CHAPTER 6 EIGENVALUES AND EIGENVECTORS SECTION 6. INTRODUCTION TO EIGENVALUES In each of Problems we first list the characteristic polynomial p( λ) A λi of the gien matrix A and then the roots of p( λ ) which are the eigenalues of A. All of the eigenalues that appear in Problems 6 are integers so each characteristic polynomial factors readily. For each eigenalue λ of the matrix A we determine the associated eigenector(s) by finding a basis j for the solution space of the linear system ( ). form in terms of the components of [ a b ]. A I We write this linear system in scalar λ j T In most cases an associated eigenector is then apparent. If A is a matrix for instance then our two scalar equations will be multiples one of the other so we can substitute a conenient numerical alue for the first component a of and then sole either equation for the second component b (or ice ersa).. Characteristic polynomial: ( ) 5 + 6 ( )( ) p λ λ λ λ λ Eigenalues: λ λ : a b a b : a b a b. Characteristic polynomial: ( ) ( + )( ) p λ λ λ λ λ Eigenalues: λ λ : 6a 6b a b : a 6b a 6b 9 Chapter 6 Copyright Pearson Education Inc. Publishing as Prentice Hall.
. Characteristic polynomial: ( ) 7 + ( )( 5) p λ λ λ λ λ Eigenalues: λ λ 5 : 6a 6b a b 5: a 6b a 6b 4. Characteristic polynomial: ( ) + ( )( ) p λ λ λ λ λ Eigenalues: λ λ 5 : a b a b : a b a b 5. Characteristic polynomial: ( ) 5 + 4 ( )( 4) p λ λ λ λ λ Eigenalues: λ λ 4 : 9a 9b 6a 6b 4: 6a 9b 6a 9b 6. Characteristic polynomial: ( ) 5 + 6 ( )( ) p λ λ λ λ λ Eigenalues: λ λ 4 : 4a 4b a b : a 4b a 4b 4 7. Characteristic polynomial: ( ) 6 + 8 ( )( 4) p λ λ λ λ λ Eigenalues: λ λ 4 Section 6. 9 Copyright Pearson Education Inc. Publishing as Prentice Hall.
: 8a 8b 6a 6b 4: 6a 8b 6a 8b 4 8. Characteristic polynomial: ( ) + + ( + )( + ) p λ λ λ λ λ Eigenalues: λ λ : 9a 6b a 8b : 8a 6b a 9b 4 9. Characteristic polynomial: ( ) 7 + ( )( 4) p λ λ λ λ λ Eigenalues: λ λ 4 : 5a b a 4b 4: 4a b a 5b 5. Characteristic polynomial: ( ) 9 + ( 4)( 5) p λ λ λ λ λ Eigenalues: λ 4 λ 5 4: 5a b a 4b 5: 4a b a 5b 5. Characteristic polynomial: ( ) 9 + ( 4)( 5) p λ λ λ λ λ Eigenalues: λ 4 λ 5 4: 5a b a 4b 5: 4a b a 5b 5 7 9 Chapter 6 Copyright Pearson Education Inc. Publishing as Prentice Hall.
. Characteristic polynomial: ( ) 7 + ( )( 4) p λ λ λ λ λ Eigenalues: λ λ 4 : a 5b 6a 9b 4: 9a 5b 6a b 5. Characteristic polynomial: ( ) + ( )( ) p λ λ λ λ λ λ λ Eigenalues: λ λ λ : : : a a b c a+ 6b+ c a a b c a+ 6b+ c a 4b c a+ 6b+ c 4. Characteristic polynomial: p λ λ λ λ λ λ λ ( ) + 7 ( )( 5) Eigenalues: λ λ λ 5 : : 5: 5a 4a 4b c a+ b+ 6c a 4a 6b c a+ b+ 4c 4a 9b c a+ b+ c Section 6. 9 Copyright Pearson Education Inc. Publishing as Prentice Hall.
5. Characteristic polynomial: p λ λ λ λ λ λ λ ( ) + ( )( ) Eigenalues: λ λ λ : : : a b a b c a+ b+ c a b a b c a+ b+ c b a 4b c a+ b+ c 6. Characteristic polynomial: p λ λ λ λ λ λ λ ( ) + 4 ( )( ) Eigenalues: λ λ λ : a c a+ b c 6a+ 6b : : c a+ b c 6a+ 6b c a c a c 6a+ 6b c 7. Characteristic polynomial: p λ λ λ λ λ λ λ ( ) + 6 + 6 ( )( )( ) Eigenalues: λ λ λ : a+ 5b c b b 94 Chapter 6 Copyright Pearson Education Inc. Publishing as Prentice Hall.
