每週問題 December 15, 2014

這是實對稱矩陣的二次型極值問題。

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).

 
參考解答:

\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}

上面令 \mathbf{y}=(y_1,\ldots,y_n)^T=U\mathbf{x}。因為 \mathbf{y}^\ast\mathbf{y}=\mathbf{x}^\ast U^\ast U\mathbf{x}=\mathbf{x}^\ast\mathbf{x},可知 \mathbf{y} 也是單位向量。所以,

\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

另一方面,

\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

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