Eigenvalues and Eigenvectors

Transcription

1 November 3, 2016

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3 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 an eigenvector of A.

4 Example Consider We have ( 2 A 1 ) = A = ( ( ) ( 2 1 ) = ). ( 4 2 ) ( 2 = 2 1 Thus λ = 2 is an eigenvalue of A and (2, 1) T is an eigenvector of A associated with the eigenvalue λ = 2. We also have ( ) ( ) ( ) ( ) A = = Thus λ = 1 is an eigenvalue of A and (3, 2) T is an eigenvector of A associated with the eigenvalue λ = 1. ).

5 Remarks 1 An eigenvalue may be zero but an eigenvector is by definition a nonzero vector. 2 If v 1 and v 2 are eigenvectors of A associated with eigenvalue λ, then for any scalars c 1 and c 2, c 1 v 1 + c 2 v 2 is also an eigenvector of A associated with eigenvalue λ if it is non-zero. 3 If λ is an eigenvalue of an n n matrix A, then the set of all eigenvectors associated with eigenvalue λ together with the zero vector 0 form a vector subspace of R n. It is called the eigenspace of A associated with eigenvalue λ.

6 Definition (Characteristic equation) Let A be an n n matrix. The polynomial equation det(a xi) = 0 of degree n is called the characteristic equation of A.

7 Theorem Let A be an n n matrix. The following statements for scalar λ are equivalent. 1 λ is an eigenvalue of A. 2 The equation (A λi)v = 0 has nontrivial solution for v. 3 Null(A λi) {0}. 4 The matrix A λi is singular. 5 λ is a root of the characteristic equation det(a xi) = 0

8 Example Find the eigenvalues and associated eigenvectors of the matrix ( ) 3 2 A =. 3 2 Solution: Solving the characteristic equation, we have det(a λi) = 0 3 λ λ = 0 λ 2 λ 12 = 0 λ = 4, 3

9 When λ = 4, (A 4I)v = 0 ( ) 1 2 v = Thus v = (2, 1) T is an eigenvector associated with λ = 4. When λ = 3, (A + 3I)v = 0 ( ) 6 2 v = Thus v = (1, 3) T is an eigenvector associated with λ = 3.

10 Example Find the eigenvalues and associated eigenvectors of the matrix ( ) 0 8 A =. 2 0 Solution: Solving the characteristic equation, we have λ 8 2 λ = 0 λ = 0 λ = ±4i

11 When λ = 4i, (A 4iI)v = 0 ( ) 4i 8 v = 0 2 4i Thus v = (2, i) T is an eigenvector associated with λ = 4i. When λ = 4i, (A + 4iI)v = 0 ( ) 4i 8 v = 0 2 4i Thus v = (2, i) T is an eigenvector associated with λ = 4i.

12 Remark For any square matrix A with real entries, the characteristic polynomial of A has real coefficients. Thus if λ = ρ + µi, where ρ, µ R, is a complex eigenvalue of A, then λ = ρ µi is also an eigenvalue of A. Furthermore, if v = a + bi is an eigenvector associated with complex eigenvalue λ, then v = a bi is an eigenvector associated with eigenvalue λ.

13 Example Find the eigenvalues and associated eigenvectors of the matrix A = Solution: Solving the characteristic equation, we have 2 λ λ = 0 (λ 2) 2 = 0 λ = 2, 2 ( ). When λ = 2, ( ) v = 0 Thus v = (1, 0) T is an eigenvector associated with λ = 2. In this example, we call only find one linearly independent eigenvector.

14 Example Find the eigenvalues and associated eigenvectors of the matrix A = Solution: Solving the characteristic equation, we have 2 λ λ 1 = λ λ(λ 1) 2 = 0 λ = 1, 1, 0

15 For λ 1 = λ 2 = 1, v = 0 Thus v 1 = (3, 1, 0) T and v 2 = ( 1, 0, 1) are two linearly independent eigenvectors associated with λ = 1. For λ 3 = 0, v = Thus v 3 = (1, 1, 1) T is an eigenvector associated with λ = 0.