: a+ 5b c b c : 5b c b b c 8. Characteristic polynomial: p λ λ λ λ λ λ λ ( ) + 6 + 6 ( )( )( ) Eigenalues: λ λ λ : : : 6a+ 7b+ c a 5b 4c a 6a+ 6b+ c a 5b 5c a 6a+ 5b+ c a 5b 6c 5 9. Characteristic polynomial: p λ λ λ λ λ λ ( ) + 5 7 + ( ) ( ) Eigenalues: λ λ λ : a+ 6b c The eigenspace of λ is -dimensional. We get the eigenector with b c and the eigenector with b c. : 6b c b c. Characteristic polynomial: p λ λ λ λ λ λ ( ) + 4 5 + ( ) ( ) Eigenalues: λ λ λ Section 6. 95 Copyright Pearson Education Inc. Publishing as Prentice Hall.
: 4a+ 6b+ c a 5b 5c The eigenspace of λ is -dimensional. We get the eigenector with b c and the eigenector with b c. : a 4a+ 5b+ c a 5b 5c 5. Characteristic polynomial: p( λ) λ + 5λ 8λ+ 4 ( λ )( λ ) Eigenalues: λ λ λ : a b+ c a b+ c c : a b+ c a b+ c The eigenspace of λ is -dimensional. We get the eigenector with b c and the eigenector with b c.. Characteristic polynomial: p λ λ λ λ λ ( ) + + ( + ) ( ) Eigenalues: λ λ λ : 6a 6b+ c 6a 6b+ c 6a 6b+ c The eigenspace of λ is -dimensional. We get the eigenector with b c and the eigenector with b c. : a 6b+ c 6a 9b+ c 6a 6b 96 Chapter 6 Copyright Pearson Education Inc. Publishing as Prentice Hall.
. Characteristic polynomial: p( λ) ( λ )( λ )( λ )( λ 4) Eigenalues: λ λ λ λ4 4 : b+ c+ d b+ c+ d c+ d d : a+ b+ c+ d c+ d c+ d d : a+ b+ c+ d b+ c+ d d d 4 4: a+ b+ c+ d b+ c+ d c+ d 4 4 4. Characteristic polynomial: p( λ) ( λ ) ( λ ) Eigenalues: λ λ λ λ4 : 4c 4c c d The eigenspace of λ is -dimensional. We note that c d but a and b are arbitrary. a+ 4c b+ 4c : 4 The eigenspace of λ is -dimensional. We get the eigenector with b c and the eigenector with b c. Section 6. 97 Copyright Pearson Education Inc. Publishing as Prentice Hall.
5. Characteristic polynomial: p( λ) ( λ ) ( λ ) Eigenalues: λ λ λ λ4 : c c c d The eigenspace of λ is -dimensional. We note that c d but a and b are arbitrary. a+ c b+ c : 4 The eigenspace of λ is -dimensional. We get the eigenector with b c and the eigenector 4 with b c. 6. Characteristic polynomial: p λ λ λ λ λ λ λ λ λ 4 ( ) 5 + 4 ( )( 4) ( + )( )( + )( ) Eigenalues: λ λ λ λ4 : 6a d 4b c 6a d : 5a d b 6a 4d : a d b c 6a 6d 98 Chapter 6 Copyright Pearson Education Inc. Publishing as Prentice Hall.
4 : a d c 6a 7d 4 7. Characteristic polynomial: p λ λ ( ) + Eigenalues: λ i λ + i i : ia+ b a+ ib i + i : ia+ b a ib i 8. Characteristic polynomial: p λ λ ( ) + 6 Eigenalues: λ 6 i λ + 6i 6: i 6ia 6b 6a+ 6ib i + 6: i 6ia 6b 6a 6ib i 9. Characteristic polynomial: p λ λ ( ) + 6 Eigenalues: λ 6 i λ + 6i 6: i 6ia b a+ 6ib i + 6: i 6ia b a 6ib i. Characteristic polynomial: p λ λ ( ) + 44 Eigenalues: λ i λ + i i : ia b a+ ib i + i : ia b a ib i Section 6. 99 Copyright Pearson Education Inc. Publishing as Prentice Hall.
. Characteristic polynomial: p λ λ ( ) + 44 Eigenalues: λ i λ + i i : ia+ 4b 6a+ ib i + i : ia+ 4b 6a ib i. Characteristic polynomial: p λ λ ( ) + 44 Eigenalues: λ i λ + i i : ia 4b 6a+ ib i + i : ia 4b 6a ib i n n n. If A λ and we assume that A λ meaning that λ n of A with associated eigenector then multiplication by A yields is an eigenalue λ λ λ λ λ n n n n n n A AA A A. Thus n λ is an eigenalue of the matrix n A with associated eigenector. 4. By the remark following Example 6 any eigenalue of an inertible matrix is nonzero. If λ is an eigenalue of the inertible matrix A with associated eigenalue then A λ A λ λ A A λ. Thus λ is an eigenalue of A with associated eigenector. T T 5. (a) Note first that ( A λi) ( A λi ) because I T I. Since the determinant of a square matrix equals the determinant of its transpose it follows that T A λi A λi. Thus the matrices A and A T hae the same characteristic polynomial and therefore hae the same eigenalues. Chapter 6 Copyright Pearson Education Inc. Publishing as Prentice Hall.
