《结构工程中的数学方法》课程教学课件（讲稿）example_inverse_power_method

2AlgebraicproblemsinmatrixformEigenvalueProblemsGeneralinversepowermethod-examplespecialeigenvalueproblem(fromslide8la,exampleforpowermethod)-44-25-42(A-2·I)·X=0A=eigenvalues are real2-24[k][k]yTK·iterationprocedurex[k]1[k][k-1](A-q·I)instead of y=C.xwithC? starting vector x = (1,1,1)selectionofg(ascloseaspossibletothetargeteigenvalue)》target:smallest eigenvalue2,=入3=q=091aMichaelBeer,EngineeringMathematics

Michael Beer, Engineering Mathematics special eigenvalue problem (from slide 81a, example for power method) General inverse power method − example 2 Algebraic problems in matrix form ● Eigenvalue Problems 91a   − −   = −    −  424 425 242 (A Ix 0 −λ⋅ ⋅ = ) A iteration procedure eigenvalues are real starting vector = ( ) [0] T x 1,1,1 ● ● ( ) − ∞ −⋅ ⋅ = ⇒ = ⇒ = k k [k] [k] [k 1] [k] [k] [k] p [k] p q y y y A Iy x y x − = ⋅ [k] [k 1] y Cx ( ) 1 q − instead of with CA I = −⋅ » target: smallest eigenvalue λi = λ3 ⇒ q = 0 ● selection of q (as close as possible to the target eigenvalue)

2AlgebraicproblemsinmatrixformEigenvalueProblemsGeneral inverse powermethod-example (cont'd)·iterative solution[1][1]4-22.5]-41yiYi[0][1]250.5-4》k=1:A·y1Y2Y2=X=24-212.0Y3LY3[2.5]1[1][1][1]y10.50.2=y; =2.51=2.5XDY.3Y1P12.5Yp2.00.8[2][1]1.667-2144Y1Y1[2][1]250.2》k=2: A.y=x40.433Y2Y2=-2420.81.200Ly3Y31.6671[2] [2]1[2]2y0.2600.433yi=1.667==y1=1.667X：Y-P[2]1.667Yp21.2000.72091bMichaelBeer,EngineeringMathematics

General inverse power method − example (cont'd) 2 Algebraic problems in matrix form Eigenvalue Problems 91b iterative solution ⋅ = [1] [0] » k = 1: Ay x ●  − −          =− ⋅ =      −          [1] 1 2 3 424 y 1 425y 1 242y 1 ∞ == ⇒ == 1 [1] [1] y y 2.5 y y 2.5 1 1 p    ⇒= = ⋅ =             1 [1] [1] [1] p 2.5 1 1 0.5 0.2 2.5 y 2.0 0.8 y x    ⇒ =             [1] 1 2 3 y 2.5 y 0.5 y 2.0 ⋅ = [2] [1] » k = 2: Ay x  − −          =− ⋅ =      −          [2] 1 2 3 424 y 1 4 2 5 y 0.2 2 4 2 y 0.8 ∞ == ⇒ == 2 [2] [2] y y 1.667 y y 1.667 1 1 p    ⇒= = ⋅ =       2 [2] [2] [2] p 1.667 1 1 0.433 0.260 1.667 y 1.200 0.720 y x    ⇒ =             [1] 1 2 3 y 1.667 y 0.433 y 1.200 Michael Beer, Engineering Mathematics

2AlgebraicproblemsinmatrixformEigenvalueProblemsGeneral inverse powermethod-example (cont'd)iterative solution[5][5]17-24-41.698Y1yi[5][4]25_40.410》k=5:A.y=x0.243Y2=Y2-2420.7311.243Y3y3[5]=yi=1.698=1.698=VYps11.698[5]1[5]yeigenvalueeigenvectorv=x0.4100.241=x[5]1.698431.2430.732remember:A.x入xtrue solution>APD(3)2 =0.5892.0.241911[5]10.73170.589, v(3)0.241~X=[5]1.6984o0.73291cMichaelBeer,EngineeringMathematics

