EIGENVALUES AND EIGENVECTORS
|
|
- Carmel Hampton
- 5 years ago
- Views:
Transcription
1 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: ( ) ( )( ) 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.
2 . 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: ( ) ( )( 4) p λ λ λ λ λ Eigenalues: λ λ 4 : 9a 9b 6a 6b 4: 6a 9b 6a 9b 6. Characteristic polynomial: ( ) ( )( ) p λ λ λ λ λ Eigenalues: λ λ 4 : 4a 4b a b : a 4b a 4b 4 7. Characteristic polynomial: ( ) ( )( 4) p λ λ λ λ λ Eigenalues: λ λ 4 Section 6. 9 Copyright Pearson Education Inc. Publishing as Prentice Hall.
3 : 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 Chapter 6 Copyright Pearson Education Inc. Publishing as Prentice Hall.
4 . 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 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 ( ) ( )( )( ) Eigenalues: λ λ λ : a+ 5b c b b 94 Chapter 6 Copyright Pearson Education Inc. Publishing as Prentice Hall.
6 : a+ 5b c b c : 5b c b b c 8. Characteristic polynomial: p λ λ λ λ λ λ λ ( ) ( )( )( ) 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 ( ) ( ) ( ) 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 ( ) ( ) ( ) Eigenalues: λ λ λ Section Copyright Pearson Education Inc. Publishing as Prentice Hall.
7 : 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.
8 . 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 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 Copyright Pearson Education Inc. Publishing as Prentice Hall.
9 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 ( ) ( )( 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.
10 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 Copyright Pearson Education Inc. Publishing as Prentice Hall.
11 . 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.
12 (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.
13 Substitution of λ and p() A I yields c 47 so the characteristic equation of A is (after multiplication by ) 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 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.
14 c + c 4 c c that we sole readily for c c. Hence the characteristic equation of A is λ λ λ λ Trying λ ± ± in turn we discoer the eigenalues λ and λ. Then diision of the quartic by ( λ 4) yields ( )( 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 Section 6. Copyright Pearson Education Inc. Publishing as Prentice Hall.
15 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.
16 : 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.
17 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.
18 . 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.
19 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.
20 : 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.
21 : 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.
22 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.
23 : 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.
24 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.
25 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.
26 . 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.
27 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.
28 : : a b a b a b a b P D P A. Characteristic polynomial: ( ) ( + )( ) p λ λ λ λ λ Eigenalues: λ λ : : 6a 6b a b a 6b a 6b P D P A 4. Characteristic polynomial: ( ) ( ) p λ λ λ λ λ Eigenalues: λ λ : : 6a 6b 4a 4b 4a 6b 4a 6b P D P A Section 6. 7 Copyright Pearson Education Inc. Publishing as Prentice Hall.
29 4. Characteristic polynomial: ( ) + ( )( ) p λ λ λ λ λ Eigenalues: λ λ : : a b a b a b a b P D P A Characteristic polynomial: ( ) + ( )( ) p λ λ λ λ λ Eigenalues: λ λ : : 4a 4b a b a 4b a 4b 4 4 P D P A 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.
30 5 5 P D P 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 ( ) ( ) ( ) Eigenalues: λ λ λ : c b c b : a b+ c b b Section 6. 9 Copyright Pearson Education Inc. Publishing as Prentice Hall.
31 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.
32 4 6 P D P 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 P 4 9 D P A 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.
33 : a 6b c a b 4c P 5 D P A 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.
34 : : : 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 A A A A A A ( + ) + 5 A A I 6. p λ λ λ ( ) + so A A+ I A A I A A A Section 6. Copyright Pearson Education Inc. Publishing as Prentice Hall.
35 A A A ( + ) + 6 A A I 7. p( λ) λ + 5λ 8λ+ 4 so A + 5A 8A+ 4I A A A A I 4 A A A A ( ) 4 A A A I 8. p( λ) λ + 4λ 5λ+ so A + 4A 5A+ I A A A A I 4 A A A A ( ) A A A I 9. p( λ) λ + 5λ 8λ+ 4 so A + 5A 8A+ 4I A A A A I 4 Chapter 6 Copyright Pearson Education Inc. Publishing as Prentice Hall.
36 A A A A ( ) 4 A A A I. p( λ) λ + 5λ 8λ+ 4 so A + 5A 8A+ 4I A A A A I 4 A A A A ( ) 4 4 A A A I. p( λ) λ λ so + + A A A 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.