(b) Consider the matrix A with characteristic equation ( λ ) and the single eigenalue λ. Then A I and it follows that the only associated T T eigenector is a multiple of [ ]. The transpose A has the same characteristic equation and eigenalue but we see similarly that its only eigenector is a T multiple of [ ]. Thus A and A T hae the same eigenalue but different eigenectors. 6. If the n n matrix A aij is either upper or lower triangular then obiously its characteristic equation is ( λ)( λ) ( λ) a a a nn. This obseration makes it clear that the eigenalues of the matrix A are its diagonal elements a a a nn. 7. If A λi ( ) n λ n + c n n λ + + cλ+ c then substitution of λ yields c A I A for the constant term in the characteristic polynomial. a b 8. The characteristic polynomial of the matrix A c d is ( a λ)( d λ) bc that is λ ( a + d) λ+ ( ad bc). Thus the coefficient of λ in the characteristic equation is ( a+ d) trace A. 9. If the characteristic equation of the n n matrix A with eigenalues λ λ λn (not necessarily distinct) is written in the factored form ( )( ) ( ) λ λ λ λ λ λ n n then it should be clear that upon multiplying out the factors the coefficient of λ will be ( λ+ λ + + λ n ). But according to Problem 8 this coefficient also equals (trace A ). Therefore λ + λ + + λn trace A a + a + ann. 4. We find that trace A and det A 6 so the characteristic polynomial of the gien matrix A is p( λ) λ + λ + cλ+ 6. Section 6. Copyright Pearson Education Inc. Publishing as Prentice Hall.
Substitution of λ and 67 47 p() A I 7 5 4 7 5 7 yields c 47 so the characteristic equation of A is (after multiplication by ) + 47 6. λ λ λ Trying λ ± ± ± (all diisors of 6) in turn we discoer the eigenalue λ. Then diision of the cubic by ( λ ) yields 9 + ( 4)( 5) λ λ λ λ so the other two eigenalues are λ 4andλ 5. We proceed to find the eigenectors associated with these three eigenalues. : 9a 67b+ 47c a c 7a 7b+ c b c 7a+ 5b 9c 4: 8a 67b+ 47c a (5/ 7) c 7a 8b+ c b c 7a+ 5b c 5: 7a 67b+ 47c a+ (/ ) c 7a 9b+ c b (/) c 7a+ 5b c 5 7 7 4. We find that trace A 8 and det A 6 so the characteristic polynomial of the gien matrix A is 4 p( λ) λ 8λ + c λ + cλ 6. Substitution of λ p() det( A I ) 4 and λ p( ) det( A+ I ) 7 yields the equations Chapter 6 Copyright Pearson Education Inc. Publishing as Prentice Hall.
c + c 4 c c that we sole readily for c c. Hence the characteristic equation of A is 8 + + 6. 4 λ λ λ λ Trying λ ± ± in turn we discoer the eigenalues λ and λ. Then diision of the quartic by ( λ 4) yields 8 + 5 ( )( 5) λ λ λ λ so the other two eigenalues are λ and λ4 5. We proceed to find the eigenectors associated with these four eigenalues. : 4a 9b 8c 8d a (/ ) d a 5b 4c+ d b a+ c d c (/ ) d 9a 9b c d : a 9b 8c 8d a d a 9b 4c+ d b (4 / ) d a+ 6c d c 9a 9b c 7d : 9a 9b 8c 8d a (/ 4) d a b 4c+ d b (/ 4) d a+ 5c d c (/ ) d 9a 9b c 8d 4 5: 7a 9b 8c 8d a d a b 4c+ d b d a+ c d c 9a 9b c d 4 4 4 Section 6. Copyright Pearson Education Inc. Publishing as Prentice Hall.
