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

2AlgebraicproblemsinmatrixformEigenvalueProblemsExample-polynomial methoddeterminationofeigenvalues[5-入445A-2·I-3 -2-3-2-2det(A - 2 ·I) = (5 - 2) · (-2 - 2) - 4 .(-3) = 022 - 3.2 +2 = (2 - 2) (2 -1) = 0入1 =2, 入2=1determinationofeigenvectors[5-入,(A-2,·I).v=0-3R2-》for2=2》for 入,=134专1.O3-3人vl) = 1 = v() =-0.75, v(1)v(2) = 1 = v(2) = -1, v(2)L-0.75176aMichaelBeer,EngineeringMathematics

Michael Beer, Engineering Mathematics determination of eigenvalues Example − polynomial method 2 Algebraic problems in matrix form ● Eigenvalue Problems 76a 5 4 3 2   =     − − A 5 4 3 2   − λ −λ⋅ =     − − −λ A I det(A I −λ⋅ = −λ ⋅ − −λ − ⋅ − = ) (5 2 4 3 0 ) ( ) ( ) ( ) ( ) 2 λ − ⋅λ+ = λ− ⋅ λ− = 3 2 2 10 ⇒ λ= λ= 1 2 2, 1 ● determination of eigenvectors » for λ = 1 2 ( ) ( ) ( ) ( ) i i i 1 i i i 2 5 4 v 0 3 2 v 0   − λ     −λ ⋅ ⋅ = ⇔   ⋅ =       − − −λ     A Iv 0 ( ) ( ) 1 1 1 2 34 0 v 34 0 v         ⋅ =      − −   (11 1 ) ( ) ( ) 1 2 1 v 1 v 0.75, 0.75   = ⇒ =− =     − v » for λ = 2 1 ( ) ( ) 2 1 2 2 44 0 v 33 0 v         ⋅ =      − −   (2 22 ) ( ) ( ) 1 2 1 v 1 v 1, 1   = ⇒ =− =     − v