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Transcript

Chapter 11 The Eigenvalue Problem 11.1 Introduction For the problem (1) Ax = lx ●S1 where A is a given n x n matrix, x is an unknown n x 1 vector, and l is an unknown scalar. If we re-express Eq. (1) as Ax = lIx (where I is an n x n identity matrix), then (A – lI)x = 0 (2) Eq. (2) is consistent because it necessarily admits the “trivial” solution x = 0. However, out interest in Eq. (2) shall be in the search for nontrivial solutions, and we anticipate that whether or not nontrivial solutions exist will depend upon the value of l. 1 Thus, the problem of interest is as follows: given the n x n matrix A, find the value(s) of l (if any) such that Eq. (2) admits nontrivial solutions, and find those nontrivial solution. The latter is called the eigenvalue problem. The l’s that lead to nontrivial solutions for x are called the eigenvalues, and the corresponding nontrivial solutions for x are called the eigenvectors. 11.2 Solution procedure and applications 11.2.1 Solution and applications The eigenvalue problem (A-lI) x = 0 (1) has the unique trivial solution x = 0 if det(A-lI) ≠0, and nontrivial solutions (in addition to the trivial solution) if and only if (2) det(A-lI) = 0 2 The latter is not a vector or matrix equation; it is an algebraic equation in l, known as the characteristic equation corresponding to the matrix A, and its left-hand side is an nth degree polynomial known as the characteristic polynomial. Let us consider Eq.(2) to have been solved for the eigenvalues l1,l2,…,lk (1 ≦ k ≦ n). Next, set l = l1 in Eq. (1). Since det(A-l1I) = 0, it is guaranteed that (A-l1I)x = 0 will have nontrivial solutions. We can find those solutions by Gauss elimination, and we designate them as e1, where the letter e is for eigenvector. The e1 solution space is called the eigenspace corresponding to the eigenvalue l1. Example 1 Determine all eigenvalues and eigenspaces of 2 2 1 A 1 3 1 1 2 2 (3) 3 2-l 2 1 det(A-l I) 1 3-l 1 l 3 7l 2 11l 5 1 2 2-l (4) =(l -5)(l -1)2 0 so the eigenvalues of A are l1 = 5 and l2 = 1, with l2 = 1 called a repeated eigenvalue. Next, find the eigenspaces: l1 = 5: Then (A - l1I)x = 0 becomes 2-5 2 1 x1 -3 2 1 x1 -3 2 1 x1 0 1 3-5 1 x 1 -2 1 x 0 1 -1 x 0 2 2 2 1 2 2-5 x3 1 2 -3 x3 0 0 0 x3 0 (5) The solution is x3 =a (arbitrary), x2 = a, x1 = a, using e in plane of x, a 1 e a a 1 a 1 (7) 1 Thus, the eigenspace corresponding to l1 = 5 is span{ 1 } 4 1 l2 = 1: Then (A - l2I)x = 0 becomes 2-1 2 1 x1 1 2 1 x1 1 2 1 x1 0 1 3-1 1 x 1 2 1 x 0 0 0 x 0 2 2 2 1 2 2-1 x3 1 2 1 x3 0 0 0 x3 0 (5) The solution is x3 =b (arbitrary), x2 = g (arbitrary), x1 = -b2g so b 2g 1 2 e g b b 0 g 1 1 0 (7) Thus, the eigenspace corresponding to l2 = 1 is 1 2 span{ 0 , 1 } 1 0 5 Example 3 Solve the coupled differential equations x x 4 y y x y (15) rt rt x ( t ) q e , y ( x ) q e Seek x, y in the form 1 2 (16) where q1, q2, r are constants that are to be determined. Putting Eq.