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