高等代数答案

第一章多项式习题解答P4411172623999FXGXXX22157FXGXXXXP4421231|9XMXXXQ余式21PMXQM021MQPQ方法二，设32011MQXPXQXMXQMQP同样。22421|XMXXPXQ余式2221MPMXQPM0220MMP221,1MPQXPQP4431用3GXX除53258FXXXX解543223303175349536673327FXXXXXXP44323232122812XXXIIXIX1281298IXII即余式98I商2252XIXIP444150,1FXXX即5432151101101511FXXXXXX当然也可以5511FXXX54151XX1P4442结果3424322328222224211FXXXXXXX4322137FXXIXIXXI432432213721575XIIIXIIIXIIXIIIXIIXIIXIXIIP45512212111GXXXXXX1313XXXXF,1FXGXX23231GXXX不可约不可约1434XXXF,1FXGX31221221102224XXXXXXXF432322426621,4222221GXXXXXFXXXXXX22,221FXGXXXP45612122XXXF2221GXXXX221112XXXXX12212XXFXXGX2，1424123YXXXXXF2124GXXXX11XFX11XGX而1112323231FXGXXXGXXX111211231133X1XXGYFXG212111333XXFXXXGX3,144234XXXXXF21GXXX,232FXGXXX211GXXX2131FGXXG32132XFXXXXGXP4572112FXGXTXTXURX2222122211111TTTUTTGXRXXXUTTTT由题意|RXXRXGX与G的公因式应为二次所以0130143322223UTTTTUTUTT0304331223UTTUTUTTXRT为一次的否则解出当1404330223TTTTTTU时32314ETT或311,03,02TTTTU只有时当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是与的组合DX所以是一个最大的公因式。DXP45，9证明,FXHXGXHXFXGXHX（的首系1）HX证设,FXHXGXHXMX由,|FXGXHXFXHX,|FXGXHXGXHX,|FXGXHXMX,FXGXHX是一个公因式。设,DXFXGXUXFXVXGX,DXHXFXGXHXUXFXHXVXGXHX而首项系数1，又是公因式得（由P45、8），它是最大公因式，且,FXGXHXFXHXGXHXP45、10已知,FXGX不全为0。证明,FXGXFXGXFXGX1证设,DXFXGX则0DX设1,FXFXDX1,GXGXDX及DXUXFXVXGX所以11DXUXFXDXVXGXDX消去得0DX111UXFXVXGXP4511证设11,0,FXGXDXFXFXDXGXGXDX111,1UXFXDXUXGXDXDXUXFXUXGX1P4512设21111111,11UFVGUFVHUUFUFVHVGUFVUGH1111UUFUVHVGUFVUGHFGH,P4513,G1IIF，固定12,1IIFGG12,1INFGGGP4514,111,1FGUFVGUVFVGFFGF同理,1GGF由12题,1FGFG令12NGGGG,1IIFG每个11,1FFG,1123,FFFG,（注反复归纳用12题）。1212,MNFFFGGGNULLNULL1推广若则,1,FXGXM，N有,1MNFXGXP45,15FXX32X22X1,GXX4X32X2X1解GXFXX12X2X1,FXX2X1X1即FX，GXX2X1令X2X10得231,23121IIFX与GX的公共根为21,P4516判断有无重因式543257248FXXXXXX34424XXXXF解4421205234XXXXXF3251325412FXFXXXXX232215492541254422FXXXXXXX324412452223XXXXXX故有重因式XF32X4843XXXF3221233FXXXXXX1311323322XXXXXF2661123311132331111XXXX131XFXFP4517有重因式（有重根）1323TXXXXFT时解TXXXF63236213TXTXXFXF如则有重因式3重因式3T13XFX如则3T21522153222TXXF此时必须415T有重因式4212XXXFP4518求多项式有重根因式的条件QPXXXF3证PXXF2323323FXXPXPXQ0P22223327323244AQXPPXQXPPPP32427PQ0P4519令4221,1|,1FXAXBXXFXXFX因为所以即12423CBXAXXBXAXXF4020AABACBBC0224BABBAA又1123DXCXBXAXXFXFX01AABACBD0BBABAA101BA2,1BAP46，20证212NXXFXXNNULL无重因式（重根）证NXFXFXN,YXFFFN,1FX,1NFXFX无重因式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“”必要性显然（见定理6推论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则是的重根根0XFX但不是的P4624若11|NNNXFXXFX则证若12FXXGXR由上节课命题10NNNFXXGXRGXRR所以|1NNXFXP46,25证明设X2X1的两个根312,1I21233111213322222100XXXXFFFF112122110110FFFF即211010FF121,XFXFXP46、26分解1NX0221COSSIN,0,12,KKKIKNN1NNULL设1101221111212122,11COSSIN,11122112COS1221112COS1NIIKMMNKKKKMKMNKKKIXXXXNNIINXXXXXXKNMXXXNKNMXXXXXNNULLNULL在中在中为奇当时1NKP46,27求有理根（1）X36X215X14FX解有理根可能为1、2、7、14。