37 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 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 Characteristic polynomial: p( λ) λ λ+ ( λ )(5λ 4) Eigenalues: λ λ 5 6 Chapter 6 Copyright Pearson Education Inc. Publishing as Prentice Hall.
38 : 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 Characteristic polynomial: p( λ) λ λ+ ( λ )(5λ 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.
39 as k. Thus the long-term distribution of population is 5% city 75% suburban Characteristic polynomial: p( λ) λ λ+ ( λ )(5λ ) 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) x S 5/8 as k. Thus the long-term distribution of population is /8 city 5/8 suburban Characteristic polynomial: p( λ) λ λ+ ( λ )(λ 7) 7 Eigenalues: λ λ a+ b 5 : a b 5 7 : a+ b a+ b Chapter 6 Copyright Pearson Education Inc. Publishing as Prentice Hall.
40 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 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.
41 : 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) x 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 Characteristic polynomial: p( λ) λ λ+ ( λ )(5λ 4) Eigenalues: λ λ 5 : 4 : 5 a+ b 5 4 a+ b 5 5 a+ b 5 4 a+ b P 4 D 4/5 P Chapter 6 Copyright Pearson Education Inc. Publishing as Prentice Hall.
42 k k xk A x 4 4/5 4 5 x F.5R F 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) Eigenalues: λ λ 7 a+ b 9 : 7 a+ b : a+ b a+ b 4 9 / P 7 D 7/ P 4 7 k k 9 / xk A x 7 7/ 4 7 x F x R as k. Thus the fox and rabbit population both die out Characteristic polynomial: p( λ) λ λ+ (λ )(4λ ) Eigenalues: λ λ 9 a+ b : 7 a+ b 9 Section 6. Copyright Pearson Education Inc. Publishing as Prentice Hall.
43 : 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 ) x 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 A PDP 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.
44 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.
and let s calculate the image of some vectors under the transformation T.
Chapter 5 Eigenvalues and Eigenvectors 5. Eigenvalues and Eigenvectors Let T : R n R n be a linear transformation. Then T can be represented by a matrix (the standard matrix), and we can write T ( v) =
More informationRemark By definition, an eigenvector must be a nonzero vector, but eigenvalue could be zero.
Sec 6 Eigenvalues and Eigenvectors Definition An eigenvector of an n n matrix A is a nonzero vector x such that A x λ x for some scalar λ A scalar λ is called an eigenvalue of A if there is a nontrivial
More informationRemark 1 By definition, an eigenvector must be a nonzero vector, but eigenvalue could be zero.
Sec 5 Eigenvectors and Eigenvalues In this chapter, vector means column vector Definition An eigenvector of an n n matrix A is a nonzero vector x such that A x λ x for some scalar λ A scalar λ is called
More informationMath Matrix Algebra
Math 44 - Matrix Algebra Review notes - 4 (Alberto Bressan, Spring 27) Review of complex numbers In this chapter we shall need to work with complex numbers z C These can be written in the form z = a+ib,
More informationMath 3191 Applied Linear Algebra
Math 9 Applied Linear Algebra Lecture 9: Diagonalization Stephen Billups University of Colorado at Denver Math 9Applied Linear Algebra p./9 Section. Diagonalization The goal here is to develop a useful
More informationDiagonalization of Matrix
of Matrix King Saud University August 29, 2018 of Matrix Table of contents 1 2 of Matrix Definition If A M n (R) and λ R. We say that λ is an eigenvalue of the matrix A if there is X R n \ {0} such that
More informationCity Suburbs. : population distribution after m years
Section 5.3 Diagonalization of Matrices Definition Example: stochastic matrix To City Suburbs From City Suburbs.85.03 = A.15.97 City.15.85 Suburbs.97.03 probability matrix of a sample person s residence
More informationHW2 - Due 01/30. Each answer must be mathematically justified. Don t forget your name.
