南京航空航天大学Matrix-Theory双语矩阵论期末考试2015

pi=<>=卩*1xdx]=pi=<>=卩*1xdx]=0v'22八2*2x-pTx-p1，3「'込X11p=<x2,石>~2+<x2,'3.31x>x=223U3=肖=2—1)(2)proj=<x2一1,u>u+<x2一1,u>u=<x2一1,~^=>~^=+<x2一1,'3x2NUAA第1页（共6页）MatrixTheory,Final,TestDate：2015年12月28日矩阵论班号：学号姓名必做题（70分）选做题（30分）总分题号12345得分PartI（必做题，共5题，70分）第1题（15分）得分7p=(1,—1,0)T3(101](110、P=00—1,P—1=0011010丿10—10丿(101]fet00'f(110、((etet—e2t0、eAt=PeJP—1=00—1（0e2tte2tfff001=f0e2t01010丿〔00e2t丿〔0—10丿*0—te2te2t丿第4题（10分）得分Solve(A—Solve(A—21)p=p32(—1—10、（0、000p=03-10丿3<1丿weobtainthat(A—2I)p3=SupposethatAgR3x3andA2—5A—61=OWhatarethepossibleminimalpolynomialsofA?Explain.Ineachcaseofpart(1),whatarethepossiblecharacteristicpolynomialsofA?Explain.Solution:(1)AnannihilatingpolynomialofAisx2—5x—6.TheminimalpolynomialofAdividesanyannihilatingpolynomialofA.Thepossibleminimalpolynomialsarex—6,x+1,andx2—5x—6(2)TheminimalpolynomialofAdividesthecharacteristicpolynomialofA.SinceAisamatrixoforder3,thecharacteristicpolynomialofAisofdegree3.TheminimalpolynomialofAandthecharacteristicpolynomialofAhavethesamelinearfactors.Casex—6,thecharacteristicpolynomialis(x—6)3Casex+1,thecharacteristicpolynomialis(x+1)3第5题（10分）得分T(120)LetA=(000丿Casex2一5x一6,thecharacteristicpolynomialis(x+1)2(x—6)or(T(120)LetA=(000丿FindtheMoore-PenroseinverseA+ofASolution:(i20)=PGP+Solution:(i20)=PGP+=(PtP)-iPt=(1,0)G+=Gt(GGt)-ir1A2<0丿A+=G+P+2(1,0)=0A00丿也可以用SVD求.PartII（选做题，每题10分）LetPLetPbethevectorspaceconsistingofallrealpolynomialsofdegreeless4than4withusualadditionandscalarmultiplication.Letx,x,xbethree123distinctrealnumbers・ForeachpairofpolynomialsfandginP,define4<f,g>=£f(x)g(x)・iii=1Determinewhether<f,g>definesaninnerproductonPornot.Explain.4第7题LetAgRnxn.Showthatifb(x)=Axistheorthogonalprojectionfrom]RntoR(A)thenAissymmetricandtheeigenvaluesofAareall1'sand0's.第8题LetAgCnxn.ShowthatxHAxisreal-valuedforallxgCnifandonlyifAisHermitian.第9题LetA,BgCnxnbeHermitianmatrices,andAbepositivedefinite.ShowthatABissimilartoBA，andissimilartoarealdiagonalmatrix.选做题得分若正面不够书写，请写在反面.Thisdoesnotdefinean第6题解答|ThisdoesnotdefineanLetf(x)=(x一x)(x一x)(x一x).Then<f,f>=0.Butf主0123innerproduct.第7题解答Foranyx,Ax—xgR(A)丄=N(At),At(Ax—x)=0.Hence,AtA=AtThus.A=AT.Fromabove,wehaveA2=A.Thiswillimplythat九2—九isanannihilatingpolynomialofA.TheeigenvalueofAmustbetherootsof九2-九=0.Thus,theeigenvaluesofAarel'sand0's.第8题解答|SeeThm7.1.1,page182.也可以用其它方法.第9题解答SinceAisnonsingular,AB=A(BA)A-i.Hence,AissimilartoBASinceAispositivedefinite,thereisanonsingularhermitianmatrixPsuchthatA=PPh.AB=PPhB=P(PhBP)P-iSincePHBPisHermitian,itissimilartoarealdiagonalmatrix.ABissimilartoPHBP,PHBPissimilartoarealdiagonalmatrix.ThusABissimilartoarealdiagonalmatrix.LetPdenotethesetofallrealpolynomialsofdegreelessthan3withdomain、、[-1，1](