第一章多項(xiàng)式習(xí)題解答P4411172623999FXGXXX22157FXGXXXXP4421231|9XMXXXQ余式21PMXQM021MQPQ方法二，設(shè)32011MQXPXQXMXQMQP同樣。22421|XMXXPXQ余式2221MPMXQPM0220MMP221,1MPQXPQP4431用3GXX除53258FXXXX解543223303175349536673327FXXXXXXP44323232122812XXXIIXIX1281298IXII即余式98I商2252XIXIP444150,1FXXX即5432151101101511FXXXXXX當(dāng)然也可以5511FXXX54151XX1P4442結(jié)果3424322328222224211FXXXXXXX4322137FXXIXIXXI432432213721575XIIIXIIIXIIXIIIXIIXIIXIXIIP45512212111GXXXXXX1313XXXXF,1FXGXX23231GXXX不可約不可約1434XXXF,1FXGX31221221102224XXXXXXXF432322426621,4222221GXXXXXFXXXXXX22,221FXGXXXP45612122XXXF2221GXXXX221112XXXXX12212XXFXXGX2，1424123YXXXXXF2124GXXXX11XFX11XGX而1112323231FXGXXXGXXX111211231133X1XXGYFXG212111333XXFXXXGX3,144234XXXXXF21GXXX,232FXGXXX211GXXX2131FGXXG32132XFXXXXGXP4572112FXGXTXTXURX2222122211111TTTUTTGXRXXXUTTTT由題意|RXXRXGX與G的公因式應(yīng)為二次所以0130143322223UTTTTUTUTT0304331223UTTUTUTTXRT為一次的否則解出當(dāng)1404330223TTTTTTU時(shí)32314ETT或311,03,02TTTTU只有時(shí)當(dāng)43331433433233232TTTTTTTTUUTUTT422468233122TTTTTTU即03422TTTU2111TP45、8|,|DXFXDXGX表明是公因式DX又已知表明任何公因式整除DXFXGX是與的組合DX所以是一個(gè)最大的公因式。DXP45，9證明,FXHXGXHXFXGXHX（的首系1）HX證設(shè),FXHXGXHXMX由,|FXGXHXFXHX,|FXGXHXGXHX,|FXGXHXMX,FXGXHX是一個(gè)公因式。設(shè),DXFXGXUXFXVXGX,DXHXFXGXHXUXFXHXVXGXHX而首項(xiàng)系數(shù)1，又是公因式得（由P45、8），它是最大公因式，且,FXGXHXFXHXGXHXP45、10已知,FXGX不全為0。證明,FXGXFXGXFXGX1證設(shè),DXFXGX則0DX設(shè)1,FXFXDX1,GXGXDX及DXUXFXVXGX所以11DXUXFXDXVXGXDX消去得0DX111UXFXVXGXP4511證設(shè)11,0,FXGXDXFXFXDXGXGXDX111,1UXFXDXUXGXDXDXUXFXUXGX1P4512設(shè)21111111,11UFVGUFVHUUFUFVHVGUFVUGH1111UUFUVHVGUFVUGHFGH,P4513,G1IIF，固定12,1IIFGG12,1INFGGGP4514,111,1FGUFVGUVFVGFFGF同理,1GGF由12題,1FGFG令12NGGGG,1IIFG每個(gè)11,1FFG,1123,FFFG,（注反復(fù)歸納用12題）。1212,MNFFFGGGNULLNULL1推廣若則,1,FXGXM，N有,1MNFXGXP45,15FXX32X22X1,GXX4X32X2X1解GXFXX12X2X1,FXX2X1X1即FX，GXX2X1令X2X10得231,23121IIFX與GX的公共根為21,P4516判斷有無(wú)重因式543257248FXXXXXX34424XXXXF解4421205234XXXXXF3251325412FXFXXXXX232215492541254422FXXXXXXX324412452223XXXXXX故有重因式XF32X4843XXXF3221233FXXXXXX1311323322XXXXXF2661123311132331111XXXX131XFXFP4517有重因式（有重根）1323TXXXXFT時(shí)解TXXXF63236213TXTXXFXF如則有重因式3重因式3T13XFX如則3T21522153222TXXF此時(shí)必須415T有重因式4212XXXFP4518求多項(xiàng)式有重根因式的條件QPXXXF3證PXXF2323323FXXPXPXQ0P22223327323