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

2AlgebraicproblemsinmatrixformEigenvalueProblemsPowermethod-examplespecialeigenvalueproblem4-2-4 52-4(A-2·I)·X=0A=eigenvalues are real24-2[k][k][k][k-1][KIky·iterationprocedureYpky=A.X=VxU二[k]Ypr· starting vector x = (1,1,1)●iterativesolution4-2-24[1][0]34521》k=1:=A·XV44-2-20.5[1][1]y1130.753 = 4=Vp,=,= 4x[1]4131481aMichaelBeer,EngineeringMathematics

Michael Beer, Engineering Mathematics special eigenvalue problem Power method − example 2 Algebraic problems in matrix form ● Eigenvalue Problems 81a   − −   = −    −  424 425 242 (A Ix 0 −λ⋅ ⋅ = ) A iteration procedure k k [k] [k] [k 1] [k] [k] [k] p [k] p y y − ∞ =⋅ ⇒ = ⇒ = y y Ax y x eigenvalues are real starting vector = ( ) [0] T x 1,1,1 iterative solution = ⋅ [1] [0] » k = 1: y Ax ● ● ●  −− −          =− ⋅ =      −          4241 2 4251 3 2421 4 ∞ = =⇒ = = 1 [1] [1] y y4y y4 3 3 p   − − ⇒= =⋅ =         1 [1] [1] [1] p 2 0.5 1 3 0.75 4 y 4 1 y x

2 Algebraicproblems in matrixformEigenvalueProblemsPower method- example (cont'd)·iterative solution4-2-0.5-4-7.5[2][1]528.5-40.75》k= 2:y=Ax426-217.5-0.882[2][2]1[2]y18.5y, =8.5==y, = 8.5XUD[2]8.5Ypz60.706-24-4-0.882-8.352[2][3]521-49.058》k=3: y=A.X24-27.1760.706[3]=9.058==9.058Yp=Y2-8.352-0.922[3] 1[3]y19.058=X=[3]9.058Yps7.1760.79281bMichaelBeer,EngineeringMathematics

Power method − example (cont'd) 2 Algebraic problems in matrix form Eigenvalue Problems 81b iterative solution = ⋅ [2] [1] » k = 2: y Ax ●  −− − −          =− ⋅ =  −     4 2 4 0.5 7.5 4 2 5 0.75 8.5 242 1 6 ∞ == ⇒ == 2 [2] [2] y y 8.5 y y 8.5 2 2 p    − − ⇒= = ⋅ =       2 [2] [2] [2] p 7.5 0.882 1 8.5 1 8.5 y 6 0.706 y x = ⋅ [3] [2] » k = 3: y Ax      −− − −      =− ⋅ = −     4 2 4 0.882 8.352 4 2 5 1 9.058 2 4 2 0.706 7.176 ∞ == ⇒ == 3 [3] [3] y y 9.058 y y 9.058 2 2 p    − − ⇒= = ⋅ =       3 [3] [3] [3] p 8.352 0.922 1 9.058 1 9.058 y 7.176 0.792 y x Michael Beer, Engineering Mathematics

2AlgebraicproblemsinmatrixformEigenvalueProblemsPower method- example (cont'd)iterative solution4-2-4-0.919-8.776[5][6]25-419.551》k = 6:y=AX42-20.7757.388[6]Yp,=yz=9.551，=9.551=-8.776-0.919[6]1[6]yeigenvalueeigenvectorv=x19.551=X=[6]9.551Yp67.3880.774remember:A.x入xy=A,X=yeX》true solution-0.919-0.9189[6][6]Y119.551,午=9.5446,XYpe0.7740.773881cMichaelBeer,EngineeringMathematics

Power method − example (cont'd) 2 Algebraic problems in matrix form Eigenvalue Problems 81c ● iterative solution = ⋅ [6] [5] » k = 6: y Ax  −− − −          =− ⋅ =  −     4 2 4 0.919 8.776 4 2 5 1 9.551 2 4 2 0.775 7.388 ∞ == ⇒ == 6 [6] [6] y y 9.551 y y 9.551 2 2 p    − − ⇒= = ⋅ =       6 [6] [6] [6] p 8.776 0.919 1 9.551 1 9.551 y 7.388 0.774 y x ● ● ● ( )   −   λ≈ = ≈ =       6 [6] [6] 1 1 p 0.919 y 9.551, 1 0.774 v x ( )   −   λ = =       1 1 0.9189 9.5446, 1 0.7738 v » true solution remember: Ax x ⋅ =λ⋅ eigenvector v = x eigenvalue =⋅= ⋅ pk y Ax x y Michael Beer, Engineering Mathematics