(16) into (15) gives rq1ert q1ert 4q2ert rq2ert q1ert q2ert expressing the result in matrix form, gives q1 1 4 q1 1 1 q r q 2 2 which is an eigenvalue problem with l = r. ●S2 6 Proceeding as above, we obtain these eigenvalues and eigenspaces: 2 -2 l1 3, e1 a ; l2 1, e2 b 1 1 (19) 2 3t -2 t x(t ) a e ; x(t ) b e 1 1 (20) or in scale form x(t ) 2a e3t 2b et y (t ) a e b e 3t t (22) 7 11.3 Symmetric Matrices 11.3.1 Eigenvalue problem of Ax = l x Theorem 11.3.1 Real eigenvalues If A is symmetric (AT = A), then all of its eigenvalues are real. Theorem 11.3.2 Dimension of eigenspace If an eigenvalue l of a symmetric matrix A is of multiplicity k, then the eigenspace corresponding to l is of dimension k. Theorem 11.3.3 Orthogonality of eigenvectors If A is symmetric, the eigenvectors corresponding to distinct eigenvalues are orthogonal. Let ej and ek be eigenvectors corresponding to distinct eigenvalues lj and lk, respectively. Thus, Aej = lj ej and Aek=lkek (1a, 1b) 8 Recall x‧y = xTy and (AB)T =BTAT for any matrices A and B that are conformable for multiplication. Then, if we dot ek into each side of (1a) and dot each side of (1b) into ej, we obtain ek‧(Aej )= ek‧(lj ej) and Aek‧ej=(lkek)‧ej e k (Ae j )=e k (λ je j ) Aek e j =(lk e k ) e j eTk Ae j =λ jeTk e j (Aek )T e j =lk eTk e j (2) eTk A T e j =lk eTk e j But AT = A by assumption so if we subtract the bottom equations on the left and right of the vertical divider, we obtain 0 = (λ j -lk )eTk e j (3) Since lj and lk were assumed to be distinct so it follows from (3) that eTkej = 0. Thus, ek‧ej = 0, as claimed. 9 Theorem 11.3.3 Orthogonal basis If an n x n matrix A is symmetric, then its eigenvectors provide an orthogonal basis for n-space. Example 2 Free vibration of a two-mass system Consider the system of two masses subjected to forces f1(t) and f2(t) and restrained laterally by springs and supported vertically by a frictionless table as shown in Fig. 1. m1 x1 (k1 k12 ) x1 k12 x2 f1 (t ) m2 x2 k12 x1 (k2 k12 ) x2 f 2 (t ) (11) 10 Let m1 = m2 = k1 = k12 = k2 = 1 and consider free vibration such that f1(t) = f2(t) = 0. Then Eq. (11) becomes x1 2 x1 x2 0 x2 x1 2 x2 0 Seek (12) x1 (t ) q1elt x2 (t ) q2elt (13) On physical grounds we expect the solution to be a vibration, and it seems more sensible to seek x1 (t ) q1 sin(t ) x2 (t ) q2 sin(t ) (14) Putting Eq. (14) into (12) gives 2 q1 2q1 q2 0 2 q2 q1 2q2 0 11 or, equivalently, q 2 -1 q1 2 1 -1 2 q q 2 2 (15) which is a matrix eigenvalue problem Aq = lq (16) with l =2 as the eigenvalue. Solving for the eigenvalues and eigenspaces, we have 1 1 (17) l1 1, e1 a ; l2 3, e2 b 1 1 Each eigenpair gives us a solution of the form (14). The first gives = (l1)1/2 = 1, and x1 (t ) 1 (18) x a sin(t 1 ) 1 x2 (t ) 12 The second gives = (l2)1/2 = 31/2, and x1 (t ) 1 x b sin( 3t 2 ) 1 x2 (t ) (19) where a, b, 1, and 2 are arbitrary and satisfies Eqs. (18) and (19). Since Eq. (12) is linear and homogeneous, it follows that the linear combination x1 (t ) 1 1 x (t ) a 1 sin(t 1 ) b 1 sin( 3t 2 ) 2 (20) is also a solution. Returning to scalar form, we have x1 (t ) a sin(t 1 ) b sin( 3t 2 ) x2 (t ) a sin(t 1 ) b sin( 3t 2 ) (21) Each eigenpair defines a vibration “mode”, the eigenvalue