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1、第七章部分习题参考答案Exercise1ShowthatanormalmatrixAisHermitianifitseigenvaluesareallreal.ProofIfAisanormalmatrix,thenthereisaunitarymatrixthatdiagonalizesA.Thatis,thereisaunitarymatrixUsuchthatwhereDisadiagonalmatrixandthediagonalelementsofDareeigenvaluesofA.Ifeigenvalu
2、esofAareallreal,thenTherefore,AisHermitian.Exercise2LetAandBbeHermitianmatricesofthesameorder.ShowthatABisHermitianifandonlyif.ProofIf,then.Hence,ABisHermitian.Conversely,ifABisHermitian,then.Therefore,.Exercise3LetAandBbeHermitianmatricesofthesameorder.Showtha
3、tAandBaresimilariftheyhavethesamecharacteristicpolynomial.ProofSincematrixAandBhavethesamecharacteristicpolynomial,theyhavethesameeigenvalues.ThereexistunitarymatricesUandVsuchthat,.Thus,.()Thatis.Hence,AandBaresimilar.Exercise4LetAbeaskew-Hermitianmatrix,i.e.,
4、,showthat(a)andareinvertible.(b)isaunitarymatrixwitheigenvaluesnotequalto.ProofofPart(a)Method1:(a)since,itfollowsthatForanyHence,ispositivedefinite.Itfollowsthatisinvertible.Hence,bothandareinvertible.Method2:5Ifissingular,thenthereexistsanonzerovectorxsuchtha
5、t.Thus,,.(1)Sinceisreal,itfollowsthat.Thatis.Since,itfollowsthat(2)Equation(1)and(2)impliesthat.Thiscontradictstheassumptionthatxisnonzero.Therefore,isinvertible.Method3:LetbeaneigenvalueofAandxbeanassociatedeigenvector..Hence,iseitherzeroorpureimaginary.1andca
6、nnotbeeigenvaluesofA.Hence,andareinvertible.Method4:Since,Aisnormal.ThereexistsaunitarymatrixUsuchthatEachispureimaginaryorzero.Sincefor,det.Hence,isinvertible.Similarly,wecanprovethatisinvertible.ProofofPart(b)Method1:Since,itfollowsthat(NotethatifPisnonsingul
7、ar.)Hence,isaunitarymatrix.Denote.Since,Hence,cannotbeaneigenvalueof.Method2:Bymethod4oftheProofofPart(a),5Theeigenvaluesofare,whichareallnotequalto.Method3:Since,itfollowsthatIfisaneigenvalueof,thenthereisanonzerovectorx,suchthat.Thatis.Itfollowsthat.Thisimpli
8、esthat.Thiscontradictionshowsthatcannotbeaneigenvalueof.Exercise6IfHisHermitian,showthatisinvertible,andisunitary.ProofLet.ThenAisskew-Hermitian.ByExercises#4,andareinvertib