Min-max theorem

Nightmare004

已于 2022-03-14 12:37:50 修改

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分类专栏：数学文章标签：线性代数矩阵

于 2022-03-14 11:07:50 首次发布

本文链接：https://blog.youkuaiyun.com/qq_39942341/article/details/123472300

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这篇博客讨论了厄尔米特矩阵的瑞利商及其与最小最大定理的关系。证明了对于厄尔米特矩阵A，其特征值λk可以通过寻找特定子空间中瑞利商的最大值和最小值来确定，从而验证了最小最大定理。同时，介绍了奇异值的概念，并指出厄尔米特矩阵的奇异值与瑞利商之间的联系。

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瑞利商

设 $A\mathbf{A}$ 是厄尔米特矩阵 $(A=AH(\mathbf{A}=\mathbf{A}^H$ )
$RA(x):Cn\{0}→RR_{\mathbf{A}}\left(\mathbf{x}\right):\mathbb{C}^n\backslash\left\{\mathbf{0}\right\}\to\mathbb{R}$ 定义为
$R_{\mathbf{A}}=\frac{\left(\mathbf{Ax},\mathbf{x}\right)}{\left(\mathbf{x},\mathbf{x}\right)}$

Min-max theorem

设 $A\mathbf{A}$ 是厄尔米特矩阵
设 $A\mathbf{A}$ 的特征值为 $λ1≤λ2≤⋯≤λn\lambda_1\le \lambda_2\le \cdots\le\lambda_n$
则
$\lambda_k=\min_{\mathbf{U}}\left\{\max_{\mathbf{x}}\left\{R_{\mathbf{A}}\left(\mathbf{x}\right)|\mathbf{x}\in\mathbf{U},\mathbf{x}\neq 0\right\}|\operatorname{dim}\left(\mathbf{U}\right)=k\right\}$
$\lambda_k=\max_{\mathbf{U}}\left\{\min_{\mathbf{x}}\left\{R_{\mathbf{A}}\left(\mathbf{x}\right)|\mathbf{x}\in\mathbf{U},\mathbf{x}\neq 0\right\}|\operatorname{dim}\left(\mathbf{U}\right)=n-k+1\right\}$

证明：
因为 $A\mathbf{A}$ 是厄尔米特矩阵，所以可以对角化
设 $A\mathbf{A}$ 两两正交的单位特征向量为 $,un\mathbf{u}_1,\cdots,\mathbf{u}_n$
即 $Aui=λiui,(ui,uj)=0,(uiui)=1\mathbf{A}\mathbf{u}_i=\lambda_i\mathbf{u}_i,\left(\mathbf{u}_i,\mathbf{u}_j\right)=0,\left(\mathbf{u}_i\mathbf{u}_i\right)=1$ 其中 $i≠ji\neq j$

先证明第一个等式
设 $dim⁡(U)=k\operatorname{dim}\left(\mathbf{U}\right)=k$
$,un})=k+n−k+1=n+1>n\operatorname{dim}\left(\mathbf{U}\right)+\operatorname{dim}\left(\operatorname{span}\left\{\mathbf{u}_k,\cdots,\mathbf{u}_n\right\}\right)=k+n-k+1=n+1>n$
所以
$,un}≠{0}\mathbf{U}\cap \operatorname{span}\left\{\mathbf{u}_k,\cdots,\mathbf{u}_n\right\}\neq\left\{\mathbf{0}\right\}$
所以存在 $v≠0\mathbf{v}\neq 0$
使得 $,un}\mathbf{v}\in \mathbf{U}\cap \operatorname{span}\left\{\mathbf{u}_k,\cdots,\mathbf{u}_n\right\}$

设 $v=∑i=knαiui\mathbf{v}=\sum_{i=k}^n \alpha_i\mathbf{u}_i$
则
$R_{\mathbf{A}}\left(\mathbf{v}\right)=\frac{\sum_{i=k}^n\lambda_i\alpha_i^2}{\sum_{i=k}^n\alpha_i^2}\ge \lambda_k$
所以
$max⁡{RA(x)∣x∈U}≥RA(v)≥λk\max \left\{R_{\mathbf{A}}\left(\mathbf{x}\right)|\mathbf{x}\in\mathbf{U}\right\}\ge R_{\mathbf{A}}\left(\mathbf{v}\right)\ge \lambda_k$

因为对于任意的 $U\mathbf{U}$ 都是成立的
所以
$\min_{\mathbf{U}}\left\{\max_{\mathbf{x}}\left\{R_{\mathbf{A}}\left(\mathbf{x}\right)|\mathbf{x}\in\mathbf{U},\mathbf{x}\neq 0\right\}|\operatorname{dim}\left(\mathbf{U}\right)=k\right\}\ge \lambda_k$

