7.4 Applications of Eigenvalues and Eigenvectors


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1 37 Chapter 7 Eigenvalues and Eigenvectors 7 Applications of Eigenvalues and Eigenvectors Model population growth using an age transition matri and an age distribution vector, and find a stable age distribution vector Use a matri equation to solve a sstem of firstorder linear differential equations Find the matri of a quadratic form and use the Principal Aes Theorem to perform a rotation of aes 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 maimum life span of a member is L ears, then the following n intervals represent the age classes, L n L n, L n n L, L n First age class Second age class nth age class The age distribution vector represents the number of population members in each age class, where n Over a period of Ln ears, the probabilit that a member of the ith age class will survive to become a member of the i th age class is given b p i, where p i, i,,, n The average number of offspring produced b a member of the ith age class is given b b i, where b i, i,,, n These numbers can be written in matri form, as follows A b p Multipling this age transition matri b the age distribution vector for a specific time period produces the age distribution vector for the net time period That is, A i i b p Number in first age class Number in second age class Number in nth age class b 3 Eample illustrates this procedure b n p n b n
2 7 Applications of Eigenvalues and Eigenvectors 373 A Population Growth Model REMARK If the pattern of growth in Eample continued for another ear, then the rabbit population would be 3 A 8 5 From the age distribution vectors,, and 3, ou can see that the percent of rabbits in each of the three age classes changes each ear To obtain a stable growth pattern, one in which the percent in each age class remains the same each ear, the n th age distribution vector must be a scalar multiple of the nth age distribution vector That is, n A n n Eample shows how to solve this problem A population of rabbits has the following characteristics a Half of the rabbits survive their first ear Of those, half survive their second ear The maimum life span is 3 ears b During the first ear, the rabbits produce no offspring The average number of offspring is during the second ear and 8 during the third ear The population now consists of rabbits in the first age class, in the second, and in the third How man rabbits will there be in each age class in ear? The current age distribution vector is and the age transition matri is A 5 5 After ear, the age distribution vector will be A 5 age < age < age age < age < age 3 Finding a Stable Age Distribution Vector Simulation Eplore this concept further with an electronic simulation available at wwwcengagebraincom Find a stable age distribution vector for the population in Eample To solve this problem, find an eigenvalue and a corresponding eigenvector such that A The characteristic polnomial of A is I A (check this), which implies that the eigenvalues are and Choosing the positive value, let Verif that the corresponding eigenvectors are of the form For instance, if t, then the initial age distribution vector would be t t t t age < age < age 3 and the age distribution vector for the net ear would be 8 age < A 5 8 age < 5 age 3 3 Notice that the ratio of the three age classes is still : :, and so the percent of the population in each age class remains the same
3 37 Chapter 7 Eigenvalues and Eigenvectors SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS (CALCULUS) A sstem of firstorder linear differential equations has the form a a a n n a a a n n n a n a n a nn n where each is a function of and i d i i t If ou let dt n and then the sstem can be written in matri form as A Solving a Sstem of Linear Differential Equations Solve the sstem of linear differential equations 3 3 From calculus, ou know that the solution of the differential equation Ce kt So, the solution of the sstem is C e t C e t 3 C 3 e t The matri form of the sstem of linear differential equations in Eample 3 is 3 3 So, the coefficients of t in the solutions i C i eit are given b the eigenvalues of the matri A If A is a diagonal matri, then the solution of A can be obtained immediatel, as in Eample 3 If A is not diagonal, then the solution requires more work First, attempt to find a matri P that diagonalies A Then, the change of variables Pw and produces Pw A APw n Pw w P APw k where P AP is a diagonal matri Eample demonstrates this procedure is A, or
