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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. L 0, First age class ΄ n΃ L 2L , Second age class ΄n n ΃ . ͑n Ϫ 1͒L , L nth age class ΄ n ΅ The age distribution vector x represents the number of population members in each age class, where x 1 Number in first age class x Number in second age class x ϭ .2 . ΄x ΅ Number in nth age class n Over a period of L͞n years, the probability that a member of the ith age class will ϩ survive to become a member of the ͑i 1͒th age class is given by pi, where ϭ Ϫ 0 Յ pi Յ 1, i 1, 2, . , n 1. The average number of offspring produced by a member of the ith age class is given by bi, where ϭ 0 Յ bi, i 1, 2, . , n. These numbers can be written in matrix form, as follows. b1 b2 b3 bnϪ1 bn . p1 0 0 0 0 A ϭ 0 p 0 . 0 0 . .2 . ΄ . ΅ 0 0 0 pnϪ1 0 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, ϭ Axi xiϩ1. Example 1 illustrates this procedure. 7.4 Applications of Eigenvalues and Eigenvectors 373 A Population Growth Model REMARK A population of rabbits has the following characteristics. If the pattern of growth in a. Half of the rabbits survive their first year. Of those, half survive their second year. Example 1 continued for The maximum life span is 3 years. another year, then the rabbit b. During the first year, the rabbits produce no offspring. The average number of population would be offspring is 6 during the second year and 8 during the third year. 168 The population now consists of 24 rabbits in the first age class, 24 in the second, ϭ ϭ and 20 in the third. How many rabbits will there be in each age class in 1 year? x3 Ax2 152 . ΄ 6΅ SOLUTION From the age distribution The current age distribution vector is vectors x1, x2, and x3, you can 24 0 Յ age < 1 see that the percent of rabbits ϭ 1 Յ age < 2 in each of the three age classes x1 24 changes each year. To obtain ΄20 ΅ 2 Յ age Յ 3 a stable growth pattern, one and the age transition matrix is in which the percent in each age class remains the same 0 6 8 ϩ each year, the ͑n 1͒th age A ϭ 0.5 0 0 . distribution vector must be ΄ 0 0.5 0΅ a scalar multiple of the nth age distribution vector. That is, After 1 year, the age distribution vector will be ϭ ϭ ␭ xnϩ1 Axn xn. Example 2 shows how to solve this 0 6 8 24 304 0 Յ age < 1 problem. ϭ ϭ ϭ Յ < x2 Ax1 0.5 0 0 24 12 . 1 age 2 ΄ 0 0.5 0΅΄20 ΅ ΄ 12 ΅ 2 Յ age Յ 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 ␭ and a corresponding eigenvector x such that Ax ϭ ␭x. The characteristic polynomial of A is Խ␭I Ϫ AԽ ϭ ͑␭ ϩ 1͒2͑␭ Ϫ 2͒ (check this), which implies that the eigenvalues are Ϫ1 and 2. Choosing the positive value, let ␭ ϭ 2. Verify that the corresponding eigenvectors are of the form Simulation x 16 t 16 Explore this concept further with 1 ϭ ϭ ϭ an electronic simulation available x x 2 4t t 4 . at www.cengagebrain.com . ΄ ΅ ΄ ΅ ΄ ΅ x 3 t 1 For instance, if t ϭ 2, then the initial age distribution vector would be 32 0 Յ age < 1 ϭ Յ < x1 8 1 age 2 ΄ 2΅ 2 Յ age Յ 3 and the age distribution vector for the next year would be 0 0 6 8 32 64 0 Յ age < 1 ϭ ϭ ϭ Յ < x2 Ax1 0.5 0 0 8 16 . 