7.4 Applications of Eigenvalues and Eigenvectors

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 L years, then the following n intervals represent the age classes.

30, n2

First age class

3 Ln, 2Ln2

Second age class

L

3

. . . sn 2 1dL ,L n

4

nth age class

The age distribution vector x represents the number of population members in each age class, where

34

x1 x x 5 .2 . . . xn

Number in first age class Number in second age class . . . Number in nth age class

Over a period of Lyn years, the probability that a member of the ith age class will survive to become a member of the si 1 1dth age class is given by pi, where 0 # pi # 1, i 5 1, 2, . . . , n 2 1. The average number of offspring produced by a member of the ith age class is given by bi, where 0 # bi, i 5 1, 2, . . . , n. These numbers can be written in matrix form, as follows.

A5

3

b1 p1 0 . . . 0

b2 0 p2 . . . 0

b3 0 0 . . . 0

. . . . . . . . . . . .

bn21 0 0 . . . pn21

bn 0 0 . . . 0

4

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, Axi 5 xi11. Example 1 illustrates this procedure.

7.4

Applications of Eigenvalues and Eigenvectors

373

A Population Growth Model

REMARK If the pattern of growth in Example 1 continued for another year, then the rabbit population would be

3 4

168 x3 5 Ax2 5 152 . 6 From the age distribution vectors x1, x2, and x3, 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 sn 1 1dth age distribution vector must be a scalar multiple of the nth age distribution vector. That is, x n11 5 Ax n 5 l x n. Example 2 shows how to solve this problem.

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

34

0 # age < 1

24 x1 5 24 20

1 # age < 2 2 # age # 3

and the age transition matrix is

3

0 A 5 0.5 0

6 0 0.5

4

8 0 . 0

After 1 year, the age distribution vector will be

3

0 x2 5 Ax1 5 0.5 0

6 0 0.5

8 0 0

43 4 3 4

24 304 24 5 12 . 20 12

0 # age < 1 1 # age < 2 2 # age # 3

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 l and a corresponding eigenvector x such that Ax 5 lx. The characteristic polynomial of A is

|lI 2 A| 5 sl 1 1d2sl 2 2d (check this), which implies that the eigenvalues are 21 and 2. Choosing the positive value, let l 5 2. Verify that the corresponding eigenvectors are of the form

Simulation Explore this concept further with an electronic simulation available at www.cengagebrain.com.

34 3 4 34

x1 16t 16 x 5 x 2 5 4t 5 t 4 . x3 t 1 For instance, if t 5 2, then the initial age distribution vector would be

34

0 # age < 1

32 x1 5 8 2

1 # age < 2 2 # age # 3

and the age distribution vector for the next year would be

3

0 x2 5 Ax1 5 0.5 0

6 0 0.5

8 0 0

43 4 3 4

32 64 8 5 16 . 2 4

0 # age < 1 1 # age < 2 2 # age # 3

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.

374

Chapter 7


SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS (CALCULUS) A system of first-order linear differential equations has the form y19 5 a11y1 1 a12 y2 1 . . . 1 a1n yn y29 5 a21y1 1 a22 y2 1 . . . 1 a2n yn . . . yn9 5 an1y1 1 an2 y2 1 . . . 1 ann yn where each yi is a function of t and yi9 5

34

y1 y y 5 .2 . . yn

and

dyi . If you let dt

34

y19 y9 y9 5 .2 . . yn9

then the system can be written in matrix form as y9 5 Ay.

Solving a System of Linear Differential Equations Solve the system of linear differential equations. y19 5 4y1 y29 5 2y2 y39 5 2y3 SOLUTION From calculus, you know that the solution of the differential equation y9 5 ky is y 5 Cekt. So, the solution of the system is y1 5 C1e4t y2 5 C2e 2t y3 5 C3e2t. The matrix form of the system of linear differential equations in Example 3 is y9 5 Ay, or

3 4 3 y19 y29 y39

4 5 0 0

0 21 0

0 0 2

43 4

y1 y2 . y3

So, the coefficients of t in the solutions yi 5 Ci el i t are given by the eigenvalues of the matrix A. If A is a diagonal matrix, then the solution of y9 5 Ay can be obtained immediately, as in Example 3. If A is not diagonal, then the solution requires more work. First, attempt to find a matrix P that diagonalizes A. Then, the change of variables y 5 Pw and y9 5 Pw9 produces Pw9 5 y9 5 Ay 5 APw

w9 5 P 21APw

where P 21AP is a diagonal matrix. Example 4 demonstrates this procedure.