244AQXPPXQXPPPP32427PQ0P4519令4221,1|,1FXAXBXXFXXFX因?yàn)樗约?2423CBXAXXBXAXXF4020AABACBBC0224BABBAA又1123DXCXBXAXXFXFX01AABACBD0BBABAA101BA2,1BAP46，20證212NXXFXXNNULL無(wú)重因式（重根）證NXFXFXN,YXFFFN,1FX,1NFXFX無(wú)重因式P46，21GX12FXFA2XAFXFXGA0又GA0/11102222XAGXFXFXFXFXXAFXGA/4/122,XAGXFXFXAXGXGXGX/是G根,且使GX的K1重根A是GX的K3重根P46，22“”必要性顯然（見定理6推論1）“”若X0是FX的T重根，TK，由定理FKX00若TK100KFX，所以矛盾P4623例如1,01MMFXXXFXMXM則是的重根根0XFX但不是的P4624若11|NNNXFXXFX則證若12FXXGXR由上節(jié)課命題10NNNFXXGXRGXRR所以|1NNXFXP46,25證明設(shè)X2X1的兩個(gè)根312,1I21233111213322222100XXXXFFFF112122110110FFFF即211010FF121,XFXFXP46、26分解1NX0221COSSIN,0,12,KKKIKNN1NNULL設(shè)1101221111212122,11COSSIN,11122112COS1221112COS1NIIKMMNKKKKMKMNKKKIXXXXNNIINXXXXXXKNMXXXNKNMXXXXXNNULLNULL在中在中為奇當(dāng)時(shí)1NKP46,27求有理根（1）X36X215X14FX解有理根可能為1、2、7、14。當(dāng)A首項(xiàng)系數(shù)則FXP為某不可約多項(xiàng)式X的方冪的充要條件是GX或者,1FG或者|MMFXGX證明11“,0,RFXPXHXHXPXHX反設(shè)不是則而111,1,|,1,PHHPXPXFG即取G則且,|,MSMFGHPX否則矛盾“”,1,1,1|RRR1FPGXPGPGFGPGPGFGX若若4881,0,PFXFXFX首項(xiàng)系數(shù)則為某不可約多項(xiàng)式的方冪|,|,/MGXHXFGHFGMFXHX由或者證明“,|,1,1,1|RRFPGHPHPHFHF設(shè)若FG,1|RRRPHPHPHFHXM111“,0,FPHHPHPHHX反設(shè)不是則而令G則|,1,1MRFGHFGFHMPHPH卻P48，補(bǔ)9證2的非零根NNMXAXB沒有重?cái)?shù)證反設(shè),NNMFXXAXBK有重根K2，0111/1/00010NNMNMMMMGXFXNXANMXKANMNXXNANMHXXNHXMXHXHXHX有重根有重根有重根無(wú)重?cái)?shù)根但重根P48、補(bǔ)100,1NFXCXFXFXN且,0MFX證明的根只能為或單位根即滿足某X1的根N證設(shè)為FX的根,由FXFXGXFXNNF0,為的根,220,NNFFX為的根,23,NNNNULL都為FX的根0,FXFX不可能有無(wú)限個(gè)根,其中必有相等者IJNNIJ不妨設(shè),210,IJNNNIJNNM令0,1MX則或或是的根P48、補(bǔ)11題/NFXFXFXAXBFX有N重根B,NAAAAAANULLNULL1212N補(bǔ)充P4812題的兩兩不同F(xiàn)XXXX證（1）1111111NIIIIIIIIIINXAXAXAXAFXFXLFAFAXAFAAAAAAAAANULLRA11110,11,1,11,1,2,1NNIJIIJJIININIIJJRJLALALAJNILXINNLXFXFXQRXFANFALXHAFARAHXRXNULLNULL，為次多項(xiàng)式設(shè)則而形式多項(xiàng)式P49、補(bǔ)13題（1）求423314052FXFXFFFF2時(shí)，至少去掉3個(gè)X,不含X3項(xiàng)了，對(duì)于I2,J2同理其它情形，至少去掉兩個(gè)X且第一行（或第二列）的兩個(gè)X只能取一個(gè)，故不含X3項(xiàng)，只剩下I1,J2時(shí)，A12本身是X項(xiàng)為12134A12A21A33A44X1XXX3,系數(shù)為11212,P971111011,NNJJJJJJDNULLNULLI故中與一樣多即號(hào),號(hào)一樣多,也即奇偶排列一樣多,N2時(shí),奇偶排列各占一半P981221111121111111NNNNNNXXXAAAPXVXAAANULLNULLNULLNULLNULLNULLNULLNULLNULL按第一行展開1211,0NNAAAVPXNNULL兩兩不同即12112101,NNPAPAPANAAANULLNULL總有兩行相同最多個(gè)根,即PX的所有根為111321332313222339813222