线性代数 05.07 用合同变换法化二次型为标准形

$\color{blue}{\S 第五章第七节用合同变换法化二次型为标准形}$

$定义10.若对方阵A作一次初等行变换,接着对所得矩阵\\ 作一次同种的初等列变换,就称对A进行一次合同变换.$

$用合同变换法化二次型为标准形的实质是: \\ 利用可逆线性变换x = Cy,把 f = x^TAx化为标准形,即\\ f = x^TAx = (Cy)^TACy = y^TC^T ACy = y^T \Lambda y \\ 只需C^TAC = \Lambda. \\ 又因 C = P_1P_2 \cdots P_s,其中P_1, P_2, \cdots, P_s均为初等方阵. \\ 所以\\ (P_1P_2 \cdots P_s)^T A P_1 P_2 \cdots P_s = \Lambda \\ 即 \\ P_s^T \cdots P_2^T P_1^T A P_1 P_2 \cdots P_s = \Lambda \quad (1) \\ 而 P_s^T \cdots P_2^T P_1^T = P_s^T \cdots P_2^T P_1^T E = C^T \quad (2) \\ 结合(1)和(2),得出将\Lambda化成对角矩阵,同时求出可逆矩阵C:$

$(A | E) \left. \begin{array}{l} A合同变换 \\ \quad \longrightarrow \\ E作行变换 \end{array} \right. (\Lambda | C^T)$

$求出C^T,作可逆线性变换x = Cy,则该变换将f化为标准形. \\ f = k_1y_1^2 + k_2y_2^2 + \cdots + k_ry_r^2.$

$例3.利用合同变换将 \\ f = 2x_1x_2 + 2x_1x_3 -2x_1x_4 - 2x_2x_3 + 2x_2x_4 + 2x_3x_4 \\ 化为标准形,并写出所用的可逆线性变换.$

$解:\\ (A | E) = \begin{pmatrix} 0 & 1 & 1 & -1 & 1 & 0 & 0 & 0 \\ 1 & 0 & -1 & 1 & 0 & 1 & 0 & 0 \\ 1 & -1 & 0 & 1 & 0 & 0 & 1 & 0 \\ -1 & 1 & 1 & 0 & 0 & 0 & 0 & 1 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 \\ 1 & 0 & -1 & 1 & 0 & 1 & 0 & 0 \\ 1 & -1 & 0 & 1 & 0 & 0 & 1 & 0 \\ -1 & 1 & 1 & 0 & 0 & 0 & 0 & 1 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 2 & 1 & 0 & 0 & 1 & 1 & 0 & 0 \\ 1 & 0 & -1 & 1 & 0 & 1 & 0 & 0 \\ 0 & -1 & 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 & 0 & 0 & 0 & 1 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 2 & 1 & 0 & 0 & 1 & 1 & 0 & 0 \\ 2 & 0 & -2 & 2 & 0 & 2 & 0 & 0 \\ 0 & -1 & 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 & 0 & 0 & 0 & 1 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 2 & 2 & 0 & 0 & 1 & 1 & 0 & 0 \\ 2 & 0 & -2 & 2 & 0 & 2 & 0 & 0 \\ 0 & -2 & 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 2 & 1 & 0 & 0 & 0 & 0 & 1 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 2 & 2 & 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & -2 & -2 & 2 & -1 & 1 & 0 & 0 \\ 0 & -2 & 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 2 & 1 & 0 & 0 & 0 & 0 & 1 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 2 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & -2 & -2 & 2 & -1 & 1 & 0 & 0 \\ 0 & -2 & 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 2 & 1 & 0 & 0 & 0 & 0 & 1 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 2 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & -2 & -2 & 2 & -1 & 1 & 0 & 0 \\ 0 & 0 & 2 & -1 & 1 & -1 & 1 & 0 \\ 0 & 2 & 1 & 0 & 0 & 0 & 0 & 1 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 2 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & -2 & 0 & 2 & -1 & 1 & 0 & 0 \\ 0 & 0 & 2 & -1 & 1 & -1 & 1 & 0 \\ 0 & 2 & -1 & 0 & 0 & 0 & 0 & 1 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 2 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & -2 & 0 & 2 & -1 & 1 & 0 & 0 \\ 0 & 0 & 2 & -1 & 1 & -1 & 1 & 0 \\ 0 & 0 & -1 & 2 & -1 & 1 & 0 & 1 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 2 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & -2 & 0 & 0 & -1 & 1 & 0 & 0 \\ 0 & 0 & 2 & -1 & 1 & -1 & 1 & 0 \\ 0 & 0 & -1 & 2 & -1 & 1 & 0 & 1 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 2 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & -2 & 0 & 0 & -1 & 1 & 0 & 0 \\ 0 & 0 & 2 & -2 & 1 & -1 & 1 & 0 \\ 0 & 0 & -2 & 8 & -2 & 2 & 0 & 2 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 2 & 0 & 0 & 0 & 1 & 1 & 0 & 0 \\ 0 & -2 & 0 & 0 & -1 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 & 1 & -1 & 1 & 0 \\ 0 & 0 & 0 & 6 & -1 & 1 & 1 & 2 \\ \end{pmatrix} \\ 于是,求得: \\ C^T = \begin{pmatrix} 1 & 1 & 0 & 0 \\ -1 & 1 & 0 & 0 \\ 1 & -1 & 1 & 0 \\ -1 & 1 & 1 & 2 \\ \end{pmatrix} , C = \begin{pmatrix} 1 & -1 & 1 & -1 \\ 1 & 1 & -1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 2 \\ \end{pmatrix} \\ 作可逆变换x = Cy,即 \\ \begin{pmatrix}x_1 \\ x_2 \\ x_3 \\x_4 \end{pmatrix} = \begin{pmatrix} 1 & -1 & 1 & -1 \\ 1 & 1 & -1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 2 \\ \end{pmatrix} \begin{pmatrix}y_1 \\ y_2 \\ y_3 \\y_4 \end{pmatrix}, \\ 把f化成标准形: \\ f = 2y_1^2 - 2y_2^2 + 2y_3^2 + 6y_4^2 .$