SECTION 6. DIAGONALIZATION OF MATRICES In Problems 8 we first find the eigenalues and associated eigenectors of the gien n n matrix A. If A has n linearly independent eigenectors then we can proceed to set up the desired diagonalizing matrix P [ n ] and diagonal matrix D such that P AP D. If you write the eigenalues in a different order on the diagonal of D then naturally the eigenector columns of P must be rearranged in the same order.. Characteristic polynomial: p λ λ λ+ λ λ ( ) 4 ( )( ) Eigenalues: λ λ : : 4a 4b a b a 4b a 4b P D. Characteristic polynomial: p λ λ λ λ λ ( ) ( ) Eigenalues: λ λ : : 6a 6b 4a 4b 4a 6b 4a 6b P D. Characteristic polynomial: p λ λ λ+ λ λ ( ) 5 6 ( )( ) Eigenalues: λ λ : a b a b 4 Chapter 6 Copyright Pearson Education Inc. Publishing as Prentice Hall.
: a b a b P D 4. Characteristic polynomial: p λ λ λ+ λ λ ( ) ( )( ) Eigenalues: λ λ : : 4a 4b a b a 4b a 4b P 4 D 4 5. Characteristic polynomial: p λ λ λ+ λ λ ( ) 4 ( )( ) Eigenalues: λ λ : : 8a 8b 6a 6b 6a 8b 6a 8b P 4 D 4 6. Characteristic polynomial: p λ λ λ+ λ λ ( ) ( )( ) Eigenalues: λ λ : : 9a 6b a 8b 8a 6b a 9b P 4 D 4 Section 6. 5 Copyright Pearson Education Inc. Publishing as Prentice Hall.
7. Characteristic polynomial: p λ λ λ+ λ λ ( ) ( )( ) Eigenalues: λ λ : : 5a b a 4b 4a b a 5b P 5 D 5 8. Characteristic polynomial: p λ λ λ+ λ λ ( ) ( )( ) Eigenalues: λ λ : : a 5b 6a 9b 9a 5b 6a b P 5 D 5 9. Characteristic polynomial: p( λ) λ λ+ ( λ ) Eigenalues: λ λ 4a b : a b Because the gien matrix A has only the single eigenector it is not diagonalizable.. Characteristic polynomial: p( λ) λ 4λ+ 4 ( λ ) Eigenalues: λ λ a b : a b Because the gien matrix A has only the single eigenector it is not diagonalizable. 6 Chapter 6 Copyright Pearson Education Inc. Publishing as Prentice Hall.
. Characteristic polynomial: p( λ) λ 4λ+ 4 ( λ ) Eigenalues: λ λ a+ b : 9a b Because the gien matrix A has only the single eigenector it is not diagonalizable.. Characteristic polynomial: p( λ) λ + λ+ ( λ+ ) Eigenalues: λ λ a+ 9b : 6a b 4 Because the gien matrix A has only the single eigenector it is not diagonalizable.. Characteristic polynomial: p( λ) λ + 5λ 8λ+ 4 ( λ )( λ ) Eigenalues: λ λ λ : b b c : b a The eigenspace of λ is -dimensional. We get the eigenector with b c and the eigenector with b c. P D 4. Characteristic polynomial: p λ λ + λ λ λ ( ) ( ) Eigenalues: λ λ λ : a b+ c a b+ c a b+ c Section 6. 7 Copyright Pearson Education Inc. Publishing as Prentice Hall.
The eigenspace of λ is -dimensional. We get the eigenector with b c and the eigenector with b c. : a b+ c a b+ c a b P D 5. Characteristic polynomial: p( λ) λ + λ λ λ( λ ) Eigenalues: λ λ λ : a b+ c a b+ c c : a b+ c a b+ c The eigenspace of λ is -dimensional. We get the eigenector with b c and the eigenector with b c. P D 6. Characteristic polynomial: p λ λ + λ λ+ λ λ ( ) 5 7 ( ) ( ) Eigenalues: λ λ λ : a b 4a 4b The eigenspace of λ is -dimensional. We get the eigenector with b c and the eigenector with b c. 8 Chapter 6 Copyright Pearson Education Inc. Publishing as Prentice Hall.
: b b 4a+ 4b c P D 7. Characteristic polynomial: p λ λ + λ + λ λ+ λ λ ( ) ( )( )( ) Eigenalues: λ λ λ : 8a 8b+ c 6a 6b+ c a b+ c : 6a 8b+ c 6a 8b+ c a b+ c : 5a 8b+ c 6a 9b+ c a b P D 8. Characteristic polynomial: p λ λ + λ λ+ λ λ λ ( ) 6 6 ( )( )( ) Eigenalues: λ λ λ : 5a 5b+ c 4a 4b+ c a b+ c : 4a 5b+ c 4a 5b+ c a b+ c Section 6. 9 Copyright Pearson Education Inc. Publishing as Prentice Hall.
