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Math 136
Assignment 10 Solutions
1.
By checking whether columns of
P
are eigenvectors of
A
, determine whether
P
diagonalizes
A
. If so, determine
P

1
, and check that
P

1
AP
is diagonal.
a)
A
=
±
42

53
²
,
P
=
±
13

11
²
.
Solution: Check whether the columns of
P
are eigenvectors of
A
by multiplying the
column vectors with the matrix
A
:
±

²±
1

1
²
=
±
2

8
²
±

3
1
²
=
±
14

12
²
We see that the columns of
P
are not eigenvectors of
A
. It follows that
P
does not
diagonalize
A
.
b)
A
=
±
31
²
,
P
=
±
1

1
²
.
Solution: Check whether the columns of
P
are eigenvectors of
A
by multiplying the
column vectors with the matrix
A
:
±
1
1
²
=
±
4
4
²
=4
±
1
1
²
±
1

1
²
=
±

2
2
²
=

2
±
1

1
²
Thus (1
,
1) is an eigenvector of
A
with eigenvalue 4, and (1
,

1) is an eigenvector of
A
with eigenvalue

2. Since the columns of
P
are not scalar multiples of each other, they
are linearly independent and hence a basis for
R
2
. Thus
P
diagonalizes
A
.
We check by calculating:
P

1
=
1
2
±
1

1
²
and
P

1
AP
=
1
2
±
1

1
1

1
²
=
±
40
0

2
²
.
2.
Let
A
and
B
be similar matrices. Prove that:
a)
A
and
B
have the same eigenvalues.
Solution: To show
A
and
B
have the same eigenvalues, we will show that they have the
same characteristic polynomial. Since
A
=
P

1
BP
we have
det(
A

λI
) = det(
P

1

λI
) = det(
P

1

λP

1
P
) = det(
P

1
(
B

λI
)
P
)
= det
P

1
det(
B

λI
) det
P
= det(
B

λI
)
.
1
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b) tr
A
= tr
B
.
Solution: Observe that
tr
AB
=
n
±
i
=1
n
±
k
=1
a
ik
b
ki
= tr
BA.
Hence,
tr
A
= tr(
P

1
BP
) = tr(
P
(
P

1
B
)) = tr
B.
c)
A
n
is similar to
B
n
for all positive integers
n
.
Solution: Since
A
=
P

1
,
A
2
=(
P

1
)(
P

1
)=
P

1
B
2
P
, and
A
2
is similar to
B
2
. To prove the statement for all
n
, we use induction. We have proved the base case
for
n
= 2, so we now assume it is true for
n
=
k
. Then
A
k
+1
=
AA
k
P

1
)(
P

1
B
k
P
P

1
B
k
+1
P,
so the statement is true for
n
=
k
+ 1. Hence the statement is true for all
n
.
3.
For each of the following matrices, determine the eigenvalues and corresponding
eigenvectors and hence determine if the matrix is diagonalizable. If it is, write the
diagonalizing matrix
P
and the resulting matrix
D
.
a)
A
=
²
4

1

25
³
Solution:
A

λI
=
²
4

λ

1


λ
³
. The characteristic polynomial is
det(
A

λI
) = (4

λ
)(5

λ
)

2=
λ
2

9
λ
+ 18 = (
λ

3)(
λ

6)
.
Thus, the eigenvalues of
A
are
λ
= 3 and
λ
= 6. Since all the eigenvalues have algebraic
multiplicity 1, we know that
A
is diagonalizable.
For
λ
= 3 we have
A

λI
=
²
1

1

22
³
∼
²
1

1
00
³
.
The general solution of (
A

λI
)
±x
=
±
0 is
x
2
(1
,
1),
x
2
∈
R
, so an eigenvector corresponding
to
λ
= 3 is (1
,
1).
λ
= 6 we have
A

λI
=
²

2

1

2

1
³
∼
²
11
/
2
³
.
The general solution of (
A

λI
)
=
±
0 is
x
2
(

1
/
2
,
1),
x
2
∈
R
, so an eigenvector
corresponding to
λ
= 6 is (1
,

2).
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This note was uploaded on 04/30/2010 for the course MATH 136 taught by Professor All during the Winter '08 term at Waterloo.
 Winter '08
 All
 Eigenvectors, Vectors

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