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372 Chapter 7 Eigenvalues and Eigenvectors
7.4 Applications of Eigenvalues and Eigenvectors
Model population growth using an age transition matrix and an
age distribution vector, and find a stable age distribution vector.
Use a matrix equation to solve a system of first-order linear
differential equations.
Find the matrix of a quadratic form and use the Principal Axes
Theorem to perform a rotation of axes for a conic and a
quadric surface.
POPULATION GROWTH
Matrices can be used to form models for population growth. The first step in this
process is to group the population into age classes of equal duration. For instance, if the
maximum life span of a member is years, then the following intervals represent the
age classes.
First age class
Second age class
th age class
The age distribution vector represents the number of population members in each
age class, where
Over a period of years, the probability that a member of the th age class will
survive to become a member of the age class is given by where
The average number of offspring produced by a member of the th age class is given by
where
These numbers can be written in matrix form, as follows.
Multiplying this age transition matrix by the age distribution vector for a specific time
period produces the age distribution vector for the next time period. That is,
Example 1 illustrates this procedure.
Axi 5 xi11.
A 5 3b1
p1
0...0
b2
0
p2
...
0
b3
0
0...0
. . .
. . .
. . .
. . .
bn21
0
0...
pn21
bn
0
0...0
40 # bi, i 5 1, 2, . . . , n.
bi,
i
0 # pi # 1, i 5 1, 2, . . . , n 2 1.
pi,si 1 1dthiLyn
x 5 3x1x2...
xn4.
x
n3sn 2 1dLn , L4
.
.
.
3Ln, 2L
n 2
30, Ln2
nL
Number in first age class
Number in second age class
Number in th age classn
.
.
.
A Population Growth Model
A population of rabbits has the following characteristics.
a. Half of the rabbits survive their first year. Of those, half survive their second year.
The maximum life span is 3 years.
b. During the first year, the rabbits produce no offspring. The average number of
offspring is 6 during the second year and 8 during the third year.
The population now consists of 24 rabbits in the first age class, 24 in the second,
and 20 in the third. How many rabbits will there be in each age class in 1 year?
SOLUTION
The current age distribution vector is
and the age transition matrix is
After 1 year, the age distribution vector will be
Finding a Stable Age Distribution Vector
Find a stable age distribution vector for the population in Example 1.
SOLUTION
To solve this problem, find an eigenvalue and a corresponding eigenvector such that
The characteristic polynomial of is
(check this), which implies that the eigenvalues are and 2. Choosing the positive
value, let Verify that the corresponding eigenvectors are of the form
For instance, if then the initial age distribution vector would be
and the age distribution vector for the next year would be
Notice that the ratio of the three age classes is still 16 : 4 : 1, and so the percent of
the population in each age class remains the same.
x2 5 Ax1 5 30
0.5
0
6
0
0.5
8
0
04 3
32
8
24 5 3
64
16
44.
x1 5 332
8
24
t 5 2,
x 5 3x1
x2
x34 5 3
16t
4t
t4 5 t3
16
4
14.
l 5 2.
21
|lI 2 A| 5 sl 1 1d2sl 2 2dAAx 5 lx.
xl
x2 5 Ax1 5 30
0.5
0
6
0
0.5
8
0
04 3
24
24
204 5 3
304
12
124.
A 5 30
0.5
0
6
0
0.5
8
0
04.
x1 5 324
24
204
7.4 Applications of Eigenvalues and Eigenvectors 373
SimulationExplore this concept further with an electronic simulation availableat www.cengagebrain.com.
REMARKIf the pattern of growth in
Example 1 continued for
another year, then the rabbit
population would be
From the age distribution
vectors and you can
see that the percent of rabbits
in each of the three age classes
changes each year. To obtain
a stable growth pattern, one
in which the percent in each
age class remains the same
each year, the age
distribution vector must be
a scalar multiple of the
age distribution vector. That is,
Example 2
shows how to solve this
problem.
xn11 5 Axn 5 lxn.
nth
sn 1 1dth
x3,x2,x1,
x3 5 Ax2 5 3168
152
64.
2 # age # 3
1 # age < 2
0 # age < 1
2 # age # 3
1 # age < 2
0 # age < 1
2 # age # 3
1 # age < 2
0 # age < 1
2 # age # 3
1 # age < 2
0 # age < 1
SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS (CALCULUS)
A system of first-order linear differential equations has the form
where each is a function of and If you let
and
then the system can be written in matrix form as
Solving a System of Linear Differential Equations
Solve the system of linear differential equations.