General inverse power method − example (cont'd) 2 Algebraic problems in matrix form Eigenvalue Problems 91c ● iterative solution » k = 5:    ⇒= = ⋅ =       5 [5] [5] [5] p 1.698 1 1 0.410 0.241 1.698 y 1.243 0.732 y x ● ● ● ( )     λ≈ = = ≈ =       5 [5] 3 3 [5] p 1 1 1 0.589, 0.241 1.698 y 0.732 v x ( )     λ = =       3 3 1 0.5892, 0.2419 0.7317 v » true solution remember: Ax x ⋅ =λ⋅ eigenvector v = x eigenvalue ⋅ = [5] [4] Ay x  − −          =− ⋅ =      −          [5] 1 2 3 424 y 1 4 2 5 y 0.243 2 4 2 y 0.731 ∞ == ⇒ == 5 [5] [5] y y 1.698 y y 1.698 1 1 p    ⇒ =             [5] 1 2 3 y 1.698 y 0.410 y 1.243 A x⋅ = A ⋅ = y ⋅x pk 1 y ⋅ pk 1 y Michael Beer, Engineering Mathematics

2AlgebraicproblemsinmatrixformEigenvalueProblemsGeneral inverse powermethod-example(cont'd)calculationof intermediateeigenvalues starting vector x - (1,1,1)selection of g (as closeas possibletothetarget eigenvalue)》target:intermediate eigenvalue ,=入2rememberthegeneralconvention2/>|22|≥|23/≥..≥|2n]= 1211>[22/≥[23 with2,=9.551(frompowermethod,slide81c)23=0.589(frompreviouscalculation,slide91c)=→ q= 0.5 (2| +2) = 5.07q=-0.5.(2/+2a)=-5.07ORpreparation of the iteration forg=5.0700-44-2-4-21-1.070525-4-4-5.07.01-3.07A-q·I=420-2-240-3.07191dMichaelBeer,EngineeringMathematics

calculation of intermediate eigenvalues General inverse power method − example (cont'd) 2 Algebraic problems in matrix form ● Eigenvalue Problems 91d starting vector = ( ) [0] T ● x 1,1,1 » target: intermediate eigenvalue λi = λ2 ● selection of q (as close as possible to the target eigenvalue) remember the general convention |λ1|>|λ2|≥|λ3|≥ . ≥|λn| ⇒ |λ1|>|λ2|≥|λ3| with λ1 = 9.551 (from power method, slide 81c) λ3 = 0.589 (from previous calculation, slide 91c) ⇒ = ⋅ λ +λ = q 0.5 ( 1 3 ) 5.07 ● preparation of the iteration for q = 5.07     − − −−−       − ⋅ =− − ⋅ =− −  − − −       4 2 4 1 0 0 1.07 2 4 q 4 2 5 5.07 0 1 0 4 3.07 5 2 4 2 0 0 1 2 4 3.07 A I OR q 0.5 =− ⋅ λ +λ =− ( 1 3 ) 5.07 Michael Beer, Engineering Mathematics

2AlgebraicproblemsinmatrixformEigenvalueProblemsGeneral inverse power method - example (cont'd)·iterativesolutionfor=5.07[1] [1]-2-4-1.07-0.381yiy1[0]5-4-3.07Y2-0.039》=1:(A-q·I)y=x=Y2D[-241-3.07LY3-0.129Ly3.17-0.381[1]1[1][1]y-0.0390.102=0.381==y,=-0.381= x[1]D0.381Yo-0.1290.339》k = 2:[2][2]-417-2-1.07-0.141Y1yi[1][2] 5-4-3.070.102-0.107(A-q.I)·y=x=Y2Y2U4-2-3.070.339-0.159Ly3[y3]0.887-0.141[2]1[2] [2]y0.159=-0.1070.673Yp,=Y3=-0.159=x[2]0.159Ypz1-0.15991eMichaelBeer,EngineeringMathematics

General inverse power method − example (cont'd) 2 Algebraic problems in matrix form Eigenvalue Problems 91e iterative solution for q = 5.07 ( −⋅ ⋅ = ) [1] [0] » k = 1: A Iy x q ● −−−          =− − ⋅ =      − −          [1] 1 2 3 1.07 2 4 y 1 4 3.07 5 y 1 2 4 3.07 y 1 ∞ = ⇒ = =− 1 [1] [1] y 0.381 y y 0.381 p 1    − ⇒ = = ⋅− =    − −   1 [1] [1] [1] p 0.381 1 1 0.039 0.102 0.381 y 0.129 0.339 y x    − ⇒ =−          −    [1] 1 2 3 y 0.381 y 0.039 y 0.129 ( −⋅ ⋅ = ) [2] [1] A Iy x q » k = 2: −−−          =− − ⋅ =      − −          [2] 1 2 3 1.07 2 4 y 1 4 3.07 5 y 0.102 2 4 3.07 y 0.339 ∞ = ⇒ = =− 2 [2] [2] y 0.159 y y 0.159 p 3    − ⇒ = = ⋅− =    − −   2 [2] [2] [2] p 0.141 0.887 1 0.107 0.673 0.159 y 0.159 1 y x    − ⇒ =−          −    [2] 1 2 3 y 0.141 y 0.107 y 0.159 Michael Beer, Engineering Mathematics

