線性代數(shù) chapter5 Eigenvalues and Eigenvectors

1、CHAPTER 5Eigenvalues and Eigenvectors IntroductionEigenvectors and Eigenvaluesnuseful through pure and applied mathematics. nused to study difference equations and continuous dynamical systemsnprovide critical information in engineering designnarise naturally in such fields as physics and chemistryC

2、hapter 5 Eigenvalues and Eigenvectors p 5.1 Eigenvectors and Eigenvaluesp 5.2 The Characteristic Equationp 5.3 Diagonalizationp 5.4 Eigenvectors and Linear Transformationsp 5.5 Complex Eigenvaluesp 5.6 Discrete Dynamical Systems5.1 Eigenvectors and Eigenvalues1 65 23-26-5p u is an eigenvector corres

3、ponding to an eigenvalue (-4)p v is not an eigenvector of A5.1 Eigenvectors and Eigenvalues,5.1 Eigenvectors and Eigenvalues333322221111000aaaaaapExample： Let By Theorem 1, The eigenvalues of A are 3, 0 an 2. The eigenvalues of B are 4 and 1. What does it mean for a matrix A to have a eigenvalue of

4、0?pA matrix A has an eigenvalue of 0, if and only if the equation has a nontrivial solution. It means A is not invertible. So the Invertible Matrix Theorem can be continued: 0 is an eigenvalue of A if and only if A is not invertible.5.1 Eigenvectors and EigenvaluesProof ：Suppose v1,vr is linearly de

5、pendent. Then there exist scalar c1,cp such thatMultiplying both sides by A and using Avk = kvk for each k(1)(2) (1) p+1 - (2) Since v1,vp is linearly independent, the weights in (3) are all zero. But the factors i- p+1 are nonzero, because the eignvalues are distinct. Hence ci = 0. It is opposite t

6、o the (1) So v1,vr is linearly independent.(3)p Eigenvectors and Difference EquationsIf A is an n*n matrix, then Xk+1 = AXk (k=0,1,2,) is a recursive description of a sequence xk in Rn. A solution of this equation is an explicit description of xk whose formula for each xk does not depend directly on

7、 A or on the preceding terms in the sequence other than the initial trem x0 How to solve? Take an eigenvector x0 and its corresponding eigenvalue and let xk= kx0 ( k=1,2,) This sequence works, because Axk=A(kx0 )= k (Ax0 )= k ( x0 )= k+1 x0 =Xk+1 Linear Combinations of solutions of the form above ar

8、e solutions, too.5.2 The Characteristic Equationp1. Determinantsp2. The Characteristic Equationp3. Similarityp4. Application to Dynamical SystemsReview The same method can be uesd to solve nn matrices5.2 The Characteristic Equationp1. Determinants Before turning to larger matrices, we summarize the

9、properties of determinants needed to study eigenvalues.5.2 The Characteristic EquationpExample ：Compute det A forpSolution： Or5.2 The Characteristic Equation So Formula for the determinant shows that A is invertible if and only if det A is nonzero. It can be added to the Invertible Matrix Theorem.5.

10、2 The Characteristic Equationp Theorem 3 Properties of Determinants Let A and B be nn matrices. a. A is invertible if and only if det A 0. b. det AB = (det A)(det B). c. det AT = det A. d. If A is triangular, then det A is the product of the entries on the main diagonal of A. e. A row replacement op

11、eration on A does not change the determinant. A row interchange changes the sign of the determinant. A row scailing also scales the determinant by the same scalar factor. 5.2 The Characteristic Equationp2. The Characteristic Equation By Theorem 3(a), we can use a determinant to determine when a matr

12、ix A - I is not invertible. The scalar equation det (A - I) = 0 is called the characteristic equation of A. A scalar is an eigenvalue of an nn matrix A if and only if satisfies the characteristic equation351Some Concept:p1. If A is an nn matrix, then det (A- I)is a polynomial of degree n called the

13、characteristic polynomial of A.p2. eg. In the upper example, the eigenvalue 5 has multiplicity 2 because (-5) occurs two times as a factor of the characteristic polynomial. pExample： The characteristic polynomial of a 66 matrix is Find the eigenvalues and their multiplicities. Solution： The eigenval

14、ues are 0 (multiplicity 4), 6 (multiplicity 1), -2 (multiplicity 1).p3. SimilarityExample： Matrices are similar. To see why, let P = ,and B= Note thatP-1=5.2 The Characteristic EquationBecauseSo we have5.2 The Characteristic Equationeg.5.2 The Characteristic Equationp4. Application to Dynamical Syst

