Math 2270 Spring 2004 Homework 20 Solutions Section 7.1

```Math 2270 Spring 2004
Homework 20 Solutions
Section 7.1) 2, 6, 10, 16, 18, 30, 31, 32
(2) We are told that A~v = λ~v . If A is invertible, we can multiply both sides of the equation
by A−1 to get ~v = λA−1~v . Since A is invertible, λ 6= 0 (Summary 7.1.4). Therefore, we
can divide by λ to see that A−1~v = λ1 ~v. Thus, ~v is an eigenvector of A−1 with associated
eigenvalue 1/λ.
(6) We are told that ~v is an eigenvector of both A and B. Let λA and λB be the associated
eigenvalues. Therefore A~v = λA~v and B~v = λB~v . Thus, AB~v = A(B~v ) = A(λB ~v) =
λB (A~v) = λB (λA~v ) = (λA λB )~v . Thus, ~v is an eigenvector of AB with associated eigenvalue
λA λB .
(10) We start with a general matrix and solve the following system.
&quot;
a b
c d
#&quot;
1
2
#
=5
&quot;
1
2
#
&quot;
⇒
a + 2b
c + 2d
#
=
&quot;
5
10
#
We have two equations a + 2b = 5 and c + 2d = 10, or a = 5 − 2b and c = 10 − 2d. The
general matrix we are looking for has the form
&quot;
5 − 2b
10 − 2d
b
d
#
where b and d are real scalars.
(16) A linear transformation in ℜ2 that rotates vectors through an angle of 180o is equivalent
to reversing the orientation of a vector, or in other words, multiplying it by −1. Therefore,
all vectors in ℜ2 except the zero vector are eigenvectors of the transformation with associated
eigenvalue λ = −1. Any basis for ℜ2 will consist of eigenvectors of the linear transformation.
(18) A linear transformation in ℜ3 that reflects vectors in the plane E has two families of
eigenvectors. All nonzero vectors on the plane will not change when reflected in the plane.
Therefore, all nonzero vectors ~v on E are eigenvectors with associated eigenvalue λ = 1. All
vectors that are perpendicular to the plane E are flipped about the origin. Therefore, all
nonzero vectors w
~ in E ⊥ are eigenvectors with associated eigenvalue λ = −1. To form a
basis for ℜ3 we can choose any two independent vectors from the plane E and one nonzero
vector from E ⊥ .
1
Math 2270 Spring 2004
Homework 20 Solutions
Problem 30 (lambda=1.2)
L2
L1
Problem 31 (lambda=1)
L2
L1
2
Math 2270 Spring 2004
Homework 20 Solutions
Problem 32 (lambda=0.9)
L2
L1
Section 7.2) 4, 12, 19, 21, 25, 28
(4)
det(λI − A) =
&quot;
λ −4
1 λ−4
#
= λ(λ − 4) + 4 = λ2 − 4λ + 4 = (λ − 2)2
The only eigenvalue is λ = 2 with algebraic multiplicity of 2.
(12)
λ−2
2
0
0

 −1
λ
+
1
0
0



det(λI − A) = 

0
0
λ−3
4 
0
0
−2
λ+3




−1
0
0
λ+1
0
0

4 
0
λ−3
4 
= (λ − 2)det 
=
 − 2det  0 λ − 3

0 &quot; −2 λ + 3 #
0 &quot; −2 λ + 3 #
λ−3
4
λ−3
4
= (λ − 2)(λ + 1)det
− 2(−1)det
−2 λ + 3
−2 # λ + 3
&quot;
λ−3
4
= ((λ − 2)(λ + 1) − 2(−1)))det
−2 λ + 3


Aside: Notice that the structure of the matrix gives the result that the eigenvalues of the
two nonzero 2 x 2 matrices inside the larger matrix are the eigenvalues of the larger matrix.
3
Math 2270 Spring 2004
Homework 20 Solutions
To see why this works in general, let ~v be composed of two column vectors ~v1 and ~v2 stacked
on top of one another. Then,
&quot;
#&quot;
A 0
0 B
~v1
~v2
#
=λ
&quot;
#
~v1
~v2
&quot;
⇒
A~v1
B~v2
#
=
&quot;
λ~v1
λ~v2
#
This does not imply that A and B must have the same eigenvalues. Instead, it is possible
that when ~v1 is nonzero, ~v2 = ~0 implying that λ is an eigenvalue of A, but not B. Or,
alternatively, when ~v2 is nonzero, ~v1 = ~0 implying that λ is an eigenvalue of B, but not A.
