线代强化NO24|二次型|二次型合同标准形的计算

二次型合同标准形的计算

$$\begin{array}{rl}& \text{解：(1) 二次型矩阵为 }\mathbf{A}=\left(\begin{array}{ccc}2& 0& 0\\ 0& 3& \alpha \\ 0& \alpha & 3\end{array}\right).\\ & \\ & |\lambda \mathbf{E}-\mathbf{A}|=\left|\begin{array}{ccc}\lambda -2& 0& 0\\ 0& \lambda -3& -\alpha \\ 0& -\alpha & \lambda -3\end{array}\right|=(\lambda -2)\left[\lambda -(\alpha +3)\right]\left[\lambda -(3-\alpha )\right]=0.\\ & \\ & \text{得特征值 }{\lambda }_1=2,{\lambda }_2=\alpha +3,{\lambda }_3=3-\alpha .\\ & \text{由标准型知特征值为}1,2,5.\text{由}\alpha >0\text{可知}3+\alpha =5,\alpha =2.\end{array}$$
$$\begin{array}{rl}& \mathbf{A}=\left(\begin{array}{ccc}2& 0& 0\\ 0& 3& 2\\ 0& 2& 3\end{array}\right),\mathbf{A}-\mathbf{E}=\left(\begin{array}{ccc}1& 0& 0\\ 0& 2& 2\\ 0& 2& 2\end{array}\right)\to \left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& 0& 0\end{array}\right),\\ & 1\text{的特征向量为}\left(\begin{array}{c}0\\ -1\\ 1\end{array}\right);\\ & \\ & \mathbf{A}-2\mathbf{E}=\left(\begin{array}{ccc}0& 0& 0\\ 0& 1& 2\\ 0& 2& 1\end{array}\right)\to \left(\begin{array}{ccc}0& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right),\\ & 2\text{的特征向量为}\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right);\\ & \\ & \mathbf{A}-5\mathbf{E}=\left(\begin{array}{ccc}-3& 0& 0\\ 0& -2& 2\\ 0& 2& -2\end{array}\right)\to \left(\begin{array}{ccc}1& 0& 0\\ 0& 1& -1\\ 0& 0& 0\end{array}\right),\\ & 5\text{的特征向量为}\left(\begin{array}{c}0\\ 1\\ 1\end{array}\right).\end{array}$$
$$\begin{array}{rl}& \text{则}\mathbf{Q}=\left(\begin{array}{ccc}0& 1& 0\\ -\frac{\sqrt{2}}{2}& 0& \frac{\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2}& 0& \frac{\sqrt{2}}{2}\end{array}\right),\text{在}\mathbf{x}=\mathbf{Q}\mathbf{y}\text{下二次型可化为标准型}\end{array}$$
补充
$$\begin{array}{rl}& f(x_1,x_2,x_3)=(x_1{x}_2{x}_3)\mathbf{A}\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)={\mathbf{x}}^T\mathbf{A}\mathbf{x}\\ & \\ & \stackrel{\mathbf{x}=\mathbf{Q}\mathbf{y}}{\to }(\mathbf{Q}\mathbf{y}{)}^T\mathbf{A}\mathbf{Q}\mathbf{y}\\ & ={\mathbf{y}}^T{\mathbf{Q}}^T\mathbf{A}\mathbf{Q}\mathbf{y}\\ & ={\mathbf{y}}^T\mathbf{\Lambda }\mathbf{y}\\ & \\ & {\mathbf{Q}}^T\mathbf{x}={\mathbf{Q}}^T\mathbf{Q}\mathbf{y}=\mathbf{y}\\ & \\ & \mathbf{x}=\mathbf{Q}\mathbf{y},\mathbf{y}={\mathbf{Q}}^T\mathbf{x}\end{array}$$