HW2 - Due 0/30 Each answer must be mathematically justified. Don t forget your name. Problem. Use the row reduction algorithm to find the inverse of the matrix 0 0, 2 3 5 if it exists. Double check your
More informationLecture 15, 16: Diagonalization
Lecture 15, 16: Diagonalization Motivation: Eigenvalues and Eigenvectors are easy to compute for diagonal matrices. Hence, we would like (if possible) to convert matrix A into a diagonal matrix. Suppose
More informationEigenvalues and Eigenvectors: An Introduction
Eigenvalues and Eigenvectors: An Introduction The eigenvalue problem is a problem of considerable theoretical interest and wide-ranging application. For example, this problem is crucial in solving systems
More informationAMS10 HW7 Solutions. All credit is given for effort. (-5 pts for any missing sections) Problem 1 (20 pts) Consider the following matrix 2 A =
AMS1 HW Solutions All credit is given for effort. (- pts for any missing sections) Problem 1 ( pts) Consider the following matrix 1 1 9 a. Calculate the eigenvalues of A. Eigenvalues are 1 1.1, 9.81,.1
More informationRecall : Eigenvalues and Eigenvectors
Recall : Eigenvalues and Eigenvectors Let A be an n n matrix. If a nonzero vector x in R n satisfies Ax λx for a scalar λ, then : The scalar λ is called an eigenvalue of A. The vector x is called an eigenvector
More information22.3. Repeated Eigenvalues and Symmetric Matrices. Introduction. Prerequisites. Learning Outcomes
Repeated Eigenvalues and Symmetric Matrices. Introduction In this Section we further develop the theory of eigenvalues and eigenvectors in two distinct directions. Firstly we look at matrices where one
More informationICS 6N Computational Linear Algebra Eigenvalues and Eigenvectors
ICS 6N Computational Linear Algebra Eigenvalues and Eigenvectors Xiaohui Xie University of California, Irvine xhx@uci.edu Xiaohui Xie (UCI) ICS 6N 1 / 34 The powers of matrix Consider the following dynamic
More informationA = 3 1. We conclude that the algebraic multiplicity of the eigenvalues are both one, that is,
65 Diagonalizable Matrices It is useful to introduce few more concepts, that are common in the literature Definition 65 The characteristic polynomial of an n n matrix A is the function p(λ) det(a λi) Example
More informationTherefore, A and B have the same characteristic polynomial and hence, the same eigenvalues.
Similar Matrices and Diagonalization Page 1 Theorem If A and B are n n matrices, which are similar, then they have the same characteristic equation and hence the same eigenvalues. Proof Let A and B be
More informationEigenvalues and Eigenvectors
Eigenvalues and Eigenvectors week -2 Fall 26 Eigenvalues and eigenvectors The most simple linear transformation from R n to R n may be the transformation of the form: T (x,,, x n ) (λ x, λ 2,, λ n x n
More informationDiagonalization. Hung-yi Lee
Diagonalization Hung-yi Lee Review If Av = λv (v is a vector, λ is a scalar) v is an eigenvector of A excluding zero vector λ is an eigenvalue of A that corresponds to v Eigenvectors corresponding to λ
More informationMAT 1302B Mathematical Methods II
MAT 1302B Mathematical Methods II Alistair Savage Mathematics and Statistics University of Ottawa Winter 2015 Lecture 19 Alistair Savage (uottawa) MAT 1302B Mathematical Methods II Winter 2015 Lecture
More informationMath 304 Fall 2018 Exam 3 Solutions 1. (18 Points, 3 Pts each part) Let A, B, C, D be square matrices of the same size such that
Math 304 Fall 2018 Exam 3 Solutions 1. (18 Points, 3 Pts each part) Let A, B, C, D be square matrices of the same size such that det(a) = 2, det(b) = 2, det(c) = 1, det(d) = 4. 2 (a) Compute det(ad)+det((b
More informationc c c c c c c c c c a 3x3 matrix C= has a determinant determined by
Linear Algebra Determinants and Eigenvalues Introduction: Many important geometric and algebraic properties of square matrices are associated with a single real number revealed by what s known as the determinant.
