Alex Gittens
February 9, 2007
ACM 104: HW 4
Exercise 1.
Find the Jordan canonical form and the corresponding Jordan canonical basis for the following
linear operators.
(a)
T
:
R
3
→
R
3
deﬁned by
Tx
=
Ax
for
x
∈
R
3
where
A
=
11

4

5
21

8

11
3

1
0
.
(b)
T
:
P
2
(
R
)
→
P
2
(
R
) deﬁned by
T
(
f
) = 2
f

f
0
.
Answer
.
(a) Since
f
(
t
) = det(
A

tI
) =

t
3
+ 3
t
2

4 =

(
t
+ 1)(
t

2)
2
,
λ
1
=

1, dim(
K
λ
1
) = 1 and
λ
2
= 2,
dim(
K
λ
2
) = 2.
To ﬁnd a basis for
K
λ
1
, note
Ax
=

x
⇒
x
=
c
1
3
0
. To ﬁnd a basis for
K
λ
2
, ﬁrst see if it can
contain two independent eigenvectors:
Ax
= 2
x
⇒
x
=
c
1
1
1
⇒
dim(
E
λ
1
) = 1
,
so it can’t. Therefore
K
λ
2
is spanned by a two cycle of generalized eigenvectors
{
(
A

2
I
)
v,v
}
such that
(
A

2
I
)
v
is an eigenvector, so
(
A

2
I
)
2
v
= 0
⇒
v
=
a
1
0
2
+
b
1
2
0
.
Taking
v
=
1
0
2
yields (
T

2
I
)
v
=

1

1

1
, so

1

1

1
,
1
0
2
is a basis for
K
λ
2
.
Therefore
β
=
1
3
0
,

1

1

1
,
1
0
2
is a Jordan canonical basis for
T
, and the corresponding Jordan canonical form of
T
is
J
=
1

1
1
3

1
0
0

1
2

1
11

4

5
21

8

11
3

1
0
1

1
1
3

1
0
0

1
2
=

1
0
0
0
2
1
0
0
2
.
1
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(b) Let
β
=
{
1
,x,x
2
}
, then
T
(1) = 2
T
(
x
) = 2
x

1
T
(
x
2
) = 2
x
2

2
x,
so
A
= [
T
]
β
=
2

1
0
0
2

2
0
0
2
,
so
λ
= 2 and dim(
K
λ
) = 3.
To ﬁnd a basis for
K
λ
, ﬁrst check the dimension of the eigenspace:
Ax
= 2
x
⇒
x
=
c
1
0
0
⇒
dim(
E
λ
) = 1
.
Note a basis of
K
λ
cannot consist of an eigenvector and a two cycle of generalized eigenvectors, because
the initial vector of the cycle would be an eigenvector, so these three vectors would not be linearly
independent. Instead, the basis of
K
λ
must be a three cycle of generalized eigenvectors
{
(
A

2
I
)
2
v,
(
A

2
I
)
v,v
}
such that the initial vector is an eigenvector:
(
A

2
I
)
3
v
= 0
⇒
v
=
a
0
0
1
+
b
0
1
0
+
c
1
0
0
.
Take
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