7.4 applications of eigenvalues and eigenvectorshomepage.ntu.edu.tw/~jryanwang/course/mathematics...

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 3x1

x2...

xn

4.

x

n3sn 2 1dLn

, L4

.

.

.

3L

n,

2L

n 2

30, L

n2

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

x3

4 5 316t

4t

t4 5 t3

16

4

14.

l 5 2.

21

|lI 2 A| 5 sl 1 1d2sl 2 2d

AAx 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

y3

4.

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

y1

y2...

yn

4yi9 5

dyi

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


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 30

1

21

04 .y19

y29

5

5

2y2

y1

y2 5 s2C1 1 C2de2t 1 2C2te2t.

y1 5 C1e2t 1 C2te

2t

l 5 2,A

A 5 3 0

24

1

44 .y19

y29

5

5 24y1 1

y2

4y2

AA

y2 5 23w1 1 w2 5 23C1e23t 1 C2e

5t.

y1 5 w1 1 w2 5 C1e23t 1 C2e

5t

3 y1

y24 5 3 1

23

1

14 3w1

w24

y 5 Pwy2y1

w2 5 C2e5t.

w1 5 C1e23t

w19 5 23w1

w29 5 5w23w19

w294 5 323

0

0

54 3w1

w24

w9 5 P21APw

P21AP 5 323

0

0

54.P21 5 314

34

214

144,P 5 3 1

23

1

14,

p2p1PAp2 5 f1 1gT.

p1 5 f1 23gTl2 5 5,l1 5 23

AA 5 33

6

2

214.P

y29 5 6y1 2 y2

y19 5 3y1 1 2y2


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 x

y4.

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 13

25

25

134.

c 5 13,a 5 13, b 5 210,

xyA 5 34

0

0

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 a

by2

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


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 13

25

25

134.

13x2 2 10xy 1 13y2 2 72 5 0.

xy

x9

xu

P 5 3cos u

sin u

2sin 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 x9

y94

P 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

by2

by2

c 4 3 x

y4 1 fd eg 3 x

y4 1 f

ax2 1 bxy 1 cy2 1 dx 1 ey 1 f


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 3l1

0

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 u

sin u

2sin 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 5

P|P| 5 1,

5 3cos u

sin u

2sin u

cos u4P 5 3 1

!2

1

!2

2 1

!2

1

!24

P,

x2 5 321

14x1 5 31

14

A


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


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 3x9

y94

1358.sin 1358 5 1y!2,cos 1358 5 21y!2

|P| 5 1. 5 3cos u

sin u

2sin 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 3

25

25

34.

3x 2 2 10xy 1 3y2 1 16!2x 2 32 5 0.

xy


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


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


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




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


Elliptic Cone




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


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


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

!2

0

2 1

!2

0

1

0

1

!2

0

1

!2

4y458

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


Figure 7.7

z'

y

x'

x

2

22

4

4

6

6

6

z

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