More informationLINEAR ALGEBRA 1, 2012-I PARTIAL EXAM 3 SOLUTIONS TO PRACTICE PROBLEMS
LINEAR ALGEBRA, -I PARTIAL EXAM SOLUTIONS TO PRACTICE PROBLEMS Problem (a) For each of the two matrices below, (i) determine whether it is diagonalizable, (ii) determine whether it is orthogonally diagonalizable,
More informationAnnouncements Monday, November 06
Announcements Monday, November 06 This week s quiz: covers Sections 5 and 52 Midterm 3, on November 7th (next Friday) Exam covers: Sections 3,32,5,52,53 and 55 Section 53 Diagonalization Motivation: Difference
More informationLinear Algebra: Matrix Eigenvalue Problems
CHAPTER8 Linear Algebra: Matrix Eigenvalue Problems Chapter 8 p1 A matrix eigenvalue problem considers the vector equation (1) Ax = λx. 8.0 Linear Algebra: Matrix Eigenvalue Problems Here A is a given
More informationDefinition (T -invariant subspace) Example. Example
Eigenvalues, Eigenvectors, Similarity, and Diagonalization We now turn our attention to linear transformations of the form T : V V. To better understand the effect of T on the vector space V, we begin
More informationStudy Guide for Linear Algebra Exam 2
Study Guide for Linear Algebra Exam 2 Term Vector Space Definition A Vector Space is a nonempty set V of objects, on which are defined two operations, called addition and multiplication by scalars (real
More informationEigenvalues and Eigenvectors
5 Eigenvalues and Eigenvectors 5.2 THE CHARACTERISTIC EQUATION DETERMINANATS nn Let A be an matrix, let U be any echelon form obtained from A by row replacements and row interchanges (without scaling),
More information235 Final exam review questions
5 Final exam review questions Paul Hacking December 4, 0 () Let A be an n n matrix and T : R n R n, T (x) = Ax the linear transformation with matrix A. What does it mean to say that a vector v R n is an
More informationMATH 1553-C MIDTERM EXAMINATION 3
MATH 553-C MIDTERM EXAMINATION 3 Name GT Email @gatech.edu Please read all instructions carefully before beginning. Please leave your GT ID card on your desk until your TA scans your exam. Each problem
More information1 :: Mathematical notation
1 :: Mathematical notation x A means x is a member of the set A. A B means the set A is contained in the set B. {a 1,..., a n } means the set hose elements are a 1,..., a n. {x A : P } means the set of
More informationMath 1553 Worksheet 5.3, 5.5
Math Worksheet, Answer yes / no / maybe In each case, A is a matrix whose entries are real a) If A is a matrix with characteristic polynomial λ(λ ), then the - eigenspace is -dimensional b) If A is an
More informationLecture 11: Eigenvalues and Eigenvectors
Lecture : Eigenvalues and Eigenvectors De nition.. Let A be a square matrix (or linear transformation). A number λ is called an eigenvalue of A if there exists a non-zero vector u such that A u λ u. ()
More informationEigenvalues and Eigenvectors
Eigenvalues and Eigenvectors Philippe B. Laval KSU Fall 2015 Philippe B. Laval (KSU) Eigenvalues and Eigenvectors Fall 2015 1 / 14 Introduction We define eigenvalues and eigenvectors. We discuss how to
More informationLinear Algebra review Powers of a diagonalizable matrix Spectral decomposition
Linear Algebra review Powers of a diagonalizable matrix Spectral decomposition Prof. Tesler Math 283 Fall 2018 Also see the separate version of this with Matlab and R commands. Prof. Tesler Diagonalizing
More informationLinear Algebra review Powers of a diagonalizable matrix Spectral decomposition
Linear Algebra review Powers of a diagonalizable matrix Spectral decomposition Prof. Tesler Math 283 Fall 2016 Also see the separate version of this with Matlab and R commands. Prof. Tesler Diagonalizing
More informationCalculating determinants for larger matrices
Day 26 Calculating determinants for larger matrices We now proceed to define det A for n n matrices A As before, we are looking for a function of A that satisfies the product formula det(ab) = det A det
More informationEigenvalues and Eigenvectors
November 3, 2016 1 Definition () The (complex) number λ is called an eigenvalue of the n n matrix A provided there exists a nonzero (complex) vector v such that Av = λv, in which case the vector v is called
More informationDiagonalization. MATH 322, Linear Algebra I. J. Robert Buchanan. Spring Department of Mathematics
Diagonalization MATH 322, Linear Algebra I J. Robert Buchanan Department of Mathematics Spring 2015 Motivation Today we consider two fundamental questions: Given an n n matrix A, does there exist a basis
More informationLinear Algebra. Matrices Operations. Consider, for example, a system of equations such as x + 2y z + 4w = 0, 3x 4y + 2z 6w = 0, x 3y 2z + w = 0.
Matrices Operations Linear Algebra Consider, for example, a system of equations such as x + 2y z + 4w = 0, 3x 4y + 2z 6w = 0, x 3y 2z + w = 0 The rectangular array 1 2 1 4 3 4 2 6 1 3 2 1 in which the
More information(the matrix with b 1 and b 2 as columns). If x is a vector in R 2, then its coordinate vector [x] B relative to B satisfies the formula.
