Solutions to PS 8 (Math 121)
J. Scholtz
January 6, 2007
Question 1: 7.1/2(a)
We are to find the basis of the generalazied eigenvectors and the corresponding Jordan form of matrix
A
:
A
=
1
1

1
3
This matrix gives us a double eigenvalue of
λ
= 2. Hence we need to find all the vectors such that:
A
=

1
1

1
1
a
b
= 0
This is satisfied by
a
(1
,
1). Therefore we need to search for another generalized eigenvector. So we need
to find a vector such that (
A

2
I
)
2
v
= 0 but (
A

2
I
)
v
= 0. The solution to the first equation is any
vector since (
A

2
I
)
2
= 0, therefore we just need to find a vector such that (
A

2
I
) = 0, hence anything
linearly independent from
a
(1
,
1), pick
v
= (

1
,
0), so that (
A

2
I
)
v
= (1
,
1). Then the Jordan form is:
J
=
2
1
0
2
Question 2: 7.1/3(a)
First, by action on the standard basis we can establish that:
A
=
2

1
0
0
2

2
0
0
2
The only (triple) eigenvalue is again
λ
= 2. We can see that the rank(
A

2
I
) = 2 and so there must be
a threecycle of generalized eigenvectors, such that (
A

2
I
)
3
v
= 0 but (
A

2
I
)
2
v
= 0. But we know
that (
A

2
I
)
3
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 Spring '08
 GUREVITCH
 Eigenvectors, Vectors

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