16 Example Find the eigenvalues and associated eigenvectors of the matrix A = Solution: Solving the characteristic equation, we have 1 λ λ 0 = λ (λ 2)(λ 1) 2 = 0 λ = 2, 1, 1

17 For λ 1 = 2, v = 0 Thus v = (0, 0, 1) T is an eigenvector associated with λ = 2. For λ 2 = λ 3 = 1, v = Thus ( 1, 2, 1) T is an eigenvectors associated with λ = 1. Note that here λ = 1 is a double root but we can only find one linearly independent eigenvector associated with λ = 1.

18 Definition (Similar matrices) Two n n matrices A and B are said to be similar if there exists an invertible matrix P such that Proposition B = P 1 AP. Similarity of square matrices is an equivalence relation, that is, 1 For any square matrix A, we have A is similar to A; 2 If A is similar to B, then B is similar to A; 3 If A is similar to B and B is similar to C, then A is similar to C.

19 Proof. 1 Since I is a non-singular matrix and A = I 1 AI, we have A is similar to A. 2 If A is similar to B, then there exists non-singular matrix P such that B = P 1 AP. Now P 1 is a non-singular matrix and (P 1 ) 1 = P. There exists non-singular matrix P 1 such that (P 1 ) 1 BP 1 = PBP 1 = A. Therefore B is similar to A. 3 If A is similar to B and B is similar to C, then there exists non-singular matrices P and Q such that B = P 1 AP and C = Q 1 BQ. Now PQ is a non-singular matrix and (PQ) 1 = Q 1 P 1. There exists non-singular matrix PQ such that (PQ) 1 A(PQ) = Q 1 (P 1 AP)Q = Q 1 BQ = C. Therefore A is similar to C.

20 Proposition 1 The only matrix similar to the zero matrix 0 is the zero matrix. 2 The only matrix similar to the identity matrix I is the identity matrix. 3 If A is similar to B, then aa is similar to ab for any real number a. 4 If A is similar to B, then A k is similar to B k for any non-negative integer k. 5 If A and B are similar non-singular matrices, then A 1 is similar to B 1. 6 If A is similar to B, then A T is similar to B T. 7 If A is similar to B, then det(a) = det(b). 8 If A is similar to B, then tr(a) = tr(b) where tr(a) = a 11 + a a nn is the trace, i.e., the sum of the entries in the diagonal, of A. 9 If A is similar to B, then A and B have the same characteristic equation.

21 Proof. (1) Suppose A is similar to 0, then there exists non-singular matrix P such that 0 = P 1 AP. Hence A = P0P 1 = 0. (4) If A is similar to B, then there exists non-singular matrix P such that B = P 1 AP. We have B k = k copies {}}{ (P 1 AP)(P 1 AP) (P 1 AP) = P 1 A(PP 1 )AP P 1 A(PP 1 )AP = P 1 AIAI IAIAP = P 1 A k P Therefore A k is similar to B k.

22 Proof. (7) If A is similar to B, then there exists non-singular matrix P such that B = P 1 AP. Thus det(b) = det(p 1 AP) = det(p 1 ) det(a) det(p) = det(a) since det(p 1 ) = det(p) 1. (8) If A is similar to B, then there exists non-singular matrix P such that B = P 1 AP. Thus tr(b) = tr((p 1 A)P) = tr(p(p 1 A)) = tr(a). (Note: It is well-known that tr(pq) = tr(qp) for any square matrices P and Q. But in general, it is not always true that tr(pqr) = tr(qpr).)

23 Definition An n n matrix A is said to be diagonalizable if there exists a nonsingular (may be complex) matrix P such that P 1 AP = D is a diagonal matrix. We say that P diagonalizes A. In other words, A is diagonalizable if it is similar to a diagonal matrix.