202011111122XYXYXXXXYPYXYXYXYXYXYXYYXYXYXXYYXYXYYXYXXYXYNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULL22334413311113114813341131624811311113PPNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULL例P9813（法一）123412341234123423410127012701271603412018100044004441230710130043600040法二10234123412341234103470113011302001020201041202220200001310123011101110011123402002016000130004法三令2412312341234321234,41,1,1,111110222222022160CFXXXXIIIDFFFIFIIIIIIII為次單位根令則行列式P9813解2211111111111111011111000111101111100011110111110001111011111000XXXXXYYYYY1,21,2111011111111111|,1111111111111111,1111111111111111XXXXXYYXFXYXXFXYFXYYYNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULL交換行再交換列解法二,設(shè)FX,Y則第一行減第二行又因?yàn)镕X,Y關(guān)于X是偶函數(shù)，即X2|FX,Y同理，Y|FX,Y，且也是偶函數(shù)，所以Y2|FX,Y2222222,4|,FXYXYFXYFXYKXYFXYXYY2而中的系數(shù)為1故有FX,YXP981322511222311323411421446921262144692126021446921262144692126XAAAAAABBBBBBCCCCCCDDDDDDNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULL性質(zhì)2133246122134272246427100001101454310076811611003427211005882941NULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULL700110010010001161100111610001001001294029400000029411294011161000性質(zhì)111111111111211112311113122222222222212111211211229814222ABCCAABABCBCPABCCAABABCBCABCCAABABCBCABCABCABCNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULL左2213121NULLNULLNULLNULLNULLNULLNULLNULLNULLNULL右右P9815求出所有代數(shù)余子式111213142122232431323334414243441214600001211260000211563000037012AAAAAAAAAAAAAAAA直接計(jì)算有1127123321641014555111213212223313233AAA直接計(jì)算有AAAAAAP99161351213512135121222012106145012142313401214012140611452112533121012165701216572103507105907105913512263310121422438400119275300817410041219135120121442300119450086311100028571351201214001101900073000285715121101214217694830011019445000103000069P9916142104211022121222012202122115213332110051568842124110220300614542601066871102202124241051564128021220668711022253021241224828511022430112413001222004151438000760036634410002017000073131117878P9917若121221121,1,2,2,11,2,3,1NNNNNNJJJJNJNJNJJJJJJNNNULLNULLNULLNULLN1NN1N中則取J取取或若則故只有兩項(xiàng)123N0，2,3,N1DX1YNULLP9917111122122111221122212111121122121111112,302NNNNNDABNDABABABABABABABBAAABBABBBABABBBBBBBABBBBBNULLNULLNULLNULLNULLNULL則則當(dāng)時(shí)第列與第列成比例121221111100199170101NIMININMINMIHHNNIIIXMXXXMMPXMXMXXXMXMMMXNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULL1IM23N2111X311N111X21X31XN11HNMNULL213244121121210002222222100991700100001000002020022111101020100002IPNNNNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULL性質(zhì)2N2（N2），且N1時(shí)。D1（左上角1）199175,213212010000200001111211122NPNNNNNNNNNNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLINULLI從最后一列開始,第N列加到第N1列,再第N1列加到第N2列第列加NN1NN122P10018從第二列起有列（第三列）乖以11IA加到第一列,則有121121121221111111111000000011NOIINNNNONOIINONIINIAAADAAAAAAAAAAAAAAAAANULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLI，（0）P10018COS112COS1COS12COS112COSAADNANANULLNULLNULL證法一，用歸納法，D1成立，COS,COS212COSCOS1COS2COSCOS2COS212COSKNDKKNKDNDDNNNANANNNNAA的第二行22112,02SAAAS其余111100000AAANULLNULLNULL12120,000IIIJIJJJJNNIAAAEEAAANULLNULL00行00AIA的第I列,且,KIIIJJAA0KIAA的第J行，JJIAAI且,SJ0JSA由于A與所有N級(jí)矩陣可換，故可換與NEEEEA1131211,NULLA的第一列全為0，的第一行只留下AAAEAE111111可解非0的第二行只留下AAAEAE121222A11其余全為0的第三行只留下AAAEAE131333A11,其余全為0的第N行只留下AAAEAENN11NNA11其余全為0所以AEAAAAA111111110011AAP200T8ACBCABAACABCBAABCABCABCACBCBACABBCAP200T9222111“,202442AABBEBEBEBE1若則得4222111“,22442BEABBEEBEBEA若則P200T10反設(shè)為列的元素行中第那么那么不防設(shè)SSAAAATSST2,0,0,0222212110NNSKKSSKSKSSSTSNKKAAAAAAAANULL20A，矛盾，0A即。P200T11AABBBAABABABABAB,“為對(duì)稱矩陣那么如果,“ABBAABABBAAB。P200T12設(shè)ABC，,BBCCCBCBAAACAAB21,21恰如,即為所求CCBBP200T13令21111211222222221233111211211111,11NNNNNNNNNNNNXXXXXXXXXDXXXDXXXXXXXXNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULL1121,NIJIJIJKIJNXNKDDAAAXS221IJJIIJNAADDDXXP200T14“0IBB取為的一個(gè)非列00,0AAXABI故有非零解而“0A0AX有非零解010200,2,N0XBXBXBNXNULL令0,21BBBBBN所以組成的列由而NULLP200T15考慮AEE的每一列去乘A的各行為0,AE0IE又AEAA0P200T16考慮齊線方程組,0NXRCXCR只含個(gè)未知量,而秩只有零解0CCRCX秩未知量個(gè)數(shù)00BCBC000BCX的各列都是適合都為0,0BB若00BCCBECBEBEP200T17設(shè)12,SANULL的行向量為（I）,B的行向量為SNULL21,（），而,21SBACNULL的行向量為（）。那么111,R222,RNULLMMRM。設(shè)1,IIRNULL為（）的極大無(wú)關(guān)組，那么秩（A）秩（）R設(shè)1,JJPNULL（）為（）的極大無(wú)關(guān)組，那么秩（A）秩（）PUU11,IIRIJPNULLNULLNULL秩（AB）秩（C）秩（）秩（）RP秩（A）秩（B）。P200T18設(shè)秩（A）R,那么，線性方程組AX0的基礎(chǔ)解系可設(shè)為RNNULL21,。設(shè)B的各列為B1,B2BMAB0說(shuō)明B的每列BJ乘以A的每行都為0，即時(shí)BJ是AX0的解。