: a 5b+ c 4a 6b+ c a b P D 9. Characteristic polynomial: p λ λ + λ λ+ λ λ λ ( ) 6 6 ( )( )( ) Eigenalues: λ λ λ : b c a+ b c 4a+ 4b : : a+ b c a+ b c 4a+ 4b c a+ b c a+ b c 4a+ 4b c P D. Characteristic polynomial: p λ λ + λ λ+ λ λ λ ( ) 5 6 ( )( 5)( 6) Eigenalues: λ λ 5 λ 6 : 6a+ 9b+ c 6a 5b c 5: a 6a+ 6b+ c 6a 5b 5c Chapter 6 Copyright Pearson Education Inc. Publishing as Prentice Hall.
6: 4a 6a+ 5b+ c 6a 5b 6c 5 P 5 D 5 6. Characteristic polynomial: p( λ) λ + λ λ+ ( λ ) Eigenalues: λ λ λ : b a b a b a The eigenspace of λ is -dimensional. We get the eigenector with b c and the eigenector with b c. Because the gien matrix A has only two linearly independent eigenectors it is not diagonalizable.. Characteristic polynomial: p( λ) λ + λ λ+ ( λ ) Eigenalues: λ λ λ : a b+ c a+ b 5a+ 7b c The eigenspace of λ is -dimensional. Because the gien matrix A has only one eigenector it is not diagonalizable.. Characteristic polynomial: p λ λ + λ λ+ λ λ ( ) 4 5 ( ) ( ) Eigenalues: λ λ λ : a+ 4b c a+ 4b c a+ b Section 6. Copyright Pearson Education Inc. Publishing as Prentice Hall.
: 4a+ 4b c a+ b c a+ b c The gien matrix A has only the two linearly independent eigenectors and and therefore is not diagonalizable. 4. Characteristic polynomial: p( λ) λ + 5λ 8λ+ 4 ( λ )( λ ) Eigenalues: λ λ λ : a b+ c a b+ c a+ b+ c : a b+ c a b+ c a+ b The gien matrix A has only the two linearly independent eigenectors and and therefore is not diagonalizable. 5. Characteristic polynomial: p( λ) ( λ+ ) ( λ ) Eigenalues: λ λ λ λ4 : a c b c The eigenspace of λ is -dimensional. We get the eigenector with c d and the eigenector with c d. : c c c d 4 The eigenspace of λ is also -dimensional. We get the eigenector with a b and the eigenector 4 with a b. Chapter 6 Copyright Pearson Education Inc. Publishing as Prentice Hall.
P D 6. Characteristic polynomial: p λ λ λ ( ) ( ) ( ) Eigenalues: λ λ λ λ4 : d d d d The eigenspace of λ is -dimensional and we hae taken adantage of the fact that we can select a b and c independently. 4 : d a d b d c P D 4 7. Characteristic polynomial: p λ λ λ ( ) ( ) ( ) Eigenalues: λ λ λ λ4 : b c d d The eigenspace of λ is -dimensional with only a single associated eigenector. Section 6. Copyright Pearson Education Inc. Publishing as Prentice Hall.
4 : b a c b d c 4 The gien matrix A has only the two linearly independent eigenectors and 4 and therefore is not diagonalizable. 8. Characteristic polynomial: p( λ) ( λ ) ( λ ) Eigenalues: λ λ λ λ4 : b+ d c+ d c+ d d The eigenspace of λ is -dimensional with only a single associated eigenector. : b a+ d c b+ d d The eigenspace of λ is also -dimensional with only a single associated eigenector. Thus the gien matrix A has only the two linearly independent eigenectors and 4 and therefore is not diagonalizable. 9. If A is similar to B and B is similar to C so A P BP and B Q CQ then A P (Q CQ)P (P Q )C(QP) (QP) C(QP) R CR with R QP so A is similar to C.. If A is similar to B so A P BP then n A ( P BP)( P BP)( P BP) ( P BP)( P BP) P B( PP ) B( PP ) B ( PP ) B( PP ) BP P BIBIB n P B P () () () IBIBP () so we see that A n is similar to B n. 4 Chapter 6 Copyright Pearson Education Inc. Publishing as Prentice Hall.
. If A is similar to B so A P BP then A (P BP) P B P so A is similar to B.. If A is similar to B so A P BP then P P P P I so A λi P A λi P P ( A λi) P λ P AP P IP B λi. Thus A and B hae the same characteristic polynomial.. If A and B are similar with A P BP then A P BP P B P P B P B. Moreoer by Problem the two matrices hae the same eigenalues and by Problem 9 in Section 6. the trace of a square matrix with real eigenalues is equal to the sum of those eigenalues. Therefore trace A (eigenalue sum) trace B. a b 4. The characteristic equation of the matrix A c d is λ ( a + d) λ+ ( ad bc) and the discriminant of this quadratic equation is a + d ad bc a d + bc ( ) 4( ) ( ) 4. (a) If > then A has two distinct eigenalues and hence has two linearly independent eigenectors and is therefore diagonalizable. (b) If < then A has no (real) eigenalues and hence no real eigenectors and therefore is not diagonalizable. (c) Finally note that for both I and A but A has only the single eigenalue therefore not diagonalizable. λ and the single eigenector [ ] and is T 5. Three eigenectors associated with three distinct eigenalues can be arranged in six different T. orders as the column ectors of the diagonalizing matrix P [ ] Section 6. 5 Copyright Pearson Education Inc. Publishing as Prentice Hall.