当A首项系数则FXP为某不可约多项式X的方幂的充要条件是GX或者,1FG或者|MMFXGX证明11“,0,RFXPXHXHXPXHX反设不是则而111,1,|,1,PHHPXPXFG即取G则且,|,MSMFGHPX否则矛盾“”,1,1,1|RRR1FPGXPGPGFGPGPGFGX若若4881,0,PFXFXFX首项系数则为某不可约多项式的方幂|,|,/MGXHXFGHFGMFXHX由或者证明“,|,1,1,1|RRFPGHPHPHFHF设若FG,1|RRRPHPHPHFHXM111“,0,FPHHPHPHHX反设不是则而令G则|,1,1MRFGHFGFHMPHPH却P48，补9证2的非零根NNMXAXB没有重数证反设,NNMFXXAXBK有重根K2，0111/1/00010NNMNMMMMGXFXNXANMXKANMNXXNANMHXXNHXMXHXHXHX有重根有重根有重根无重数根但重根P48、补100,1NFXCXFXFXN且,0MFX证明的根只能为或单位根即满足某X1的根N证设为FX的根,由FXFXGXFXNNF0,为的根,220,NNFFX为的根,23,NNNNULL都为FX的根0,FXFX不可能有无限个根,其中必有相等者IJNNIJ不妨设,210,IJNNNIJNNM令0,1MX则或或是的根P48、补11题/NFXFXFXAXBFX有N重根B,NAAAAAANULLNULL1212N补充P4812题的两两不同FXXXX证（1）1111111NIIIIIIIIIINXAXAXAXAFXFXLFAFAXAFAAAAAAAAANULLRA11110,11,1,11,1,2,1NNIJIIJJIININIIJJRJLALALAJNILXINNLXFXFXQRXFANFALXHAFARAHXRXNULLNULL，为次多项式设则而形式多项式P49、补13题（1）求423314052FXFXFFFF2时，至少去掉3个X,不含X3项了，对于I2,J2同理其它情形，至少去掉两个X且第一行（或第二列）的两个X只能取一个，故不含X3项，只剩下I1,J2时，A12本身是X项为12134A12A21A33A44X1XXX3,系数为11212,P971111011,NNJJJJJJDNULLNULLI故中与一样多即号,号一样多,也即奇偶排列一样多,N2时,奇偶排列各占一半P981221111121111111NNNNNNXXXAAAPXVXAAANULLNULLNULLNULLNULLNULLNULLNULLNULL按第一行展开1211,0NNAAAVPXNNULL两两不同即12112101,NNPAPAPANAAANULLNULL总有两行相同最多个根,即PX的所有根为111321332313222339813222202011111122XYXYXXXXYPYXYXYXYXYXYXYYXYXYXXYYXYXYYXYXXYXYNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULL22334413311113114813341131624811311113PPNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULL例P9813（法一）123412341234123423410127012701271603412018100044004441230710130043600040法二10234123412341234103470113011302001020201041202220200001310123011101110011123402002016000130004法三令2412312341234321234,41,1,1,111110222222022160CFXXXXIIIDFFFIFIIIIIIII为次单位根令则行列式P9813解2211111111111111011111000111101111100011110111110001111011111000XXXXXYYYYY1,21,2111011111111111|,1111111111111111,1111111111111111XXXXXYYXFXYXXFXYFXYYYNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULL交换行再交换列解法二,设FX,Y则第一行减第二行又因为FX,Y关于X是偶函数，即X2|FX,Y同理，Y|FX,Y，且也是偶函数，所以Y2|FX,Y2222222,4|,FXYXYFXYFXYKXYFXYXYY2而中的系数为1故有FX,YXP981322511222311323411421446921262144692126021446921262144692126XAAAAAABBBBBBCCCCCCDDDDDDNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULL性质2133246122134272246427100001101454310076811611003427211005882941NULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULL700110010010001161100111610001001001294029400000029411294011161000性质111111111111211112311113122222222222212111211211229814222ABCCAABABCBCPABCCAABABCBCABCCAABABCBCABCABCABCNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULL左2213121NULLNULLNULLNULLNULLNULLNULLNULLNULLNULL右右P9815求出所有代数余子式111213142122232431323334414243441214600001211260000211563000037012AAAAAAAAAAAAAAAA直接计算有1127123321641014555111213212223313233AAA直接计算有AAAAAAP99161351213512135121222012106145012142313401214012140611452112533121012165701216572103507105907105913512263310121422438400119275300817410041219135120121442300119450086311100028571351201214001101900073000285715121101214217694830011019445000103000069P9916142104211022121222012202122115213332110051568842124110220300614542601066871102202124241051564128021220668711022253021241224828511022430112413001222004151438000760036634410002017000073131117878P9917若121221121,1,2,2,11,2,3,1NNNNNNJJJJNJNJNJJJJJJNNNULLNULLNULLNULLN1NN1N中则取J取取或若则故只有两项123N0，2,3,N1DX1YNULLP9917111122122111221122212111121122121111112,302NNNNNDABNDABABABABABABABBAAABBABBBABABBBBBBBABBBBBNULLNULLNULLNULLNULLNULL则则当时第列与第列成比例121221111100199170101NIMININMINMIHHNNIIIXMXXXMMPXMXMXXXMXMMMXNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULL1IM23N2111X311N111X21X31XN11HNMNULL213244121121210002222222100991700100001000002020022111101020100002IPNNNNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULL性质2N2（N2），且N1时。D1（左上角1）199175,213212010000200001111211122NPNNNNNNNNNNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLINULLI从最后一列开始,第N列加到第N1列,再第N1列加到第N2列第列加NN1NN122P10018从第二列起有列（第三列）乖以11IA加到第一列,则有121121121221111111111000000011NOIINNNNONOIINONIINIAAADAAAAAAAAAAAAAAAAANULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLI，（0）P10018COS112COS1COS12COS112COSAADNANANULLNULLNULL证法一，用归纳法，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与所有N级矩阵可换，故可换与NEEEEA1131211,NULLA的第一列全为0，的第一行只留下AAAEAE111111可解非0的第二行只留下AAAEAE121222A11其余全为0的第三行只留下AAAEAE131333A11,其余全为0的第N行只留下AAAEAENN11NNA11其余全为0所以AEAAAAA111111110011AAP200T8ACBCABAACABCBAABCABCABCACBCBACABBCAP200T9222111“,202442AABBEBEBEBE1若则得4222111“,22442BEABBEEBEBEA若则P200T10反设为列的元素行中第那么那么不防设SSAAAATSST2,0,0,0222212110NNSKKSSKSKSSSTSNKKAAAAAAAANULL20A，矛盾，0A即。P200T11AABBBAABABABABAB,“为对称矩阵那么如果,“ABBAABABBAAB。