12,JNRBNULL1212,NNBBBRNULLNULL秩（B）秩1212,NNRBBBNRNULLNULL秩ABRNRN秩秩KKP200T19若AK0212121KKEAEAAAEAAAAAAANULLNULLNULL0KEAEE121KAAAEAENULLP201T20,1DBAACAACBDA11231112100110,AAAA12134XXAXXX令112312243132AXAXAXAXAXAXAX131311322231123123130001111231AXAXAXEXAAXAXEAXEX存在而1122442120AXAXXAAX411221311,1,1233X1110331203321133A4613513411203602121011322332313322212312111AAAAAAAAAA即143153164A46135134111AAAA123423123111102612241234231211111026AAAAA113221312121113443120EOAAAAAAEAAAA131143211323152AAAAAAB4111121213431412341000231201000031111000523201AAEOAAAAAAEOBA132261710002262617212013010017520130021111010210053320141535法1EA42AA411法2100102200101202000112200000011111100011110100111100101111000111110011220011002200100102201001200200112200010120201001022000022222111140002200440002024040200240041111400011002200010120201001200240001111040011111|00401111400041111EAA41411A334310751001061106110100,5421043726501523320317122003AAE1075100101201430120033234106004330703310007512140100325800104130691110001594399159A151219325841306911159439915971357012300120001A111131357120101230112001A10002100721038113182100320057181316A1111210321857211816133216A1210021032003257345768111221122222001110001221000103140100031401002761001000111000122100010321001212210001120120010314010003051100001110000011100000130112000211121205110022273503005222311001012221110001122271011100062637553101006263310010122211100011222A113720735101933663336求A1A方法1令B12100NAANULLB1EC1ECC2C3C4A12B112B112C12C212C312C4111112481632111124816111248112412方法2A2100002100210021002BDOCA01111BBOCOC111241021182411024BDC1111124816321111024816111002481100024100002P201T21設(shè),KXKA123400RRYYACXYYYC則令才能乘,XYYX與3412AYAYXYCYCY而2143YCYAYXYCYA11114440,00,0,00CYYAYYXXYYXEAYYCY若則13331222KRAYYAEYACYYCEYC11100CXAP201T22將00NAXA121NAAAANULL由21題，（見上面）1111112111000000000NNNAAAXAAANULLNULL0P202T232153315211254635462231321122108X解02112001122011111111201101122011111101321000213101012161100103230002121110123212010132021611216111BAXAXB，則XA1B，故11111210000111112100,001110120000010012ABNULLNULLNULLNULLNULLNULLNULLNULLNULLNULL所以1111011111101XAB2NULLNULLNULLNULLNULL4,111022110A11213112223213233321621222AAAAAAAAAA421112142281111021242266211222458XP202T24EAAAA11EAAAA1111AA若對(duì)稱即則1111,AAAAAA若反對(duì)稱即則11111,AAAAAAA若AKKAAAN由那么,1NAAAAA0A于是A不可逆。