In both cases, λ will be an eigenvalue of the larger matrix.
Now, continuing with our previous calculation:
= ((λ − 2)(λ + 1) − 2(−1)) &middot; ((λ − 3)(λ + 3) − 4(−2))
= λ2 − λ
λ2 − 1 = λ(λ − 1)(λ − 1)(λ + 1) = λ(λ − 1)2 (λ + 1)
The eigenvalues are λ = 0 with multiplicity of one, λ = 1 with multiplicity of two, and
λ = −1 with multiplicity of one.
(19) The characteristic polynomial of a 2 x 2 matrix is fA (λ) = λ2 − tr(A)λ + det(A). Using
the quadratic formula we know that zeros of this polynomial are
q
1
λ=
tr(A) &plusmn; (tr(A))2 − 4det(A)
2
If det(A) &lt; 0, then the discriminant ((tr(A))2 − 4det(A)) is always positive. In this case
there are two distinct (real) values of λ that are zeros of fA (λ).
(21) If λ1 , λ2 , . . ., λn are eigenvalues of A, then the characteristic polynomial must factor as (λ − λ1 )(λ − λ2 ) &middot; &middot; &middot; (λ − λn ). If we multiply out this polynomial we see that the
constant term will be the product of the eigenvalues altered by a possible negative sign:
(−1)n λ1 λ2 &middot; &middot; &middot; λn . From Fact 7.2.5, we know that the constant term of the characteristic
polynomial is (−1)n det(A). Therefore det(A) = λ1 λ2 &middot; &middot; &middot; λn . Multiplying out the coefficient
of the λn−1 term and using Fact 7.2.5, we find that tr(A) = λ1 + λ2 + &middot; &middot; &middot; λn .
(25) First we verify that
&quot;
a c
b d
#&quot;
c
b
#
=λ
&quot;
c
b
#
&quot;
c
b
#
and
&quot;
⇒
1
−1
#
&quot;
are eigenvectors of A.
ac + bc
bc + bd
4
#
=
&quot;
c(a + b)
b(c + d)
#
=
&quot;
c
b
#
=
&quot;
λc
λb
#
Math 2270 Spring 2004
Homework 20 Solutions
Problem 25
span[c,b]
span[1,-1]
Therefore,
&quot;
c
b
#
&quot;
1
−1
#
a c
b d
#&quot;
&quot;
is an eigenvector with associated eigenvalue λ = 1.
=λ
&quot;
1
−1
#
⇒
&quot;
a−c
b−d
#
=
&quot;
a−c
(1 − a) − (1 − c)
#
=
&quot;
a−c
−(a − c)
#
=
&quot;
λ
−λ
#
1
Therefore,
is an eigenvector with associated eigenvalue λ = a − c. We are told that
−1
a, b, c and d are positive numbers such that a + b = c + d = 1. Therefore, 0 &lt; a, b, c, d &lt; 1
and |a − c| &lt; 1.
(28) The problem tells us that 20% of the usual Wipf’s customers switch to Migros, so
80% stay. Also, 10% of the Migros customers switch back to Wipf’s. Therefore w(t + 1) =
.8w(t) + .1m(t). Likewise, 20% of the Wipf’s customers switch to Migros and only 90% of
the Migros customers stay at Migros. Therefore m(t + 1) = .2w(t) + .9m(t).
A=
&quot;
.8 .1
.2 .9
#
This is a transition matrix since its columns sum to 1 and its entries are all positive. We can
use the information from problem (25) to find the eigenvalues and
of the matrix
&quot; eigenvectors
#
.1
A. In this case a = .8, b = .2, c = .1 and d = .9. Therefore, ~v1 =
is an eigenvector with
.2
5
#
Math 2270 Spring 2004
Homework 20 Solutions
&quot;
1
associated eigenvalue λ1 = 1 and ~v2 =
−1
λ2 = a − c = .7. The initial state vector is
~x0 =
&quot;
1200
0
#
#
= 4000
is an eigenvector with associated eigenvalue
&quot;
.1
.2
#
+ 800
&quot;
1
−1
#
Therefore, the closed formulas are
w(t) = 4000(.1)(1)t + 800(1)(.7)t = 400 + 800(.7)t
m(t) = 4000(.2)(1)t + 800(−1)(.7)t = 800 − 800(.7)t
Looking at these two equations we see that as t → ∞, the number of customers shopping
at Wipf’s decreases and approaches 400 and the number of customers shopping at Migros
increases and approaches 800. Therefore, Wipf’s will always have more than 250 customers
per week and will not go out of business.
6
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