2AlgebraicproblemsinmatrixformEigenvalueProblemsGeneral inversepowermethod-example(cont'd)iterative solution[3][3]》k=3:-2-4-0.322[-1.070.887yiyi[2][3]5-4-3.070.673(A-q·I)·y=x =-0.011Y2Y2D1-24-3.07LY3-0.130[y3]17-0.322[3]1[3] [3]y-0.0110.034=0.322==y1=-0.322=x[3]0.322Yps0.404-0.130[2][3]alternating iteration with 2 = 2.45, 22 =-1.22, 22 =1.96typical ifi)gisinthemiddle betweenthenexteigenvaluesi)two eigenvalues are very closeto oneanother (alternation only in x)ii)starting vectornot suitableAs=5.07=0.5.(2+2,itisverylikelythatcasei)hascausedtheproblem.Inthis case,eigenvalue22is negative.alternative selection:q=-5.0791fMichael Beer,EngineeringMathematics

General inverse power method − example (cont'd) 2 Algebraic problems in matrix form Eigenvalue Problems 91f ● iterative solution alternating iteration with ( −⋅ ⋅ = ) [3] [2] A Iy x q » k = 3: −−−          =− − ⋅ =  − −      [3] 1 2 3 1.07 2 4 y 0.887 4 3.07 5 y 0.673 2 4 3.07 y 1 ∞ = ⇒ = =− 3 [3] [3] y 0.322 y y 0.322 p 1    − ⇒ = = ⋅− =    − −   3 [3] [3] [3] p 0.322 1 1 0.011 0.034 0.322 y 0.130 0.404 y x    − ⇒ =−          −    [3] 1 2 3 y 0.322 y 0.011 y 0.130 λ = λ =− λ = [1] [2] [3] 22 2 2.45, 1.22, 1.96 As , it is very likely that case i) has caused the problem. In this case, eigenvalue λ2 is negative. alternative selection: q = –5.07 typical if i) q is in the middle between the next eigenvalues ii) two eigenvalues are very close to one another (alternation only in x) iii) starting vector not suitable q 5.07 0.5 = = ⋅ λ +λ ( 1 3 ) Michael Beer, Engineering Mathematics

2AlgebraicproblemsinmatrixformEigenvalueProblemsGeneral inverse power method - example (cont'd)preparationof the iteration g=-5.070-2-4109.07-2-44552100-47.07-4A-q·I=(-5.07)00-244217.07-2iterativesolution forg=-5.07[1][1]-4-20.1919.07yiyi[0][1]5-47.070.186y2》 k = 1: (A-q.I) y=x =Y2U4[-27.0710.091LY3LY3170.191EY1[1][1]0.1860.5340.191y1=0.191UXYpy-[1]0.191Yp.0.0910.47691gMichaelBeer,EngineeringMathematics

General inverse power method − example (cont'd) 2 Algebraic problems in matrix form Eigenvalue Problems 91g ● preparation of the iteration q = –5.07 ( )     − − − −       − ⋅ = − −− ⋅ = −  − −       4 2 4 1 0 0 9.07 2 4 q 4 2 5 5.07 0 1 0 4 7.07 5 2 4 2 0 0 1 2 4 7.07 A I ● iterative solution for q = –5.07 ( −⋅ ⋅ = ) [1] [0] » k = 1: A Iy x q  − −          = − ⋅ =      −          [1] 1 2 3 9.07 2 4 y 1 4 7.07 5 y 1 2 4 7.07 y 1 ∞ = ⇒ == 1 [1] [1] y 0.191 y y 0.191 p 1    ⇒= = ⋅ =       1 [1] [1] [1] p 0.191 1 1 0.186 0.534 0.191 y 0.091 0.476 y x    ⇒ =             [1] 1 2 3 y 0.191 y 0.186 y 0.091 Michael Beer, Engineering Mathematics

2AlgebraicproblemsinmatrixformEigenvalueProblemsGeneral inverse power method - example (cont'd)·iterativesolutionforq=-5.07》k=2:[2][2]-2-419.070.155y1yi[1][2]5-47.070.534(A-q·I)·y=x=0.141Y2=Y2-247.070.4760.031Ly3Ly3110.155By1[2][2]1210.1410.910=yi=0.155=0.155XUYp2[2]0.155Yp20.0310.200》k=3:[3][3]17-2-49.070.132Y1Y1[2][3] 5-47.070.2610.910(A-q·I)·y=x=Y2Y2U4-27.070.200-0.082LY3Y30.5060.132[3]1[3][3]y31Yp=y,=0.2610.2610.261= x[3]0.261Yp-0.082-0.31491hMichael Beer,EngineeringMathematics