P202T25若A,B上三角形，則IJAA，,0,IJIJIJBBIJAB0當(dāng)時(shí)0時(shí)，IJ當(dāng)1111100NININIJIKKJIKKJIKKJKJIKKKKIKKICABABABBACAB為上三角,IJIJABAABB若為下三角形則,0,IJIJIJAB,0的秩符號(hào)差則可逆使其中都是的齊次一次式1211221122“”,NNNNFXXXAXAXAXBXBXBX設(shè)12121212,0NNAAAAABBBBB若線性無(wú)關(guān),不妨設(shè)令212211222111YYFXYXBXBXBYXAXAXAYIINNNN則02,2221212211符號(hào)差秩為則令CZZFZYZZYZZYII1112222121,0,1NNNNYAXAXAXYXKAYXFKY若線性相關(guān),不妨設(shè)及令則秩為233123123996247161030243071990,128340,2067267228816196035880104122421412141100201640,2067228816196035880PAPPPAAPPPAXAX故A正定，二次型12,NFXXX正定這里順便發(fā)現(xiàn)一個(gè)等式2111112111112111112111112210002100012100012100012P2337判別1212111,NNNIIIIFXXXXXX是否正定。證121021121211210211A由斜行列式,1110111112,42222KKKKKKNKPPPPPPPP12KNKKPP2121212121111KKPP212121122KKKKPP212111101,2,2KKKPKNXXXCFAN正定正定,2123322123112811213510,10,541545044110,055405TPATPPTPTTTTTTTTA2即又因?yàn)闊o(wú)公共解0即對(duì)任何都有主子式大于T0233121290,KKPAAAIIIIIIKNULLNULL正定的主子式全大于證明此時(shí)的順序主子式也大于0。所以A正定（定理）任取的行，列作成一個(gè)階主子式111212122212,KKKKKIIIIIIIIIIIIIIIIIIAAAAAABPBAAA設(shè)12,KIIIFXAXXXX作一個(gè)關(guān)于的二次型1212,0,0,0,0,0KKIIIIIIGXXXFXXX1212,KKIIIIIIXXXXXBX12121,0,0,0,00,0,0KKKIIIIIIIIBGXXGCCCFCC是的矩陣，因?yàn)槿谓oX0BKB為的正定矩陣，有P233,10,，IJNNAA證設(shè)TEAL那么的第個(gè)順序主子式KKKKKKKKKKKRBTBTATAAAATAAAATTP1,1212222111211是一個(gè)T的多項(xiàng)式函數(shù),且NULLLIM,0KTPTM,KKKNTNPTMNULL當(dāng)后，恒有012012MIN,000NNNNNNTNPTPTPTNULLNULLNULLNULL取則當(dāng)，恒有正定ATEP23311A正定,證明A1正定證可逆使存在正定可逆CCCCAAEEACCEACC11即111111,CACEGCGCCCGAGEAEA1NULL取那么則正定234577720012,1000,000,0,0,000,000PTEATPTEATEATEAXEAXXEAXXXXXXXAXXX作12FFFXABX也有任0,0XFXAXXBX所以F正定,即AB正定P23414PRAXXF慣性指數(shù)秩0證設(shè)XCY,使22122221RPPYYYYYAXXF“充分性若PR,則負(fù)系數(shù)平方項(xiàng)不出現(xiàn)“任取1100022100010,0NRCXYCXFXCFXXXAXYYF0必有在的值為半正定“必要性,反設(shè)”1200000,0RRRNRCCPRYXCYCNULL半正定即于的所有主子式大于或等AXXFCA0P2341證首先12,XX,線性無(wú)關(guān),反設(shè)線性相關(guān),不妨設(shè)有21,XXRKKXX,12221211XAXKXAKXKXAX0與220XAXNULLNULLNULLNULLNULL作一個(gè)被為元二次型那么任取必有是一個(gè)元正定二次型必有112222112121122222211112,RRRRRNRNRNNRRSYZYZGFYYYCCYZZYYGOGOZCOEOEYZFYYXCYCCZ使令可則且12CZCCCFRA2Z逆NULL可逆的正慣性指數(shù)秩P234補(bǔ)3先講補(bǔ)2,221221QPPPLLLLF證設