6. The fact that the matrices A and B hae the same eigenalues (with the same multiplicities) implies that they are both similar to the same diagonal matrix D haing these eigenalues as its diagonal elements. But two matrices that are similar to a third matrix are (by Problem 9) similar to one another. 7. If A PDP with P the eigenector matrix of A and D its diagonal matrix of eigenalues then A ( PDP )( PDP ) PD( P P) DP PD P. Thus the same (eigenector) matrix P diagonalizes A but the resulting diagonal (eigenalue) matrix D is the square of the one for A. The diagonal elements of D are the eigenalues of A and the diagonal elements of D are the eigenalues of A so the former are the squares of the latter. 8. If the n n matrix A has n linearly independent eigenectors associated with the single eigenalue λ then A PDP with D λi so A P( λi) P λpp λi D. 9. Let the n n matrix A hae k n distinct eigenalues λ λ λ k. Then the definition of algebraic multiplicity and the fact that all solutions of the nth degree polynomial equation A λi are real imply that the sum of the multiplicities of the eigenalues equals n p p p n + + + k. Now Theorem 4 in this section implies that A is diagonalizable if and only if q + q + + q n k where q i denotes the geometric multiplicity of λ i ( i k). But because pi qi for each i k the two equations displayed aboe can both be satisfied if and only if p q for each i. i i SECTION 6. APPLICATIONS INVOLVING POWERS OF MATRICES In Problems we first find the eigensystem of the gien matrix A so as to determine its eigenector matrix P and its diagonal eigenalue matrix D. Then we calculate the matrix power A 5 PD 5 P.. Characteristic polynomial: p λ λ λ λ λ ( ) + ( )( ) Eigenalues: λ λ 6 Chapter 6 Copyright Pearson Education Inc. Publishing as Prentice Hall.
: : a b a b a b a b P D P 5 6 6 A. Characteristic polynomial: ( ) ( + )( ) p λ λ λ λ λ Eigenalues: λ λ : : 6a 6b a b a 6b a 6b P D P 5 65 66 A 4. Characteristic polynomial: ( ) ( ) p λ λ λ λ λ Eigenalues: λ λ : : 6a 6b 4a 4b 4a 6b 4a 6b P D P 5 96 96 A 64 64 Section 6. 7 Copyright Pearson Education Inc. Publishing as Prentice Hall.
4. Characteristic polynomial: ( ) + ( )( ) p λ λ λ λ λ Eigenalues: λ λ : : a b a b a b a b P D P 5 94 9 A 6 6 5. Characteristic polynomial: ( ) + ( )( ) p λ λ λ λ λ Eigenalues: λ λ : : 4a 4b a b a 4b a 4b 4 4 P D P 5 4 4 5 4 A 9 9 4 6. Characteristic polynomial: ( ) + ( )( ) p λ λ λ λ λ Eigenalues: λ λ : 5a b a 4b : 4a b a 5b 5 8 Chapter 6 Copyright Pearson Education Inc. Publishing as Prentice Hall.
5 5 P D P 5 5 5 56 A 6 7. Characteristic polynomial: p( λ) ( λ )( λ ) Eigenalues: λ λ λ : b b c : b a P D P 9 5 A 8. Characteristic polynomial: p λ λ λ λ λ λ ( ) + 4 5 + ( ) ( ) Eigenalues: λ λ λ : c b c b : a b+ c b b Section 6. 9 Copyright Pearson Education Inc. Publishing as Prentice Hall.
P D P 6 5 A 6 9. Characteristic polynomial: p( λ) ( λ )( λ ) Eigenalues: λ λ λ : c b b c : a b+ c P D P 9 5 A. Characteristic polynomial: p( λ) ( λ )( λ ) Eigenalues: λ λ λ : a b+ c a b+ c c : a b+ c a b+ c Chapter 6 Copyright Pearson Education Inc. Publishing as Prentice Hall.
4 6 P D P 4 6 94 9 5 A 6 6. Characteristic polynomial: p λ λ λ λ λ λ ( ) + ( + )( ) Eigenalues: λ λ λ : : : a 6a+ 6b+ c a 5b 5c a 6a+ 5b+ c a 5b 6c 6a+ 4b+ c a 5b 7c 5 4 87 6 5 P 4 9 D P 5 87 6 5 A 4 9 78 5 5 87 95 5 6. Characteristic polynomial: p( λ) ( λ)( λ ) ( λ+ )( λ ) Eigenalues: λ λ λ : a 6b c a b 4c c Section 6. Copyright Pearson Education Inc. Publishing as Prentice Hall.
