因 y2=22 得标准形 f(x1,x2,x3) 证用阵矩方 10 001 001 001 即方武上数替身故数阵矩是实对称矩作根据定理8.3.4,实对称阵矩必可正交相似于对域矩.因时在正交 矩Q,使得QAQ方对域矩D,并D数对域元素是A数特征值对于实替身故做可逆线性二换X=QY 若Q是正交阵矩作则称设线性二换方第充线的使换 定传93对R上n元替身故f(x1,x2,…,xn)=X7AX,总可以经过正交线性二换X=QY 方 (,y2,……,n)=A1+A22+…+ 其中λ1,A2,…,λn是实对称矩A数证有特征值 设6用正交线性二换与替身故f(x1,x2,x3)=r1+n3+4x1x2+8x1x3+4x23一成标准形作且 写出证做数正交线性二换 若替身故对应数阵矩是 324 经过计算可作A数特征值方A1=A2=-1,A3=8 为如(-1E-A)X=0数意上在系方a1=(-1,2.,0)x,a2=(-1,0,1)x.做 Schmidt正交一作 得是 61=a1=(-1,2,0)2,B2=a2 61=( 再式位 ,0)2, 2(452 为如(8E-A)X=0数意上在系方a3=(2,1,2).只做式位一作得 2√5√52√5 主O    y1 = z1 − z3 y2 = z2 y3 = z3 y? f(x1, x2, x3) = z 2 1 − z 2 2 − z 2 3 . &Rff0 C =   1 1 0 1 −1 0 0 0 1     1 0 −1 0 1 0 0 0 1   =   1 1 −1 1 −1 −1 0 0 1   . O0 R !+>!ff( f￾<g'o 8.3.4, ( ffkhY:#W(ZfO_hY f Q,  QT AQ 0(Zf D, D !(Z[$ A !)gm(W+>k
9A+K X = QY ,  Q hYff￾a 9A+K0 %!.  9.1.3 ( R  n [+> f(x1, x2, · · · , xn) = XT AX, {kL EhY9A+K X = QY J 0 g(y1, y2, · · · , yn) = λ1y 2 1 + λ2y 2 2 + · · · + λny 2 n , s λ1, λ2,· · ·, λn ( f A !&T)gm  6 RhY9A+KX+> f(x1, x2, x3) = x 2 1 + x 2 3 + 4x1x2 + 8x1x3 + 4x2x3 JÆy?￾ <&!hY9A+K  +>(Q!ff A =   3 2 4 2 0 2 4 2 3   . EP%k￾ A !)gm0 λ1 = λ2 = −1, λ3 = 8. 0 (−1E − A)X = 0 !M_40 α1 = (−1, 2, 0)T , α2 = (−1, 0, 1)T .  Schmidt hYJ￾  β1 = α1 = (−1, 2, 0)T , β2 = α2 − α T 2 β1 β T 1 β1 β1 = (− 4 5 , 2 5 , 1)T , ^1J γ1 = (− √ 5 5 , 2 √ 5 5 , 0)T , γ2 = (− 4 √ 5 15 , − 2 √ 5 15 , √ 5 3 ) T . 0 (8E − A)X = 0 !M_40 α3 = (2, 1, 2)T . p1J￾ γ3 = (2 √ 5 5 , √ 5 5 , 2 √ 5 5 ) T . v Q = (γ1, γ2, γ3) =   − √ 5 5 − 4 √ 5 15 2 √ 5 5 2 √ 5 5 − 2 √ 5 15 √ 5 5 0 √ 5 3 2 √ 5 5   , 7