SOLUTION
From calculus, you know that the solution of the differential equation is
So, the solution of the system is
The matrix form of the system of linear differential equations in Example 3 is or
So, the coefficients of in the solutions are given by the eigenvalues of the
matrix
If is a diagonal matrix, then the solution of
can be obtained immediately, as in Example 3. If is not diagonal, then the solution
requires more work. First, attempt to find a matrix that diagonalizes Then, the
change of variables and produces
where is a diagonal matrix. Example 4 demonstrates this procedure.P 21AP
w9 5 P 21APwPw9 5 y9 5 Ay 5 APw
y9 5 Pw9y 5 Pw
A.P
A
y9 5 Ay
A
A.
yi 5 Ci el itt
3y19
y29
y39
4 5 34
0
0
0
21
0
0
0
24 3
y1
y2
y34.
y9 5 Ay,
y3 5 C3e2t.
y2 5 C2e2 t
y1 5 C1e4t
y 5 Cekt.
y9 5 ky
y39 5 2y3
y29 5 2y2
y19 5 4y1
y9 5 Ay.
y9 5 3y19
y29...
yn94y 5 3
y1y2...
yn4
yi9 5dyi
dt.tyi
yn9 5 an1y1 1 an2y2 1. . . 1 annyn
.
.
.
y29 5 a21y1 1 a22y2 1. . . 1 a2nyn
y19 5 a11y1 1 a12y2 1. . . 1 a1nyn
374 Chapter 7 Eigenvalues and Eigenvectors
Solving a System of Linear Differential Equations
Solve the system of linear differential equations.
SOLUTION
First, find a matrix that diagonalizes The eigenvalues of are
and with corresponding eigenvectors and
Diagonalize using the matrix whose columns consist of and
to obtain
and
The system has the following form.
The solution of this system of equations is
To return to the original variables and , use the substitution and write
which implies that the solution is
If has eigenvalues with multiplicity greater than 1 or if has complex
eigenvalues, then the technique for solving the system must be modified.
1. Eigenvalues with multiplicity greater than 1: The coefficient matrix of the system
is
The only eigenvalue of is and the solution of the system is
2. Complex eigenvalues: The coefficient matrix of the system
is
The eigenvalues of are and and the solution of the system is
Try checking these solutions by differentiating and substituting into the original
systems of equations.
y2 5 2C2 cos t 1 C1 sin t.
y1 5 C1 cos t 1 C2 sin t
l2 5 2i, l1 5 i A
A 5 30121
04 .y19
y29
5
5
2y2y1
y2 5 s2C1 1 C2de2t 1 2C2te2t. y1 5 C1e
2t 1 C2te2t
l 5 2,A
A 5 3 024
1
44 .y19
y29
5
5 24y1 1
y24y2
AA
y2 5 23w1 1 w2 5 23C1e23t 1 C2e
5t.
y1 5 w1 1 w2 5 C1e23t 1 C2e
5t
3 y1y24 5 31
23
1
14 3w1
w24
y 5 Pwy2y1
w2 5 C2e5t.
w1 5 C1e23t
w19 5 23w1w29 5 5w2
3w19w294 5 323
0
0
54 3w1
w24
w9 5 P21APw
P21AP 5 32300
54.P21 5 314
34
214
144,P 5 3 123 114,
p2p1PAp2 5 f1 1gT.p1 5 f1 23gTl2 5 5,l1 5 23
AA 5 3362
214.P
y29 5 6y1 2 y2
y19 5 3y1 1 2y2
7.4 Applications of Eigenvalues and Eigenvectors 375
QUADRATIC FORMS
Eigenvalues and eigenvectors can be used to solve the rotation of axes problem
introduced in Section 4.8. Recall that classifying the graph of the quadratic equation
Quadratic equation
is fairly straightforward as long as the equation has no -term (that is, ). If the
equation has an -term, however, then the classification is accomplished most easily
by first performing a rotation of axes that eliminates the -term. The resulting
equation (relative to the new -axes) will then be of the form
You will see that the coefficients and are eigenvalues of the matrix
The expression
Quadratic form
is called the quadratic form associated with the quadratic equation
and the matrix is called the matrix of the quadratic form. Note that the matrix is
symmetric. Moreover, the matrix will be diagonal if and only if its corresponding
quadratic form has no -term, as illustrated in Example 5.
Finding the Matrix of a Quadratic Form
Find the matrix of the quadratic form associated with each quadratic equation.
a. b.
SOLUTION
a. Because and the matrix is
Diagonal matrix (no -term)
b. Because and the matrix is
Nondiagonal matrix ( -term)
In standard form, the equation is
which is the equation of the ellipse shown in Figure 7.3. Although it is not apparent
by inspection, the graph of the equation is similar.