P200T12设ABC，,BBCCCBCBAAACAAB21,21恰如,即为所求CCBBP200T13令21111211222222221233111211211111,11NNNNNNNNNNNNXXXXXXXXXDXXXDXXXXXXXXNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULLNULL1121,NIJIJIJKIJNXNKDDAAAXS221IJJIIJNAADDDXXP200T14“0IBB取为的一个非列00,0AAXABI故有非零解而“0A0AX有非零解010200,2,N0XBXBXBNXNULL令0,21BBBBBN所以组成的列由而NULLP200T15考虑AEE的每一列去乘A的各行为0,AE0IE又AEAA0P200T16考虑齐线方程组,0NXRCXCR只含个未知量,而秩只有零解0CCRCX秩未知量个数00BCBC000BCX的各列都是适合都为0,0BB若00BCCBECBEBEP200T17设12,SANULL的行向量为（I）,B的行向量为SNULL21,（），而,21SBACNULL的行向量为（）。那么111,R222,RNULLMMRM。设1,IIRNULL为（）的极大无关组，那么秩（A）秩（）R设1,JJPNULL（）为（）的极大无关组，那么秩（A）秩（）PUU11,IIRIJPNULLNULLNULL秩（AB）秩（C）秩（）秩（）RP秩（A）秩（B）。P200T18设秩（A）R,那么，线性方程组AX0的基础解系可设为RNNULL21,。设B的各列为B1,B2BMAB0说明B的每列BJ乘以A的每行都为0，即时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设,KXKA123400RRYYACXYYYC则令才能乘,XYYX与3412AYAYXYCYCY而2143YCYAYXYCYA11114440,00,0,00CYYAYYXXYYXEAYYCY若则13331222KRAYYAEYACYYCEYC11100CXAP201T22将00NAXA121NAAAANULL由21题，（见上面）1111112111000000000NNNAAAXAAANULLNULL0P202T232153315211254635462231321122108X解02112001122011111111201101122011111101321000213101012161100103230002121110123212010132021611216111BAXAXB，则XA1B，故11111210000111112100,001110120000010012ABNULLNULLNULLNULLNULLNULLNULLNULLNULLNULL所以1111011111101XAB2NULLNULLNULLNULLNULL4,111022110A11213112223213233321621222AAAAAAAAAA421112142281111021242266211222458XP202T24EAAAA11EAAAA1111AA若对称即则1111,AAAAAA若反对称即则11111,AAAAAAA若AKKAAAN由那么,1NAAAAA0A于是A不可逆。P202T25若A,B上三角形，则IJAA，,0,IJIJIJBBIJAB0当时0时，IJ当1111100NININIJIKKJIKKJIKKJKJIKKKKIKKICABABABBACAB为上三角,IJIJABAABB若为下三角形则,0,IJIJIJAB,0的秩符号差则可逆使其中都是的齐次一次式1211221122“”,NNNNFXXXAXAXAXBXBXBX设12121212,0NNAAAAABBBBB若线性无关,不妨设令212211222111YYFXYXBXBXBYXAXAXAYIINNNN则02,2221212211符号差秩为则令CZZFZYZZYZZYII1112222121,0,1NNNNYAXAXAXYXKAYXFKY若线性相关,不妨设及令则秩为233123123996247161030243071990,128340,2067267228816196035880104122421412141100201640,2067228816196035880PAPPPAAPPPAXAX故A正定，二次型12,NFXXX正定这里顺便发现一个等式2111112111112111112111112210002100012100012100012P2337判别1212111,NNNIIIIFXXXXXX是否正定。证121021121211210211A由斜行列式,1110111112,42222KKKKKKNKPPPPPPPP12KNKKPP2121212121111KKPP212121122KKKKPP212111101,2,2KKKPKNXXXCFAN正定正定,2123322123112811213510,10,541545044110,055405TPATPPTPTTTTTTTTA2即又因为无公共解0即对任何都有主子式大于T0233121290,KKPAAAIIIIIIKNULLNULL正定的主子式全大于证明此时的顺序主子式也大于0。所以A正定（定理）任取的行，列作成一个阶主子式111212122212,KKKKKIIIIIIIIIIIIIIIIIIAAAAAABPBAAA设12,KIIIFXAXXXX作一个关于的二次型1212,0,0,0,0,0KKIIIIIIGXXXFXXX1212,KKIIIIIIXXXXXBX12121,0,0,0,00,0,0KKKIIIIIIIIBGXXGCCCFCC是的矩阵，因为任给X0BKB为的正定矩阵，有P233,10,，IJNNAA证设TEAL那么的第个顺序主子式KKKKKKKKKKKRBTBTATAAAATAAAATTP1,1212222111211是一个T的多项式函数,且NULLLIM,0KTPTM,KKKNTNPTMNULL当后，恒有012012MIN,000NNNNNNTNPTPTPTNULLNULLNULLNULL取则当，恒有正定ATEP23311A正定,证明A