ENSC 483 HW#4 Solutions by M. Saif
1
Simon Fraser University
School of Engineering Science
ENSC 483  Modern Control Systems
1. Find the modal matrix and the Jordan Canonical Form of the following matrices
A
=

1
0

1
1
1
3 0
0
1
0
0
0
0 0
2
1
2

1

1

6 0

2
0

1
2
1
3 0
0
0
0
0
1
0 0
0
0
0
0
0
1 0

1

1
0
1
2
4 1
B
=

1

2
3
0

2
3
0
2
3
0
0
2

4
0
1

2
1

1

4

1
0
0

1
0
0
0
0
0
0
0
0
5

7

1
2

4
1

2

7

1
0

1
1
0

1
0
0
0
1
0
0
3

4
0
1

3
1

1

4

1
0
1

1
0
1

1

1

1

1
0
0

1
1
0
0
0
0

1
1
0
0
0
0
0
0
0
0
0

1
0
0
4

5
0
2

3
1

2

5

2
Hint:
A
has all of its eigenvalues located at
λ
= 1, whereas eigenvalues of
B
are all at
λ
=

1.
Solution:
•
Consider matrix
A
:
λ
1
=
λ
2
=
···
=
λ
7
= 1
.
ρ
(
A

I
) = 3 =
⇒
γ
(
A

I
) = 4
⇒
dim
N
1
= 4
v
1
v
4
v
6
v
7
ρ
(
A

I
)
2
= 1 =
⇒
γ
(
B
+
I
)
2
= 6
⇒
dim
N
2
= 6
v
2
v
5
ρ
(
B

I
)
3
= 0 =
⇒
γ
(
B

I
)
3
= 7
⇒
dim
N
3
= 7
v
3
Based on the above information the largest Jordan Block is of
3
×
3
size, and there are four JBs in
all. So the size of the blocks are
3
×
3
,
2
×
2
,
1
×
1
, and
1
×
1
. Now to ﬁnd a chain of generalized
eigenvectors of length three, pick
v
3
such that
(
A

I
)
3
v
3
= 0
, and
(
A

I
)
2
v
3
6
= 0
. One such
vector is
v
3
=
0
1
0
0
0
0
0
;
then
v
2
= (
A

I
)
v
3
=
0
0
1
0
0
0

1
;
v
1
= (
A

I
)
v
2
=

1
0
1

1
0
0
0
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2
Now pick a chain of generalized eigenvectors of length 2. So pick a
v
5
∈ N
2
,v
5
/
∈ N
1
and
independent of
v
2
such that
(
A

I
)
2
v
5
= 0
but
(
A

I
)
v
5
6
= 0
.
v
5
=
0
0
0
1
0
0
0
;
then
v
4
= (
A

I
)
v
5
=
1
0

1
1
0
0
1
Now ﬁnd two more linearly independent (from one another as well as
v
1
and
v
4
eigenvectors in
N
1
. These are
v
7
=
0
0

1

2
1
0
0
;
and
v
8
=
1
3
1
0
0
1
0
Now
M
=
£
v
1
v
2
v
3
v
4
v
5
v
6
v
7
/
, and
ˆ
A
=
M

1
AM
=
1 1 0 0 0 0 0
0 1 1 0 0 0 0
0 0 1 0 0 0 0
0 0 0 1 1 0 0
0 0 0 0 1 0 0
0 0 0 0 0 1 0
0 0 0 0 0 0 1
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 Spring '07
 MehrdadSaif
 Linear Algebra, Characteristic polynomial, M. Saif

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