$$\begin{array}{rl}& \text{解：(2) 求}f(x_1,x_2,x_3)=2x_1^2+3x_2^2+3x_3^2+4x_2{x}_3\text{在约束条件}{x}_1^2+x_2^2+x_3^2=1\text{的最值}.\\ & \\ & \text{转化为}f(x_1,x_2,x_3)={\mathbf{x}}^T\mathbf{A}\mathbf{x}\stackrel{\mathbf{x}=\mathbf{Q}\mathbf{y}}{=}{\mathbf{y}}^T{\mathbf{Q}}^T\mathbf{A}\mathbf{Q}\mathbf{y}=y_1^2+2y_2^2+5y_3^2.\\ & \\ & \text{在}{\mathbf{x}}^T\mathbf{x}=(\mathbf{Q}\mathbf{y}{)}^T(\mathbf{Q}\mathbf{y})={\mathbf{y}}^T{\mathbf{Q}}^T\mathbf{Q}\mathbf{y}={\mathbf{y}}^T\mathbf{y}=1,y_1^2+y_2^2+y_3^2=1\text{下的最值}.\\ & \\ & f(y_1,y_2,y_3)=y_1^2+2y_2^2+5y_3^2\le 5y_1^2+5y_2^2+5y_3^2=5(y_1^2+y_2^2+y_3^2)=5,\\ & \text{当}{y}_1=0,y_2=0,y_3=1\text{时，取等号，}\text{故最大值为}5.\\ & \\ & f(y_1,y_2,y_3)=y_1^2+2y_2^2+5y_3^2\ge y_1^2+y_2^2+y_3^2=1,\\ & \text{当}{y}_1=1,y_2=0,y_3=0\text{时，取等号，}\text{故最小值为}1.\end{array}$$
$$\text{【小结】①函数的最大值和最小值一定是函数值；②正交变换只改变向量的方向，不改变向量的长度。}$$

$$\begin{array}{rl}& \text{解：(1) 法1设}f(x_1,x_2,x_3)=2\left(a_1^2{x}_1^2+a_2^2{x}_2^2+a_3^2{x}_3^2+2a_1{a}_2{x}_1{x}_2+2a_1{a}_3{x}_1{x}_3+2a_2{a}_3{x}_2{x}_3\right)\\ & +\left(b_1^2{x}_1^2+b_2^2{x}_2^2+b_3^2{x}_3^2+2b_1{b}_2{x}_1{x}_2+2b_1{b}_3{x}_1{x}_3+2b_2{b}_3{x}_2{x}_3\right)\\ & =\left(2a_1^2+b_1^2\right)x_1^2+\left(2a_2^2+b_2^2\right)x_2^2+\left(2a_3^2+b_3^2\right)x_3^2\\ & +\left(4a_1{a}_2+2b_1{b}_2\right)x_1{x}_2+\left(4a_1{a}_3+2b_1{b}_3\right)x_1{x}_3+\left(4a_2{a}_3+2b_2{b}_3\right)x_2{x}_3\\ & \\ & \text{矩阵为}\left(\begin{array}{ccc}2a_1^2+b_1^2& 2a_1{a}_2+b_1{b}_2& 2a_1{a}_3+b_1{b}_3\\ 2a_1{a}_2+b_1{b}_2& 2a_2^2+b_2^2& 2a_2{a}_3+b_2{b}_3\\ 2a_1{a}_3+b_1{b}_3& 2a_2{a}_3+b_2{b}_3& 2a_3^2+b_3^2\end{array}\right)=2{\mathbf{\alpha \alpha }}^T+{\mathbf{\beta \beta }}^T,\\ & \text{则二次型}f\text{对应的矩阵为}2{\mathbf{\alpha \alpha }}^T+{\mathbf{\beta \beta }}^T.\end{array}$$
$$\begin{array}{rl}& \text{法2：设}\mathbf{x}=\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right),{\mathbf{x}}^T\mathbf{\alpha }={\mathbf{\alpha }}^T\mathbf{x}=a_1{x}_1+a_2{x}_2+a_3{x}_3,\\ & {\mathbf{x}}^T\mathbf{\beta }={\mathbf{\beta }}^T\mathbf{x}=b_1{x}_1+b_2{x}_2+b_3{x}_3,\\ & \\ & f(x_1,x_2,x_3)=2{\mathbf{x}}^T\mathbf{\alpha }{\mathbf{\alpha }}^T\mathbf{x}+{\mathbf{x}}^T\mathbf{\beta }{\mathbf{\beta }}^T\mathbf{x}={\mathbf{x}}^T\left(2\mathbf{\alpha }{\mathbf{\alpha }}^T+\mathbf{\beta }{\mathbf{\beta }}^T\right)\mathbf{x},\\ & \\ & {\left(2\mathbf{\alpha }{\mathbf{\alpha }}^T+\mathbf{\beta }{\mathbf{\beta }}^T\right)}^T=2{\left({\mathbf{\alpha }}^T\right)}^T{\mathbf{\alpha }}^T+{\left({\mathbf{\beta }}^T\right)}^T{\mathbf{\beta }}^T=2\mathbf{\alpha }{\mathbf{\alpha }}^T+\mathbf{\beta }{\mathbf{\beta }}^T,\\ & \text{故}2\mathbf{\alpha }{\mathbf{\alpha }}^T+\mathbf{\beta }{\mathbf{\beta }}^T\text{为实对称矩阵}.