则 Q=Q-IAQ 008 所以,作正交线性替换X=QY,二次型f(x1,x2,x3)化为标准形 y2-v2+8y3 例7设f(x1,x2,x2,…,xn)=XAX是实二次型,若detA<0,证明:必存在一组实数a1,a an,使f(a1,a2,……,an)<0. 证明设C为可逆阵使CAC为对角阵B,由于detA<0,则dtB<0.不妨设B的主对角元素 前r个为负,则r为奇数设a=(1,,1(r个1),0,…,0),又令(a1,a2,……,an)=Ca,则 f(e n)=(Ca)A(Ca)=a Ba<o 注意到,对称矩阵在合同下的标准形不唯一.为了深入讨论问题,下节将考虑规范形 习题 1.写出下列二次型的矩阵,并求二次型的秩 (1)f(x1,x2,x3)=2x-n2+4x1x3-2x2x3; 2.求下列矩阵的二次型 222 3.设f(x1,x2,x3)=x+42+43-4x1x2+2ax1x3+2bx2x3的秩为1,求a,b的值 4.求f(x1,x2,x3)=(ax1+bx2+cr3)2的矩阵和秩 5.用矩阵初等变换的方法将二次型化为标准形 (1)f(x1,x2,x3)=2+2n2+2n3-2x1x2+4x2x3; (2)f(x1,x2,x3)=a-32+ 6.用配方法将二次型化为标准形 (1)f(x1,x2,x3)=x1x2+2x1x3; (2)f(x1,x2,x3)=2r1-4n2+4x1x3-2x2x3 7.用正交线性替换的方法将下列实二次型化为标准形.a Q T AQ = Q −1AQ =   −1 0 0 0 −1 0 0 0 8   . &L￾￾hY9A+K X = QY , +> f(x1, x2, x3) J0y? −y 2 1 − y 2 2 + 8y 2 3 .  7  f(x1, x2, x2, · · · , xn) = XT AX +>￾ detA < 0, i~_J}! a1, a2, . . ., an,  f(a1, a2, . . . , an) < 0. &  C 0k
f C T AC 0(Zf B, SW detA < 0, a detB < 0. 1 B !v(Z[$ r 904￾a r 0! α = (1, . . . , 1(r91), 0, . . . , 0)T , Vv (a1, a2, . . . , an) T = Cα, a f(a1, a2, . . . , an) = (Cα) T A(Cα) = α T Bα < 0. wM￾( ff_H-5!y?/J0s(x2*￾5℄XiwB.?  1. <5t+>!ff￾ +>!r (1) f(x1, x2, x3) = 2x 2 1 − x 2 2 + 4x1x3 − 2x2x3; (2) f(x1, x2, x3) = 2x1x2 + 2x1x3 + 2x1x4 − x3x4. 2. 5tff!+> (1)   1 1 3 0 1 3 0 −5 0 −5 3   ; (2)   −1 −2 4 −2 2 2 4 2 1   . 3.  f(x1, x2, x3) = x 2 1 + 4x 2 2 + 4x 2 3 − 4x1x2 + 2ax1x3 + 2bx2x3 !r0 1 ￾ a, b !m 4.  f(x1, x2, x3) = (ax1 + bx2 + cx3) 2 !ffGr 5. Rff"K!0,X+>J0y? (1) f(x1, x2, x3) = x 2 1 + 2x 2 2 + 2x 2 3 − 2x1x2 + 4x2x3; (2) f(x1, x2, x3) = x 2 1 − 3x 2 2 + x 2 3 + x1x3. 6. R0,X+>J0y? (1) f(x1, x2, x3) = x1x2 + 2x1x3; (2) f(x1, x2, x3) = 2x 2 1 − 4x 2 2 + 4x1x3 − 2x2x3. 7. RhY9A+K!0,X5t+>J0y? 8