(完整版)矩阵的特征值与特征向量-英文文献

1、6.2 Definitions and examplesDEFINITION6.1.1 (Eigenvalue,eigenvector) Let A be a complex square matrix. Then ifis a complex number and X a non ero complex column vector satisfying AX X , we call X an eigenvector of A, while is called an eigenvalue of A. We also say that X is an eigenvector correspond

2、ing to the eigenvalue ?.So in the above example P1 and P2 are eigenvectors corresponding to 1 and 2 ,respectively. We shall give an algorithm which starts from the eigenvalues ofa htAand constructs a rotation matrix A such that P AP is diagonal.hbAs noted above, if is an eigenvalue of an n n matrix

3、A, with corresponding eigenvector X , then (A In)X 0, with X 0, so det(A In) 0 and there are at most n distinct eigenvalues of A.Conversely if det(A I n)0 , then (A I n )X 0 has a non-rivial solution X andso， is an eigenvalue of A with X a corresponding eigenvector.DEFINITION6.1.2 (Characteristicpol

4、ynomial,equation)The polynomial det(A In) is called the characteristic polynomial of A and is oftendenoted by chA( ). The equation det(A In)0 is called the characteristic equation ofA. Hence the eigenvalues of A are the roots of the characteristic polynomial of A.abFor a 2 2 matrix A, it is easily v

5、erified that the characteristic polynomial iscd2(traceA) det A, where traceA a dis the sum of the diagonal elements of A.EXAMPLE6.2.1 Find the eigenvalues of A21and find all eigen-vectors.Solution. The characteristic equation of Ais430, or( 1)(3)0.Hence 1 or 3. The eigenvector equation(AIn)X 0 reduc

6、es toor12Taking 1 giveswhich has solution x1 are the vectorsTaking 3 giveswhich has solution xsponding to(2 )x y 0x (2 )y 0y , y arbitrary. Consequently the eigenvectors corresponding towith y 0.y , y arbitrary. Consequently the eigenvectors corre-3 are the vectorsy , with yy0.Our n ext result has w

7、ide applicability:THEOREM6.2.1 Let A be a 2 2 matrix having distinct eigenvalues1 and 2 and corresponding eigenvectors X1 and X2. Let P be the matrix whose columns are X1 andX2, respectively. Then P is non-singular and P 1APProof. Suppose AX11X1 and AX22X2. We show that the system of homogeneousequa

8、ti onsxX1 yX20has only the trivial soluti on. Then by theorem 2.5.10 the matrix P% . x2 is non-singular.So assumexX1yX20.(6.3)Then A(xX1yX2) A0 0, so x(AX1)y(g)0. Hencex 1X1y 2X20.(6.4)Multipl ying equati on 6.3 byand subtract ing from equati on 6.4 gives1(21) yX 20.Hence y 0, as ( 21)0 and X20. The

9、n from equation 6.3, xX10and hence x 0.x X202Then the equati onsAX11X1 and AX22X2 giveAPA xv x2Ax仁Ax20X1. X2EXAMPLE6.2.2 Let A2 112 be the matrix of example 6.2.1. Then1and X2are eigenvectors corresponding to eigenvalues11 and 3, respectively. Hence if P1 1 1,we have P AP1 11 0There are two0 32Pdiag

10、 ( 1, ?)Pan dAn(P 0P 1)nP10 P1immediate applications of theorem 6.1.1.The first is to the calculation of An: If1P AP diag ( 1, 2), then AThe second application is to solving a system of linear differential equationswhere Adxdtax bydydtxc dya bis a matrix of real or complex numbers and x and y c dare

11、 functions of t. The system can be written in matrix form as X AX ,where XxanddxdtdydtWe make the substitution XPY , where Y.Then 为 and % are also functions x2of tandX? ? 1PY AX A(PY), soY (P AP)YHencex11x1 and y11y1.These differential equations are well nown to have the solutions xixi (0) e lt and唱

12、yiyOeJ where 论(0) is the value of 论 when t 0.kx, where k is a constant, thend / ktkt kt dy kt kt.(e x) ke x eke x e kx 0.dtdtHence e ktx is constant, so e ktx e k0x(0)x(0) . Hence x x(0)e kt.HoweverX1(0)%(0)p 1 x(0)y(0),so this determines x1(0) and y1(0) in11terms of x(0) and y(0) . Hence ultimately

13、 x and y are determined as explicit functions of t, using the equation X PY .23EXAMPLE6.1.3 Let A. Use the eige nvalue method to45roots 11 and 22. We find corresponding eigenvectors X1derive an explicit formula forAn and also solve the system of differentialequati onsdx2x 3y dtdy dt4x5y,given x 7 an

14、d y 13 when t 0.Solution. The characteristic polynomial of A.3is32 which has distinet3 1 31and X2. Hence if P, we haveP AP diag ( 1, 2)4 1 4An (Pdiag( 1, 2)P 1)n Pdiag ( 1)n,( 2)n)P 113 ( 1)n0431 40( 2)n 11n 131043Hence( 1)n140 2131)n3 2n4 2n(1)n3 2n4 2n3 3 2n3 4 2nTo solve the differential equation

15、 system, make the substitution X PYThe nx 捲 3yy x1 4y1. The system the n becomesX1xi? so y2y1为 x1 (0)e t andY1%(o)e2t.Now .%(0)门1 x(0)43 711t2tP.S0X111e andY12e , Hence%（o）y(0)11136为11et3(6e 2t)11et2t18e , y111e t4(6e2t)11e t24e For amore complicated example we solve a system of in homoge neous recu

16、rre nee relati ons.EXAMPLE6.2.4 Solve the system of recurre nee relatio nsX 2xn yn 1 yn 1Xn 2yn 2 given that x0 0 and y00 .Solution. The system can be written in matrix form asX n 1AXn B ,2 1 1Where Aand B.1 2 2It is the n an easy in duct ion to prove thatXnAnX0 (An 1 L A I2)B.(6.5)Also it is easy t

17、o verify by the eige nvalue method thatAn 113n13n 1u $v2 1 3n 1 3n 22111 1where Uand V.Hence111 1j.(3n1 LAn 1 L I2Xn (R V)(3n4Then equati on 6.5 giveswhich simplifies toXn2nyn2n1 4 ?5 3n.4Hence xn (2n 13n)；4 andyn(2n53n),4.1REMARK6.2.1 If (A 12) existed (that is, if det(A 12)0 , orequivalently, if 1

18、 is not an eigenvalue of A), then we could have used the formulan 1n1An 1 L A I2(AnI2)(A I2) 1.(6.6)However the eigenvalues of A are 1 and 3 in the above problem, so formula 6.6 cannot be used there.Our discussion of eigenvalues and eigenvectors has been limited to 2 2 matrices. The discussion is mo

19、re complicated for matrices of size greater than two and is best left to a second course in linear algebra. Nevertheless the following result is a useful generalization of theorem 6.2.1. The reader is referred to 28, page 350 for a proof.THEOREM6.2.2 Let AA be an n n matrix having distinct eigenvalues 1丄 nandcorresponding eigenvectors X1,LXn. Let P be the matrix whose columns are respectively X1,L Xn. Then P is noningular and100010200P 1APMMMM00LnAno ther useful result which cover