用配方法求二次型f=2x1x2-x1x3+x1x4-x2x3+x2x4-2x3x4的标准形．
f=2x1x2-x1x3+x1x4-x2x3+x2x4-2x3x4 =2x1x2-(x1+x2)x3+(x1+x2)x4-2x3x4． 为了产生平方项，需要先作可逆线性变换 (x1 x2 x3 x4) = (1 1 0 0 1 -1 0 0 0 0 1 0 0 0 0 1) (y1 y2 y3 y4)， 即 {x1=y1+y2 {x2=y1-y2 {x3=y3 {x4=y4 将上式代入原二次型得 f=2y12-2y22-2y1y3+2y1y4-2y3y4 =2(y12-y1y3+y1y4)-2y22-2y3y4 =2[y1-(1/2)y3+(1/2)y4]2-(1/2)y32-(1/2)y24+y3y4-2y22-2y3y4 =2[y1-(1/2)y3+(1/2)y4]2-2y22-1/2(y3+y4)2． 再经可逆线性变换 {z1=y1-(1/2)y3+(1/2)y4 {z2=y2 {z3=y3+y4 {z4=y4 就可得到标准形f=2z-2z-(1/2)z．