接着证明反过来的
设 $,uk}\mathbf{V}=\operatorname{span}\left\{\mathbf{u}_1,\cdots,\mathbf{u}_k\right\}$
则
$\max \left\{R_{\mathbf{A}}\left(\mathbf{x}\right)|\mathbf{x}\in\mathbf{V}\right\}\le\lambda_k$
于是
$\min_{\mathbf{U}}\left\{\max_{\mathbf{x}}\left\{R_{\mathbf{A}}\left(\mathbf{x}\right)|\mathbf{x}\in\mathbf{U},\mathbf{x}\neq 0\right\}|\operatorname{dim}\left(\mathbf{U}\right)=k\right\}\le \max \left\{R_{\mathbf{A}}\left(\mathbf{x}\right)|\mathbf{x}\in\mathbf{V}\right\}\le\lambda_k$
所以
$\lambda_k=\min_{\mathbf{U}}\left\{\max_{\mathbf{x}}\left\{R_{\mathbf{A}}\left(\mathbf{x}\right)|\mathbf{x}\in\mathbf{U},\mathbf{x}\neq 0\right\}|\operatorname{dim}\left(\mathbf{U}\right)=k\right\}$

接着证明第二个等式
设 $dim⁡(U)=n−k+1\operatorname{dim}\left(\mathbf{U}\right)=n-k+1$
$,uk})=n−k+1+k=n+1>n\operatorname{dim}\left(\mathbf{U}\right)+\operatorname{dim}\left(\operatorname{span}\left\{\mathbf{u}_1,\cdots,\mathbf{u}_k\right\}\right)=n-k+1+k=n+1>n$
所以
$,uk}≠{0}\mathbf{U}\cap \operatorname{span}\left\{\mathbf{u}_1,\cdots,\mathbf{u}_k\right\}\neq\left\{\mathbf{0}\right\}$
所以存在 $v≠0\mathbf{v}\neq 0$
使得 $,un}\mathbf{v}\in \mathbf{U}\cap \operatorname{span}\left\{\mathbf{u}_k,\cdots,\mathbf{u}_n\right\}$
设 $v=∑i=1kαiui\mathbf{v}=\sum_{i=1}^k \alpha_i\mathbf{u}_i$
则
$R_{\mathbf{A}}\left(\mathbf{v}\right)=\frac{\sum_{i=1}^k\lambda_i\alpha_i^2}{\sum_{i=1}^k\alpha_i^2}\le \lambda_k$
所以
$min⁡{RA(x)∣x∈U}≤RA(v)≤λk\min \left\{R_{\mathbf{A}}\left(\mathbf{x}\right)|\mathbf{x}\in\mathbf{U}\right\}\le R_{\mathbf{A}}\left(\mathbf{v}\right)\le \lambda_k$
因为对于任意的 $U\mathbf{U}$ 都是成立的
所以
$\max_{\mathbf{U}}\left\{\min_{\mathbf{x}}\left\{R_{\mathbf{A}}\left(\mathbf{x}\right)|\mathbf{x}\in\mathbf{U},\mathbf{x}\neq 0\right\}|\operatorname{dim}\left(\mathbf{U}\right)=n-k+1\right\}\le \lambda_k$

接着证明反过来的
设 $,un}\mathbf{V}=\operatorname{span}\left\{\mathbf{u}_k,\cdots,\mathbf{u}_n\right\}$
则
$\min \left\{R_{\mathbf{A}}\left(\mathbf{x}\right)|\mathbf{x}\in\mathbf{V}\right\}\ge\lambda_k$
于是
$\max_{\mathbf{U}}\left\{\min_{\mathbf{x}}\left\{R_{\mathbf{A}}\left(\mathbf{x}\right)|\mathbf{x}\in\mathbf{U},\mathbf{x}\neq 0\right\}|\operatorname{dim}\left(\mathbf{U}\right)=n-k+1\right\}\ge \min \left\{R_{\mathbf{A}}\left(\mathbf{x}\right)|\mathbf{x}\in\mathbf{V}\right\}\ge\lambda_k$
所以
$\lambda_k=\max_{\mathbf{U}}\left\{\min_{\mathbf{x}}\left\{R_{\mathbf{A}}\left(\mathbf{x}\right)|\mathbf{x}\in\mathbf{U},\mathbf{x}\neq 0\right\}|\operatorname{dim}\left(\mathbf{U}\right)=n-k+1\right\}$

奇异值版

设 $σk\sigma_k$ 是 $A\mathbf{A}$ 的奇异值
则
$\sigma_k=\min_{\mathbf{S}:\operatorname{dim}\left(\mathbf{S}\right)=k}\max_{\mathbf{x}\in\mathbf{S},\|\mathbf{x}\|=1}\|\mathbf{Ax}\|$
$\sigma_k=\min_{\mathbf{S}:\operatorname{dim}\left(\mathbf{S}\right)=n-k+1}\max_{\mathbf{x}\in\mathbf{S},\|\mathbf{x}\|=1}\|\mathbf{Ax}\|$