1 age 2 ΄ 0 0.5 0΅΄ 2΅ ΄ 4΅ 2 Յ age Յ 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 Eigenvalues and Eigenvectors SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS (CALCULUS) A system of first-order linear differential equations has the form Ј ϭ ϩ ϩ . ϩ y1 a11 y1 a12 y2 a1nyn Ј ϭ ϩ ϩ . ϩ y2 a21 y1 a22 y2 a2nyn . Ј ϭ ϩ ϩ . ϩ yn an1y1 an2 y2 ann yn dy where each y is a function of t and y Ј ϭ i. If you let i i dt Ј y1 y1 y y Ј y ϭ .2 and yЈ ϭ .2 . ΄y ΅ ΄y Ј΅ n n then the system can be written in matrix form as yЈ ϭ Ay. Solving a System of Linear Differential Equations Solve the system of linear differential equations. Ј ϭ y1 4y1 Ј ϭ Ϫ y2 y2 Ј ϭ y3 2y3 SOLUTION From calculus, you know that the solution of the differential equation yЈ ϭ ky is y ϭ Ce kt . So, the solution of the system is ϭ 4t y1 C1e ϭ Ϫt y2 C2e ϭ 2t y3 C3e . The matrix form of the system of linear differential equations in Example 3 is yЈ ϭ Ay, or Ј y1 4 0 0 y1 Ј ϭ Ϫ y2 0 1 0 y2 . ΄ Ј ΅ ΄ ΅΄ ΅ y3 0 0 2 y3 ␭ ϭ it So, the coefficients of t in the solutions yi Ci e are given by the eigenvalues of the matrix A. If A is a diagonal matrix, then the solution of yЈ ϭ 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 ϭ Pw and yЈ ϭ PwЈ produces PwЈ ϭ yЈ ϭ Ay ϭ AP w wЈ ϭ P Ϫ1AP w where P Ϫ1AP is a diagonal matrix. Example 4 demonstrates this procedure. 7.4 Applications of Eigenvalues and Eigenvectors 375 Solving a System of Linear Differential Equations Solve the system of linear differential equations. Ј ϭ ϩ y1 3y1 2y2 Ј ϭ Ϫ y2 6y1 y2 SOLUTION 3 2 First, find a matrix P that diagonalizes A ϭ . The eigenvalues of A are ΄6 Ϫ1΅ ␭ ϭ Ϫ ␭ ϭ ϭ Ϫ T 1 3 and 2 5, with corresponding eigenvectors p1 ͓1 3͔ and ϭ T p2 ͓1 1 ͔ . Diagonalize A using the matrix P whose columns consist of p1 and p2 to obtain 1 Ϫ1 1 1 4 4 Ϫ3 0 P ϭ , PϪ1 ϭ , and P Ϫ1AP ϭ . ΄Ϫ3 1΅ 3 1 ΄ 0 5΅ ΄4 4΅ The system wЈ ϭ P Ϫ1AP w has the following form. w Ј Ϫ3 0 w w Ј ϭ Ϫ 3w 1 ϭ 1 1 1 Ј Ј ϭ ΄w2 ΅ ΄ 0 5΅΄w2΅ w2 5w2 The solution of this system of equations is ϭ Ϫ3t w1 C1e ϭ 5t w2 C2e . ϭ To return to the original variables y1 and y2, use the substitution y Pw and write y 1 1 w 1 ϭ 1 Ϫ ΄ y2΅ ΄ 3 1΅΄w2΅ which implies that the solution is ϭ ϩ ϭ Ϫ3t ϩ 5t y1 w1 w2 C1e C2e ϭ Ϫ ϩ ϭ Ϫ Ϫ3t ϩ 5t y2 3w1 w2 3C1e C2e . 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 Ј ϭ 0 1 y1 y2 ϭ Ј ϭ Ϫ ϩ is A . y2 4y1 4y2 ΄Ϫ4 4΅ The only eigenvalue of A is ␭ ϭ 2, and the solution of the system is ϭ 2t ϩ 2t y1 C1e C2te ϭ ϩ 2t ϩ 2t y2 ͑2C1 C2͒e 2C2te . 2. Complex eigenvalues: The coefficient matrix of the system Ј ϭ Ϫ 0 Ϫ1 y1 y2 ϭ Ј ϭ is A . y2 y1 ΄1 0΅ ␭ ϭ ␭ ϭ Ϫ The eigenvalues of A are 1 i and 2 i, and the solution of the system is ϭ ϩ y1 C1 cos t C2 sin t ϭ Ϫ ϩ y2 C2 cos t C1 sin t. Try checking these solutions by differentiating and substituting into the original systems of equations. 376 Chapter 7 Eigenvalues and Eigenvectors 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 ϩ bxy ϩ cy 2 ϩ dx ϩ ey ϩ f ϭ 0 Quadratic equation is fairly straightforward as long as the equation has no xy -term (that is,b ϭ 0 ).
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