2x2対称行列の固有値の差

$$\begin{eqnarray} S_2&=&\begin{bmatrix} x & y \\ y & z \end{bmatrix} \;\cdots\;S:対称行列 \\det\left(S-\lambda I\right) &=&\begin{bmatrix} x - \lambda & y \\ y & z - \lambda \end{bmatrix} \\&=&\left(x - \lambda\right)\left(z - \lambda\right)-y^2 \\&=&\lambda^2-(x+z)\lambda+xz-y^2 \\\lambda_{1,2} &=&\frac{-\left\{ -(x+z) \right\}\pm\sqrt{\left\{-(x+z)\right\}^2-4\left(xz-y^2\right)}}{2} \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/11/blog-post.html}{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}} \\&=&\frac{1}{2}\left(x+z \pm \sqrt{\left\{-(x+z)\right\}^2-4\left(xz-y^2\right)}\right) \\&=&\frac{1}{2}\left(x+z \pm \sqrt{x^2+2xz+z^2-4xz+4y^2}\right) \\&=&\frac{1}{2}\left(x+z \pm \sqrt{x^2-2xz+z^2+4y^2}\right) \\&=&\frac{1}{2}\left(x+z \pm \sqrt{(x-z)^2+4y^2}\right) \\\lambda_{1}-\lambda_{2}&=&\frac{1}{2}\left(x+z + \sqrt{(x-z)^2+4y^2}\right) - \frac{1}{2}\left(x+z - \sqrt{(x-z)^2+4y^2}\right) \;\cdots\;\lambda_{1}\gt\lambda_{2}とする \\&=&\frac{1}{2}\left(x+z + \sqrt{(x-z)^2+4y^2} - x -z + \sqrt{(x-z)^2+4y^2}\right) \\&=&\frac{1}{2}\left(2\sqrt{(x-z)^2+4y^2}\right) \\&=&\sqrt{(x-z)^2+4y^2} \end{eqnarray}$$