## 每週問題 January 20, 2014

Let $A$ be an $n\times n$ matrix and $B=(A+A^{\ast})/2$. Let $\mu_{\max}$ and $\mu_{\min}$ be the largest eigenvalue and the smallest eigenvalue of $B$, respectively. Show that every eigenvalue $\lambda$ of $A$ satisfies $\mu_{\min}\le\text{Re}(\lambda)\le\mu_{\max}$.

\displaystyle \begin{aligned} \mathbf{x}^\ast B\mathbf{x}&=\mathbf{x}^\ast\left(\frac{A+A^\ast}{2}\right)\mathbf{x}=\frac{1}{2}\left(\mathbf{x}^\ast A\mathbf{x}+\mathbf{x}^\ast A^\ast\mathbf{x}\right)\\ &=\frac{1}{2}\left(\mathbf{x}^\ast A\mathbf{x}+(\mathbf{x}^\ast A\mathbf{x})^\ast\right)=\frac{1}{2}\left(\mathbf{x}^\ast A\mathbf{x}+\overline{\mathbf{x}^\ast A\mathbf{x}}\right)\\ &=\text{Re}(\mathbf{x}^\ast A\mathbf{x}) .\end{aligned}

$\displaystyle \mu_{\min}\le\frac{\mathbf{x}^\ast B\mathbf{x}}{\mathbf{x}^{\ast}\mathbf{x}}\le\mu_{\max}$

$\displaystyle \mu_{\min}\le\frac{\text{Re}(\mathbf{x}^\ast A\mathbf{x})}{\mathbf{x}^{\ast}\mathbf{x}}\le\mu_{\max}$

$\mathbf{x}$$A$ 的特徵向量，對應特徵值 $\lambda$，則 $\text{Re}(\mathbf{x}^\ast A\mathbf{x})=\text{Re}(\lambda\mathbf{x}^\ast\mathbf{x})=\text{Re}(\lambda)\mathbf{x}^\ast\mathbf{x}$，代入上式即證得所求。