15、emsExample： Let . Analyze the long-term behavior of a dynamical system by xk+1 = Axk (k = 0,1,2,) with x0 = Sol.97.05.03.95.A4 .6 . It is readily checked that eigenvectors corresponding to =1 and =0.92 are multiples of and Next step is write the given x0 in terms of v1 and v2. there exist weights c1

16、 and c2 such that (3)In fact Because v1 and v2 in (3) are eigenvectors of A, with Av1=v1, and Av2=.92v2, we easily compute each xk And so on. In general Using c1 and c2 from (4) This explicit formula for xk gives the solution of the difference equation xk+1 = Axk. As k, (.92)k tends to zero and xk t

17、ends to 5.3 DiagonalizationpIn many cases, we can get the eigenvalue and eigenvector of a matrix A form A= PDP-1.pIn this section, we learn how to compute Ak quickly for large values of k. It is a fundamental idea in several applications of linear algebra.nnaaa00000022115.3 Diagonalization5.3 Diagon

18、alizationA square matrix A is diagonalizable if A is similar to a diagonal matrix, that is, if Proof ：First, if P is any n*n matrix with columns v1,vn, and if D is any diagonal matrix with diagonal entries 1,n, thenWhile Now suppose A is diagonalizable and A = PDP-1, then right-multiplying this rela

19、tion by P, we have AP = PD, In this case,(1) and (2) imply thatEquating columns, we find that Since P is invertible, its columns v1,vn must be linearly independent, Also, since these columns are nonzero, (4) shows that 1,n are eigenvalues and v1,vn are corresponding eigenvectors. Finally, given any

20、n eigenvectors v1,vn use them to construct the columns of P and use corresponding eigenvalues 1,n to construct D, by (1)-(3), AP=PD. If the eigenvectors are linearly independent, then P is invertible, and AP = PD implies that A = PDP-1.5.3 Diagonalization5.3 Diagonalization The Theorem 6 provides a

21、sufficient condition for a matrix to be diagonalizable.Proof：Let v1,vn be eigenvectors corresponding to the n distinct eigenvalues of a matrix A, then v1,vn is linearly independent, by Theorem 2 (Section 5.1) Hence A is diagonalizable, by Theorem 5.5.3 Diagonalization261If an nn matrix A has the eig

22、envalues which are not distinct, we can use the Theorem 7 to build P that makes P automatically invertible.5.3 DiagonalizationpExample : Diagonalize the following matrix, if possible.pSolution : A is a triangular matrix, the eigenvalues are 5 and -3, each with multiplicity 2. Using the method of Sec

23、tion 5.1, By Theorem 7, the set v1,v4 is linearly independent. So the matrix P = v1 v4 is invertible, and A = PDP-1, where5.4 Eigenvectors and Linear Transformationsp1. The Matrix of a Linear Transformationp2. Linear Transformations from V into Vp3. Linear Transformations on Rn5.4 Eigenvectors and L

24、inear Transformationsp1. The Matrix of a Linear Transformation Suppose V - n-dimensional vector space, W - m-dimensional vector space, T - any linear transformation from V to W and C - the bases for V and W. x - the coordinate vector for any x in V (in Rn ) T(x)C - the coordinate vector of its image

25、 ( in Rm) The connection between and is easy to find. Let b1,bn be the basis for V. If then Because T is linear，then Using the basis C in W, we have=M: the matrix for T relative to the bases and CExample： Suppose = b1,b2 is a basis for V and C = c1,c2,c3 is a basis for W. Let T: VW be a linear trans

26、formation with the property thatFind the matrix M for T relative to and C.Solution：The C-coordinate vectors of the images of b1 and b2 are5.4 Eigenvectors and Linear Transformationsp2. Linear Transformations from V into V When W = V, C = , M is called the matrix for T relative to , or -matrix for T,

27、 denoted by T The B-matrix for T: VV satisfiespExample： The mapping T: P2P2 is a linear transformation. a. Find the B-matrix for T, when B is the basis 1,t,t2. b. Verify that for each p in P2 Solution：a. Compute the images of the basis vectors:pb. For a general polynomialWe have5.4 Eigenvectors and

28、Linear Transformationsp3. Linear Transformations on Rn Theorem 8 Diagonal Matrix Representation Suppose A = PDP-1, where D is a diagonal nn matrix. If is the basis for Rn formed from the columns of P, then D is the -matrix for the transformation x Ax.Proof：Denote the columns of P by b1, bn, so thatP