1正定证可逆使存在正定可逆CCCCAAEEACCEACC11即111111,CACEGCGCCCGAGEAEA1NULL取那么则正定234577720012,1000,000,0,0,000,000PTEATPTEATEATEAXEAXXEAXXXXXXXAXXX作12FFFXABX也有任0,0XFXAXXBX所以F正定,即AB正定P23414PRAXXF惯性指数秩0证设XCY,使22122221RPPYYYYYAXXF“充分性若PR,则负系数平方项不出现“任取1100022100010,0NRCXYCXFXCFXXXAXYYF0必有在的值为半正定“必要性,反设”1200000,0RRRNRCCPRYXCYCNULL半正定即于的所有主子式大于或等AXXFCA0P2341证首先12,XX,线性无关,反设线性相关,不妨设有21,XXRKKXX,12221211XAXKXAKXKXAX0与220XAXNULLNULLNULLNULLNULL作一个被为元二次型那么任取必有是一个元正定二次型必有112222112121122222211112,RRRRRNRNRNNRRSYZYZGFYYYCCYZZYYGOGOZCOEOEYZFYYXCYCCZ使令可则且12CZCCCFRA2Z逆NULL可逆的正惯性指数秩P234补3先讲补2,221221QPPPLLLLF证设NINIIIXAXAXAL2211并设XCY,C可逆使22122221QPPPYYYYYF那么反设110000PPNLLPPYY作线性方程组XCYXBXBYNINIII11共有PNPNPPN不全为矛盾,所有PP同理,负惯性指数QQ另推论如本例形式二次型,例秩RQP1112212211111211111221212221212211111211214,AAEXTATATAAOEAAXAAAXAEXEOAAAAAXAOEXEAAXAXAA235P补证明0设1112111121221111121121111212111210,AXAXAAAAXAAAAAAAAAA则取EOAAET即合要求12111P235,补5,若N1,显然A0若N2,A0,显然1100000,10101AAAO则成立【】由于A2仍为反对称故归纳假设A20001100110222TAT2,22101010110011000STTAECPPTTCAC取则证毕P235补,6由习题第10题57772,一定存在120,CTC当时C1EA永为正定取120000000000MAX,0,020,CCCCEACEA0XXCEAXCXXXAXXCEAXCXXXAXCXOXOXAXOCXOXOXXAXCXX推出所以D正定,即A正定,证毕P236补8,1,11100001AYAYAEAYFAYAYYYYYAY0,46765111值故对任一组题见习题第正定已知正定YPIAJCIA11120,0,0NYAFYYYAYAYAF是负定二次型2设1120,0NNNNAAAABBAA112111NNNNNNN1ABBAAAAAAQP3NNNNNNNNNNNNNNNNAAAAAAAPAAAQAPAACPAACA,122111122122332111111即则如此下去仍然证定4212222121111,IINKIKNNNNIIKIIINIKIIIATTTETAAT2ATATTTT作则正定且23612120000NNPD0ACCACDIDCADDDANULLNULL补9（必要性）可逆,1212111111,0,0,0,KKKKKKKIIIIIIIIIIIIIIIIGXXXFXXXAAGXXGAAA作则半正定,的矩阵为0011AA11121222211121NNNNNKNKKNNNNAAAAAADAAAAAA中的系数KA这样取到在0中主对角线上任取NK项中的的系数项所在的行和列,得一个K级子式含入DK,由于是KNKA即令中只能取所有的常数项的子式故的系数CDKCKN0,NKKDKDKD中的这正是的一个级主子式,要是的系数中的一元,故A为的所有K阶主子式之和,如1112NNAAAA等110,RKKKKAEMAAAEKBB现在考虑任意它的阶顺序主子式,为的右上角的M阶方阵,作0,00,0LIM0KIIKKIKKKBAIAABAEAEXXAEXXAEZXAXA由于的系数是的一切阶主子式之和,而的主子式仍为的主子式,由充分条件,因此正定,故对任何边续性所以半正定第六章线性空间习题解答P2671设,MNMNMMN证明N,XMXNMXMNMMNXMNXMMNMMNMMNM证XNXMNNMNXMNXNXMNXNMNN又或P2672证LMNMLNMLMNMLNM12XXMXNLXMXNXLXMNXMLXXMXNLXMXNXLXMNXMLX证左且且或或右反过来。证毕证左或或且且右。证毕1P2673不做成,因为2个N次多项式相加不一定是N次多项式,如122UUXXXXX做成,因为1122,FAGAHAHXFXGXKFAHAHXKFX为多项式为多项式,做成因为实对称反对称,上三角下三角之和之倍数仍为实对称反对称,上三角下三角故做成线性空间不做成,设|V为平面上不平行的向量不做成,违反定义351100，但这里。取即得矛盾。P26732111112121212211121,AKKKBKABAKAABBAABABANULL解显然V非空10以及2个封闭的代数运算02验证先设03RTKBARBABA,332221及21212112312123123123123231231223231231,2,AABBAARAAABBABAAAAAABBBAARAAABBBBAAAAA12312323121311111211121111111211111,300,0,00,00,4,0,001511,1111,2AAABBBAAAAAARABAABAABAABABAAABAABNULLNULLNU