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王品超上一道关于矩阵不等式的证明及其推广

热度 4已有 862 次阅读2017-11-25 19:00 |个人分类:奇妙数学

1.设$A$为2阶正定矩阵,$0<\left|\overrightarrow\alpha\right|\leq1$,则有$$\frac{\left(\alpha^TA\alpha\right)\left(\alpha^TA^{‐1}\alpha\right)}{\alpha^T\alpha}\leq\frac{\displaystyle\left(\lambda_1+\lambda_2\right)^2}{4\lambda_1\lambda_2},其中\lambda_1 , \lambda_2是A特征值$$
证明:由条件知,存在正交矩阵$T$使$T^TAT=\begin{pmatrix}\lambda_1&0\\0&\lambda_2\end{pmatrix},\;\;\;\;\;\;\lambda_i>0,i=1,2$
令$T^T\alpha=\begin{pmatrix}y_1\\y_2\end{pmatrix}$,则$$\begin{array}{l}\frac{\left(\alpha^TA\alpha\right)\left(\alpha^TA^{‐1}\alpha\right)}{\alpha^T\alpha}=\frac{(\lambda_1y_1^2+\lambda_2y_2^2)(\lambda_1^{‐1}y_1^2+\lambda_2^{‐1}y_2^2)}{y_1^2+y_2^2}=\frac{y_1^4+y_2^4+\left(\frac{\lambda_2}{\lambda_1}+\frac{\lambda_1}{\lambda_2}\right)y_1^2y_2^2}{y_1^2+y_2^2}=\frac{(y_1^2+y_2^2)^2-2y_1^2y_2^2+\left(\frac{\lambda_2}{\lambda_1}+\frac{\lambda_1}{\lambda_2}\right)y_1^2y_2^2}{y_1^2+y_2^2}\\=y_1^2+y_2^2+\frac{\left(\sqrt{\frac{\lambda_2}{\lambda_1}}-\sqrt{\frac{\lambda_1}{\lambda_2}}\right)^2y_1^2y_2^2}{y_1^2+y_2^2}=(y_1^2+y_2^2)\left[1+\frac{\left(\sqrt{\frac{\lambda_2}{\lambda_1}}-\sqrt{\frac{\lambda_1}{\lambda_2}}\right)^2y_1^2y_2^2}{(y_1^2+y_2^2)^2}\right]\leq1+\frac{\left(\sqrt{\frac{\lambda_2}{\lambda_1}}-\sqrt{\frac{\lambda_1}{\lambda_2}}\right)^2y_1^2y_2^2}{(y_1^2+y_2^2)^2}\end{array}$$
因为$y_1^2+y_2^2\geq2\vert y_1y_2\vert,故\frac12\geq\frac{\vert y_1y_2\vert}{y_1^2+y_2^2},即\frac14\geq\frac{y_1^2y_2^2}{(y_1^2+y_2^2)^2}$,$$\Rightarrow \frac{\left(\alpha^TA\alpha\right)\left(\alpha^TA^{‐1}\alpha\right)}{\alpha^T\alpha}\leq1+\frac14\left(\sqrt{\frac{\lambda_2}{\lambda_1}}-\sqrt{\frac{\lambda_1}{\lambda_2}}\right)^2=\frac{\displaystyle\left(\lambda_1+\lambda_2\right)^2}{4\lambda_1\lambda_2}$$


更进一步,可以得到更强的不等式
  2.设$A$为$n(n\geq 2)$阶正定矩阵,$0<\left|\overrightarrow\alpha\right|\leq1$,则有$$\frac{\left(\alpha^TA\alpha\right)\left(\alpha^TA^{‐1}\alpha\right)}{\alpha^T\alpha}\leq\frac{\displaystyle(n-1)\left(\lambda_1+\lambda_2\right)^2}{2n\lambda_1\lambda_2},其中\lambda_1 , \lambda_2分别是A最小及最大特征值$$