24 Theorem Let A be an n n matrix. Then A is diagonalizable if and only if A has n linearly independent eigenvectors. Proof. Let P be an n n matrix and write P = [ v 1 v 2 v n ]. where v 1, v 2,, v n are the column vectors of P. First observe that P is nonsingular if and only if v 1, v 2,, v n are linearly independent.

25 Furthermore λ P 1 0 λ 2 0 AP = D =. is a diagonal matrix λ n λ λ 2 0 AP = PD for some diagonal matrix D = λ n A [ ] [ ] v 1 v 2 v n = λ1v 1 λ 2v 2 λ nv n. [ ] [ ] Av 1 Av 2 Av n = λ1v 1 λ 2v 2 λ nv n. v k is an eigenvector of A associated with eigenvalue λ k for k = 1, 2,, n. Therefore P diagonalizes A if and only if v 1, v 2,, v n are linearly independent eigenvectors of A.

26 Example Diagonalize the matrix A = ( ). Solution: We have seen in Example 8 that A has eigenvalues λ 1 = 4 and λ 2 = 3 associated with linearly independent eigenvectors v 1 = (2, 1) T and v 2 = (1, 3) T respectively. Thus the matrix ( ) 2 1 P = 1 3 diagonalizes A and P 1 AP = = ( ) 1 ( ( ) ) ( )

27 Example Diagonalize the matrix A = ( ). Solution: We have seen in Example 10 that A has eigenvalues λ 1 = 4i and λ 2 = 4i associated with linearly independent eigenvectors v 1 = (2, i) T and v 2 = (2, i) T respectively. Thus the matrix ( ) 2 2 P = i i diagonalizes A and P 1 AP = ( 4i 0 0 4i ).

28 Example We have seen in Example 14 that ( ) 2 3 A = 0 2 has only one linearly independent eigenvector. Therefore it is not diagonalizable.

29 Example Diagonalize the matrix A = Solution: We have seen in Example 10 that A has eigenvalues λ 1 = λ 2 = 1 and λ 3 = 0. For λ 1 = λ 2 = 1, there are two linearly independent eigenvectors v 1 = (3, 1, 0) T and v 2 = ( 1, 0, 1) T. For λ 3 = 0, there associated one linearly independent eigenvector v 3 = (1, 1, 1) T. The three vectors v 1, v 2, v 3 are linearly independent eigenvectors. Thus the matrix diagonalizes A and P = P 1 AP =

30 Example Show that the matrix A = ( ) is not diagonalizable. Solution: One can show that there is at most one linearly independent eigenvector. Alternatively, one can argue in the following way. The characteristic equation of A is (r 1) 2 = 0. Thus λ = 1 is the only eigenvalue of A. Hence if A is diagonalizable by P, then P 1 AP = I. But then A = PIP 1 = I which leads to a contradiction.

31 Theorem Suppose that eigenvectors v 1, v 2,, v k are associated with the distinct eigenvalues λ 1, λ 2,, λ k of a matrix A. Then v 1, v 2,, v k are linearly independent. Proof. We prove the theorem by induction on k. The theorem is obviously true when k = 1. Now assume that the theorem is true for any set of k 1 eigenvectors. Suppose c 1 v 1 + c 2 v c k v k = 0. Multiplying A λ k I to the left on both sides, we have c 1 (A λ k I)v 1 + c 2 (A λ k I)v c k 1 (A λ k I)v k 1 + c k (A λ k I)v k = 0 c 1 (λ 1 λ k )v 1 + c 2 (λ 2 λ k )v c k 1 (λ k 1 λ k )v k 1 = 0

32 Note that (A λ k I)v k = 0 since v k is an eigenvector associated with λ k. From the induction hypothesis, v 1, v 2,, v k 1 are linearly independent. Thus c 1 (λ 1 λ k ) = c 2 (λ 2 λ k ) = = c k 1 (λ k 1 λ k ) = 0. Since λ 1, λ 2,, λ k are distinct, λ 1 λ k, λ 2 λ k,, λ k 1 λ k are all nonzero. Hence c 1 = c 2 = = c k 1 = 0. It follows then that c k is also equal to zero because v k is a nonzero vector. Therefore v 1, v 2,, v k are linearly independent.