: a 6b c a b 4c 5 5 5 5 5 P 5 D P 5 5 5 5 5 5 A 5 5 5 5. Characteristic polynomial: p λ λ λ λ λ λ ( ) + ( + )( ) Eigenalues: λ λ λ : : a b+ c a b+ c 4a 4b+ c a b+ c a b+ c 4a 4b+ c : b+ c a b+ c 4a 4b P D P A 4. Characteristic polynomial: p λ λ λ λ λ λ ( ) + ( + )( ) Eigenalues: λ λ λ Chapter 6 Copyright Pearson Education Inc. Publishing as Prentice Hall.
: : : 6a 5b c a b c 4a 4b c 5a 5b c a b c 4a 4b c 4a 5b c a b c 4a 4b 4c P D P A 5. p λ λ λ ( ) + so A A+ I 5 4 A A I 9 8 5 4 9 8 A A A 9 8 4 9 8 6 6 A A A 9 8 45 44 5 4 4 ( + ) + 5 A A I 6. p λ λ λ ( ) + so A A+ I 6 6 6 6 6 6 7 6 4 7 A A I A A A Section 6. Copyright Pearson Education Inc. Publishing as Prentice Hall.
4 6 7 6 76 5 A A A 4 7 6 59 6 ( + ) + 6 A A I 7. p( λ) λ + 5λ 8λ+ 4 so A + 5A 8A+ 4I 9 4 5 8 4 8 + 4 8 45 5 8 + 4 6 6 A A A A I 4 A A A A ( 5 + 8 ) 4 A A A I 8. p( λ) λ + 4λ 5λ+ so A + 4A 5A+ I 6 4 7 4 5 + 6 4 4 8 5 4 5 + 6 A A A A I 4 A A A A ( 4 + 5 ) A A A I 9. p( λ) λ + 5λ 8λ+ 4 so A + 5A 8A+ 4I 9 7 4 5 8 + 4 8 4 8 A A A A I 4 Chapter 6 Copyright Pearson Education Inc. Publishing as Prentice Hall.
45 5 5 8 + 4 6 6 4 A A A A ( 5 + 8 ) 4 A A A I. p( λ) λ + 5λ 8λ+ 4 so A + 5A 8A+ 4I 9 7 6 5 5 8 4 4 7 + 4 8 46 45 5 5 8 + 4 9 5 6 A A A A I 4 A A A A ( 5 + 8 ) 4 4 A A A I. p( λ) λ λ so + + A A 78 5 6 5 95 5 6 5 6 A 78 5 95 5 6 A A A A 4 Because λ is an eigenalue A is singular and A does not exist.. p( λ) λ λ λ so + + + + A A A I 6 4 A I A A + A I A Section 6. 5 Copyright Pearson Education Inc. Publishing as Prentice Hall.
4 A A + A A I 6 4 A A + A+ I A. p( λ) λ λ so + + A A 4 4 A A A A A 4 Because λ is an eigenalue A is singular and A does not exist. 4. p( λ) λ λ so + + A A 5 5 4 4 A A A A A 4 Because λ is an eigenalue A is singular and A does not exist. In Problems 5 we first find the eigensystem of the gien transition matrix A so as to determine its eigenector matrix P and its diagonal eigenalue matrix D. Then we determine how the matrix power A k PD k P behaes as k. For simpler calculations of eigenalues and eigenectors we write the entries of A in fractional rather than decimal form. 9 4 5. Characteristic polynomial: p( λ) λ λ+ ( λ )(5λ 4) 5 5 5 4 Eigenalues: λ λ 5 6 Chapter 6 Copyright Pearson Education Inc. Publishing as Prentice Hall.
: a+ b a b 4 : 5 a+ b a+ b P D 4/5 P k k xk A x 4/5 x C / ( C S) + x S / as k. Thus the long-term distribution of population is 5% city 5% suburban. 9 4 6. Characteristic polynomial: p( λ) λ λ+ ( λ )(5λ 4) 5 5 5 4 Eigenalues: λ λ 5 a+ b : a b 4 : 5 a+ b a+ b P D 4/5 P 4 k k xk A x 4/5 4 x C /4 ( C S) 4 + x 4 S /4 Section 6. 7 Copyright Pearson Education Inc. Publishing as Prentice Hall.
as k. Thus the long-term distribution of population is 5% city 75% suburban. 8 7. Characteristic polynomial: p( λ) λ λ+ ( λ )(5λ ) 5 5 5 Eigenalues: λ λ 5 a+ b 4 : a b 5 4 : 5 a+ b a+ b 4 4 P 5 D /5 P 8 5 x k k A x 5 /5 8 5 x k C /8 ( C S) 5 8 5 x + 8 5 5 S 5/8 as k. Thus the long-term distribution of population is /8 city 5/8 suburban. 7 7 8. Characteristic polynomial: p( λ) λ λ+ ( λ )(λ 7) 7 Eigenalues: λ λ a+ b 5 : a b 5 7 : a+ b a+ b 5 5 8 Chapter 6 Copyright Pearson Education Inc. Publishing as Prentice Hall.
P D 7/ P k k xk A x 7/ x C / ( C S) + x S / as k. Thus the long-term distribution of population is / city / suburban. 7 7 9. Characteristic polynomial: p( λ) λ λ+ ( λ )(λ 7) 7 Eigenalues: λ λ a+ b : a b 7 : a+ b a+ b P D 7/ P x k k A x 7/ x k C / ( C S) x + S / as k. Thus the long-term distribution of population is / city / suburban.. Characteristic polynomial: p( λ) λ λ+ ( λ )(λ ) Eigenalues: λ λ Section 6. 9 Copyright Pearson Education Inc. Publishing as Prentice Hall.
: a+ b 5 a b 5 4 : a+ b a+ b 5 5 P 4 D / P 7 4 k k xk A x 4 / 7 4 x C /7 ( C S) 4 + 7 4 x 7 4 4 S 4/7 as k. Thus the long-term distribution of population is /7 city 4/7 suburban. In the following three problems just as in Problems 5 we first write the elements of A in fractional rather than decimal form with r 4/5 in Problem r 7/4 in Problem and r 7/ in Problem. 9 4. Characteristic polynomial: p( λ) λ λ+ ( λ )(5λ 4) 5 5 5 4 Eigenalues: λ λ 5 : 4 : 5 a+ b 5 4 a+ b 5 5 a+ b 5 4 a+ b 5 5 5 5 5 P 4 D 4/5 P 4 5 5 4 5 Chapter 6 Copyright Pearson Education Inc. Publishing as Prentice Hall.
k k 5 5 5 xk A x 4 4/5 4 5 x 5 5 5 5 F.5R F 4 4 5 x 8 R R.8F as k. Thus the fox-rabbit population approaches a stable situation with.5r F foxes and R.8F rabbits. 9. Characteristic polynomial: p( λ) λ λ+ (λ 9)(λ 7) 5 4 4 9 7 Eigenalues: λ λ 7 a+ b 9 : 7 a+ b 7 4 4 7 : a+ b 4 7 7 a+ b 4 9 / P 7 D 7/ P 4 7 k k 9 / xk A x 7 7/ 4 7 x F 7 4 7 x R as k. Thus the fox and rabbit population both die out. 9 6. Characteristic polynomial: p( λ) λ λ+ (λ )(4λ ) 5 8 8 9 7 Eigenalues: λ λ 9 a+ b : 7 a+ b 9 Section 6. Copyright Pearson Education Inc. Publishing as Prentice Hall.
: 4 a+ b 7 9 a+ b / P 9 D /4 P 6 9 k k / xk A x 9 /4 6 9 x k k F (.5 ) 6 9 9 x 6 7 9 R k (.5 ) ( R F 6 9 when k is sufficiently large. Thus the fox and rabbit populations are both increasing at 5% per year with foxes for each 9 rabbits. 4. 5 7 5 4 A PDP 4 7 4 56 4 If n is een then D n I so n n A A A IA A. Thus n n A PD P PIP I. If n is odd then 99 A A and A I. 5. The fact that each λ so λ ± implies that n n A PD P PIP I. D n I if n is een in which case 6. We find immediately that forth. 4 A Iso A A A IA A A A A A I and so 7. We find immediately that so forth. 4 A Iso A A A IA A A A A A I and 8. If A then + + I B B so it follows that n n n n n n A ( I+ B) I + ni B+ n( n ) I B + I+ nb. Chapter 6 Copyright Pearson Education Inc. Publishing as Prentice Hall.
9. The characteristic equation of A is ( p λ)( q λ) ( p)( q) λ ( p+ q) λ+ ( p+ q ) ( λ )[ λ ( p+ q )] so the eigenalues of A are λ and λ p+ q. 4. The fact that the column sums of A are each implies that the row sums of the transpose matrix A T are each so it follows readily that A T. Thus λ is an eigenalue of A T. But A and A T hae the same eigenalues (by Problem 5 in Section 6.). Section 6. Copyright Pearson Education Inc. Publishing as Prentice Hall.