7.4


375

Solving a System of Linear Differential Equations Solve the system of linear differential equations. y19 5 3y1 1 2y2 y29 5 6y1 2 y2 SOLUTION

36

P5

3

4

1 23

1 , P21 5 1

3 4 1 4

214

3 4

1 4

,

and

3

4

2 . The eigenvalues of A are 21 l1 5 23 and l2 5 5, with corresponding eigenvectors p 1 5 f1 23g T and p 2 5 f1 1g T. Diagonalize A using the matrix P whose columns consist of p 1 and p 2 to obtain First, find a matrix P that diagonalizes A 5

P 21AP 5

3

23 0

4

0 . 5

The system w9 5 P 21APw has the following form.

3 w 94 5 3 w19 2

23 0

4 3w 4

0 5

w19 5 23w1 w29 5 5w2

w1 2

The solution of this system of equations is w1 5 C1e23t w2 5 C2e5t. To return to the original variables y1 and y2, use the substitution y 5 Pw and write

3 y 4 5 323 y1

1

2

4 3w 4

1 1

w1 2

which implies that the solution is y1 5 w1 1 w2 5 C1e23t 1 C2e5t y2 5 23w1 1 w2 5 23C1e23t 1 C2e5t. If A has eigenvalues with multiplicity greater than 1 or if A 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 y19 5 y2 y29 5 24y1 1 4y2

A5

is

324 0

4

1 . 4

The only eigenvalue of A is l 5 2, and the solution of the system is y1 5 C1e2t 1 C2te2t y2 5 s2C1 1 C2de2t 1 2C2te2t. 2. Complex eigenvalues: The coefficient matrix of the system y19 5 2y2 is y29 5 y1

A5

31 0

4

21 . 0

The eigenvalues of A are l1 5 i and l2 5 2i, and the solution of the system is y1 5 C1 cos t 1 C2 sin t y2 5 2C2 cos t 1 C1 sin t. Try checking these solutions by differentiating and substituting into the original systems of equations.

376

Chapter 7


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 ax 2 1 bxy 1 cy 2 1 dx 1 ey 1 f 5 0

Quadratic equation

is fairly straightforward as long as the equation has no xy-term (that is, b 5 0). If the equation has an xy-term, however, then the classification is accomplished most easily by first performing a rotation of axes that eliminates the xy-term. The resulting equation (relative to the new x9 y9-axes) will then be of the form a9sx9 d2 1 c9sy9 d2 1 d9x9 1 e9y9 1 f9 5 0. You will see that the coefficients a9 and c9 are eigenvalues of the matrix A5

3by2

4

a by2 . c

The expression ax 2 1 bxy 1 cy 2

Quadratic form

is called the quadratic form associated with the quadratic equation ax 2 1 bxy 1 cy 2 1 dx 1 ey 1 f 5 0 and the matrix A is called the matrix of the quadratic form. Note that the matrix A is symmetric. Moreover, the matrix A will be diagonal if and only if its corresponding quadratic form has no xy-term, as illustrated in Example 5.

Finding the Matrix of a Quadratic Form

y 3

Find the matrix of the quadratic form associated with each quadratic equation. a. 4x 2 1 9y 2 2 36 5 0 b. 13x 2 2 10xy 1 13y 2 2 72 5 0

1 x

−2

−1

1

2

SOLUTION a. Because a 5 4, b 5 0, and c 5 9, the matrix is

−1

A5 2

y2

x + =1 2 2 3 2

−3

(x′)2 y

3

2

+

3

y′

(y′) 2 2

2

=1

45° x

−3

−1

1

−2 −3

13x 2 − 10xy + 13y 2 − 72 = 0

Figure 7.4

3

4

0 . 9

Diagonal matrix (no xy-term)

325 13

4

25 . 13

Nondiagonal matrix (xy -term)

In standard form, the equation 4x 2 1 9y 2 2 36 5 0 is

x′

1

4

b. Because a 5 13, b 5 210, and c 5 13, the matrix is A5

Figure 7.3

30

x2 y2 1 251 2 3 2 which is the equation of the ellipse shown in Figure 7.3. Although it is not apparent by inspection, the graph of the equation 13x 2 2 10xy 1 13y 2 2 72 5 0 is similar. In fact, when you rotate the x- and y-axes counterclockwise 458 to form a new x9y9-coordinate system, this equation takes the form

sx9 d2 s y9 d2 1 2 51 32 2 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 X5

3 y4. x

7.4


377

Then the quadratic expression ax 2 1 bxy 1 cy 2 1 dx 1 ey 1 f can be written in matrix form as follows. X TAX 1 fd

eg X 1 f 5 fx yg

3by2

4 3 y4 1 fd

a by2 c

x

eg

3 y4 1 f x

5 ax 2 1 bxy 1 cy 2 1 dx 1 ey 1 f If b 5 0, then no rotation is necessary. But if b Þ 0, then because A is symmetric, you can apply Theorem 7.10 to conclude that there exists an orthogonal matrix P such that P TAP 5 D is diagonal. So, if you let P T X 5 X9 5