\text{则二次型}f\text{对应的矩阵为}2\mathbf{\alpha }{\mathbf{\alpha }}^T+\mathbf{\beta }{\mathbf{\beta }}^T.\end{array}$$
$$\begin{array}{rl}& \\ & (\text{2})r(2\mathbf{\alpha }{\mathbf{\alpha }}^T+\mathbf{\beta }{\mathbf{\beta }}^T)\le r(2\mathbf{\alpha }{\mathbf{\alpha }}^T)+r(\mathbf{\beta }{\mathbf{\beta }}^T)=1+1=2,\\ & \text{故}0\text{为}2\mathbf{\alpha }{\mathbf{\alpha }}^T+\mathbf{\beta }{\mathbf{\beta }}^T\text{的一个特征值}.\\ & {\mathbf{\alpha }}^T\mathbf{\beta }={\mathbf{\beta }}^T\mathbf{\alpha }=0,{\mathbf{\alpha }}^T\mathbf{\alpha }={\mathbf{\beta }}^T\mathbf{\beta }=1,\\ & \\ & (2\mathbf{\alpha }{\mathbf{\alpha }}^T+\mathbf{\beta }{\mathbf{\beta }}^T)\mathbf{\beta }=2\mathbf{\alpha }{\mathbf{\alpha }}^T\mathbf{\beta }+\mathbf{\beta }{\mathbf{\beta }}^T\mathbf{\beta }=\mathbf{\beta },1\text{为一个特征值}.\\ & \\ & (2\mathbf{\alpha }{\mathbf{\alpha }}^T+\mathbf{\beta }{\mathbf{\beta }}^T)\mathbf{\alpha }=2\mathbf{\alpha }{\mathbf{\alpha }}^T\mathbf{\alpha }+\mathbf{\beta }{\mathbf{\beta }}^T\mathbf{\alpha }=2\mathbf{\alpha },2\text{为一个特征值}.\\ & \\ & \text{则}f\text{在正交变换下的标准形为}2y_1^2+y_2^2.\end{array}$$
$$\begin{array}{rl}& \text{改：}f\text{的矩阵为}3\mathbf{\alpha }{\mathbf{\alpha }}^T+2\mathbf{\beta }{\mathbf{\beta }}^T+\mathbf{\gamma }{\mathbf{\gamma }}^T,\\ & \mathbf{\alpha },\mathbf{\beta },\mathbf{\gamma }\text{正交且均为单位向量，求正交变换}\mathbf{x}=\mathbf{Q}\mathbf{y}\text{，使得}f\text{化为标准形}.\\ & \\ & \text{解：}\left(3\mathbf{\alpha }{\mathbf{\alpha }}^T+2\mathbf{\beta }{\mathbf{\beta }}^T+\mathbf{\gamma }{\mathbf{\gamma }}^T\right)\mathbf{\alpha }=3\mathbf{\alpha }{\mathbf{\alpha }}^T\mathbf{\alpha }+2\mathbf{\beta }{\mathbf{\beta }}^T\mathbf{\alpha }+\mathbf{\gamma }{\mathbf{\gamma }}^T\mathbf{\alpha }=3\mathbf{\alpha },3,\mathbf{\alpha }.\\ & \\ & \left(3\mathbf{\alpha }{\mathbf{\alpha }}^T+2\mathbf{\beta }{\mathbf{\beta }}^T+\mathbf{\gamma }{\mathbf{\gamma }}^T\right)\mathbf{\beta }=3\mathbf{\alpha }{\mathbf{\alpha }}^T\mathbf{\beta }+2\mathbf{\beta }{\mathbf{\beta }}^T\mathbf{\beta }+\mathbf{\gamma }{\mathbf{\gamma }}^T\mathbf{\beta }=2\mathbf{\beta },2,\mathbf{\beta }.