## 每週問題 December 15, 2014

Let $A$ be an $n\times n$ Hermitian matrix. Let $\lambda_{\min}(A)$ and $\lambda_{\max}(A)$ be the smallest and largest eigenvalues of $A$, respectively. For every unit vector $\mathbf{x}\in\mathbb{C}^n$, show that

$\displaystyle \lambda_{\min}(A)\le\mathbf{x}^\ast A\mathbf{x}\le\lambda_{\max}(A)$.

$\lambda_1,\ldots,\lambda_n$ 為 Hermitian 矩陣 $A$ 的特徵值，且 $\lambda_1\le\lambda_2\le\cdots\le\lambda_n$。Hermitian 矩陣 $A$ 可么正對角化為 $A=U^\ast DU$，其中 $U$ 是么正矩陣 (unitary matrix) 滿足 $U^\ast=U^{-1}$$D=\text{diag}(\lambda_1,\ldots,\lambda_n)$。對於任一單位向量 $\mathbf{x}$

\displaystyle\begin{aligned} \mathbf{x}^\ast A\mathbf{x}&=\mathbf{x}^\ast U^\ast DU\mathbf{x}=\mathbf{y}^\ast D\mathbf{y}\\ &=\lambda_1\vert y_1\vert^2+\lambda_2\vert y_2\vert^2+\cdots+\lambda_n\vert y_n\vert^2,\end{aligned}

$\displaystyle \mathbf{x}^\ast A\mathbf{x}\ge\lambda_1\vert y_1\vert^2+\lambda_1\vert y_2\vert^2+\cdots+\lambda_1\vert y_n\vert^2=\lambda_1(\vert y_1\vert^2+\vert y_2\vert^2+\cdots+\vert y_n\vert^2)=\lambda_1$

$\displaystyle \mathbf{x}^\ast A\mathbf{x}\le\lambda_n\vert y_1\vert^2+\lambda_n\vert y_2\vert^2+\cdots+\lambda_n\vert y_n\vert^2=\lambda_n(\vert y_1\vert^2+\vert y_2\vert^2+\cdots+\vert y_n\vert^2)=\lambda_n$