29、 is the change-of-coordinates matrix , whereExample： Define T: R2 R2 by T(x) = Ax, where Find a basis for R2 with the property that the -matrix for T is a diagonal matrix.By Theorem 8, D is the -matrix for T when = p1, p2p1 p2Solution：5.4 Eigenvectors and Linear Transformationsp Similarity of Matrix

30、 Representations If A is similar to a matrix C, with A=PCP-1, then C is the -matrix for the transformation x|Ax when the basis is formed from the columns of P. The factorization A = PCP-1 is shown below. p Conversely, if T: Rn Rn is defined by T(x) = Ax, and if is any basis for Rn, then the -matrix

31、for T is similar to A. In fact , the proof of theorem 8 show that if P is the matrix whose columns come from the vectors in B, then . Thus, the set of all matrices similar to a matrix A coincide with the set of all matrix representations of the transformation x-Ax.Example： Let The characteristic pol

32、ynomial of A is (+2)2, but the eigenspace for the eigenvalue -2 is only one-dimensional; so A is not diagonalizable. However, the basis = b1,b2 has the property that the -matrix for the transformation x|Ax is a triangular matrix called the Jordan Form of A. Find this -matrix . Sol：If P = b1, b2, the

33、n the -matrix is P-1AP, compute5.5 Complex EigenvaluesIntroductionpIf the characteristic equation of an nn matrix involves a polynomial of degree n, the equation always has exactly n roots, counting multiplicities.pComplex eigenvalues is essential to uncover “hidden” information about certain matric

34、es.5.5 Complex EigenvaluespCn - n-dimension complex space (n維復(fù)空間)p - complex scalar (complex) eigenvalue （復(fù)）特征值p x - nonzero vector x in Cn (complex) eigenvector （復(fù)）特征向量 satisfies det(A-I) = 0 if and only if Ax = x.5.5 Complex EigenvaluespDefinition： A complex scalar satisfies det(A-I) = 0 if and on

35、ly if there is a nonzero vector x in Cn such that Ax = x. We call complex eigenvalue and x a complex eigenvector corresponding to .pExample： If , then the linear transformation x -Ax on R2 rotates the plane counterclockwise through a quarter-term. The action of A is periodic, since after four quarte

36、r-term, a vector is back where it started. Obviously, no nonzero vector is mapped into a multiple of itself, so A has no eigenvectors in R2 and hence no real eigenvalues. In fact, the characteristic equation of A is 2 +1 = 0. The only roots are complex: = i and = -i . However, if we permit A to act

37、on C2 Then Thus i and -i are eigenvalues, with and pExample： Let Find the eigenvalues of A, and find a basis for eacheigenspace.pSolution：The characteristic equation of A ispFor Since 0.8-0.6i is an eigenvalue, the system has a nontrivial solution: So the solution is pChoose x2 = 5, x1 = -2-4i. So a

38、 basis for the eigenspace corresponding to is pFor , using the same way，we haveExample：One way to see how multiplication by the A in Example 1 affects pionts is to plot an arbitrary initial piont-say, x0 =(2,0)- and then to plot sucessive of this point under repeated multiplications by A, that is, p

39、lot The figure shows x0,x8 are large dotsThe smaller dots are the locations of x9,x100.5.5 Complex EigenvaluespReal and Imaginary Parts of Vectors 1. The complex conjugate of a complex vector x in Cn is the vector in Cn whose entries are the complex conjugate of the entries in x. （ Cn的復(fù)向量x的共軛向量 也是Cn

40、的向量，它的分量是x對應(yīng)分量的共軛復(fù)數(shù)。） 2. The real and imaginary parts of a complex vector x are the vectors Rex and Imx formed from the real and imaginary parts of the entries of x.pExample： suppose then If B is an m*n matrix with possibly complex entries, then denotes the matrix whose entries are the complex conju

41、gates of the entries in B. Properties of conjugates for complex numbers: pEigenvalues and Eigenvectors of a Real Matrix That Axts on Cn： If A be an nxn matrix whose entries are real. If is an eigenvalue of A and x is a corresponding eigenvectors in Cn , then Hence is also an eigenvalue of A ,with a

42、corresponding eigenvector. This shows when A is real, its complex eigenvalues occur in conjugate pairs. p Example: If , where a and b are real and not both zero, then the eigenvalues of C are Also, if then Where is the angle between the positive x-axis and the ray form (0,0) through (a, b). The angl

43、e is called the argument of Thus the transformation xCx may be viewed as the composition of a rotation through the angle and a scaling by |. 5.5 Complex EigenvaluesA rotation followed by a scalingExample: Let and let P be 2*2 real matrixAnd let because C is a pure rotation. From then5.5 Complex Eige