4 Solving a Sstem of Linear Differential Equations Solve the sstem of linear differential equations 3 First, find a matri P 3 that diagonalies A The eigenvalues of A are 3 and 5, with corresponding eigenvectors p 3 T and p T Diagonalie A using the matri P whose columns consist of p and p to obtain 3 and P AP 5 P P 3, 3 The sstem w w 3 has the following form The solution of this sstem of equations is w C e 3t w C e 5t To return to the original variables and, use the substitution Pw and write 3 which implies that the solution is w w C e3t C e 5t 3w w 3C e3t C e 5t If A has eigenvalues with multiplicit greater than or if A has comple eigenvalues, then the technique for solving the sstem must be modified Eigenvalues with multiplicit greater than : The coefficient matri of the sstem is The onl eigenvalue of A is and the solution of the sstem is C et C te t C C e t C te t Comple eigenvalues: The coefficient matri of the sstem 7 Applications of Eigenvalues and Eigenvectors 375 w P APw is 5 w w w w A The eigenvalues of A are i and i, and the solution of the sstem is C cos t C sin t, C cos t C sin t, A w 3w w 5w Tr checking these solutions b differentiating and substituting into the original sstems of equations
5 37 Chapter 7 Eigenvalues and Eigenvectors QUADRATIC FORMS Eigenvalues and eigenvectors can be used to solve the rotation of aes problem introduced in Section 8 Recall that classifing the graph of the quadratic equation a b c d e f Quadratic equation is fairl straightforward as long as the equation has no term (that is, b ) If the equation has an term, however, then the classification is accomplished most easil b first performing a rotation of aes that eliminates the term The resulting equation (relative to the new aes) will then be of the form You will see that the coefficients and are eigenvalues of the matri A a b The epression b c a b c a c d e f a Quadratic form is called the quadratic form associated with the quadratic equation a b c d e f c and the matri A is called the matri of the quadratic form Note that the matri A is smmetric Moreover, the matri A will be diagonal if and onl if its corresponding quadratic form has no term, as illustrated in Eample 5 Finding the Matri of a Quadratic Form Figure = Find the matri of the quadratic form associated with each quadratic equation a 9 3 b a Because a, b, and c 9, the matri is A b Because a 3, b, and c 3, the matri is A Diagonal matri (no term) Nondiagonal matri ( term) Figure ( ) ( ) 3 + = = In standard form, the equation 9 3 is which is the equation of the ellipse shown in Figure 73 Although it is not apparent b inspection, the graph of the equation is similar In fact, when ou rotate the  and aes counterclockwise 5 to form a new coordinate sstem, this equation takes the form 3 3 which is the equation of the ellipse shown in Figure 7 To see how to use the matri of a quadratic form to perform a rotation of aes, let X
6 Then the quadratic epression a b c d e f matri form as follows can be written in If b, then no rotation is necessar But if b, then because A is smmetric, ou can appl Theorem 7 to conclude that there eists an orthogonal matri P such that P T AP D is diagonal So, if ou let then it follows that X PX, and X T AX PX T APX X T P T APX X T DX The choice of the matri P must be made with care Because P is orthogonal, its determinant will be ± It can be shown (see Eercise 5) that if P is chosen so that, then P will be of the form P X T AX d ex f a b P T X X P cos sin 7 Applications of Eigenvalues and Eigenvectors 377 sin cos b c d e a b c d e f f where gives the angle of rotation of the conic measured from the positive ais to the positive ais This leads to the Principal Aes Theorem REMARK Note that the matri product d epx has the form d cos e sin d sin e cos Principal Aes Theorem For a conic whose equation is a b c d e f, the rotation given b X PX eliminates the term when P is an orthogonal matri, with, that diagonalies A That is, P P T AP where and are eigenvalues of A The equation of the rotated conic is given b d epx f Rotation of a Conic Perform a rotation of aes to eliminate the term in the quadratic equation The matri of the quadratic form associated with this equation is A Because the characteristic polnomial of A is 8 8 (check this), it follows that the eigenvalues of A are 8 and 8 So, the equation of the rotated conic is which, when written in the standard form 3 is the equation of an ellipse (See Figure 7)
7 378 Chapter 7 Eigenvalues and Eigenvectors In Eample, the eigenvectors of the matri A are and which ou can normalie to form the columns of P, as follows cos sin P sin cos Note first that P, which implies that P is a rotation Moreover, because cos 5 sin 5, the angle of rotation is 5, as shown in Figure 7 The orthogonal matri P specified in the Principal Aes 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 Eample, an of the following choices of P would have worked 8, , 8 8, 8 35 For an of these choices of P, the graph of the rotated conic will, of course, be the same (See Figure 75) ( ) ( ) 3 + = ( ) ( ) + 3 = ( ) ( ) + 3 = Figure 75 The following summaries the steps used to appl the Principal Aes Theorem Form the matri A and find its eigenvalues and Find eigenvectors corresponding to and Normalie these eigenvectors to form the columns of P P 3 If, then multipl one of the columns of P b to obtain a matri of the form P cos sin sin cos The angle represents the angle of rotation of the conic 5 The equation of the rotated conic is d e PX f
