每週問題 February 13, 2017

證明遍歷定理 (ergodic theorem)。

Let A be a unitary matrix, i.e., A^\ast=A^{-1}. Prove that

\displaystyle \lim_{m\to\infty}\frac{1}{m}\sum_{k=0}^{m-1}A^k\mathbf{x}=P\mathbf{x},

where P is the Hermitian projection matrix onto N(I-A^\ast).

 
參考解答:

An\times n 階么正 (unitary) 矩陣。根據正交補餘性質 C(I-A)^\perp=N(I-A^\ast),其中 C(I-A) 代表 I-A 的行空間 (column space),每一 \mathbf{x}\in\mathbb{C}^n 可唯一分解為 \mathbf{x}=\mathbf{y}+\mathbf{z},其中 \mathbf{y}\in C(I-A)\mathbf{z}\in N(I-A^\ast)。因為 \mathbf{y}\in C(I-A),存在 \mathbf{u}\in\mathbb{C}^n 使得 \mathbf{y}=(I-A)\mathbf{u}。當 m\to\infty,使用 \Vert A\mathbf{u}\Vert^2=\mathbf{u}^\ast A^\ast A\mathbf{u}=\mathbf{u}^\ast\mathbf{u}=\Vert\mathbf{u}\Vert^2,可得

\displaystyle \begin{aligned}\left\|\frac{1}{m}\sum_{k=0}^{m-1}A^k\mathbf{y}\right\|&=\left\|\frac{1}{m}\sum_{k=0}^{m-1}(A^k\mathbf{u}-A^{k+1}\mathbf{u})\right\|=\frac{1}{m}\left\|\mathbf{u}-A^{m}\mathbf{u}\right\|\\ &\le\frac{1}{m}(\Vert\mathbf{u}\Vert+\Vert A^m\mathbf{u}\Vert)=\frac{2\Vert\mathbf{u}\Vert}{m}\to 0, \end{aligned}

可知 \lim_{m\to\infty}\frac{1}{m}\sum_{k=0}^{m-1}A^k\mathbf{y}=0。因為 \mathbf{z}\in N(I-A^\ast),即 (I-A^\ast)\mathbf{z}=\mathbf{0},可得 \mathbf{z}=A^\ast\mathbf{z}=A^{-1}\mathbf{z},或 A\mathbf{z}=\mathbf{z},即有

\displaystyle \lim_{m\to\infty}\frac{1}{m}\sum_{k=0}^{m-1}A^k\mathbf{z}=\lim_{m\to\infty}\frac{1}{m}\sum_{k=0}^{m-1}\mathbf{z}=\mathbf{z}

合併以上結果即得證。

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This entry was posted in pow 內積空間, 每週問題 and tagged , . Bookmark the permalink.

1 Response to 每週問題 February 13, 2017

  1. levinc417 says:

    酷! 原來這定理也能用線代証。(我是在高等機率學過它)

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