4 Diagonalization 4 Change of basis Let B (b,b ) be an ordered basis for R and let B b b (the matrix with b and b as columns) If x is a vector in R, then its coordinate vector x B relative to B satisfies
More informationEigenvalues for Triangular Matrices. ENGI 7825: Linear Algebra Review Finding Eigenvalues and Diagonalization
Eigenvalues for Triangular Matrices ENGI 78: Linear Algebra Review Finding Eigenvalues and Diagonalization Adapted from Notes Developed by Martin Scharlemann The eigenvalues for a triangular matrix are
More informationMAC Module 12 Eigenvalues and Eigenvectors. Learning Objectives. Upon completing this module, you should be able to:
MAC Module Eigenvalues and Eigenvectors Learning Objectives Upon completing this module, you should be able to: Solve the eigenvalue problem by finding the eigenvalues and the corresponding eigenvectors
More informationMAC Module 12 Eigenvalues and Eigenvectors
MAC 23 Module 2 Eigenvalues and Eigenvectors Learning Objectives Upon completing this module, you should be able to:. Solve the eigenvalue problem by finding the eigenvalues and the corresponding eigenvectors
More informationMAT1302F Mathematical Methods II Lecture 19
MAT302F Mathematical Methods II Lecture 9 Aaron Christie 2 April 205 Eigenvectors, Eigenvalues, and Diagonalization Now that the basic theory of eigenvalues and eigenvectors is in place most importantly
More informationChapter 5 Eigenvalues and Eigenvectors
Chapter 5 Eigenvalues and Eigenvectors Outline 5.1 Eigenvalues and Eigenvectors 5.2 Diagonalization 5.3 Complex Vector Spaces 2 5.1 Eigenvalues and Eigenvectors Eigenvalue and Eigenvector If A is a n n
More informationHomework For each of the following matrices, find the minimal polynomial and determine whether the matrix is diagonalizable.
Math 5327 Fall 2018 Homework 7 1. For each of the following matrices, find the minimal polynomial and determine whether the matrix is diagonalizable. 3 1 0 (a) A = 1 2 0 1 1 0 x 3 1 0 Solution: 1 x 2 0
More informationQuestion: Given an n x n matrix A, how do we find its eigenvalues? Idea: Suppose c is an eigenvalue of A, then what is the determinant of A-cI?
Section 5. The Characteristic Polynomial Question: Given an n x n matrix A, how do we find its eigenvalues? Idea: Suppose c is an eigenvalue of A, then what is the determinant of A-cI? Property The eigenvalues
More informationSOLUTIONS: ASSIGNMENT Use Gaussian elimination to find the determinant of the matrix. = det. = det = 1 ( 2) 3 6 = 36. v 4.
SOLUTIONS: ASSIGNMENT 9 66 Use Gaussian elimination to find the determinant of the matrix det 1 1 4 4 1 1 1 1 8 8 = det = det 0 7 9 0 0 0 6 = 1 ( ) 3 6 = 36 = det = det 0 0 6 1 0 0 0 6 61 Consider a 4
More informationEcon Slides from Lecture 7
Econ 205 Sobel Econ 205 - Slides from Lecture 7 Joel Sobel August 31, 2010 Linear Algebra: Main Theory A linear combination of a collection of vectors {x 1,..., x k } is a vector of the form k λ ix i for
More informationEigenvalue and Eigenvector Homework
Eigenvalue and Eigenvector Homework Olena Bormashenko November 4, 2 For each of the matrices A below, do the following:. Find the characteristic polynomial of A, and use it to find all the eigenvalues
More informationMath 315: Linear Algebra Solutions to Assignment 7
Math 5: Linear Algebra s to Assignment 7 # Find the eigenvalues of the following matrices. (a.) 4 0 0 0 (b.) 0 0 9 5 4. (a.) The characteristic polynomial det(λi A) = (λ )(λ )(λ ), so the eigenvalues are
More informationEigenvalues and Eigenvectors
5 Eigenvalues and Eigenvectors 5.2 THE CHARACTERISTIC EQUATION DETERMINANATS n n Let A be an matrix, let U be any echelon form obtained from A by row replacements and row interchanges (without scaling),
More informationRepeated Eigenvalues and Symmetric Matrices
Repeated Eigenvalues and Symmetric Matrices. Introduction In this Section we further develop the theory of eigenvalues and eigenvectors in two distinct directions. Firstly we look at matrices where one
More informationTMA Calculus 3. Lecture 21, April 3. Toke Meier Carlsen Norwegian University of Science and Technology Spring 2013