33 Theorem If the n n matrix A has n distinct eigenvalues, then it is diagonalizable. Theorem Let A be an n n matrix and λ 1, λ 2,, λ k be the roots of the characteristic equation of A of multiplicity n 1, n 2,, n k respectively, i.e., the characteristic equation of A is (x λ 1 ) n 1 (x λ 2 ) n2 (x λ n ) n k = 0. Then A is diagonalizable if and only if for each 1 i k, there exists n i linearly independent eigenvectors associated with eigenvalue λ i.

34 Remark Let A be a square matrix. A (complex) number λ is an eigenvalue of A if and only if λ is a root of the characteristic equation of A. The multiplicity of λ being a root of the characteristic equation is called the algebraic multiplicity of λ. The dimension of the eigenspace associated to eigenvalue λ, that is, the maximum number of linearly independent eigenvectors associated with eigenvalue λ, is called the geometric multiplicity of λ. It can be shown that for each eigenvalue λ, we have 1 m g m a where m g and m a are the geometric and algebraic multiplicity of λ respectively. Therefore we have A is diagonalizable if and only if the algebraic multiplicity and the geometric multiplicity are equal for all eigenvalues of A.

35 Theorem A square matrix is diagonlizable if and only if its minimal polynomial has no repeated factor. Theorem All eigenvalues of a symmetric matrix are real. Theorem Any symmetric matrix is diagonalizable (by orthogonal matrix).

36 Exercise Diagonalize the following matrices. ( ) ( ) (a) (b) (c) (d) Exercise Show that that following matrices arenot diagonalizable. ( ) (a) (b) (c)

37 Suggested answer ( ) ( ) (a) P =, D =, ( ) ( ) 1 + i 1 i 1 + i 0 (b) P =, D =, i (c) P = 1 2 3, D = 0 2 0, (d) P = , D =

38 Suggested answer 2(a) The characteristic equation of the matrix is (λ 2) 2 = 0. There is only one eigenvalue λ = 2. For eigenvalue λ = 2, the eigenspace is span((1, 1) T ) which is of dimension 1. Therefore the matrix is not diagonalizable. 2(b) There are two eigenvalues 2 and 1. The algebraic multiplicity of eigenvalue λ = 1 is 2 but the geometric multiplicity of it is 1. Therefore the matrix is not diagonalizable. 2(c) There are two eigenvalues 2 and 1. The algebraic multiplicity of eigenvalue λ = 2 is 2 but the geometric multiplicity of it is 1. Therefore the matrix is not diagonalizable.

39 Let A be an n n matrix and P be a matrix diagonalizes A, i.e., P 1 AP = D is a diagonal matrix. Then A k = ( PDP 1) k = PD k P 1

40 Example Find A 5 if A = ( ). Solution: From Example 26, ( ) P AP = D = where P = 0 3 Thus ( ). A 5 = PD 5 P 1 ( ) ( ) 5 ( ) = ( ) ( ) ( ) = ( ) =

41 Example Find A 5 if A = Solution: Diagonalizing A, we have P 1 AP = D = where P = Thus A 5 = = =

42 Example Consider a metropolitan area with a constant total population of 1 million individuals. This area consists of a city and its suburbs, and we want to analyze the changing urban and suburban populations. Let C k denote the city population and S k the suburban population after k years. Suppose that each year 15% of the people in the city move to the suburbs, whereas 10% of the people in the suburbs move to the city. Then it follows that C k+1 = 0.85C k + 0.1S k S k+1 = 0.15C k + 0.9S k Find the urban and suburban populations after a long time.

