18.06 Problem Set 6
Due Wednesday, 9 April 2008 at 4 pm in 2106.
Problem 1:
a) Do problem 2 from section 6.1 (pg. 283) in the book.
b) Do problem 9 from section 6.1 (pg. 284).
Solution
(5+5 points)
a) To find the eigenvalues of
A
, we take the determinant of
A

λI
, where
λ
is
a variable:
det(
A

λI
) = det
1

λ
4
2
3

λ
= (1

λ
)(3

λ
)

8
This simplifies to
λ
2

4
λ

5, which has roots 5
,

1. (A shortcut to find this equation
for a 2 by 2 matrix is to take
λ
2

tr
(
A
)
λ
+ det(
A
).) We then find the eigenvectors
with eigenvalue 1 by taking the nullspace of
A

(5)
I
: the nullspace of
A

5
I
=

4
4
2

2
is given by the basis [1
,
1]
T
. Similarly,
A

(

1)
I
=
2
4
2
4
has nullspace given by the basis [

2
,
1]
T
. In sum, the eigenvalue 5 leads to eigen
vector (1
,
1), and the eigenvalue

1 leads to the eigenvector (

2
,
1).
We now consider
A
+
I
. The equation giving the eigenvalues is
λ
2

6
λ
= 0.
This has roots 0
,
6. Note that (
A
+
I
)

0
I
and (
A
+
I
)

6
I
are exactly the matrices
we had as before. So they have the same nullspaces. That is,
A
+
I
has the same
eigenvectors as
A
, but the eigenvalues have increased by 1.
b) Part 1: if
Ax
=
λx
, then
x
is also an eigenvector of
A
2
with eigenvalue
λ
2
,
because
A
2
x
=
A
(
λx
)
=
λ
(
Ax
)
=
λ
(
λx
)
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Part 2: Supposing that
A
is invertible,
x
is an eigenvector of
A

1
with eigenvalue
λ

1
: we just take the equation
Ax
=
λx
and multiply both sides by
λ

1
A

1
to get
λ

1
x
=
λ

1
A

1
(
λx
) =
A

1
x
.
Part 3: If
Ax
=
λx
, then (
A
+
I
)
x
=
λx
+
x
= (
λ
+ 1)
x
.
Problem 2:
Do problem 13 from section 6.1 (pg. 285) in the book.
Solution
(10 points)
a) We have
Pu
=
uu
T
u
=
u
because
u
T
u
=
u
·
u
= 1. So
u
is an eigenvector
with eigenvalue 1.
b) If
v
is perpendicular to
u
then 0 =
u
·
v
=
u
T
v
. So
Pv
=
uu
T
v
= 0, showing
that
v
has eigenvalue 0.
c) By part b, it suffices to find three independent vectors perpendicular to
u
(and in fact this will be equivalent).
Using the fact that the row space and null
space are perpendicular, we need to find three independent vectors in the nullspace
of [1
/
6
,
1
/
6
,
3
/
6
,
5
/
6].
We obtain special solutions [

1
,
1
,
0
,
0]
T
, [

3
,
0
,
1
,
0]
T
, and
[

5
,
0
,
0
,
1]
T
.
Problem 3:
Consider the matrix
M
=
2
2
1
1

14

6

9

7

2

1

2

1
8
1
7
4
a) One eigenvector is
x
1
= (1
,
1
,
0
,

3). What is the corresponding eigenvalue?
b) Note that det(
M
) = 0. Use this information to find another eigenvalue
λ
2

how do you know this must be an eigenvalue?
c) A third eigenvalue is
λ
3
=

1. Write down (but don’t solve) a linear system
that can be solved to find
x
3
.
d) What is the fourth eigenvalue? (Hint: use the trace.)
Solution
(10 points)
a) Since
x
1
is an eigenvector, we have
Mx
1
=
λ
1
x
1
for the corresponding eigen
value
λ
1
. So we just calculate
Mx
1
= [1
,
1
,
0
,

3]
T
. Thus
λ
1
= 1.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 CHANILLO
 Linear Algebra, Eigenvectors, Matrices, Singular value decomposition, linearly independent eigenvectors

Click to edit the document details