実対称行列を回転行列(実直交行列)と実対角行列で表現する

$$\begin{eqnarray} S_{2}&=&\begin{bmatrix} x & y \\ y & z \end{bmatrix} \;\cdots\;S_{2}:x,y,zを実数とする2\times2の実対称行列 \\&&\:\cdots\;\small{実対称行列:S_{2}=S_{2}^T(対称行列),S_{2}=S_{2}^{*}(実行列(各要素の複素共役が変わらない,虚部が0))} \\ \\R(\theta)&=&\begin{bmatrix} \cos\left(\theta\right) & -\sin\left(\theta\right) \\ \sin\left(\theta\right) & \cos\left(\theta\right) \end{bmatrix} \\R(\theta)^T&=&\begin{bmatrix} \cos\left(\theta\right) & \sin\left(\theta\right) \\ -\sin\left(\theta\right) & \cos\left(\theta\right) \end{bmatrix} =\begin{bmatrix} \cos\left(-\theta\right) & -\sin\left(-\theta\right) \\ \sin\left(-\theta\right) & \cos\left(-\theta\right) \end{bmatrix} =R(-\theta)\;\cdots\;逆回転 \\R(\theta)R(\theta)^T&=&R(\theta)^TR(\theta)=I \;\cdots\;R(\theta):直交行列(特に各要素の複素共役が変わらないので実直交行列) \\ \\\Lambda&=&\begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{bmatrix} \;\cdots\;\lambda_1,\lambda_2はS_{2}の固有値 \\&&\;\cdots\;対角行列(特に各要素の複素共役が変わらない場合，実対角行列) \end{eqnarray}$$ $$\begin{eqnarray} S_{2}&=&R(\theta) \Lambda R(\theta)^T \\&=&\begin{bmatrix} \cos\left(\theta\right) & -\sin\left(\theta\right) \\ \sin\left(\theta\right) & \cos\left(\theta\right) \end{bmatrix} \begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{bmatrix} \begin{bmatrix} \cos\left(\theta\right) & \sin\left(\theta\right) \\ -\sin\left(\theta\right) & \cos\left(\theta\right) \end{bmatrix} \\&=&\begin{bmatrix} \lambda_1\cos\left(\theta\right) & -\lambda_2\sin\left(\theta\right) \\ \lambda_1\sin\left(\theta\right) & \lambda_2\cos\left(\theta\right) \end{bmatrix} \begin{bmatrix} \cos\left(\theta\right) & \sin\left(\theta\right) \\ -\sin\left(\theta\right) & \cos\left(\theta\right) \end{bmatrix} \\&=&\begin{bmatrix} \lambda_1\cos^2\left(\theta\right) +\lambda_2\sin^2\left(\theta\right) & \left(\lambda_1-\lambda_2\right)\cos\left(\theta\right)\sin\left(\theta\right) \\ \left(\lambda_1-\lambda_2\right)\cos\left(\theta\right)\sin\left(\theta\right) & \lambda_1\sin^2\left(\theta\right) +\lambda_2\cos^2\left(\theta\right) \end{bmatrix} \\ \\x&=&\lambda_1\cos^2\left(\theta\right) +\lambda_2\sin^2\left(\theta\right) \\&=&\frac{1}{2}2\left\{\lambda_1\cos^2\left(\theta\right) +\lambda_2\sin^2\left(\theta\right)\right\} \\&=&\frac{1}{2}\left\{2\lambda_1\cos^2\left(\theta\right) +2\lambda_2\sin^2\left(\theta\right)\right\} \\&=&\frac{1}{2}\left\{\lambda_1\left(\cos^2\left(\theta\right)+1-\sin^2\left(\theta\right)\right) +\lambda_2\left(\sin^2\left(\theta\right)+1-\cos^2\left(\theta\right)\right)\right\} \;\cdots\;\cos^2\left(\theta\right)=1-\sin^2\left(\theta\right),\;\sin^2\left(\theta\right)=1-\cos^2\left(\theta\right) \\&=&\frac{1}{2}\left\{\lambda_1\left(\cos^2\left(\theta\right)+1-\sin^2\left(\theta\right)\right) +\lambda_2\left(1-\left(\cos^2\left(\theta\right)-\sin^2\left(\theta\right)\right)\right)\right\} \\&=&\frac{1}{2}\left\{\lambda_1\left(\cos\left(2\theta\right)+1\right) +\lambda_2\left(1-\cos\left(2\theta\right)\right)\right\} \;\cdots\;\cos\left(2\theta\right)=\cos^2\left(\theta\right)-\sin^2\left(\theta\right) \\&=&\frac{1}{2}\left\{ \lambda_1 \cos\left(2\theta\right)+\lambda_1+\lambda_2-\lambda_2\cos\left(2\theta\right)\right\} \\&=&\frac{1}{2}\left\{ \left( \lambda_1-\lambda_2\right) \cos\left(2\theta\right)+\lambda_1+\lambda_2\right\} \\y&=&\left(\lambda_1-\lambda_2\right)\cos\left(\theta\right)\sin\left(\theta\right) \\&=&\frac{1}{2}2\left(\lambda_1-\lambda_2\right)\cos\left(\theta\right)\sin\left(\theta\right) \\&=&\frac{1}{2}\left(\lambda_1-\lambda_2\right)\sin\left(2\theta\right) \;\cdots\;\sin\left(2\theta\right)=2\cos\left(\theta\right)\sin\left(\theta\right) \\z&=&\lambda_1\sin^2\left(\theta\right) +\lambda_2\cos^2\left(\theta\right) \\&=&\frac{1}{2}2\left\{\lambda_1\sin^2\left(\theta\right) +\lambda_2\cos^2\left(\theta\right)\right\} \\&=&\frac{1}{2}\left\{2\lambda_1\sin^2\left(\theta\right) +2\lambda_2\cos^2\left(\theta\right)\right\} \\&=&\frac{1}{2}\left\{\lambda_1\left(\sin^2\left(\theta\right)+1-\cos^2\left(\theta\right)\right) +\lambda_2\left(\cos^2\left(\theta\right)+1-\sin^2\left(\theta\right)\right)\right\} \;\cdots\;\sin^2\left(\theta\right)=1-\cos^2\left(\theta\right),\;\cos^2\left(\theta\right)=1-\sin^2\left(\theta\right) \\&=&\frac{1}{2}\left\{\lambda_1\left(1-\left(\cos^2\left(\theta\right)-\sin^2\left(\theta\right)\right)\right) +\lambda_2\left(\cos^2\left(\theta\right)+1-\sin^2\left(\theta\right)\right)\right\} \\&=&\frac{1}{2}\left\{\lambda_1\left(1-\cos\left(2\theta\right)\right) +\lambda_2\left(\cos\left(2\theta\right)+1\right)\right\} \;\cdots\;\cos\left(2\theta\right)=\cos^2\left(\theta\right)-\sin^2\left(\theta\right) \\&=&\frac{1}{2}\left\{\lambda_1-\lambda_1\cos\left(2\theta\right) + \lambda_2\cos\left(2\theta\right)+\lambda_2\right\} \\&=&\frac{1}{2}\left\{ -\left( \lambda_1-\lambda_2\right) \cos\left(2\theta\right)+\lambda_1+\lambda_2\right\} \end{eqnarray}$$ $$ \left\{ \begin{eqnarray} x&=&\frac{1}{2}\left\{ \left( \lambda_1-\lambda_2\right) \cos\left(2\theta\right)+\lambda_1+\lambda_2\right\} \\y&=&\frac{1}{2}\left(\lambda_1-\lambda_2\right)\sin\left(2\theta\right) \\z&=&\frac{1}{2}\left\{ -\left( \lambda_1-\lambda_2\right) \cos\left(2\theta\right)+\lambda_1+\lambda_2\right\} \end{eqnarray} \right. $$ $$\begin{eqnarray} 2x&=& \left( \lambda_1-\lambda_2\right) \cos\left(2\theta\right)+\lambda_1+\lambda_2 \\2z&=&-\left( \lambda_1-\lambda_2\right) \cos\left(2\theta\right)+\lambda_1+\lambda_2 \\2x-2z&=&\left( \lambda_1-\lambda_2\right) \cos\left(2\theta\right)+\lambda_1+\lambda_2 - \left\{-\left( \lambda_1-\lambda_2\right) \cos\left(2\theta\right)+\lambda_1+\lambda_2\right\} \\2(x-z)&=&\left( \lambda_1-\lambda_2\right) \cos\left(2\theta\right)+\lambda_1+\lambda_2 + \left( \lambda_1-\lambda_2\right) \cos\left(2\theta\right)-\lambda_1-\lambda_2 \\2(x-z)&=&2\left( \lambda_1-\lambda_2\right) \cos\left(2\theta\right) \\\cos\left(2\theta\right)&=&\frac{x-z}{\lambda_1-\lambda_2}=\frac{x-z}{\sqrt{(x-z)^2+4y^2}} \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/11/2x2.html}{\lambda_{1}-\lambda_{2}=\sqrt{(x-z)^2+4y^2}} \\\sin\left(2\theta\right)&=&\frac{2y}{\lambda_1-\lambda_2}=\frac{2y}{\sqrt{(x-z)^2+4y^2}} \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/11/2x2.html}{\lambda_{1}-\lambda_{2}=\sqrt{(x-z)^2+4y^2}} \end{eqnarray}$$ \(\lambda_1,\lambda_2\)だけでなく\(\theta\)も\(x,y,z\)により一意に決まる．