Math 102  Homework 7 (selected problems)
David Lipshutz
Problem 1.
(Strang, 5.2: #4)
If a 3 by 3 upper triangular matrix has diagonal entries 1, 2, 7, how do you know it can be
diagonalized? What is Λ?
Proof.
Its eigenvalues must be 1, 2 and 7 since
A

λI
where
λ
= 1
,
2 or 7 is upper triangular
with a 0 along the diagonal.
So
A

λI
is singular, thus
λ
is an eigenvalue.
Since the
eigenvalues are all distinct,
A
can be diagonalized.
Λ =
1
0
0
0
2
0
0
0
7
Problem 2.
(Strang, 5.2: #8)
Suppose
A
=
uv
T
is a column times a row (a rank1 matrix).
(a) By multiplying
A
times
u
, show that
u
is an eigenvector. What is
λ
?
(b) What are the other eigenvalues of
A
?
(c) Compute trace(
A
) from the sum on the diagonal and the sum of
λ
’s.
Proof.
(a)
Au
=
uv
T
u
= (
u
T
vu
T
)
T
= ((
u
T
v
)
u
T
)
T
=
λu
where
λ
=
u
T
v
.
(b) The other eigenvalues are zero since
A
is a rank1 matrix.
(c) trace(
A
) =
∑
n
i
=1
u
i
v
i
=
u
T
v
=
λ
Problem 3.
(Strang, 5.2: #30)
Find Λ and
S
to diagonalize
A
in Problem 29. What is the limit of Λ
k
as
k
→ ∞
? What is
the limit of
S
Λ
S

1
? In the columns of this limiting matrix what do you see?
Proof.
det(
A

λI
) = det
"
.
6

λ
.
4
.
4
.
6

λ
#
=
λ
2

1
.
2
λ
+
.
2 = (
λ

.
2)(
λ

1)
We get
A
(1
,
1)
T
= (1
,
1)
T
and
A
(1
,

1)
T
=
.
2(1
,

1)
T
. So
A
=
"
1
1
1

1
# "
1
0
0
.
2
# "
1
/
2
1
/
2
1
/
2

1
/
2
#
1
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lim
k
→∞
Λ
k
= lim
k
→∞
"
1
k
0
0
.
2
k
#
=
"
1
0
0
0
#
So
lim
k
→∞
SA
k
S
=
"
1
1
1

1
# "
1
0
0
0
# "
1
/
2
1
/
2
1
/
2

1
/
2
#
=
"
1
/
2
1
/
2
1
/
2
1
/
2
#
The columns give you the equilibrium distribution of
A
.
Problem 4.
(Strang, 5.2: #34)
Suppose that
A
=
S
Λ
S

1
. Take determinants to prove that det
A
=
λ
1
λ
2
· · ·
λ
n
= product
of
λ
’s.
Proof.
det
A
= det
S
det Λ(det
S
)

1
= det Λ =
λ
1
λ
2
· · ·
λ
n
This quick proof only works when
A
is diagonalizable.
Problem 5.
(Strang, 5.2: #40)
Substitue
A
=
S
Λ
S

1
into the product (
A

λ
1
I
)(
A

λ
2
I
)
· · ·
(
A

λ
n
I
) and explain why
this produces the zero matrix.
Proof.
(
A

λ
1
I
)
· · ·
(
A

λ
n
I
) = (
S
Λ
S

1

λ
1
I
)(
S
Λ
S

1

λ
2
I
)
· · ·
(
S
Λ
S

1

λ
n
I
)
=
S
(Λ

λ
1
I
)
S

1
S
(Λ

λ
2
I
)
S

1
· · ·
S
(Λ

λ
n
I
)
S

1
=
S
(Λ

λ
1
I
)(Λ

λ
2
I
)
· · ·
(Λ

λ
n
I
)
S

1
Even if
A
is not diagonalizable, it can be written as
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 Fall '08
 SZYPOWSK
 Math, Matrices, Steady State, Characteristic polynomial, Eigenvalue, eigenvector and eigenspace, Strang

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