1、會計學1LinearAlgebra第一頁，編輯于星期六：十八點 二十九分。2Chapter Overview第1頁/共26頁第二頁，編輯于星期六：十八點 二十九分。3Chapter 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 existenc

2、e of such a factorization and look at a number of examples.第2頁/共26頁第三頁，編輯于星期六：十八點 二十九分。4 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/共26頁第四頁，編輯于星期六：十八點 二十九分

3、。5Let r = dim(Span(x1, x2, xk), and assume that r k.( Implies that there are at most r linearly independent 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 z

4、eros such thatc1x1+ + crxr + cr+1xr+1 = 0(1)6.3 Diagonalization6.3 Diagonalization第4頁/共26頁第五頁，編輯于星期六：十八點 二十九分。6(2)Note 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

5、+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+ + cr(r r+1)xr = 0.6.3 Diagonalization6.3 Diagonalization第5頁/共26頁第六頁，編輯于星期六：十八點 二十九分。7Since 1, 2, , k are distinct eigenvalues, then x1, , , xr becomes linearly dependent.So it is a contr

6、adiction to the assumption thatr = dim Span(x1, , , xk) k.Hence r = k and x1, , , xk are linearly independent.6.3 Diagonalization6.3 Diagonalization第6頁/共26頁第七頁，編輯于星期六：十八點 二十九分。8An 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

7、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 matrix X is not unique.6.3 Diagonalization6.3 Diagonalization第7頁/共26頁第八頁，編輯于星期六：十八點 二十九分。96.3 Diagonalization6.3 DiagonalizationAn n n matrix A is

8、 diagonalizable if and only if A has n linearly independent eigenvectors.第8頁/共26頁第九頁，編輯于星期六：十八點 二十九分。10Suppose that A has n linearly independent eigenvectors x1, x2 , , xn . Let i be the eigenvalue of A corresponding to xi for each i (some of the is may be equal). Let X = (x1, x2, , , xn), then jxj

9、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 nonsingular and hence.1DXAX第9頁/共26頁第十頁，編輯于星期六：十八點 二十九分。11#Conversely, suppose that A is diagonalizable, the

10、n 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, A has n linearly independent engienvectors.6.3 Diagonalization6.3 Diagonalization(

11、)xxjjjjjjAd第10頁/共26頁第十一頁，編輯于星期六：十八點 二十九分。12 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 not unique. Reorderingthe columns of a given diagonalizing mat

12、rix X ormultiplying them by nonzero scalars will produce anew diagonalizing matrix.6.3 Diagonalization6.3 Diagonalization第11頁/共26頁第十二頁，編輯于星期六：十八點 二十九分。13 If A is n n and A has n distinct eigenvalues, then Ahas n linearly independent eigenvectors, and henceA is diagonalizable. If the eigenvalues are

13、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 Diagonalization第12頁/共26頁第十三頁，編輯于星期六：十八點 二十九分。14 I

14、f 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/共26頁第十四頁，編輯于星期六：十八點 二十九分。15Factor the matrix A into XDX-1, where D is diagonal:6.3 Diagonalization6.3 Diagonalization2325A第

15、14頁/共26頁第十五頁，編輯于星期六：十八點 二十九分。16The 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/共26頁第十六頁，編輯于星期六：十八點 二十九分。17Factor the matrix A

16、 into XDX-1, where D is diagonal:6.3 Diagonalization6.3 Diagonalization3122 02211A第16頁/共26頁第十七頁，編輯于星期六：十八點 二十九分。186.3 Diagonalization6.3 DiagonalizationFactor the matrix A into XDX-1, where D is diagonal:1 10 1A第17頁/共26頁第十八頁，編輯于星期六：十八點 二十九分。19 If an n n matrix A has fewer than n linearly independent

17、 eigenvectors, we say that A is e. A defective matrix is not diagonalizable.The matrixhas 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/共26頁第十九頁，

18、編輯于星期六：十八點 二十九分。206.3 Diagonalization6.3 Diagonalization Consider the eigenvectors of the matrices200200040and140102362AB 第19頁/共26頁第二十頁，編輯于星期六：十八點 二十九分。21eigenvalues: 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 multi

19、plicity 26.3 Diagonalization6.3 Diagonalization第20頁/共26頁第二十一頁，編輯于星期六：十八點 二十九分。22 Given a scalar a, the exponential ea can be expressed in terms of a power 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 Di

20、agonalization6.3 Diagonalization231112!3!aeaaa 23112!3!AeIAAA12nD1211111lim2!1!lim1!nDmmmkkmmknkeIDDDmekek第21頁/共26頁第二十二頁，編輯于星期六：十八點 二十九分。23 If the n n matrix A is diagonalizable, 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/共26頁第二十三頁，編輯于星期六：十八點 二十九分。24 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