General inverse power method − example (cont'd) 2 Algebraic problems in matrix form Eigenvalue Problems 91h ● iterative solution for q = –5.07 ( −⋅ ⋅ = ) [2] [1] A Iy x q » k = 2:  − −          = − ⋅ =  −     [2] 1 2 3 9.07 2 4 y 1 4 7.07 5 y 0.534 2 4 7.07 y 0.476 ∞ = ⇒ == 2 [2] [2] y 0.155 y y 0.155 p 1    ⇒= = ⋅ =       2 [2] [2] [2] p 0.155 1 1 0.141 0.910 0.155 y 0.031 0.200 y x    ⇒ =             [2] 1 2 3 y 0.155 y 0.141 y 0.031 ( −⋅ ⋅ = ) [3] [2] A Iy x q » k = 3:  − −          = − ⋅ =  −     [3] 1 2 3 9.07 2 4 y 1 4 7.07 5 y 0.910 2 4 7.07 y 0.200 ∞ = ⇒ == 3 [3] [3] y 0.261 y y 0.261 p 2    ⇒= = ⋅ =    − −    3 [3] [3] [3] p 0.132 0.506 1 0.261 1 0.261 y 0.082 0.314 y x    ⇒ =          −    [3] 1 2 3 y 0.132 y 0.261 y 0.082 Michael Beer, Engineering Mathematics

2AlgebraicproblemsinmatrixformEigenvalueProblemsGeneral inverse power method - example (cont'd)·iterativesolutionfor=-5.07[10][10]》k=10:-2-49.070.3620.127Y1Yi[9][10]57.07-4-0.858Y2-0.292(A-qI).y =x==Y24-217.070.342Y3[y3]0.1270.371[10]1[10][10]y-0.292-0.854=0.342=Y,=0.342XD-YPio[10]0.342y10.342P10(A-q·I)eigenvalueeigenvectorv=x1(A-q·I)remember:A×入X1A.x-g.xAX二UX全91iMichaelBeer,EngineeringMathematics

General inverse power method − example (cont'd) 2 Algebraic problems in matrix form Eigenvalue Problems 91i ● iterative solution for q = –5.07 ● ● ● remember: Ax x ⋅ =λ⋅ eigenvector v = x eigenvalue ( −⋅ ⋅ = ) [10] [9] A Iy x q  − −          = − ⋅ =−  −     [10] 1 2 3 9.07 2 4 y 0.362 4 7.07 5 y 0.858 2 4 7.07 y 1 ∞ = ⇒ == 10 [10] [10] y 0.342 y y 0.342 p 3    ⇒ = = ⋅− = −       10 [10] [10] [10] p 0.127 0.371 1 0.292 0.854 0.342 y 0.342 1 y x    ⇒ =−             [10] 1 2 3 y 0.127 y 0.292 y 0.342 » k = 10: (A I − ⋅ q ) ⋅ x pk 1 y ⋅ pk 1 y ⋅ = y ( −⋅ ⋅ = ⋅ ) pk 1 q y A Ix x ⋅ −⋅ = ⋅ ⇒ pk 1 q y Ax x x   ⇒ ⋅= + ⋅       pk 1 q y A x x Michael Beer, Engineering Mathematics

2AlgebraicproblemsinmatrixformEigenvalueProblemsGeneral inverse power method-example (cont'd)·iterativesolutionforq=-5.070.1270.371[10]1[10]y-0.292-0.854x=[10]0.342Ypio0.34210.371[10]11(-5.07) = -2.146, v(2) ~ x-0.854[10]0.342y1P》truesolution0.376520.8452=-2.1338191jMichaelBeer,EngineeringMathematics

General inverse power method − example (cont'd) 2 Algebraic problems in matrix form Eigenvalue Problems 91j ● iterative solution for q = –5.07 ( ) ( )     λ ≈ + = + − =− ≈ = −      5 [10] 2 2 [10] p 0.371 1 1 q 5.07 2.146, 0.854 0.342 y 1 v x ( )     λ =− = −      2 2 0.3765 2.1338, 0.8452 1 v » true solution       = = ⋅− = −    10 [10] [10] [10] p 0.127 0.371 1 0.292 0.854 0.342 y 0.342 1 y x Michael Beer, Engineering Mathematics