## 實對稱矩陣特徵值變化界定的典型問題

$f(x_1,x_2,\ldots,x_n)=\begin{bmatrix} x_1&x_2&\cdots&x_n \end{bmatrix}\begin{bmatrix} 0&\frac{1}{2}&&&\\ \frac{1}{2}&0&\frac{1}{2}&&\\ &\ddots&\ddots&\ddots&\\ &&\frac{1}{2}&0&\frac{1}{2}\\ &&&\frac{1}{2}&0 \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ \vdots\\ x_n \end{bmatrix}=\mathbf{x}^TA\mathbf{x}$

$\displaystyle\max_{\Vert\mathbf{x}\Vert=1}f(\mathbf{x})=\max_{\Vert\mathbf{x}\Vert=1}\mathbf{x}^TA\mathbf{x}$

$\displaystyle \max_{\Vert\mathbf{x}\Vert=1}\mathbf{x}^TA\mathbf{x}=\max_{j=1,\ldots,n}\lambda_j=\lambda_{\max}$

$L(\mathbf{x},\lambda)=\mathbf{x}^TA\mathbf{x}+\lambda(1-\mathbf{x}^T\mathbf{x})$

\displaystyle\begin{aligned} \frac{\partial L}{\partial \mathbf{x}}&=2(A\mathbf{x}-\lambda\mathbf{x})=\mathbf{0}\\ \frac{\partial L}{\partial \lambda}&=1-\mathbf{x}^T\mathbf{x}=0.\end{aligned}

$f(\mathbf{x})=\mathbf{x}^T(A\mathbf{x})=\mathbf{x}^T(\lambda\mathbf{x})=\lambda\Vert\mathbf{x}\Vert^2=\lambda$

$B=\begin{bmatrix} b&a&&&\\ c&b&a&&\\ &\ddots&\ddots&\ddots&\\ &&c&b&a\\ &&&c&b \end{bmatrix}$

$\displaystyle \mu_j=b+2a\sqrt{\frac{c}{a}}\cos\left(\frac{j\pi}{n+1}\right),~~j=1,2,\ldots,n$

$\displaystyle \lambda_j=\cos\left(\frac{j\pi}{n+1}\right),~~j=1,\ldots,n$

$f(x_1,x_2,\ldots,x_n)=\begin{bmatrix} x_1&x_2&\cdots&x_n \end{bmatrix}\begin{bmatrix} 0&\frac{1}{2}&&&\frac{1}{2}\\ \frac{1}{2}&0&\frac{1}{2}&&\\ &\ddots&\ddots&\ddots&\\ &&\frac{1}{2}&0&\frac{1}{2}\\ \frac{1}{2}&&&\frac{1}{2}&0 \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ \vdots\\ x_n \end{bmatrix}=\mathbf{x}^TA\mathbf{x}$

$\displaystyle\max_{\Vert\mathbf{x}\Vert=1\atop\mathbf{x}\perp\mathbf{e}}f(\mathbf{x})=\max_{\Vert\mathbf{x}\Vert=1\atop\mathbf{x}\perp\mathbf{e}}\mathbf{x}^TA\mathbf{x}$

$\displaystyle \max_{\Vert\mathbf{x}\Vert=1\atop\mathbf{x}\perp\mathbf{e}}\mathbf{x}^TA\mathbf{x}=\max_{\Vert\mathbf{x}\Vert=1\atop\mathbf{x}\in\mathrm{span}\{\mathbf{u}_2,\ldots,\mathbf{u}_n\}}\mathbf{x}^TA\mathbf{x}=\max_{j=2,\ldots,n}\lambda_j$

$C=\begin{bmatrix} c_0&c_1&c_2&\cdots&c_{n-1}\\ c_{n-1}&c_0&c_1&\cdots&c_{n-2}\\ c_{n-2}&c_{n-1}&c_0&\cdots&c_{n-3}\\ \vdots&\vdots&\ddots&\ddots&\vdots\\ c_1&c_2&\cdots&c_{n-1}&c_0 \end{bmatrix}$

$\displaystyle \mu_j=\sum_{k=0}^{n-1}c_ke^{-2\pi i(j-1)k/n},~~j=1,\ldots,n$

\displaystyle\begin{aligned} \lambda_j&=\frac{1}{2}\left(e^{-2\pi i(j-1)/n}+e^{-2\pi i(j-1)(n-1)/n}\right)\\ &=\frac{1}{2}\left(e^{-2\pi i(j-1)/n}+e^{2\pi i(j-1)/n}\right)\\ &=\cos\left(\frac{2\pi(j-1)}{n}\right),~~~j=1,\ldots,n.\end{aligned}

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