线性代数中用配方法化二次型,如果没有平方项,这个作出平方项是随便设出来的吗?

2025-03-15 07:39:48
推荐回答(2个)
回答1:

二次型中没有平方项,只有交叉项,先利用平方差公式构造可逆线性变换, 化二次型为含平方项的二次型。令 x1=y1+y2,x2=y1-y2,x3=y3,x4=y4,代入就有平方项了,之后中按有平方项的方法做就行了。

令 x1=y1+y2

x2=y1-y2

x3=y3

代入后 f = y1^2 + 2y3y1 - y2^2

之后按有平方项的方法配方

扩展资料:

把二次型f所化得的标准二次型的平方项的系数中,正的个数和负的个数分别称为f的正惯性指数和负惯性指数.

正负惯性指数之和=f的秩

用矩阵的语言来表述即:与一个给定的实对称矩阵A合同的对角矩阵的对角线元素中,正的个数和负的个数是由A确定的,把这两个数分别称为A的正惯性指数和负惯性指数,合同于A的规范对角矩阵是唯一的,其中的自然数p,q就是A的正,负惯性指数。

参考资料来源:百度百科-正惯性指数

回答2:

二次型中没有平方项, 只有交叉项. 先利用平方差公式构造可逆线性变换, 化二次型为含平方项的二次型。令 x1=y1+y2,x2=y1-y2,x3=y3,x4=y4,代入就有平方项了,之后中按有平方项的方法做就行了。

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