gives the vibrational frequency ( = l1/2) and the eigenvector gives the mode shape or configuration. The frequencies 13 are called the eigenfrequencies, or natural frequencies. The first term in Eq. (20) is called the low mode because it occurs at the lower of the two natural frequencies, and the second term is called the high mode. ●S3 Case 1: x1 (0) 0, x2 (0) 0, x1(0) 1, x2 (0) 1 b 0, a 1, 1 0 Case 2 : x1 (0) 1, x2 (0) 1, x1(0) 0, x2 (0) 0 a 0, b 1, 2 2 Case 3: x1 (0) 1, x2 (0) 0, x1(0) 0, x2 (0) 1, motion containing both modes Case 1(b =0) Case 2(a = 0) Case 3(mixed) 14 11.4 Diagonalization Find the solution of Ax = c is tedious, where A is n x n matrix, if n is large; find the result of Am is tedious if m Is large; find the solution of (1) x’(t) = Ax(t) is generally tedious. But it is simple if A is diagonal. Find a matrix Q such that the variables x1 ,..., xn can be converted to x1 ,..., xn . and x Qx x Qx (2) where Q is a constant matrix and can be expressed as x1 q11 xn qn1 q1n x1 qnn xn (3) 15 Qx AQx (4) Choosing Q to be invertible. Then multiplying (4) by Q-1 gives Q1Qx Q 1 AQx or x Q AQx 1 (5) Given a matrix A, the idea is to find a Q matrix so that Q 1 AQ D (6) is diagonal because then the differential equations within Eq. (5) will be uncoupled. If there does exist such a Q we say that A is diagonalizable and that Q diagonalizes A. Two questions:(1) Given A, does there exist such a Q? (2) How do we find it? 16 Theorem 11.4.1 Diagonalization Let A be n x n 1. A is diagonalizable if and only if it has n LI eigenvalues. 2. If A has n LI eigenvectors e1,…,en and we make these the columns of Q, so that Q =[e1,…, en], then Q-1AQ = D is diagonal and the jth diagonal element of D is the jth eigenvalue of A. Proof: Let Q =[e1,…,en]. (1)Prove: If A is diagonalizable, then it has n LI eigenvectors. If A is diagonalizable, then there is an invertible matrix Q such that d1 0 0 0 d 0 2 (7) Q 1 AQ D 17 0 d n Pre-multplying both sides of Eq. (7) by Q gives AQ = QD q11 AQ QD qn1 d1q11 d1qn1 d1 0 q1n 0 d 2 qnn 0 d n q1n dq 1 1 d n qnn 0 0 dn (8) dnqn where the vector qj simply denotes the jth column of Q. Alternatively AQ = A[q1,q2,…, qn] = [Aq1, Aq2,…,Aqn] Comparing (8) and (9), we have Aq1=d1q1,…, Aqn=dnqn (9) (10) 18 (2) If A has n LI eigenvectors, then it is diagonalizable. Let Q to be made up of columns which are the eigenvectors of A, so Q = [e1, …, en]. AQ [Ae1 , l1e11 l1en1 , Ae n ] [l1e1 , ln e1n e11 ln enn en1 , ln e n ] l1 0 e1n 0 l2 enn 0 0 0 QD ln (11) Q is invertible since its columns are LI, thus Q-1AQ Q-1QD=D 19 Theorem 11.4.2 Distinct eigenvalues, LI eigenvectors If n x n matrix A has distinct eigenvalues l1,…, ln, then the corresponding eigenvectors e1,…, en are LI. Theorem 11.4.3 Diagonalizability If an n x n has n distinct eigenvalues, then it is diagonalizable. Theorem 11.4.4 Symmetric matrices Every symmetric matrix is diagonalizable. 20 Problems for Chapter 11 Exercise 11.2 1. (a)、(b) 3.(a)、(e)、(i)、(l) 5.(a)、(c)、(d); 6.(b)；11；16.(c); 18.(b); Exercise 11.4 1.(a)、(c)、(g)、 (j)、 3.(c)； 4; Exercise 11.3 1.(a)、(d)、(g) 、(i) 10. 21