43 Solution: Let x k = (C k, S k ) T be the population vector after k years. Then x k = Ax k 1 = A 2 x k 2 = = A k x 0, where A = ( ). Solving the characteristic equation, we have 0.85 λ λ = 0 λ λ = 0 λ = 1, 0.75 Hence the eigenvalues are λ 1 = 1 and λ 2 = By solving A λi = 0, the associated eigenvectors are v 1 = (2, 3) T and v 2 = ( 1, 1) T respectively.

44 Thus where When k is very large P 1 AP = D = P = ( ( ). ), A k = PD k P 1 ( ) ( ) k ( ) = ( ) ( ) ( ) = k ( ) ( ) ( ) = 1 ( )

45 Therefore x k = A k x ( 0 ) ( C S 0 ( ) 0.4 = (C 0 + S 0 ) 0.6 ( ) 0.4 = 0.6 ) That mean whatever the initial distribution of population is, the long-term distribution consists of 40% in the city and 60% in the suburbs.

46 An n n matrix is called a stochastic matrix if it has nonnegative entries and the sum of the elements in each column is one. A Markov process is a stochastic process having the property that given the present state, future states are independent of the past states. A Markov process can be described by a Morkov chain which consists of a sequence of vectors x k, k = 0, 1, 2,, satisfying x k = Ax k 1 = Ax 0. The vector x k is called the state vector and A is called the transition matrix.

47 PageRank The PageRank algorithm was developed by Larry Page and Sergey Brin and is used to rank the web sites in the Google search engine. It is a probability distribution used to represent the likelihood that a person randomly clicking links will arrive at any particular page. The linkage of the web sites in the web can be represented by a linkage matrix A. The probability distribution of a person to arrive the web sites are given by A k x 0 for a sufficiently large k and is independent of the initial distribution x 0.

48 Example Consider a small web consisting of three pages P, Q and R, where page P links to the pages Q and R, page Q links to page R and page R links to page P and Q. Assume that a person has a probability of 0.5 to stay on each page and the probability of going to other pages are evenly distributed to the pages which are linked to it. Find the page rank of the three web pages. R P Q

49 Solution: Let p k, q k and r k be the number of people arrive the web pages P, Q and R respectively after k iteration. Then p k p k 1 q k = q k 1. r k r k 1 Thus where A = x k = Ax k 1 = = A k x 0, and x 0 = p 0 q 0 r 0 are the linkage matrix and the initial state.

50 Solving the characteristic equation of A, we have 0.5 λ λ λ = 0 λ 3 1.5λ λ = 0 λ = 1 or 0.25 For λ 1 = 1, we solve v = 0 and v 1 = (2, 3, 4) T is an eigenvector of A associated with λ 1 = 1. For λ 2 = λ 3 = 0.25, we solve v = 0 and there associates only one linearly independent eigenvector v 2 = (1, 0, 1) T.

51 Thus A is not diagonalizable. However we have P = where P 1 AP = J = (J is called the Jordan normal form of A.) When k is sufficiently large, we have A k = PJ k P k = /9 2/9 2/9 = 3/9 3/9 3/9 4/9 4/9 4/ /9 1/9 1/9 4/9 4/9 5/9 1/12 1/6 1/12. 1

52 Thus after sufficiently many iteration, the number of people arrive the web pages are given by 2/9 2/9 2/9 p 0 2/9 A k x 0 3/9 3/9 3/9 q 0 = (p 0 + q 0 + r 0 ) 3/9. 4/9 4/9 4/9 r 0 4/9 Note that the ratio does not depend on the initial state. The PageRank of the web pages P, Q and R are 2/9, 3/9 and 4/9 respectively.

