LinearAlgebraPPT課件

1、1Chapter Overview第1頁(yè)/共26頁(yè)2Chapter 6 EigenvaluesChapter 6 Eigenvalues We consider the problem of factoring an n n matrix A into a product of the form XDX-1, where D is diagonal; We will give a necessary and sufficient condition for the existence of such a factorization and look at a number of example

2、s.第2頁(yè)/共26頁(yè)3 If 1, 2, , k are distinct eigenvalues of an n n matrix A with corresponding eigenvectors x1, x2, , xk, then x1, x2, , xk are linearly independent.6.3 Diagonalization6.3 Diagonalization第3頁(yè)/共26頁(yè)4Let r = dim(Span(x1, x2, xk), and assume that r k.( Implies that there are at most r linearly i

3、ndependent vectors in x1, x2, xk ). Assume that x1, x2, xr are linearly independent without loss of generality.Since x1, x2, xr, xr+1 are linearly dependent, there exists scalars c1, , cr, cr+1 not all zeros such thatc1x1+ + crxr + cr+1xr+1 = 0(1)6.3 Diagonalization6.3 Diagonalization第4頁(yè)/共26頁(yè)5(2)Not

4、e that cr+1 cannot be 0 (if cr+1 =0, then x1, , , xr are l.d.). So cr+1xr+1 0, and hence c1, , , cr cannot all be 0.Multiplying both sides of (1) by A, we havec1Ax1+ + crAxr + cr+1Axr+1 = 0 c11x1+ + crrxr + cr+1r+1xr+1 = 0.Multiplying (1) by r+1 and subtracting it from (2), we obtainc1(1 r+1)x1+ + c

5、r(r r+1)xr = 0.6.3 Diagonalization6.3 Diagonalization第5頁(yè)/共26頁(yè)6Since 1, 2, , k are distinct eigenvalues, then x1, , , xr becomes linearly dependent.So it is a contradiction to the assumption thatr = dim Span(x1, , , xk) k.Hence r = k and x1, , , xk are linearly independent.6.3 Diagonalization6.3 Diag

6、onalization第6頁(yè)/共26頁(yè)7An n n matrix A is said to be if there exists a nonsingular matrix X and a diagonal matrix D such thatX-1AX = Dwe say that X A. If A is diagonalizable by X, the column vectors of X areeigenvectors of A, and the diagonal elements of D arethe eigenvalues of A. The diagonalizing mat

7、rix X is not unique.6.3 Diagonalization6.3 Diagonalization第7頁(yè)/共26頁(yè)86.3 Diagonalization6.3 DiagonalizationAn n n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors.第8頁(yè)/共26頁(yè)9Suppose that A has n linearly independent eigenvectors x1, x2 , , xn . Let i be the eigenvalue

8、of A corresponding to xi for each i (some of the is may be equal). Let X = (x1, x2, , , xn), then jxj is the jth column vector of AX. Therefore,1211221212(,)(,)(,)xxxxxxx xxnnnnnAXAAAXD6.3 Diagonalization6.3 DiagonalizationSince X has n linearly independent column vectors, it follows that X is nonsi

9、ngular and hence.1DXAX第9頁(yè)/共26頁(yè)10#Conversely, suppose that A is diagonalizable, then there exists a nonsingular matrix X such that AX = XD.If x1, x2, , , xn are the column vectors of X, thenThus, for each j, j is an eigenvalue of A and xj ( 0) is an eigenvector belonging to j.Since X is nonsingular,

10、A has n linearly independent engienvectors.6.3 Diagonalization6.3 Diagonalization()xxjjjjjjAd第10頁(yè)/共26頁(yè)11 If A is diagonalizable, then the column vectors of the diagonalizing matrix X areeigenvectors of A the diagonal elements of D are the correspondingeigenvalues of A The diagonalizing matrix X is n

11、ot unique. Reorderingthe columns of a given diagonalizing matrix X ormultiplying them by nonzero scalars will produce anew diagonalizing matrix.6.3 Diagonalization6.3 Diagonalization第11頁(yè)/共26頁(yè)12 If A is n n and A has n distinct eigenvalues, then Ahas n linearly independent eigenvectors, and henceA is

12、 diagonalizable. If the eigenvalues are not distinct, then A may or maynot be diagonalizable depending on whether A has nlinearly independent eigenvectors. A has n linearly independent eigenvectors ifdimension of N(A- I) = multiplicity of for all repeated eigenvalues.6.3 Diagonalization6.3 Diagonali

13、zation第12頁(yè)/共26頁(yè)13 If A is diagonalizable, then A can be factored into a product XDX-1. Therefore, A2 = (XDX1)(XDX1) = XD2X1in general,6.3 Diagonalization6.3 Diagonalization111200kkkkknAXD XXX第13頁(yè)/共26頁(yè)14Factor the matrix A into XDX-1, where D is diagonal:6.3 Diagonalization6.3 Diagonalization2325A第14

14、頁(yè)/共26頁(yè)15The eigenvalues of A are 1=1 and 2= 4. Corresponding to 1 and 2, we have eigenvectors x1=(3,1)T and x2=(1,2)T. Let3 11 2Xthen1121233 1101 3251 2045XAXand1213 11023551 204132555XDXA6.3 Diagonalization6.3 Diagonalization第15頁(yè)/共26頁(yè)16Factor the matrix A into XDX-1, where D is diagonal:6.3 Diagona

15、lization6.3 Diagonalization3122 02211A第16頁(yè)/共26頁(yè)176.3 Diagonalization6.3 DiagonalizationFactor the matrix A into XDX-1, where D is diagonal:1 10 1A第17頁(yè)/共26頁(yè)18 If an n n matrix A has fewer than n linearly independent eigenvectors, we say that A is e. A defective matrix is not diagonalizable.The matrix

16、has two distinct eigenvalues 1 = 4, 2 = 3 = 2. The eigenspace of 1 is Span(e2) and the eigenspace of 2 and 3 is Span(e3). A is a defective matrix.6.3 Diagonalization6.3 Diagonalization2 0 00 4 01 0 2A第18頁(yè)/共26頁(yè)196.3 Diagonalization6.3 Diagonalization Consider the eigenvectors of the matrices200200040

17、and140102362AB 第19頁(yè)/共26頁(yè)20eigenvalues: 1=4, 2= 3=2eigenspaces: 1:e2, 2, 3:e3 =2 has geometric multiplicity 1 1=4, 2= 3=2 1:x1=(0,1,3)T, 2, 3:x2=(2,1,0)T, e3 geometric multiplicity 26.3 Diagonalization6.3 Diagonalization第20頁(yè)/共26頁(yè)21 Given a scalar a, the exponential ea can be expressed in terms of a p

18、ower series Given an n n matrix A, we can define the matrix exponential eA in terms of the convergent power series In case of a diagonal matrix:6.3 Diagonalization6.3 Diagonalization231112!3!aeaaa 23112!3!AeIAAA12nD1211111lim2!1!lim1!nDmmmkkmmknkeIDDDmekek第21頁(yè)/共26頁(yè)22 If the n n matrix A is diagonali

19、zable, then2311112!3!ADeXIDDDXXe X1for1,2,.kkAXD Xkcompute eA for2613A11211110131,0( 2,1) ,( 3,1)1223013326611121 320TTADAXDXeeeeXe Xeee xx6.3 Diagonalization6.3 Diagonalization第22頁(yè)/共26頁(yè)23 The matrix exponential can be applied to the The solution to the initial value problem is simply If A is a diagonal matrix:6.3 Diagonalization6.3 Dia