(shè)NINIIIXAXAXAL2211并設(shè)XCY,C可逆使22122221QPPPYYYYYF那么反設(shè)110000PPNLLPPYY作線性方程組XCYXBXBYNINIII11共有PNPNPPN不全為矛盾,所有PP同理,負(fù)慣性指數(shù)QQ另推論如本例形式二次型,例秩RQP1112212211111211111221212221212211111211214,AAEXTATATAAOEAAXAAAXAEXEOAAAAAXAOEXEAAXAXAA235P補(bǔ)證明0設(shè)1112111121221111121121111212111210,AXAXAAAAXAAAAAAAAAA則取EOAAET即合要求12111P235,補(bǔ)5,若N1,顯然A0若N2,A0,顯然1100000,10101AAAO則成立【】由于A2仍為反對(duì)稱故歸納假設(shè)A20001100110222TAT2,22101010110011000STTAECPPTTCAC取則證畢P235補(bǔ),6由習(xí)題第10題57772,一定存在120,CTC當(dāng)時(shí)C1EA永為正定取120000000000MAX,0,020,CCCCEACEA0XXCEAXCXXXAXXCEAXCXXXAXCXOXOXAXOCXOXOXXAXCXX推出所以D正定,即A正定,證畢P236補(bǔ)8,1,11100001AYAYAEAYFAYAYYYYYAY0,46765111值故對(duì)任一組題見習(xí)題第正定已知正定YPIAJCIA11120,0,0NYAFYYYAYAYAF是負(fù)定二次型2設(shè)1120,0NNNNAAAABBAA112111NNNNNNN1ABBAAAAAAQP3NNNNNNNNNNNNNNNNAAAAAAAPAAAQAPAACPAACA,122111122122332111111即則如此下去仍然證定4212222121111,IINKIKNNNNIIKIIINIKIIIATTTETAAT2ATATTTT作則正定且23612120000NNPD0ACCACDIDCADDDANULLNULL補(bǔ)9（必要性）可逆,1212111111,0,0,0,KKKKKKKIIIIIIIIIIIIIIIIGXXXFXXXAAGXXGAAA作則半正定,的矩陣為0011AA11121222211121NNNNNKNKKNNNNAAAAAADAAAAAA中的系數(shù)KA這樣取到在0中主對(duì)角線上任取NK項(xiàng)中的的系數(shù)項(xiàng)所在的行和列,得一個(gè)K級(jí)子式含入DK,由于是KNKA即令中只能取所有的常數(shù)項(xiàng)的子式故的系數(shù)CDKCKN0,NKKDKDKD中的這正是的一個(gè)級(jí)主子式,要是的系數(shù)中的一元,故A為的所有K階主子式之和,如1112NNAAAA等110,RKKKKAEMAAAEKBB現(xiàn)在考慮任意它的階順序主子式,為的右上角的M階方陣,作0,00,0LIM0KIIKKIKKKBAIAABAEAEXXAEXXAEZXAXA由于的系數(shù)是的一切階主子式之和,而的主子式仍為的主子式,由充分條件,因此正定,故對(duì)任何邊續(xù)性所以半正定第六章線性空間習(xí)題解答P2671設(shè),MNMNMMN證明N,XMXNMXMNMMNXMNXMMNMMNMMNM證XNXMNNMNXMNXNXMNXNMNN又或P2672證LMNMLNMLMNMLNM12XXMXNLXMXNXLXMNXMLXXMXNLXMXNXLXMNXMLX證左且且或或右反過(guò)來(lái)。證畢證左或或且且右。證畢1P2673不做成,因?yàn)?個(gè)N次多項(xiàng)式相加不一定是N次多項(xiàng)式,如122UUXXXXX做成,因?yàn)?122,FAGAHAHXFXGXKFAHAHXKFX為多項(xiàng)式為多項(xiàng)式,做成因?yàn)閷?shí)對(duì)稱反對(duì)稱,上三角下三角之和之倍數(shù)仍為實(shí)對(duì)稱反對(duì)稱,上三角下三角故做成線性空間不做成,設(shè)|V為平面上不平行的向量不做成,違反定義351100，但這里。取即得矛盾。P26732111112121212211121,AKKKBKABAKAABBAABABANULL解顯然V非空10以及2個(gè)封閉的代數(shù)運(yùn)算02驗(yàn)證先設(shè)03RTKBARBABA,332221及21212112312123123123123231231223231231,2,AABBAARAAABBABAAAAAABBBAARAAABBBBAAAAA12312323121311111211121111111211111,300,0,00,00,4,0,001511,1111,2AAABBBAAAAAARABAABAABAABABAAABAABNULLNULLNU