1
) = dim(
K
λ
2
) = 2).
For
λ
1
= 2, we have
A

λ
1
I
=
0
1
0
0
0
0
1
0
0
0
1
0
0
1

1
1
So dim(
E
λ
1
) = 4

rank
(
A

λ
1
I
) = 1. We have
(
A

λ
1
I
)
2
=
0
0
1
0
0
0
1
0
0
0
1
0
0
1

1
1
Thus we can choose
v
=
0
1
0

1
such that (
A

λ
1
I
)
v
=
1
0
0
0
6
= 0
and (
A

λ
2
I
)
2
v
= 0. We have
β
1
=