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

2AlgebraicproblemsinmatrixformEigenvalueProblemsExample-multiple eigenvaluesdeterminationofeigenvalues40000[5-入4500-300-2 - 入-3-2A-2.I4004055-入0400-300-2-2-3-2det(A-2.1)=24-6.23+13.22-12.2+4=0(α-2) (2-1)° =0 =212 = 2, 234 = 1determinationofeigenvectors40ve0[5-入0v900-30-2- 入(A-2,I).v=0vo00045-入LOva00-3-2-入77aMichael Beer,Engineering Mathematics

Michael Beer, Engineering Mathematics λ= λ= 1,2 2, 1 3,4 determination of eigenvalues Example − multiple eigenvalues 2 Algebraic problems in matrix form ● Eigenvalue Problems 77a 5400 3 20 0 0054 00 32     − − =   − − A 5 400 32 0 0 0 05 4 0 0 32   − λ   − − −λ −λ⋅ = − λ   − − −λ A I ( ) 43 2 det A I −λ⋅ = λ − ⋅λ + ⋅λ − ⋅λ+ = 6 13 12 4 0 ( ) ( ) 2 2 λ− ⋅ λ− = 2 10 ⇒ ● determination of eigenvectors ( ) ( ) ( ) ( ) ( ) ( ) i 1 i i 2 i i 3 i 4 5 400 0 v 32 0 0 0 v 0 05 4 0 v 0 0 32 0 v  − λ          − − −λ      −λ ⋅ ⋅ = ⇔ ⋅ =  − λ           − − −λ      A Iv 0

2AlgebraicproblemsinmatrixformEigenvalueProblemsExample-multipleeigenvalues (cont'd)determination of eigenvectors》for入1= 入2 = 240030000-400-3-0.7500o034010000-0.75as v(1)andv(2)belong to the same eigenvalue,all linear combinations of v(1) and v(2) are also eigenvectors.0.5110.75-0.375-0.75110.2-0.750.75-0.15》for 2=2=1 accordingly77bMichaelBeer,EngineeringMathematics

determination of eigenvectors Example − multiple eigenvalues (cont'd) 2 Algebraic problems in matrix form ● Eigenvalue Problems 77b ( ) ( ) ( ) ( ) i 1 i 2 i 3 i 4 3400 0 v 3 40 0 0 v 0034 0 v 00 34 0 v           − −      ⋅ =            − −       » for λ =λ = 1 2 2 (1 2 ) ( ) 1 0 0.75 0 , 0 1 0 0.75     − ⇒= =   − v v ! as v(1) and v(2) belong to the same eigenvalue, all linear combinations of v(1) and v(2) are also eigenvectors. 1 0.5 1 0.75 0.375 0.75 , , , . 1 1 0.2 0.75 0.75 0.15         −− − ⇒= = =     −−− vv v » for accordingly λ =λ = 3 4 1 Michael Beer, Engineering Mathematics