44、nvaluespTHEOREM 9 Let A be a real 2X2 matrix with a complex eigenvalue and an associated eigenvector v in .Then , where , andExample: The matrix has eigenvalues and 1.07 Any vector w0 in the x1x2-plane is rotated by A into another point in the plane. Any vector x0 not in the plane has its x3-coordin

45、ate multiplied by 1.07. The iterates(迭代) of the points w0= (2,0,0) and x0 = (2,0,1) under multiplication by A are show in figure 5.6 Discrete Dynamical SystemspA Predator-Prey SystempGraphical Description of SolutionspChange of VariablepComplex EigenvaluespSurvival of the Spotted Owls Assume that A

46、is diagonalizable, with n linearly independent eigenvectors, v1,vn, and corresponding eigenvalues, 1, n. For convenience, assume the eigenvectors are arranged so that | 1 | | 2 | | n | . Since V1,Vn is a basis for Rn, any initial vector x0 can be written uniquely as x0 = c1v1 + cnvn This eigenvector

47、 decomposition of x0 determines what happens to the sequence xk. The next calculation generalizes the simple case examined in Example 5 of Section 5.2. Since the vi are eigenvectors. x1 = Ax0 = c1Av1+ cnAvn = c11v1+ cnnvn In general, Xk = c1(1)kv1+cn(n)kvn (k=0,1,2)p A Predator-Prey SystemExample :

48、Denote the owl and wood rat population at time k by , where k is the time in months, Ok is the number of owls in the region studied, and Rk is the number of rats (measured in thousands). Suppose Ok+1 = (.5)Ok + (.4) Rk Rk+1 = -p*Ok + (1.1)RkWhere p is a positive parameter to be specified. The (.5)Ok

49、 in the first equation says that with no wood rats for food, only half of owls will survive each month, while the (1.1)Rk in the second equation says that with no owls as predators, the rat population will grow by 10% per month. If rats are plentiful, the (.4)Rk will tend to make the owl population

50、rise, while p*Ok measures the deaths of the rats due to predator by owlskkkROx Determine the evolution of this system when the predation parameter p is .104.Solution: When p=1.04, the eigenvalues of the coefficient matrix A to be 1=1.02 and 2=.58. corresponding eigenvectors are An initial x0 can be

51、written as x0 =c1v1 + c2v2. then for k0,As k, we have (.58)k 0. Assume c10, then, for all sufficiently large k, xk c1(1.02)Kv1, and we writeAnd so for large k,This says that eventually both entries of xk grow by a factor of almost 1.02 each month, a 2% monthly growth rate. And the entries in xk are

52、nearly in the ratio as 10 to 13. That is, for every 10 owls there are about 13 thousand rats.Example 1: illustrates two general facts about a dynamical system xk+1 = Axk in which A is n*n, its eigenvalues satisfy |1|1 and 1 |j| for j=2,.,n, and v1 is an eigenvector corresponding to 1. if x0 is given

53、 by (1), with c10, then for all sufficiently large k,So 1 determines the eventual growth rate of system.The ratio of any two entries in xk is nearly the same as the ratio of the corresponding entries in v1. 5.6 Discrete Dynamical Systemsp Graphical Description of Solutions When A is 2*2, algebraic c

54、alculations can be supplemented by a geometric description of a systems evolution. We look xk+1 = Axk as a description of what happens to an initial point x0 in R2 as it is transformed repeatedly by the mapping x|Ax. The graph of x0,x1, is called a trajectory (軌跡軌跡) of the dynamical system.Example:

55、plot several trajectories of the dynamical system xk+1 = Axk, whenSolution: The eigenvalues of A are .8 and .64, with eigenvectors and , if thenxk0, when k because (.8)k and (.64)k both approach 0 as k The origin is called an attractor(吸引子吸引子) of the dynamical system .Example: Plot several typical s

56、olution of the equation xk+1 = Axk, whereSolution: The eigenvalues of A are 1.44 and 1.2, if ,then Both eigenvalues of A are large then 1 in magnitude, and o is called a repellor(排斥子排斥子) of a dynamical system.Example:Plot several typical solution of the equation yk+1 = Dyk, whereSolution :The eigenv

57、alues of D are 2 and 0.5. if then 0 is called a saddle point (鞍點鞍點) because the origin attracts solutions from some directions and repels then in other directions. 5.6 Discrete Dynamical Systemsp Change of Variable A is a n*n matrix, which eigenvectors form a basis v1, vn for Rn. Let P=v1,vn, and let D be the diago