TMA4115 - Calculus 3 Lecture 21, April 3 Toke Meier Carlsen Norwegian University of Science and Technology Spring 2013 www.ntnu.no TMA4115 - Calculus 3, Lecture 21 Review of last week s lecture Last week
More informationComputationally, diagonal matrices are the easiest to work with. With this idea in mind, we introduce similarity:
Diagonalization We have seen that diagonal and triangular matrices are much easier to work with than are most matrices For example, determinants and eigenvalues are easy to compute, and multiplication
More information22m:033 Notes: 7.1 Diagonalization of Symmetric Matrices
m:33 Notes: 7. Diagonalization of Symmetric Matrices Dennis Roseman University of Iowa Iowa City, IA http://www.math.uiowa.edu/ roseman May 3, Symmetric matrices Definition. A symmetric matrix is a matrix
More informationMATH 221, Spring Homework 10 Solutions
MATH 22, Spring 28 - Homework Solutions Due Tuesday, May Section 52 Page 279, Problem 2: 4 λ A λi = and the characteristic polynomial is det(a λi) = ( 4 λ)( λ) ( )(6) = λ 6 λ 2 +λ+2 The solutions to the
More informationReview of Linear Algebra
Review of Linear Algebra Definitions An m n (read "m by n") matrix, is a rectangular array of entries, where m is the number of rows and n the number of columns. 2 Definitions (Con t) A is square if m=
More informationLecture 12: Diagonalization
Lecture : Diagonalization A square matrix D is called diagonal if all but diagonal entries are zero: a a D a n 5 n n. () Diagonal matrices are the simplest matrices that are basically equivalent to vectors
More informationDimension. Eigenvalue and eigenvector
Dimension. Eigenvalue and eigenvector Math 112, week 9 Goals: Bases, dimension, rank-nullity theorem. Eigenvalue and eigenvector. Suggested Textbook Readings: Sections 4.5, 4.6, 5.1, 5.2 Week 9: Dimension,
More informationDM554 Linear and Integer Programming. Lecture 9. Diagonalization. Marco Chiarandini
DM554 Linear and Integer Programming Lecture 9 Marco Chiarandini Department of Mathematics & Computer Science University of Southern Denmark Outline 1. More on 2. 3. 2 Resume Linear transformations and
More informationFundamentals of Engineering Analysis (650163)
Philadelphia University Faculty of Engineering Communications and Electronics Engineering Fundamentals of Engineering Analysis (6563) Part Dr. Omar R Daoud Matrices: Introduction DEFINITION A matrix is
More informationEigenvalues, Eigenvectors, and Diagonalization
Math 240 TA: Shuyi Weng Winter 207 February 23, 207 Eigenvalues, Eigenvectors, and Diagonalization The concepts of eigenvalues, eigenvectors, and diagonalization are best studied with examples. We will
More informationa 11 a 12 a 11 a 12 a 13 a 21 a 22 a 23 . a 31 a 32 a 33 a 12 a 21 a 23 a 31 a = = = = 12
24 8 Matrices Determinant of 2 2 matrix Given a 2 2 matrix [ ] a a A = 2 a 2 a 22 the real number a a 22 a 2 a 2 is determinant and denoted by det(a) = a a 2 a 2 a 22 Example 8 Find determinant of 2 2
More information4. Linear transformations as a vector space 17
4 Linear transformations as a vector space 17 d) 1 2 0 0 1 2 0 0 1 0 0 0 1 2 3 4 32 Let a linear transformation in R 2 be the reflection in the line = x 2 Find its matrix 33 For each linear transformation
More information1. Linear systems of equations. Chapters 7-8: Linear Algebra. Solution(s) of a linear system of equations (continued)
1 A linear system of equations of the form Sections 75, 78 & 81 a 11 x 1 + a 12 x 2 + + a 1n x n = b 1 a 21 x 1 + a 22 x 2 + + a 2n x n = b 2 a m1 x 1 + a m2 x 2 + + a mn x n = b m can be written in matrix
More informationSymmetric and anti symmetric matrices
Symmetric and anti symmetric matrices In linear algebra, a symmetric matrix is a square matrix that is equal to its transpose. Formally, matrix A is symmetric if. A = A Because equal matrices have equal
More informationConceptual Questions for Review
Conceptual Questions for Review Chapter 1 1.1 Which vectors are linear combinations of v = (3, 1) and w = (4, 3)? 1.2 Compare the dot product of v = (3, 1) and w = (4, 3) to the product of their lengths.