53 Fibonacci sequence The Fibonacci sequence is defined by { F k+2 = F k+1 + F k, for k 0 F 0 = 0, F 1 = 1. The recursive equation can be written as ( ) ( ) ( ) Fk Fk+1 =. F k F k If we let then A = ( ) and x k = ( Fk+1 x k+1 = ( Ax k ), for k 0 1 x 0 =. 0 F k ),

54 It follows that x k = A k x 0. To find A k, we can diagonalize A and obtain ( Hence A k = = ( ) 1 ( ( ( ) ) k+1 ( ) k ( ( ) ( ) k ( ) k+1 ( ) k ( ) ( = ( 1+ 5 ) k ) k ( ) k 1 ( ) k ) k 1 ) ) 1

55 Now we have x k = A k x 0 = 1 5 ( F k = 1 5 ( ( ) k+1 ( ) k ( 1 + ) k ( ) k ) k ) k 5. 2, We also have F k+1 ((1 + 5)/2) k+1 ((1 5)/2) k+1 lim = lim k F k k ((1 + 5)/2) k ((1 = )/2) k 2 which links the Fibonacci sequence with the celebrated golden ratio.

56 One of the most notable and important theorems in linear algebra is the. Definition Let A be an n n matrix. The minimal polynomial of A is a nonzero polynomial m(x) of minimum degree with leading coefficient 1 satisfying m(a) = 0. Theorem () Let A be an n n matrix and p(x) = det(a xi) be its characteristic polynomial. Then p(a) = 0. In other words, the minimal polynomial of A always divides the characteristic polynomial of A.

57 Proof Let B = A xi and p(x) = det(b) = c nx n + c n 1x n c 1x + c 0 be the characteristic polynomial of A. Consider B = A xi as an n n matrix whose entries are polynomial in x. Write the adjoint adj(b) of B as a polynomial of degree n 1 in x with matrix coefficients adj(b) = B n 1x n B 1x + B 0 where the coefficients B i are n n constant matrices. On one hand, we have det(b)i = (c nx n + c n 1x n c 1x + c 0)I On the other hand, we have = (c ni)x n + (c n 1I)x n (c 1I)x + c 0I Badj(B) = (A xi)(b n 1x n B 1x + B 0) = B n 1x n + (AB n 1 B n 2)x n (AB 1 B 0)x + AB 0

58 Proof. Comparing the coefficients of we get det(b)i = Badj(B), c ni = B n 1 c n 1 I = AB n 1 B n 2. c 1 I = AB 1 B 0 c 0 I = AB 0 If we multiply the first equation by A n, second by A n 1,, the last by I and add up the resulting equations, we obtain p(a) = c na n + c n 1A n c 1A + c 0I = A n B n 1 + (A n B n 1 A n 1 B n 2) + + (A 2 B 1 AB 0) + AB 0 = 0

59 Example Let The characteristic equation is A = p(x) = x 3 + 7x 2 16x So A 3 + 7A 2 16A + 12I = 0.

60 Example Now We have A 2 = A 3 = 7A 2 16A + 12I = =

61 Example Multiplying by A gives A 4 = 7A 3 16A A = 7(7A 2 16A + 12) 16A A = 33A 2 100A + 84I = =

62 Example Thus we can use Cayley-Hamilton theorem to express power of A in terms of A and A 2. We can also use Cayley-Hamilton theorem to find A 1 as follow 12I = A 3 7A A A 1 = 1 12 (A2 7A + 16I) = =

63 Example Find the minimal polynomial of A = Solution: The characteristic polynomial of A is p(x) = x(x 1) 2. It can be proved that every root of the characteristic polynomial must also be a root of the minimal polynomial. Thus the minimal polynomial must be one of the following polynomials x(x 1) or x(x 1) 2. By direct calculation A(A I) = = 0. Hence the minimal polynomial of A is x(x 1).

64 Definition Let A be an n n matrix. The matrix exponential of A is defined as exp(a) = k=0 A k k!. Theorem (Properties of matrix exponential) Let A and B be two n n matrix and a, b be any scalars. Then 1 exp(0) = I; 2 exp( A) = exp(a) 1 ; 3 exp((a + b)a) = exp(aa) exp(ba); 4 If AB = BA, then exp(a + B) = exp(a) exp(b); 5 If A is nonsingular, then exp(a 1 BA) = A 1 exp(b)a; 6 det(exp(a)) = e tr(a). (tr(a) = a 11 + a a nn is the trace of A.)

