1 A complex matrix is hermitian if or ij ji is said to be hermitian positive de64257nite if Ax for all 0 Remark is hermitian positive de64257nite if and only if its eigenvalues are all positive If is hermitian positive de64257nite and LU is the LU ID: 22868
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=266664u110000u22000000um�1m�10000umm3777752666641u12 u11u1m�1 u11u1m u1101u2m�1 u22u2m u2200001um�1m um�1m�10001377775=W.Sinceu11�0,,umm�0,wecanwrite=D2whereD=266664p u110000p u22000000p um�1m�10000p umm377775.SowehaveA=LU=LW=LD2WwhereLisunit-triangularandWisunitupper-triangular.SinceA=A,wehaveLD2W=(LD2W)=W(D2)L=W(D2)L.NotethatWisunitlower-triangular.BytheuniquenessofLUfactorization,wehaveL=W.SoA=LDDL=(LD)(LD).LetR=DL.ThenRisupper-triangularandA=RR.Lemma2.4.SupposeAAisinvertible.ThenAA=RRwhereRisuupper-triangular.Proof:OnecancheckeasilythatAAishermitian(b/c(AA)=A(A)=AA).Sincex(AA)x=(Ax)(Ax)=jjAxjj20andx(AA)x=0ifAx=0.NotethatAAisinvertible.SoAx=0impliesAAx=0andx=0.AlgorithmforCholeskyFactorizationforaHermitianpositivedef-initematrixStep1.FindaLUdecompositionofA=LU.Step2.FactorU=D2WwhereWisaunitupper-triangularmatrixandDisadiagonalmatrix.Step3.A=RRwhereR=DW.Example2.5.Determineifthefollowingmatrixishermitianpositivedenite.AlsonditsCholesskyfactorizationifpossible.A=26641212331323775,B=266412228020243775.Solution:(1)Fromtherowreduction,wehavethefollowing.A=26641212331323775^�2r1+r2;�r1+r3241210�1101135^1r2+r3241210�1100235.SoA=241002101�1135241210�1100235Nowu22=�10.SoAisnotpositivedenite.2 2.Fromtherowreduction,wehavethefollowing.AA=26642�20�24�20�243775^r1+r226642�2002�20�243775^1r2+r326642�2002�20023775.SoAA=2664100�1100�11377526642�2002�20023775=2664100�1100�1137752664p 2000p 2000p 23775226641�1001�10013775SoR=2664p 2000p 2000p 2377526641�1001�10013775=2664p 2�p 200p 2�p 200p 23775andAA=RR.ComputeA2664211�23775=26641�100�111111�1�137752664211�23775=26641�243775.NowsolveRy=A2664211�23775()2664p 200�p 2p 200�p 2p 2377524y1y2y335=26641�243775()y1=1 p 2,y2=�2+p 2y1 p 2=�1 p 2andy3=4+p 2y2 p 2=3 p 2Last,wesolveRx=y()2664p 2�p 200p 2�p 200p 2377524x1x2x335=2641 p 2�1 p 23 p 2375()x3=3 2,x2=�1 p 2+p 2x3 p 2=1andx1=1 p 2+p 2x2 p 2=3 2.Hencethesolutionisx=243 213 235.4