\\ & \\ & \left(3\mathbf{\alpha }{\mathbf{\alpha }}^T+2\mathbf{\beta }{\mathbf{\beta }}^T+\mathbf{\gamma }{\mathbf{\gamma }}^T\right)\mathbf{\gamma }=3\mathbf{\alpha }{\mathbf{\alpha }}^T\mathbf{\gamma }+2\mathbf{\beta }{\mathbf{\beta }}^T\mathbf{\gamma }+\mathbf{\gamma }{\mathbf{\gamma }}^T\mathbf{\gamma }=\mathbf{\gamma },1,\mathbf{\gamma }.\\ & \\ & \mathbf{Q}=(\mathbf{\alpha },\mathbf{\beta },\mathbf{\gamma }),3y_1^2+2y_2^2+y_3^2.\end{array}$$
$$\begin{array}{rl}& \text{【小结】将二次型}{\mathbf{x}}^T\mathbf{A}\mathbf{x}\text{经过正交变换}\mathbf{x}=\mathbf{Q}\mathbf{y}\text{化为标准形}{\mathbf{y}}^T\mathbf{\Lambda }\mathbf{y}\text{之后，由于}\\ & \mathbf{\Lambda }={\mathbf{Q}}^T\mathbf{A}\mathbf{Q}={\mathbf{Q}}^{-1}\mathbf{A}\mathbf{Q},\\ & \text{因此矩阵}\mathbf{A}\text{与}\mathbf{\Lambda }\text{之间既是合同关系也是相似关系。也就是说}\mathbf{\Lambda }\text{也}\\ & \text{是}\mathbf{A}\text{的相似标准形。因此}\mathbf{\Lambda }\text{的对角元均为}\mathbf{A}\text{的特征值，而}\mathbf{Q}\text{的列向量就是所对应的特}\\ & \text{征向量。因此，考研对二次型的考查与特征值特征向量结合得比较紧密。}\end{array}$$

选A
根据标准形得出P为2，1，-1，则Q为2，-1，1，选A

$$\begin{array}{rl}& \text{解：}(1)f\text{的矩阵为}\left(\begin{array}{cc}1& -2\\ -2& 4\end{array}\right),g\text{的矩阵为}\left(\begin{array}{cc}a& 2\\ 2& b\end{array}\right).\\ & \\ & \text{因为正交变换不改变矩阵的特征值，故}\\ & \{\begin{array}{l}a+b=5\\ ab-4=0\end{array}\text{则}\{\begin{array}{l}a=4\\ b=1\end{array}\text{或}\{\begin{array}{l}a=1\\ b=4\end{array}.\end{array}$$
$$\begin{array}{rl}& (2)法1\left(\begin{array}{cc}1& -2\\ -2& 4\end{array}\right)\text{的}0\text{的特征向量为}\left(\begin{array}{c}2\\ 1\end{array}\right),5\text{的特征向量为}\left(\begin{array}{c}1\\ -2\end{array}\right).\\ & \\ & \left(\begin{array}{cc}4& 2\\ 2& 1\end{array}\right)\text{的}0\text{的特征向量为}\left(\begin{array}{c}1\\ -2\end{array}\right),5\text{的特征向量为}\left(\begin{array}{c}2\\ 1\end{array}\right).\\ & \\ & \text{则正交矩阵}{\mathbf{Q}}_1=\left(\begin{array}{cc}\frac{2}{\sqrt{5}}& \frac{1}{\sqrt{5}}\\ \frac{1}{\sqrt{5}}& -\frac{2}{\sqrt{5}}\end{array}\right),{\mathbf{Q}}_2=\left(\begin{array}{cc}\frac{2}{\sqrt{5}}& \frac{2}{\sqrt{5}}\\ -\frac{2}{\sqrt{5}}& \frac{1}{\sqrt{5}}\end{array}\right).\\ & \\ & f={\mathbf{x}}^T\mathbf{A}\mathbf{x}=({\mathbf{Q}}_1\mathbf{y}{)}^T\mathbf{A}({\mathbf{Q}}_1\mathbf{y})={\mathbf{y}}^T{\mathbf{Q}}_1^T\mathbf{A}{\mathbf{Q}}_1\mathbf{y}={\mathbf{y}}^T\mathbf{\Lambda }\mathbf{y},\\ & \\ & g={\mathbf{x}}^T\mathbf{B}\mathbf{x}=({\mathbf{Q}}_2\mathbf{y}{)}^T\mathbf{B}({\mathbf{Q}}_2\mathbf{y})={\mathbf{y}}^T{\mathbf{Q}}_2^T\mathbf{B}{\mathbf{Q}}_2\mathbf{y}={\mathbf{y}}^T\mathbf{\Lambda }\mathbf{y},\\ & \\ & \text{即}{\mathbf{Q}}_1^T\mathbf{A}{\mathbf{Q}}_1={\mathbf{Q}}_2^T\mathbf{B}{\mathbf{Q}}_2.\end{array}$$