More informationft-uiowa-math2550 Assignment NOTRequiredJustHWformatOfQuizReviewForExam3part2 due 12/31/2014 at 07:10pm CST
me me ft-uiowa-math2550 Assignment NOTRequiredJustHWformatOfQuizReviewForExam3part2 due 12/31/2014 at 07:10pm CST 1. (1 pt) local/library/ui/eigentf.pg A is n n an matrices.. There are an infinite number
More informationDiagonalization. P. Danziger. u B = A 1. B u S.
7., 8., 8.2 Diagonalization P. Danziger Change of Basis Given a basis of R n, B {v,..., v n }, we have seen that the matrix whose columns consist of these vectors can be thought of as a change of basis
More informationChapter 3. Determinants and Eigenvalues
Chapter 3. Determinants and Eigenvalues 3.1. Determinants With each square matrix we can associate a real number called the determinant of the matrix. Determinants have important applications to the theory
More informationLecture 15 Review of Matrix Theory III. Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science - Bangalore
Lecture 15 Review of Matrix Theory III Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science - Bangalore Matrix An m n matrix is a rectangular or square array of
More informationMath Camp Notes: Linear Algebra II
Math Camp Notes: Linear Algebra II Eigenvalues Let A be a square matrix. An eigenvalue is a number λ which when subtracted from the diagonal elements of the matrix A creates a singular matrix. In other
More information4. Determinants.
4. Determinants 4.1. Determinants; Cofactor Expansion Determinants of 2 2 and 3 3 Matrices 2 2 determinant 4.1. Determinants; Cofactor Expansion Determinants of 2 2 and 3 3 Matrices 3 3 determinant 4.1.
More informationGlossary of Linear Algebra Terms. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB
Glossary of Linear Algebra Terms Basis (for a subspace) A linearly independent set of vectors that spans the space Basic Variable A variable in a linear system that corresponds to a pivot column in the
More informationDIAGONALIZATION. In order to see the implications of this definition, let us consider the following example Example 1. Consider the matrix
DIAGONALIZATION Definition We say that a matrix A of size n n is diagonalizable if there is a basis of R n consisting of eigenvectors of A ie if there are n linearly independent vectors v v n such that
More informationDifferential Geometry of Surfaces
Differential Geometry of urfaces Jordan mith and Carlo équin C Diision, UC Berkeley Introduction These are notes on differential geometry of surfaces ased on reading Greiner et al. n. d.. Differential
More informationPROBLEM SET. Problems on Eigenvalues and Diagonalization. Math 3351, Fall Oct. 20, 2010 ANSWERS
PROBLEM SET Problems on Eigenvalues and Diagonalization Math 335, Fall 2 Oct. 2, 2 ANSWERS i Problem. In each part, find the characteristic polynomial of the matrix and the eigenvalues of the matrix by
More informationChapter 3 Transformations
Chapter 3 Transformations An Introduction to Optimization Spring, 2014 Wei-Ta Chu 1 Linear Transformations A function is called a linear transformation if 1. for every and 2. for every If we fix the bases
More informationMatrices and Linear Algebra
Contents Quantitative methods for Economics and Business University of Ferrara Academic year 2017-2018 Contents 1 Basics 2 3 4 5 Contents 1 Basics 2 3 4 5 Contents 1 Basics 2 3 4 5 Contents 1 Basics 2
More informationTBP MATH33A Review Sheet. November 24, 2018
TBP MATH33A Review Sheet November 24, 2018 General Transformation Matrices: Function Scaling by k Orthogonal projection onto line L Implementation If we want to scale I 2 by k, we use the following: [
More informationTopic 1: Matrix diagonalization
Topic : Matrix diagonalization Review of Matrices and Determinants Definition A matrix is a rectangular array of real numbers a a a m a A = a a m a n a n a nm The matrix is said to be of order n m if it
More informationExercise Set 7.2. Skills
Orthogonally diagonalizable matrix Spectral decomposition (or eigenvalue decomposition) Schur decomposition Subdiagonal Upper Hessenburg form Upper Hessenburg decomposition Skills Be able to recognize
More informationGeneralized Eigenvectors and Jordan Form
Generalized Eigenvectors and Jordan Form We have seen that an n n matrix A is diagonalizable precisely when the dimensions of its eigenspaces sum to n. So if A is not diagonalizable, there is at least
More informationMath 205, Summer I, Week 4b:
