PDF-Theorem20.3.LetAbeannnmatrix.ThenAisdiagonalisableifandonlyifwecan nd
Author : karlyn-bohler | Published Date : 2016-05-27
Example205IsitpossibletodiagonaliseA01216101211AIftheanswerisyesthendiagonaliseAWealreadysawthatv11613isaneigenvectorwitheigenvalue0v2121isaneigenvectorwitheigenvalue4and
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Theorem20.3.LetAbeannnmatrix.ThenAisdiagonalisableifandonlyifwecan nd: Transcript
Example205IsitpossibletodiagonaliseA01216101211AIftheanswerisyesthendiagonaliseAWealreadysawthatv11613isaneigenvectorwitheigenvalue0v2121isaneigenvectorwitheigenvalue4and. Nullspace LetA=(aij)beanmnmatrix.De nition.ThenullspaceofthematrixA,denotedA),isthesetofalln-dimensionalcolumnvectorssuchthat A0. 0BBB@a11a12a13:::a1na21a22a23:::a2n...............am1am2am3:::amn1CCC TheQCQPproblemConsideraquadraticallyconstrainedquadraticprogram:(QCQP)z=minf0(x)s:t:fi(x)di;i=1;:::;qx0;Axb;wherefi(x)=xTQix+cTix,i=0;1;:::;q,eachQiisannnsymmetricmatrix,andAisanmnmatrix.LetF=fx WIGNERENSEMBLEH=(hjk)isahermitianNNmatrix,N 1.hjk=1pN(xjk+iyjk);(jk);hjj=s2Nxjjwherexjk;yjk(jk)andxjjareindependentwithdistributionsxjk;yjkd:=e g(x)dx;Normalization:Exjk=0,Ex2jk=12.Example:g(x)=x2i Thm.[B] LetX1;X2;;Xkbeeigenvectorscorrespondingto distinct eigenvalues1;2;;kofA.ThenfX1;X2;;Xkgis linearlyindependent . Proof.AssumethatfX1;X2;&#
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