De64257nition 2 Computation and Properties 3 Chains brPage 3br Generalized Eigenvectors Math 240 De64257nition Computation and Properties Chains Motivation Defective matrices cannot be diagonalized because they do not possess enough eigenvectors to ID: 24036
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GeneralizedEigenvectorsMath240DenitionComputationandPropertiesChains Motivation Defectivematricescannotbediagonalizedbecausetheydonotpossessenougheigenvectorstomakeabasis.Howcanwecorrectthisdefect? ExampleThematrixA=1101isdefective. 1.Onlyeigenvalueis=1. 2.A�I=0100 3.Singleeigenvectorv=(1;0). 4.Wecoulduseu=(0;1)tocompleteabasis. 5.Noticethat(A�I)u=vand(A�I)2u=0. Maybewejustdidn'tmultiplybyA�Ienoughtimes. GeneralizedEigenvectorsMath240DenitionComputationandPropertiesChains Computinggeneralizedeigenvectors ExampleDeterminegeneralizedeigenvectorsforthematrixA=2411001200335: 1.Characteristicpolynomialis(3�)(1�)2. 2.Eigenvaluesare=1;3. 3.Eigenvectorsare1=3:v1=(1;2;2);2=1:v2=(1;0;0): 4.Finalgeneralizedeigenvectorwillavectorv36=0suchthat(A�2I)2v3=0but(A�2I)v36=0: Pickv3=(0;1;0). Notethat(A�2I)v3=v2. GeneralizedEigenvectorsMath240DenitionComputationandPropertiesChains Computinggeneralizedeigenvectors ExampleDeterminegeneralizedeigenvectorsforthematrixA=241201120�1135: 1.Singleeigenvalueof=1. 2.Singleeigenvectorv1=(�2;0;1). 3.Lookat(A�I)2=24204000�10�235tondgeneralizedeigenvectorv2=(0;1;0). 4.Finally,(A�I)3=0,sowegetv3=(1;0;0). GeneralizedEigenvectorsMath240DenitionComputationandPropertiesChains Computinggeneralizedeigenvectors ExampleDeterminegeneralizedeigenvectorsforthematrixA=241201120�1135: 1.Fromlasttime,wehaveeigenvalue=1andeigenvectorv1=(�2;0;1). 2.Solve(A�I)v2=v1togetv2=(0;�1;0). 3.Solve(A�I)v3=v2togetv3=(�1;0;0). GeneralizedEigenvectorsMath240DenitionComputationandPropertiesChains Jordancanonicalform What'stheanalogueofdiagonalizationfordefectivematrices? Thatis,iffv1;v2;:::;vngarethelinearlyindependentgeneralizedeigenvectorsofA,whatdoesthematrixS�1ASlooklike,whereS=v1v2vn?