3 y9 4 x9

then it follows that X 5 PX9, and X TAX 5 sPX9 dTAsPX9 d 5 sX9 dTP TAPX9 5 sX9 dTDX9. The choice of the matrix P must be made with care. Because P is orthogonal, its determinant will be ± 1. It can be shown (see Exercise 65) that if P is chosen so that P 5 1, then P will be of the form

||

P5

3 sin u cos u

2sin u cos u

4

where u gives the angle of rotation of the conic measured from the positive x-axis to the positive x9-axis. This leads to the Principal Axes Theorem.

Principal Axes Theorem

REMARK Note that the matrix product fd eg PX9 has the form

sd cos u 1 e sin udx9 1 s2d sin u 1 e cos udy9.

For a conic whose equation is ax 2 1 bxy 1 cy 2 1 dx 1 ey 1 f 5 0, the rotation given by X 5 PX9 eliminates the xy-term when P is an orthogonal matrix, with P 5 1, that diagonalizes A. That is,

||

PTAP 5

30

l1

0 l2

4

where l1 and l 2 are eigenvalues of A. The equation of the rotated conic is given by

l1sx9 d2 1 l 2s y9 d2 1 fd

eg PX9 1 f 5 0.

Rotation of a Conic Perform a rotation of axes to eliminate the xy-term in the quadratic equation 13x 2 2 10xy 1 13y 2 2 72 5 0. SOLUTION The matrix of the quadratic form associated with this equation is A5

325 13

4

25 . 13

Because the characteristic polynomial of A is sl 2 8dsl 2 18d (check this), it follows that the eigenvalues of A are l1 5 8 and l2 5 18. So, the equation of the rotated conic is 8sx9 d2 1 18s y9 d2 2 72 5 0 which, when written in the standard form

sx9 d2 s y9 d2 1 2 51 32 2 is the equation of an ellipse. (See Figure 7.4.)

378

Chapter 7

Eigenvalues and Eigenvectors In Example 6, the eigenvectors of the matrix A are

314 1

x1 5

x2 5

and

3 14 21

which you can normalize to form the columns of P, as follows.

3

P5

4

2 1 !2 cos u 5 1 sin u !2

1 !2

1 !2

3

2sin u cos u

4

||

Note first that P 5 1, which implies that P is a rotation. Moreover, because cos 458 5 1y!2 5 sin 458 , the angle of rotation is 458, as shown in Figure 7.4. The orthogonal matrix P specified in the Principal Axes Theorem is not unique. Its entries depend on the ordering of the eigenvalues l1 and l2 and on the subsequent choice of eigenvectors x1 and x2. For instance, in the solution of Example 6, any of the following choices of P would have worked. x1

3

2 2

x2

x1

1

1

!2

!2

1

2

!2

1 !2

4 3

l 1 5 8, l 2 5 18 u 5 2258

2

1 !2

1 !2

x2

2 2

x1

1 !2

1 !2

4 3

x2

1

1

!2

!2

4

1 1 ! !2 2 l 1 5 18, l 2 5 8 u 5 3158

l 1 5 18, l 2 5 8 u 5 1358

2

For any of these choices of P, the graph of the rotated conic will, of course, be the same. (See Figure 7.5.) y

y

y

3

x′

3

225°

135°

315°

x −3

1

x

3

−3

−2 x′

y′

−3

(x′)2 3

2

+

(y′) 2 2

2

y′

3

y′

1

x −3 −2

−3

−3

2

2

+

(y ′)2 3

2

=1

3

1

−2

(x ′)2

=1

3

(x′)2 2

2

x′

+

(y′)2 3

2

=1

Figure 7.5

The following summarizes the steps used to apply the Principal Axes Theorem. 1. Form the matrix A and find its eigenvalues l1 and l 2. 2. Find eigenvectors corresponding to l1 and l 2. Normalize these eigenvectors to form the columns of P.

||

3. If P 5 21, then multiply one of the columns of P by 21 to obtain a matrix of the form P5

3 sin u

4

cos u 2sin u . cos u

4. The angle u represents the angle of rotation of the conic. 5. The equation of the rotated conic is l 1sx9 d2 1 l 2s y9 d2 1 fd

eg PX9 1 f 5 0.

7.4


379

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 xy-term in the quadratic equation 3x 2 2 10xy 1 3y 2 1 16!2x 2 32 5 0. SOLUTION The matrix of the quadratic form associated with this equation is A5

325

4

25 . 3

3

The eigenvalues of A are

l1 5 8 and l2 5 22 with corresponding eigenvectors of x1 5 s21, 1d and

x2 5 s21, 21d.