In fact, when you rotate the - and -axes counterclockwise to form a new
-coordinate system, this equation takes the form
which is the equation of the ellipse shown in Figure 7.4.
To see how to use the matrix of a quadratic form to perform a rotation of axes, let
X 5 3 xy4.
sx9d2
321
sy9d2
225 1
x9y9
458yx
13x2 2 10xy 1 13y2 2 72 5 0
x2
321
y2
225 1
4x2 1 9y2 2 36 5 0
xyA 5 3 1325
25
134.c 5 13,a 5 13, b 5 210,
xyA 5 3400
94.c 5 9,a 5 4, b 5 0,
13x2 2 10xy 1 13y2 2 72 5 04x2 1 9y2 2 36 5 0
xy
A
AA
dx 1 ey 1 f 5 0ax2 1 bxy 1 cy2 1
ax2 1 bxy 1 cy2
A 5 3 aby2 by2
c4.c9a9
a9sx9d2 1 c9sy9 d2 1 d9x9 1 e9y9 1 f9 5 0.
x9y9
xy
xy
b 5 0xy
ax 2 1 bxy 1 cy2 1 dx 1 ey 1 f 5 0
376 Chapter 7 Eigenvalues and Eigenvectors
Figure 7.3
−2 −1
−1
2
−3
1
3
x
y
1
x2 y2
32
22
+ = 1
Figure 7.4
−1−3 3
−3
−2
1
3y ′ x ′
x
45°
y
1
13x2 − 10xy + 13y2 − 72 = 0
(x ′)2 (y ′)2
32
22
+ = 1
Then the quadratic expression can be written in
matrix form as follows.
If then no rotation is necessary. But if then because is symmetric, you
can apply Theorem 7.10 to conclude that there exists an orthogonal matrix such that
is diagonal. So, if you let
then it follows that and
The choice of the matrix must be made with care. Because is orthogonal, its
determinant will be It can be shown (see Exercise 65) that if is chosen so that
then will be of the form
where gives the angle of rotation of the conic measured from the positive -axis to the
positive -axis. This leads to the Principal Axes Theorem.
Rotation of a Conic
Perform a rotation of axes to eliminate the -term in the quadratic equation
SOLUTION
The matrix of the quadratic form associated with this equation is
Because the characteristic polynomial of is (check this), it follows
that the eigenvalues of are and So, the equation of the rotated conic is
which, when written in the standard form
is the equation of an ellipse. (See Figure 7.4.)
sx9d2
321
sy9 d2
225 1
8sx9 d2 1 18sy9 d2 2 72 5 0
l2 5 18.l1 5 8A
sl 2 8dsl 2 18dA
A 5 3 1325
25
134.
13x2 2 10xy 1 13y2 2 72 5 0.
xy
x9
xu
P 5 3cos usin u2sin u
cos u4P|P| 5 1,
P±1.
PP
5 sX9 dTDX9.5 sX9 dTPTAPX9X TAX 5 sPX9dTAsPX9dX 5 PX9,
P TX 5 X9 5 3 x9y94P TAP 5 D
P
Ab Þ 0,b 5 0,
5 ax2 1 bxy 1 cy2 1 dx 1 ey 1 f
X TAX 1 fd egX 1 f 5 fx yg 3 a by2by2
c 4 3x
y4 1 fd eg 3x
y4 1 f
ax2 1 bxy 1 cy2 1 dx 1 ey 1 f
7.4 Applications of Eigenvalues and Eigenvectors 377
REMARKNote that the matrix product
has the form
1 s2d sin u 1 e cos udy9.
sd cos u 1 e sin udx9
fd egPX9
Principal Axes Theorem
For a conic whose equation is the
rotation given by eliminates the -term when is an orthogonal
matrix, with that diagonalizes That is,
where and are eigenvalues of The equation of the rotated conic is
given by
l1sx9 d2 1 l2sy9 d2 1 fd egPX9 1 f 5 0.
A.l2l1
PTAP 5 3l10 0
l24
A.|P| 5 1,PxyX 5 PX9
ax2 1 bxy 1 cy2 1 dx 1 ey 1 f 5 0,
In Example 6, the eigenvectors of the matrix are
and
which you can normalize to form the columns of as follows.
Note first that which implies that is a rotation. Moreover, because
cos the angle of rotation is as shown in Figure 7.4.
The orthogonal matrix specified in the Principal Axes Theorem is not unique. Its
entries depend on the ordering of the eigenvalues and and on the subsequent
choice of eigenvectors and For instance, in the solution of Example 6, any of the
following choices of would have worked.