8 7 Applications of Eigenvalues and Eigenvectors 379 Eample 7 shows how to appl the Principal Aes Theorem to rotate a conic whose center has been translated awa from the origin Rotation of a Conic 8 Figure ( ) ( + ) = Perform a rotation of aes to eliminate the term in the quadratic equation The matri of the quadratic form associated with this equation is A 3 5 The eigenvalues of A are 8 and with corresponding eigenvectors of, and This implies that the matri P is P cos sin where Because cos 35 and sin 35, the angle of rotation is Finall, from the matri product d epx the equation of the rotated conic is 8 3 In standard form, the equation 5 3 sin cos,, is the equation of a hperbola Its graph is shown in Figure 7 P 35 Quadratic forms can also be used to anale equations of quadric surfaces in R 3, which are the threedimensional analogs of conic sections The equation of a quadric surface in is a seconddegree polnomial of the form R 3 a b c d e f g h i j There are si basic tpes of quadric surfaces: ellipsoids, hperboloids of one sheet, hperboloids of two sheets, elliptic cones, elliptic paraboloids, and hperbolic paraboloids The intersection of a surface with a plane, called the trace of the surface in the plane, is useful to help visualie the graph of the surface in R 3 The si basic tpes of quadric surfaces, together with their traces, are shown on the net two pages
9 38 Chapter 7 Eigenvalues and Eigenvectors Ellipsoid a b c trace trace Parallel to plane Parallel to plane Parallel to plane The surface is a sphere when a b c trace Hperboloid of One Sheet a b c Hperbola Hperbola Parallel to plane Parallel to plane Parallel to plane The ais of the hperboloid corresponds to the variable whose coefficient is negative trace trace trace Hperboloid of Two Sheets c a b trace trace Hperbola Hperbola Parallel to plane Parallel to plane Parallel to plane The ais of the hperboloid corresponds to the variable whose coefficient is positive There is no trace in the coordinate plane perpendicular to this ais parallel to plane no trace
10 7 Applications of Eigenvalues and Eigenvectors 38 Elliptic Cone trace a b c Hperbola Hperbola Parallel to plane Parallel to plane Parallel to plane The ais of the cone corresponds to the variable whose coefficient is negative The traces in the coordinate planes parallel to this ais are intersecting lines trace (one point) parallel to plane trace Elliptic Paraboloid trace trace a b Parabola Parabola Parallel to plane Parallel to plane Parallel to plane The ais of the paraboloid corresponds to the variable raised to the first power parallel to plane trace (one point) Hperbolic Paraboloid trace b a Hperbola Parabola Parabola Parallel to plane Parallel to plane Parallel to plane The ais of the paraboloid corresponds to the variable raised to the first power trace parallel to plane
11 38 Chapter 7 Eigenvalues and Eigenvectors 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, Brail, is in the shape of a hperboloid of one sheet It was designed b Pritker Prie winning architect Oscar Niemeer, and dedicated in 97 The siteen identical curved steel columns, weighing 9 tons each, are intended to represent two hands reaching up to the sk Pieced together between the columns, in the meterwide and 3meterhigh triangular gaps formed b the columns, is semitransparent stained glass, which allows light inside for nearl the entire height of the columns The quadratic form of the equation a b c d e f g h i j Quadric surface is defined as a b c d e f Quadratic form The corresponding matri is A a d e d b f e f c In its threedimensional version, the Principal Aes Theorem relates the eigenvalues and eigenvectors of A to the equation of the rotated surface, as shown in Eample 8 Rotation of a Quadric Surface Perform a rotation of aes to eliminate the term in the quadratic equation ' ' Figure 77 The matri A associated with this quadratic equation is A 5 5 which has eigenvalues of, and So, in the rotated sstem, the quadratic equation is, 9 9 3, which in standard form is 3 The graph of this equation is an ellipsoid As shown in Figure 77, the aes represent a counterclockwise rotation of 5 about the ais Moreover, the orthogonal matri 3 P whose columns are the eigenvectors of A, has the propert that P T AP is diagonal ostill, /Used under license from Shutterstockcom
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