证明:由条件知,存在正交矩阵$T$使$T^TAT=\begin{pmatrix}\lambda_1&&0\\&\ddots&\\0&&\lambda_m\end{pmatrix},\;\lambda_i>0,i=1,2,\cdots,n$,不妨设$\lambda_1\leq\lambda_2\leq\cdots\leq\lambda_n$。
令$T^T\alpha=\begin{pmatrix}y_1\\\vdots\\y_n\end{pmatrix}$,有$$\frac{\left(\alpha^TA\alpha\right)\left(\alpha^TA^{‐1}\alpha\right)}{\alpha^T\alpha}=\frac{(\alpha^TT)(T^TAT)(T^T\alpha)(\alpha^TT)(T^TA^{‐1}T)(T^T\alpha)}{(\alpha^TT)(T^T\alpha)}=\frac{(\lambda_1y_1^2+\cdots+\lambda_ny_n^2)(\lambda_1^{‐1}y_1^2+\cdots+\lambda_n^{‐1}y_n^2)}{y_1^2+\cdots+y_n^2}$$
令$\triangle=y_1^2+\cdots+y_n^2$,$$则原式=\frac{\displaystyle\sum_{i=1}^ny_i^4+\sum_{1\leq i<j\leq n}\left(\frac{\displaystyle\lambda_i}{\displaystyle\lambda_j}+\frac{\displaystyle\lambda_j}{\displaystyle\lambda_i}\right)y_i^2y_j^2}\triangle=\frac{\triangle^2+{\displaystyle\sum_{1\leq i<j\leq n}}\left(\sqrt{\frac{\lambda_i}{\lambda_j}}-\sqrt{\frac{\lambda_j}{\lambda_i}}\right)^2y_i^2y_j^2}\triangle$$
由$$\left(\sqrt{\frac{\lambda_1}{\lambda_n}}-\sqrt{\frac{\lambda_n}{\lambda_1}}\right)^2-\left(\sqrt{\frac{\lambda_i}{\lambda_j}}-\sqrt{\frac{\lambda_j}{\lambda_i}}\right)^2=\frac{\lambda_1^2+\lambda_n^2}{\lambda_1\lambda_n}-\frac{\lambda_i^2+\lambda_j^2}{\lambda_i\lambda_j}=\frac{(\lambda_n\lambda_j-\lambda_i\lambda_1)(\lambda_i\lambda_n-\lambda_1\lambda_j)}{\lambda_1\lambda_n\lambda_i\lambda_j}$$
因为$\lambda_n\lambda_j-\lambda_i\lambda_1 \geq 0 ,\lambda_i\lambda_n-\lambda_1\lambda_j \geq 0$,则$\left(\sqrt{\frac{\lambda_1}{\lambda_n}}-\sqrt{\frac{\lambda_n}{\lambda_1}}\right)^2 \geq \left(\sqrt{\frac{\lambda_i}{\lambda_j}}-\sqrt{\frac{\lambda_j}{\lambda_i}}\right)^2$
$$\Rightarrow 原式\leq \triangle+\frac{({\displaystyle\sum_{1\leq i<j\leq n}}y_i^2y_j^2)\left(\sqrt{\frac{\lambda_1}{\lambda_n}}-\sqrt{\frac{\lambda_n}{\lambda_1}}\right)^2}\triangle=\triangle\left[1+\frac{({\displaystyle\sum_{1\leq i<j\leq n}}y_i^2y_j^2)\left(\sqrt{\frac{\lambda_1}{\lambda_n}}-\sqrt{\frac{\lambda_n}{\lambda_1}}\right)^2}{\triangle^2}\right]$$
由$$\triangle^2=(y_1^2+\cdots+y_n^2)^2=y_1^4+\cdots+y_n^4+2\sum_{1\leq i<j\leq n}y_i^2y_j^2$$
$$=\left(\frac{y_1^2}{\sqrt{n-1}}-\frac{y_2^2}{\sqrt{n-1}}\right)^2+\cdots+\left(\frac{y_1^2}{\sqrt{n-1}}-\frac{y_n^2}{\sqrt{n-1}}\right)^2+\cdots+\left(\frac{y_{n-1}^2}{\sqrt{n-1}}-\frac{y_n^2}{\sqrt{n-1}}\right)^2+\frac2{n-1}\sum_{1\leq i<j\leq n}y_i^2y_j^2+2\sum_{1\leq i<j\leq n}y_i^2y_j^2\geq\frac{2n}{n-1}\sum_{1\leq i<j\leq n}y_i^2y_j^2$$
$$\Rightarrow \frac{n-1}{2n}\geq\frac1{\triangle^2}\sum_{1\leq i<j\leq n}y_i^2y_j^2,且因为\triangle\leq1$$
$\Rightarrow 原式\leq1+\frac{n-1}{2n}\left(\sqrt{\frac{\lambda_1}{\lambda_n}}-\sqrt{\frac{\lambda_n}{\lambda_1}}\right)^2=\frac{(n-1)\lambda_1^2+2\lambda_1\lambda_n+(n-1)\lambda_n^2}{2n\lambda_1\lambda_n}\\\leq\frac{(n-1)\lambda_1^2+2(n-1)\lambda_1\lambda_n+(n-1)\lambda_n^2}{2n\lambda_1\lambda_n}=\frac{(n-1)(\lambda_1+\lambda_n)^2}{2n\lambda_1\lambda_n}$


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