For any of these choices of the graph of the rotated conic will, of course, be the
same. (See Figure 7.5.)
Figure 7.5
The following summarizes the steps used to apply the Principal Axes Theorem.
1. Form the matrix and find its eigenvalues and
2. Find eigenvectors corresponding to and Normalize these eigenvectors to
form the columns of
3. If then multiply one of the columns of by to obtain a matrix of
the form
4. The angle represents the angle of rotation of the conic.
5. The equation of the rotated conic is l1sx9 d2 1 l2sy9 d2 1 fd eg PX9 1 f 5 0.
u
P 5 3cos usin u2sin u
cos u4.
21P|P| 5 21,P.
l2.l1
l2.l1A
−3 3
−3
−2
3 y ′
x ′
x
315°
y
1
(x ′)2 (y ′)2
22
32
+ = 1
−3 3
−3
−2
3
y ′
x ′
x
135°
y
1
(x ′)2 (y ′)2
22
32
+ = 1
−3 31
−3
−2
3
y ′x ′
x
225°
y
(x ′)2 (y ′)2
32
22
+ = 1
P,
u 5 3158u 5 1358u 5 2258
l1 5 18, l2 5 8l1 5 18, l 2 5 8l1 5 8, l 2 5 18
3 1
!2
2 1
!2
1
!2 1
!2432
1
!2 1
!2
2 1
!2
2 1
!2432
1
!2
2 1
!2
1
!2
2 1
!24
x2x1x2x1x2x1
P
x2.x1
l2 l1
P
458,1y!2 5 sin 458 ,458 5P|P| 5 1,
5 3cos usin u2sin u
cos u4P 5 3 1
!2 1
!2
2 1
!2 1
!24
P,
x2 5 32114x1 5 31
14 A
378 Chapter 7 Eigenvalues and Eigenvectors
Example 7 shows how to apply the Principal Axes Theorem to rotate a conic whose
center has been translated away from the origin.
Rotation of a Conic
Perform a rotation of axes to eliminate the -term in the quadratic equation
SOLUTION
The matrix of the quadratic form associated with this equation is
The eigenvalues of are
and
with corresponding eigenvectors of
and
This implies that the matrix is
where
Because and the angle of rotation is
Finally, from the matrix product
the equation of the rotated conic is
In standard form, the equation
is the equation of a hyperbola. Its graph is shown in Figure 7.6.
Quadratic forms can also be used to analyze equations of quadric surfaces in
which are the three-dimensional analogs of conic sections. The equation of a quadric
surface in is a second-degree polynomial of the form
There are six basic types of quadric surfaces: ellipsoids, hyperboloids of one
sheet, hyperboloids of two sheets, elliptic cones, elliptic paraboloids, and hyperbolic
paraboloids. The intersection of a surface with a plane, called the trace of the surface
in the plane, is useful to help visualize the graph of the surface in The six basic types
of quadric surfaces, together with their traces, are shown on the next two pages.
R3.
ax2 1 by 2 1 cz2 1 dxy 1 exz 1 fyz 1 gx 1 hy 1 iz 1 j 5 0.
R3
R3,
sx9 2 1d2
122
sy9 1 4d2
225 1
8sx9 d2 2 2sy9d2 2 16x9 2 16y9 2 32 5 0.
5 216x9 2 16y9
fd egPX9 5 f16!2 0g 32 1
!2 1
!2
2 1
!2
2 1
!24 3x9y94
1358.sin 1358 5 1y!2,cos 1358 5 21y!2
|P| 5 1. 5 3cos usin u2sin u
cos u4,
P 5 32 1
!2 1
!2
2 1
!2
2 1
!24
P
x2 5 s21, 21d.x1 5 s21, 1d
l2 5 22l1 5 8
A
A 5 3 325
25
34.
3x 2 2 10xy 1 3y2 1 16!2x 2 32 5 0.
xy
7.4 Applications of Eigenvalues and Eigenvectors 379
Figure 7.6
−4 −2 4 62 8
−2
4
6
8
10
x
x ′
y ′
135°
y
(x ′ − 1) (y ′ + 4)
12
22
− = 12 2
y
x
z
y
x
z
xz-trace yz-trace
xy-trace
y
x
z
y
x
z
xz-trace yz-trace
xy-trace
x y
zz
x y
yz-trace xz-trace
no xy-traceparallel to
xy-plane
380 Chapter 7 Eigenvalues and Eigenvectors
Ellipsoid
Ellipse Parallel to xy-plane
Ellipse Parallel to xz-plane
Ellipse Parallel to yz-plane
The surface is a sphere when
a 5 b 5 c Þ 0.
