合同矩阵——线性代数VIP免费

合同矩阵合同矩阵一.实对称方阵的合同变换.T12(,,...,)XAXnfxxx设P是一个n阶可逆方阵,令:X=PY,则:XTAX=(PY)TA(PY)=YT(PTAP)Y12(,,...,)ngyyy实二次型:核心问题:对于给定的实对称方阵A，如何寻找一个可逆方阵P，使得PTAP成为对角方阵?这样的对角方阵具有什么特点呢?1,11,21,31,2,12,22,32,3,13,23,33,,1,2,3,.........A...nnnnnnnnaaaaaaaaaaaaaaaa(1)若a1,101,11a1,11a设A是一个n阶实对称方阵.1,21,31,1,11,11,12,12,22,32,1,13,13,23,33,1,1,1,2,3,1,11............nnnnnnnnaaaaaaaaaaaaaaaaaaaaa2,22,32,3,23,33,,2,3,100...00...0...0...nnnnnnbbbbbbbbb(2)若a1,1=0,而ak,k01,21,31,2,12,22,32,3,13,23,33,,1,2,3,0.........A...nnnnnnnnaaaaaaaaaaaaaaa无妨说a3,303,13,23,33,2,12,22,32,1,21,31,,1,2,3,......0......nnnnnnnnaaaaaaaaaaaaaaa3,33,23,13,2,32,22,12,1,31,21,,3,2,1,......0......nnnnnnnnaaaaaaaaaaaaaaa13rr13cc(3)若a1,1=a2,2=…=an,n=0,而ai,j01,21,31,2,12,32,3,13,23,,1,2,30...0...0...A...0nnnnnnaaaaaaaaaaaa无妨说,an,301,21,31,2,12,32,3,1,13,2,2,33,,1,2,30...0...0...0...0nnnnnnnnnaaaaaaaaaaaaaaa1,21,31,1,2,12,32,2,3,1,13,2,2,33,,1,2,30...0...2......0nnnnnnnnnnnaaaaaaaaaaaaaaaaa3nrr3ncc2,22,32,3,23,33,,2,3,100...00...0...0...nnnnnnbbbbbbbbb3,33,,3,100...0010...000...00...nnnnccccEEOst其中s+t=R(A)，称s为该二次型的正惯性指数；称t为该二次型的负惯性指数；s–t为该二次型的符号差。即,我们用了一系列初等方阵P1,P2,…,Pl,使得:PlPl-1…P2P1AP1TP2T…Pl-1TPlTEEOst令P=P1TP2T…Pl-1TPlT=PTAP那么如何在变换过程中自动记录P呢？AETTPAPP例1.利用合同变换把对称方阵111A123136化为标准形.解：111100123010136001100100012110025101100100010110,001121111P012001即T3PAPE例2.利用合同变换把对称矩阵21341121A32334134化为标准形解：213410001121010032330010413400011121010012341000233300101434000112rr12cc21;rr312;rr41rr21;cc312;cc41cc100001000113110001110210031301011000010001001100002211100026320110000100010011000020111000042311100001000100110000101/21/21/20000113/21/21/212121212011/21111/23/2P001/21/20001/2令:则2T2EPAPE例3.利用合同变换把对称方阵011A103130化为标准形.解：01110010301013000121211010301023000120011001/221/21/2002211120011001/201/21/200063111001/21/200101/21/200013/61/61/61001/21/200103/61/61/60011/21/201/23/61/2P1/21/61/201/60...