65 Theorem If D = λ λ is a diagonal matrix, then λ n e λ e λ2 0 exp(d) = e λn Moreover, for any n n matrix A, if there exists nonsingular matrix P such that P 1 AP = D is a diagonal matrix. Then exp(a) = P exp(d)p 1.

66 Example Find exp(at) where A = ( Solution: Diagonalizing A, we have ( ) 1 ( ). ) ( ) = ( Therefore ( ) ( ) ( 1 2 e 2t exp(at) = e 5t 3 1 = 1 ( ) e 2t + 6e 5t 2e 2t + 2e 5t 7 3e 2t + 3e 5t 6e 2t + e 5t ). ) 1

67 Example Find exp(at) where A = Solution: First compute Therefore A 2 = and A 3 = exp(at) = I + At A2 t 2 = = t 3t + 2t t t t 2

68 Example Find exp(at) where A = ( Solution: ( ) 4t t exp(at) = exp 0 4t (( 4t 0 = exp 0 4t ( 4t 0 = exp 0 4t = = ( 4t 0 (since 0 4t ( e 4t 0 0 e 4t ( ) e 4t te 4t 0 e 4t ). ) ( 0 t ) exp ) ( I + )) ( ) 0 t 0 0 ) ( ) 0 t = 0 0 ( )) 0 t 0 0 ( 0 t 0 0 ) ( 4t 0 0 4t ) )

69 Example Find exp(at) where A = ( ). Solution: Solving the characteristic equation 5 λ λ = 0 (5 λ)(3 λ) + 1 = 0 λ 2 8λ + 16 = 0 (λ 4) 2 = 0 λ = 4, 4 we see that A has only one eigenvalue λ = 4.

70 det(a 4I)v = 0 ( ) 1 1 v = we can find only one linearly independent eigenvector v = (1, 1) T. Thus A is not diagonalizable. To find exp(at), we may take ( ) 1 1 T = 1 0 and let (J is called the Jordan normal form of A.) J = T 1 AT ( ) 1 ( ) ( ) = ( ) ( ) ( ) = ( ) 4 1 = 0 4

71 Consider det(a 4I)v = 0 ( ) 1 1 v = we can find only one linearly independent eigenvector v = (1, 1) T. Thus A is not diagonalizable. To find exp(at), we may use the so called generalized eigenvector. A vector v is called a generalized eigenvector of rank 2 associated with eigenvalue λ if it satisfies { (A λi)v 0, (A λi) 2 v = 0.

72 Now if we take v 0 = then ( 1 0 ), v 1 = (A 4I)v 0 = ( ) ( 1 0 ) = ( 1 1 ), ( ) ( ) ( ) (A 4I)v 0 = = 0, 1 1 ( 0 ) 1 ( ) (A 4I) 2 v 0 = (A 4I)v 1 = = So v 0 is a generalized eigenvector of rank 2 associated with eigenvalue λ = 4.

73 We may let and T = [v 1 v 0 ] = ( ), J = T 1 AT ( ) 1 ( ) ( ) = ( ) ( ) ( ) = ( ) 4 1 = 0 4 (J is called the Jordan normal form of A.)

74 Then exp(at) = exp(tjt 1 t) = T exp(jt)t 1 ( ) ( 1 1 e 4t te = 4t ) ( e 4t 1 1 ( e = 4t + te 4t te 4t ) te 4t e 4t te 4t ) (By above Example)

75 Exercise Find exp(at) for each of the following matrices A. ( ) ( )

76 Answer ( 1 1 3e 5t + 4e 2t 3e 5t 3e 2t 7 4e 5t 4e 2t 4e 5t + 3e 2t ( ) t t e 2t t 1 t t t 6t t e 3t te 3t 1 2 t2 e 3t 0 e 3t te 3t 0 0 e 3t )