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1、RepeatedEigenvaluesWenowconsiderthecaseofrepeatedeigenvalues.Inthecasewhenwehaveenoughlinearlyindependenteigenvectors,wewillnothavetodoanythingspecial,butwewillhavetodealwiththepossibilitythatwemaynothaveenoughlinearlyindependenteigenvectors.1RepeatedEigenvalues:Al
2、gebraicandGeomet-ricMultiplicityWeknowthatifwehaveasystemofnfirstorderequations,weneednvectorvaluedsolutions.Wealsoknowthatwegetatleastonenewlinearlyindependenteigenvector(andthussolution)pereigenvalueofthematrix.However,wehavealreadyseenthatitispossibletohavelessth
3、anneigenvaluesandstillhavenlinearlyindependentvectors.Inthefollowingexample,wesolveainwhichthematrixhasonlyoneeigenvalue1,buthastwolinearlyindependenteigenvectors.Inthiscase,wedonothavetodoanythingspecial.Example:Solve!′20x=x02Itisnotdifficulttoseethatthecharacterist
4、icpolynomialis(λ−2)2=0,sowehavearepeatedeigenvalueλ=2.Itisalsonotdifficulttoseethatwhenwesolveforeigenvectors,wegetbothx1andxasfreevariables,sothatwehaveeigenvectors(1,0)Tand(0,1)T.Thusthegeneral2solutionis!!12t12tc1e+c2e00Inparticular,itcanbeshownthatthisisthecasewh
5、eneverarealmatrixAinx′=Axissymmetric.(AsymmetricmatrixisoneforwhichAT=A.)Here’s1anotherexample:Example:Solve011′x=101x110Itisnotdifficulttoseethatthecharacteristicpolynomialis(λ−2)(λ+1)2=0,sowehavearepeatedeigenvalueλ=−1.Itisalsonotdifficulttoseethatwhenwesolvefo
6、reigenvectorsassociatedwithλ=−1,wegetbothx2andx3asfreevariables:111v10111v2=0111v30sothatwehaveeigenvectors(−1,1,0)Tand(−1,0,1)T.However,ifwehaveaneigenvaluewithmultiplicitynwhichhaslessthannlinearlyindependenteigenvectors,wewillnothaveenoughsolut
7、ions.Withthisinmind,wedefinethealgebraicmultiplicityofaneigenvaluetobethenumberoftimesitisarootofthecharacteristicequation.Wedefinethegeometricmultiplicityofaneigenvaluetobethenumberoflinearlyindependenteigenvectorsfortheeigenvalue.Thus,inExample1above,λ=2isaneigenva
8、luewithbothalgebraicandgeometricmultiplicityoftwo.However,itispossiblefortheretobelessthannlinearlyinde-pendenteigenvectorsifaneigenvalueisrepeat