practice problem_solution 外文学习材料

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1、MatrixTheory{PracticeProblems{Solutions1.Let23100A=41015010(1)FindtheJordanformforA.(2)Assumek3.ComputeAkAk2andA2I.(3)ComputeeA.Rt(4)Compute(IA2)eAsds:0(5)FindthecharacteristicpolynomialandminimalpolynomialforA.Solution:(1)TheJordanformforAis23110J=40105001(2)+(5)

2、Thecharacteristicpolynomialandminimalpolynomialareboth(1)2(+1).Hence(AI)2(A+I)=0)A(A2I)=A2I)Ak(A2I)=(A2I)forallk1:Here23000D=A2I=41005satisesD2=0100(3)Fromabove,Ak=Ak2+Dforallk2.A2=I+D;A3=A+D;A4=A2+D=I+2D


A2k=I+kD;A2k+1=A+kDSoX11X11X11eA=An=(I+kD)+(A+kD)n!(

3、2k)!(2k+1)!n=0k=0k=0!!!X11X11X1kk=I+A++D(2k)!(2k+1)!(2k)!(2k+1)!k=0k=0k=0e1+e1e1e1e1+e1=I+A+D224213e00111111=43eee+eee5422111111e+eeee+e422(4)SinceAk(A2I)=A2IwehaveA2I=Ak(A2I)forallk1.HenceZts=t(IA2)eAsds=(A2I)A3eAs=D(eAtI)0s=0NoticethatD2=0;DA=D.H

4、enceX11X11DeAt=DAntn=Dtn=etDn!n!n=0n=0FinallyZt(IA2)eAsds=(et1)D:02.AssumeA;B2M4(C)andtheminimalpolynomialsareq(t)=(t1)2(t2);q(t)=(t1)(t2)2:ABThenwhatistheminimalpolynomialforthefollowingmatrix:AAB0B12Solution:Noticethat11A011AAB=010B010BA0AABSoand

5、aresimilarandtheyhavethesameminimalpolynomial0B0Bq(t)=(t1)2(t2)2:3.SupposeA2M(C)andsatisesA2A2I=0.CanAbediagonalised?Ifyes,giveyourreason.Otherwise,giveancounterexample.Solution:Yes,Acanbediagonalised.HereAsatisestheequationA2A2I=0,whichmeanstheminimalpolynomialofA

6、isq(t)=t2t2=(t2)(t+1).SoAhastwoeigenvalues2;1.Thepowerofeachfactor(t2);(t+1)intheAminimalpolynomialis1,sointheJordanform,eachJordanblockshouldhavedimension11,whichmeanstheJordanformisadiagonalmatrix.4.LetR[x]4bethespaceofrealpolynomialswithdegreelessthan3:R[x]=ff(t):

7、f(t)=a+at+at2+at3;a2Rg:40123i(1)Letf1;(t1);(t1)2;(t1)3gbeabasisofR[x].Findcoecientsoff(t)=a+at+at2+at3underthisbasis.40123(2)DeneZ1hf;gi=f(t)g(t)dt:0ShowthatthisisaninnerproductonR[x]4.(c)FindthematrixAsuchthatforf(t)=a+at+at2+at3,g(t)=b+bt+bt2+bt3,01230123hf;gi=[b;b;

8、b;b]A[a;a;a;a]T01230123Solution:(1)Iff(t)=a+at+at2+at3=b+b(t1)+b(t1)2+b(t1)3,thenbydir