用配方法将下列二次型化为标准型，并写出相应的可逆线性变换

Question

为了看起来清楚些，用 X,Y,Z,W 替代常用的 X1,X2,X3,X4

1. f(X,Y,Z)= X^2 + XY + 2 Z^2

2. f(X,Y,Z)= X^2 + 2XY + 4XZ + Y^2 - 2YZ + 2 Z^2

3. f(X,Y,Z)= XY + 4XZ - 2YZ

4. f(X,Y,Z,W)= XY-ZW
注意是用配方法

Answer 1

1. f(X,Y,Z)= X^2+XY+2Z^2=(X+1/2Y)^2-Y^2/4+2Z^2，X=x-1/2y，Y=y，Z=z。2. f(X,Y,Z)= X^2+2XY+4XZ+Y^2-2YZ + 2Z^2=(X+Y+2Z)^2+9/2Y^2-2(Z+3/2Y)^2，X=x+2y-2z，Y=y，Z=z-3/2y。3. f(X,Y,Z)=XY+4XZ-2YZ=(X+Y)^2/4-(X-Y)^2/4+2(X+Y)Z+2(X-Y)Z-(X+Y)Z+(X-Y)Z=(X+Y)^2/4-(X-Y)^2/4+(X+Y)Z+3(X-Y)Z=(X/2+Y/2+Z)^2-(X-Y)^2/4+3(X-Y)Z-Z^2=(X/2+Y/2+Z)^2-(X/2-Y/2-3Z)^2+8Z^2，X=x+y-2z，Y=x-y-4z，Z=z。4. f(X,Y,Z,W)= XY-ZW=(X+Y)^2/4-(X-Y)^2/4-(Z+W)^2/4+(Z-W)^2/4，X=x/2+y/2，Y=x/2-y/2，Z=z/2+w/2，W=z/2-w/2。

Answer 2

问题其实就是找一个可逆C,使得x'Ax=(Cy)'A(Cy)=y'C'ACy=y'(C'AC)y=y'Λy,即使C'AC=Λ为对角阵.
C要可逆,因为考察的原表达式是关于x的,考察原表达式必然要考察所有的x取值,就要求对于所有的x,都对应一个y,也就要求变换C不降秩,C可逆,y=C‘x.