This implies that the matrix P is

P5

5

3

2

1 !2

1 !2

u 3cos sin u

2 2

1 !2

1 !2

4

4

2sin u , where P 5 1. cos u

||

Because cos 1358 5 21y!2 and sin 1358 5 1y!2, the angle of rotation is 1358. Finally, from the matrix product

y 10

fd

8

egPX9 5 f16!2

0g

6 x′

4

3

2

1 !2

1 !2

2

1 !2

1 2 !2

4

3xy99 4

5 216x9 2 16y9

135°

the equation of the rotated conic is x

−4

−2 y′

2

4

6

8

−2

8sx9 d2 2 2s y9 d2 2 16x9 2 16y9 2 32 5 0. In standard form, the equation

(x′ − 1) 2 1

Figure 7.6

2

−

(y′ + 4) 2 2

2

=1

sx9 2 1d2 s y9 1 4d2 2 51 12 22 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 R 3, which are the three-dimensional analogs of conic sections. The equation of a quadric surface in R 3 is a second-degree polynomial of the form ax 2 1 by 2 1 cz 2 1 dxy 1 exz 1 fyz 1 gx 1 hy 1 iz 1 j 5 0. 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 R 3. The six basic types of quadric surfaces, together with their traces, are shown on the next two pages.

380

Chapter 7


Ellipsoid x2 y2 z2 1 1 51 a 2 b 2 c2

z

Trace

Plane

Ellipse Ellipse Ellipse

Parallel to xy-plane Parallel to xz-plane Parallel to yz-plane

z

xz-trace

yz-trace

y

y

The surface is a sphere when a 5 b 5 c Þ 0.

x

x

xy-trace

z z

Hyperboloid of One Sheet x2 y2 z2 1 22 251 2 a b c

y

Trace

Plane

Ellipse Hyperbola Hyperbola


xy-trace

y

x

x

The axis of the hyperboloid corresponds to the variable whose coefficient is negative. yz-trace

xz-trace

Hyperboloid of Two Sheets

z

z2 c2

x

y

2

x2 a2

2

y2 b2

yz-trace

z

xz-trace

51

Trace

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.

parallel to xy-plane x

no xy-trace y

7.4

381


z

z

xz-trace

Elliptic Cone x2 y2 z2 1 22 250 2 a b c

y

Trace

Plane



x

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.

z5

parallel to xy-plane

yz-trace

z

y

yz-trace

Plane

Ellipse Parabola Parabola


parallel to xy-plane

The axis of the paraboloid corresponds to the variable raised to the first power.

y

y

x

Hyperbolic Paraboloid z5

xz-trace

x2 y2 1 2 2 a b

Trace

z

x

y x

Elliptic Paraboloid

z

x

xy-trace (one point)

yz-trace

xy-trace (one point)

z

y2 x2 2 2 2 b a

Trace

Plane

Hyperbola Parabola Parabola


y x

The axis of the paraboloid corresponds to the variable raised to the first power.

parallel to xy-plane xz-trace

382

Chapter 7


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 ax 2 1 by 2 1 cz 2 1 dxy 1 exz 1 fyz 1 gx 1 hy 1 iz 1 j 5 0

Quadric surface

is defined as ax 2 1 by 2 1 cz 2 1 dxy 1 exz 1 fyz.

Quadratic form

The corresponding matrix is

3 4 a

A5

d 2 e 2

d 2

e 2 f . 2

b

f 2

c

In its three-dimensional version, the Principal Axes Theorem relates the eigenvalues and eigenvectors of A 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 xz-term in the quadratic equation 5x 2 1 4y 2 1 5z 2 1 8xz 2 36 5 0. z

SOLUTION The matrix A associated with this quadratic equation is

6 z'

3

5 A5 0 4

4 2 2 6

2 4 6

x

x'

Figure 7.7

y

0 4 0

4 0 5

4

which has eigenvalues of l 1 5 1, l 2 5 4, and l 3 5 9. So, in the rotated x9y9z9-system, the quadratic equation is sx9 d2 1 4s y9 d2 1 9sz9 d2 2 36 5 0, which in standard form is

sx9 d2 s y9 d2 sz9 d2 1 2 1 2 5 1. 62 3 2 The graph of this equation is an ellipsoid. As shown in Figure 7.7, the x9y9z9-axes represent a counterclockwise rotation of 458 about the y-axis. Moreover, the orthogonal matrix

P5

3

1 !2

0 1 2 !2

0

1 !2

1 0

0 1 !2

4

whose columns are the eigenvectors of A, has the property that P TAP is diagonal. ostill, 2010/Used under license from Shutterstock.com

7.4 Applications of Eigenvalues and Eigenvectors

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