Week 8:
1.
6.1 1st order systems of DE (briefly!)
2.
5.4 Eigenvalues and Eigenvectors
3.
5.5 Eigenspaces and Diagonalization
—————
A vector
v
= 0 in
R
n
(or in
C
n
)
is an
eigenvector with eigenvalue
λ
of
an
n
by
n
matrix
A
if
Av
=
λv.
We rewrite the vector equation as (
A

λI
n
)
v
= 0
,
which is a homogeneous system with coef
matrix (
A

λI
)
,
and we want
λ
so that the system has a nontrivial solution.
We see that the eigenvalues are the roots
of the
characteristic polynomial
P
(
λ
) = 0
,
where
P
(
λ
) = det(
A

λI
)
.
To find the eigenvectors we find the distinct
roots
λ
=
λ
i
,
and for each
i
solve (
A

λ
i
I
)
v
= 0
.
—————
Problem 1.
Find the eigenvalues and eigenvectors
of
A
=
3

1

5

1
¶
Solution.
P
(
λ
) =
det
(
A

λI
2
) =
fl
fl
fl
fl
3

λ

1

5

1

λ
fl
fl
fl
fl
= (

1

λ
)(3

λ
)

5 =
λ
2

2
λ

3

5
=
λ
2

2
λ

8 = (
λ

4)(
λ
+ 2)
,
so
λ
1
= 4
, λ
2
=

2
.
For
λ
1
= 4
,
we reduce the coef matrix of the system (
A

4
I
)
x
= 0
,
2
A

4
I
=
3

4

1

5

1

4
¶
=

1

1

5

5
¶
→
1
1
0
0
¶
,
so for
x
2
=
a
we get
x
1
=

a,
and the eigenvectors for
λ
1
= 4 are the
vectors (
x
1
, x
2
) =
a
(

1
,
1) with
a
= 0
,
from the space with basis (

1
,
1)
.
—————
For
λ
2
=

2
,
we start over with coef
A
+ 2
I
=
3 + 2

1

5

1 + 2
¶
→
5

1
0
0
¶
.
To find a spanning set without fractions, we
anticipate that finding
x
1
will use division
by 5, and take
x
2
= 5
b.
Then 5
x
1

x
2
= 0 gives
5
x
1
= 5
b,
so
x
1
=
b,
and (
x
1
, x
2
) =
b
(1
,
5)
,
spanned by (1
,
5) [or, if you must, (
1
5
,
1)].
Problem 2.
Find the eigenvalues and eigenvectors
of
A
=
10

12
8
0
2
0

8
12

6
.
Solution.
Since
A
is 3by3
,
the characteristic polynomial
P
(
λ
) is cubic,
and we can make this problem difficult by neglecting to think
strategically. How should
P
(
λ
) be given; for example,
do we need the coef of
λ
2
? We’re looking for the roots, so
we need the factors (
λ

λ
i
). For that objective
there is only one good method of computing the determinant;
expansion on the 2nd row. Look why the 2nd row exp is best:
we get
P
(
λ
) = (2

λ
)
fl
fl
fl
fl
10

λ
8

8

6

λ
fl
fl
fl
fl
= (2

λ
)[(10

λ
)(

6

λ
) + 64] = (2

λ
)[
λ
2

4
λ

60 + 64] =

(
λ

2)
3
.
3
Can we agree that
P
(
λ
) =

λ
3
+ 12
λ
2

24
λ
+ 8
,
[which we would have gotten from the other methods]
is not as useful as having a factor [since the other factors only
need a quadratic; not solving a cubic]?