Plane Trace
x2
a21
y2
b21
z2
c25 1
Hyperboloid of One Sheet
Ellipse Parallel to xy-plane
Hyperbola Parallel to xz-plane
Hyperbola Parallel to yz-plane
The axis of the hyperboloid
corresponds to the variable whose
coefficient is negative.
Plane Trace
x2
a21
y2
b22
z2
c25 1
Hyperboloid of Two Sheets
Ellipse Parallel to xy-plane
Hyperbola Parallel to xz-plane
Hyperbola Parallel to yz-plane
The axis of the hyperboloid
corresponds to the variable whose
coefficient is positive. There is
no trace in the coordinate plane
perpendicular to this axis.
Plane Trace
z2
c22
x2
a22
y2
b25 1
x
y
z
x
y
z xz-trace
yz-trace
xy-trace
(one point)
parallel to
xy-plane
xy
z
xy
zyz-trace xz-trace
parallel to
xy-plane
xy-trace(one point)
x
y
z
x
y
zyz-trace
xz-trace
parallel to
xy-plane
7.4 Applications of Eigenvalues and Eigenvectors 381
Elliptic Cone
Ellipse Parallel to xy-plane
Hyperbola Parallel to xz-plane
Hyperbola Parallel to yz-plane
The axis of the cone corresponds
to the variable whose coefficient
is negative. The traces in the
coordinate planes parallel to this
axis are intersecting lines.
Plane Trace
x2
a21
y2
b22
z2
c25 0
Elliptic Paraboloid
Ellipse Parallel to xy-plane
Parabola Parallel to xz-plane
Parabola Parallel to yz-plane
The axis of the paraboloid
corresponds to the variable raised
to the first power.
Plane Trace
z 5x2
a21
y2
b2
Hyperbolic Paraboloid
Hyperbola Parallel to xy-plane
Parabola Parallel to xz-plane
Parabola Parallel to yz-plane
The axis of the paraboloid
corresponds to the variable raised
to the first power.
Plane Trace
z 5y2
b22
x2
a2
LINEAR ALGEBRA APPLIED
Some of the world’s most unusual architecture makes use
of quadric surfaces. For instance, Catedral Metropolitana
Nossa Senhora Aparecida, a cathedral located in Brasilia,
Brazil, is in the shape of a hyperboloid of one sheet. It was
designed by Pritzker Prize winning architect Oscar Niemeyer,
and dedicated in 1970. The sixteen identical curved steel
columns, weighing 90 tons each, are intended to represent
two hands reaching up to the sky. Pieced together between
the columns, in the 10-meter-wide and 30-meter-high
triangular gaps formed by the columns, is semitransparent
stained glass, which allows light inside for nearly the entire
height of the columns.
The quadratic form of the equation
Quadric surface
is defined as
Quadratic form
The corresponding matrix is
In its three-dimensional version, the Principal Axes Theorem relates the eigenvalues
and eigenvectors of to the equation of the rotated surface, as shown in Example 8.
Rotation of a Quadric Surface
Perform a rotation of axes to eliminate the -term in the quadratic equation
SOLUTION
The matrix associated with this quadratic equation is
which has eigenvalues of and So, in the rotated -system,
the quadratic equation is which in standard form is
The graph of this equation is an ellipsoid. As shown in Figure 7.7, the -axes represent
a counterclockwise rotation of about the -axis. Moreover, the orthogonal matrix
whose columns are the eigenvectors of has the property that is diagonal.P TAPA,
P 5 3 1
!20
2 1
!2
0
1
0
1
!20
1
!24
y458
x9y9z9
sx9d2
621
sy9d2
321
sz9d2
225 1.
sx9d2 1 4sy9d2 1 9sz9d2 2 36 5 0,x9y9z9l3 5 9.l1 5 1, l 2 5 4,
A 5 35
0
4
0
4
0
4
0
54
A
5x2 1 4y2 1 5z2 1 8xz 2 36 5 0.
xz
A
A 5 3a
d
2
e
2
d
2
b
f
2
e
2
f
2
c4 .
ax2 1 by2 1 cz2 1 dxy 1 exz 1 fyz.
ax2 1 by2 1 cz2 1 dxy 1 exz 1 fyz 1 gx 1 hy 1 iz 1 j 5 0
382 Chapter 7 Eigenvalues and Eigenvectors
Figure 7.7
z'
y
x'
x
2
22
4
4
6
6
6
z
ostill, 2010/Used under license from Shutterstock.com