Math 205, Summer I, 2016 Week 4b: Chapter 5, Sections 6, 7 and 8 (5.5 is NOT on the syllabus) 5.6 Eigenvalues and Eigenvectors 5.7 Eigenspaces, nondefective matrices 5.8 Diagonalization [*** See next slide
More informationEigenvalues for Triangular Matrices. ENGI 7825: Linear Algebra Review Finding Eigenvalues and Diagonalization
Eigenvalues for Triangular Matrices ENGI 78: Linear Algebra Review Finding Eigenvalues and Diagonalization Adapted from Notes Developed by Martin Scharlemann June 7, 04 The eigenvalues for a triangular
More information1. General Vector Spaces
1.1. Vector space axioms. 1. General Vector Spaces Definition 1.1. Let V be a nonempty set of objects on which the operations of addition and scalar multiplication are defined. By addition we mean a rule
More informationMATH 20F: LINEAR ALGEBRA LECTURE B00 (T. KEMP)
MATH 20F: LINEAR ALGEBRA LECTURE B00 (T KEMP) Definition 01 If T (x) = Ax is a linear transformation from R n to R m then Nul (T ) = {x R n : T (x) = 0} = Nul (A) Ran (T ) = {Ax R m : x R n } = {b R m
More informationEigenvalues and Eigenvectors
Contents Eigenvalues and Eigenvectors. Basic Concepts. Applications of Eigenvalues and Eigenvectors 8.3 Repeated Eigenvalues and Symmetric Matrices 3.4 Numerical Determination of Eigenvalues and Eigenvectors
More informationA matrix is a rectangular array of. objects arranged in rows and columns. The objects are called the entries. is called the size of the matrix, and
Section 5.5. Matrices and Vectors A matrix is a rectangular array of objects arranged in rows and columns. The objects are called the entries. A matrix with m rows and n columns is called an m n matrix.
More informationLinear Algebra Primer
Linear Algebra Primer David Doria daviddoria@gmail.com Wednesday 3 rd December, 2008 Contents Why is it called Linear Algebra? 4 2 What is a Matrix? 4 2. Input and Output.....................................
More information1. In this problem, if the statement is always true, circle T; otherwise, circle F.
Math 1553, Extra Practice for Midterm 3 (sections 45-65) Solutions 1 In this problem, if the statement is always true, circle T; otherwise, circle F a) T F If A is a square matrix and the homogeneous equation
More informationMATH 369 Linear Algebra
Assignment # Problem # A father and his two sons are together 00 years old. The father is twice as old as his older son and 30 years older than his younger son. How old is each person? Problem # 2 Determine
More informationElementary Linear Algebra
Matrices J MUSCAT Elementary Linear Algebra Matrices Definition Dr J Muscat 2002 A matrix is a rectangular array of numbers, arranged in rows and columns a a 2 a 3 a n a 2 a 22 a 23 a 2n A = a m a mn We
More informationUnit 5: Matrix diagonalization
Unit 5: Matrix diagonalization Juan Luis Melero and Eduardo Eyras October 2018 1 Contents 1 Matrix diagonalization 3 1.1 Definitions............................. 3 1.1.1 Similar matrix.......................
More informationALGEBRA QUALIFYING EXAM PROBLEMS LINEAR ALGEBRA
ALGEBRA QUALIFYING EXAM PROBLEMS LINEAR ALGEBRA Kent State University Department of Mathematical Sciences Compiled and Maintained by Donald L. White Version: August 29, 2017 CONTENTS LINEAR ALGEBRA AND
More informationA matrix is a rectangular array of. objects arranged in rows and columns. The objects are called the entries. is called the size of the matrix, and
Section 5.5. Matrices and Vectors A matrix is a rectangular array of objects arranged in rows and columns. The objects are called the entries. A matrix with m rows and n columns is called an m n matrix.
More informationLinear Algebra II Lecture 13
Linear Algebra II Lecture 13 Xi Chen 1 1 University of Alberta November 14, 2014 Outline 1 2 If v is an eigenvector of T : V V corresponding to λ, then v is an eigenvector of T m corresponding to λ m since
More informationMath 489AB Exercises for Chapter 2 Fall Section 2.3
Math 489AB Exercises for Chapter 2 Fall 2008 Section 2.3 2.3.3. Let A M n (R). Then the eigenvalues of A are the roots of the characteristic polynomial p A (t). Since A is real, p A (t) is a polynomial
More informationMATH 315 Linear Algebra Homework #1 Assigned: August 20, 2018
Homework #1 Assigned: August 20, 2018 Review the following subjects involving systems of equations and matrices from Calculus II. Linear systems of equations Converting systems to matrix form Pivot entry
More information