2x2対称行列の固有値の差

$$\begin{array}{rcl}{S}_2& =& \left[\begin{array}{cc}x& y\\ y& z\end{array}\right]\cdots S:\mathrm{対}\mathrm{称}\mathrm{行}\mathrm{列}\\ det(S-\lambda I)& =& \left[\begin{array}{cc}x-\lambda & y\\ y& z-\lambda \end{array}\right]\\ & =& (x-\lambda )(z-\lambda )-y^2\\ & =& {\lambda }^2-(x+z)\lambda +xz-y^2\\ {\lambda }_{1,2}& =& \frac{-\{-(x+z)\}\pm \sqrt{{\{-(x+z)\}}^2-4(xz-y^2)}}{2}\cdots x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ & =& \frac{1}{2}(x+z\pm \sqrt{{\{-(x+z)\}}^2-4(xz-y^2)})\\ & =& \frac{1}{2}(x+z\pm \sqrt{x^2+2xz+z^2-4xz+4y^2})\\ & =& \frac{1}{2}(x+z\pm \sqrt{x^2-2xz+z^2+4y^2})\\ & =& \frac{1}{2}(x+z\pm \sqrt{(x-z{)}^2+4y^2})\\ {\lambda }_1-{\lambda }_2& =& \frac{1}{2}(x+z+\sqrt{(x-z{)}^2+4y^2})-\frac{1}{2}(x+z-\sqrt{(x-z{)}^2+4y^2})\cdots {\lambda }_1>{\lambda }_2\mathrm{と}\mathrm{す}\mathrm{る}\\ & =& \frac{1}{2}(x+z+\sqrt{(x-z{)}^2+4y^2}-x-z+\sqrt{(x-z{)}^2+4y^2})\\ & =& \frac{1}{2}\left(2\sqrt{(x-z{)}^2+4y^2}\right)\\ & =& \sqrt{(x-z{)}^2+4y^2}\end{array}$$