$$\begin{array}{rl}f& ={\mathbf{X}}^T\mathbf{A}\mathbf{X}\stackrel{\mathbf{X}=\mathbf{Q}\mathbf{Y}}{\to }{\mathbf{Y}}^T{\mathbf{Q}}^T\mathbf{A}\mathbf{Q}\mathbf{Y}\\ & ={\mathbf{Y}}^T\mathbf{B}\mathbf{Y},\\ {\mathbf{Q}}^T\mathbf{A}\mathbf{Q}& =\mathbf{B}.\end{array}$$
$$\mathbf{B}={\mathbf{Q}}_2{\mathbf{Q}}_1^T\mathbf{A}{\mathbf{Q}}_1{\mathbf{Q}}_2^T={\left({\mathbf{Q}}_1,{\mathbf{Q}}_2^T\right)}^T\mathbf{A}{\left({\mathbf{Q}}_1,{\mathbf{Q}}_2\right)}^T$$
$$\text{则}\mathbf{Q}={\mathbf{Q}}_1{\mathbf{Q}}_2^T=\frac{1}{5}\left(\begin{array}{cc}2& 1\\ 1& -2\end{array}\right)\left(\begin{array}{cc}1& -2\\ 2& 1\end{array}\right)=\frac{1}{5}\left(\begin{array}{cc}4& -3\\ -3& -4\end{array}\right)$$
$$\begin{array}{rl}& \text{法2}：\\ & g(y_1,y_2)=4y_1^2+4y_1{y}_2+y_2^2\\ & f(x_1,x_2)=x_1^2-4x_1{x}_2+4x_2^2\\ & x_1=y_2,x_2=-y_1,\text{即}\left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\left(\begin{array}{c}{y}_1\\ y_2\end{array}\right)\\ & \text{可化为}4y_1^2+4y_1{y}_2+y_2^2,\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\text{为正交矩阵}\\ & \text{则}\mathbf{Q}=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\end{array}$$
$$\begin{array}{rl}& \text{【小结】将二次型}{\mathbf{x}}^T\mathbf{A}\mathbf{x}\text{经过正交变换}\mathbf{x}=\mathbf{Q}\mathbf{y}\text{化为了二次型}{\mathbf{y}}^T\mathbf{B}\mathbf{y}\text{之后，与正交相似}\\ & \text{对角化类似，由于}\mathbf{B}={\mathbf{Q}}^T\mathbf{A}\mathbf{Q}={\mathbf{Q}}^{-1}\mathbf{A}\mathbf{Q}\text{，可知矩阵}\mathbf{A}\text{与}\mathbf{B}\text{之间既是合同关系也是相似关}\\ & \text{系，所以}\mathbf{A}\text{与}\mathbf{B}\text{的特征值是相同的。}\\ & \\ & \text{要计算正交矩阵}\mathbf{Q}\text{，可以按照如下步骤，分别求出正交矩阵}{\mathbf{Q}}_1,{\mathbf{Q}}_2\text{使得}{\mathbf{Q}}_1^T\mathbf{A}{\mathbf{Q}}_1={\mathbf{Q}}_2^T\mathbf{B}{\mathbf{Q}}_2\text{为}\\ & \text{对角矩阵，由矩阵的运算法则，有}{\mathbf{Q}}_2{\mathbf{Q}}_1^T\mathbf{A}{\mathbf{Q}}_1{\mathbf{Q}}_2^T={\left({\mathbf{Q}}_1{\mathbf{Q}}_2^T\right)}^T\mathbf{A}{\mathbf{Q}}_1{\mathbf{Q}}_2^T=\mathbf{B}\text{，然后再令}\mathbf{Q}={\mathbf{Q}}_1{\mathbf{Q}}_2^T\text{。}\end{array}$$