$例4.设二次型 \\ f = 4x_1^2 + 3x_2^2 + 3x_3^2 + 2x_2x_3 \\ 1) 用正交变换化f为标准形;\\ 2) 用配方法化f为标准形;\\ 3) 用合同法化f为标准形.$

$解: 1) 用正交变换化f为标准形. ①把f写成矩阵形式: \\ f(x_1, x_2, x_3) = \begin{pmatrix} 4 & 0 & 0 \\ 0 & 3 & 1 \\ 0 & 1 & 3 \\ \end{pmatrix} \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix} \\ ②通过 |A - \lambda E | = 0,求A的全部特征值. \\ |A - \lambda E | = \begin{vmatrix} 4 - \lambda & 0 & 0 \\ 0 & 3 - \lambda & 1 \\ 0 & 1 & 3 - \lambda \\ \end{vmatrix} \\ = (4 - \lambda)[(3 - \lambda)^2 - 1] \\ = -(\lambda - 2)(\lambda - 4)^2 = 0 解得A的特征值为: \lambda_1 = 2, \lambda_2 = \lambda_3 = 4. \\ ③通过(A - \lambda E) x = 0, 求A的特征向量. \\ 当\lambda_1 = 2时，(A - 2E) x = 0,由 \\ A - 2E = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \\ \end{pmatrix} \sim \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ \end{pmatrix} \\ 解得基础解系 \xi_1 = \begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix}, 单位化 p_1 = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix} \\ 当\lambda_2 = \lambda_3 = 4时, (A - 4E) x = 0,由 \\ A - 4E = \begin{pmatrix} 0 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 1 & -1 \\ \end{pmatrix} \sim \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ \end{pmatrix} \\ 解得基础解系: \\ \xi_2 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \xi_3 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, 因为\xi_2与\xi_3正交,直接单位化得: \\ p_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, p_2 = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, \\ ④ 把p_1,p_2,p_3拼成正交变换矩阵P, 作正交变换x = Py,即\\ P = \begin{pmatrix} p_1 , p_2, p_3 \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ -\dfrac{1}{\sqrt{2}} & 0 & \dfrac{1}{\sqrt{2}} \\ \dfrac{1}{\sqrt{2}} & 0 & \dfrac{1}{\sqrt{2}} \\ \end{pmatrix} \\ \begin{pmatrix} x_1 \\ x_2 \\x _3 \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ -\dfrac{1}{\sqrt{2}} & 0 & \dfrac{1}{\sqrt{2}} \\ \dfrac{1}{\sqrt{2}} & 0 & \dfrac{1}{\sqrt{2}} \\ \end{pmatrix}\begin{pmatrix} y_1 \\ y_2 \\y _3 \end{pmatrix} \\ ⑤ 把f化为标准形: \\ f = 2y_1^2 + 4y_2^2 + 4y_3^2 \\ 2) 用配方法化f为标准形 \\ f = 4x_1^2 + 3x_2^2 + 3x_3^2 + 2x_2x_3 \\ = 4x_1^2 + 2(x_2 + x_3)^2 + (x_2 - x_3)^2$

$令 \left \{ \begin{array}{l} y_1 = x_1 \\ y_2 = x_2 + x_3 \\ y_3 = x_2 - x_3 \end{array} \right.$

$即\left \{ \begin{array}{l} x_1 = y_1 \\ x_2 = \dfrac{1}{2}y_2 + \dfrac{1}{2} y_3 \\ x_3 = \dfrac{1}{2}y_2 - \dfrac{1}{2}y_3 \end{array} \right.$

$即 \begin{pmatrix} x_1 \ x_2 \ x_3 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \dfrac{1}{2} & \dfrac{1}{2} \\ 0 & \dfrac{1}{2} & -\dfrac{1}{2} \\ \end{pmatrix} \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix} \\ f为标准形为: f = 4y_1^2 + 2y_2^2 + y_3^2 \\ 3) 用合同法化f为标准形. \\ (A | E) = \begin{pmatrix} 4 & 0 & 0 & 1 & 0 & 0 \\ 0 & 3 & 1 & 0 & 1 & 0 \\ 0 & 1 & 3 & 0 & 0 & 1 \\ \end{pmatrix} \sim \begin{pmatrix} 4 & 0 & 0 & 1 & 0 & 0 \\ 0 & 3 & 1 & 0 & 1 & 0 \\ 0 & 3 & 9 & 0 & 0 & 3 \\ \end{pmatrix} \sim \begin{pmatrix} 4 & 0 & 0 & 1 & 0 & 0 \\ 0 & 3 & 3 & 0 & 1 & 0 \\ 0 & 3 & 27 & 0 & 0 & 3 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 4 & 0 & 0 & 1 & 0 & 0 \\ 0 & 3 & 3 & 0 & 1 & 0 \\ 0 & 0 & 24 & 0 & -1 & 3 \\ \end{pmatrix} \\ \sim \begin{pmatrix} 4 & 0 & 0 & 1 & 0 & 0 \\ 0 & 3 & 0 & 0 & 1 & 0 \\ 0 & 0 & 24 & 0 & -1 & 3 \\ \end{pmatrix} \\ 于是,得 \\ A = \begin{pmatrix} 4 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 24 \\ \end{pmatrix} \\ C^T = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -1 & 3 \\ \end{pmatrix}, C = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 3 \\ \end{pmatrix}, \\ 从而c = Cy, 即 \\ \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 3 \\ \end{pmatrix} \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix}, \\ 则该变换将f化成标准形为: \\ f = 4y_1^2 + 3y_2^2 + 24y_3^2$

原文链接：https://blog.csdn.net/longji/article/details/78917465

§ 第 五 章 第 七 节 用 合 同 变 换 法 化 二 次 型 为 标 准 形 \color{blue}{\S 第五章 第七节 用合同变换法化二次型为标准形}

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