每週問題 December 5, 2016

給定正整數 k,證明任一矩陣 A 可分解為 A=B(B^\ast B)^k

Let A be any n\times n complex matrix. Show that for each positive integer k there exists a unique matrix B such that A=B(B^\ast B)^k.

 
參考解答:

我們先證明存在性。考慮 n\times n 階矩陣 A 的奇異值分解 A=U\Sigma V^\ast,其中 U^\ast=U^{-1}V^\ast=V^{-1},且 \Sigma=\hbox{diag}(\sigma_1,\ldots,\sigma_n),每一奇異值 \sigma_i\ge 0。令 B=U\Sigma^{\frac{1}{2k+1}}V^\ast,其中 k 是正整數。計算可得

\displaystyle \begin{aligned} B^\ast B&=\left(U\Sigma^{\frac{1}{2k+1}}V^\ast\right)^\ast\left(U\Sigma^{\frac{1}{2k+1}}V^\ast\right)\\ &=V\Sigma^{\frac{1}{2k+1}}U^\ast U\Sigma^{\frac{1}{2k+1}}V^\ast\\ &=V\Sigma^{\frac{2}{2k+1}}V^\ast,\end{aligned}

因此

\displaystyle \begin{aligned} B(B^\ast B)^k&=U\Sigma^{\frac{1}{2k+1}}V^\ast\left(V\Sigma^{\frac{2}{2k+1}}V^\ast\right)^k\\ &=U\Sigma^{\frac{1}{2k+1}}V^\ast V\Sigma^{\frac{2k}{2k+1}}V^\ast\\ &=U\Sigma V^\ast=A. \end{aligned}

 
接著證明唯一性。假設 A=B(B^\ast B)^k=C(C^\ast C)^k,則 A^\ast A=(B^\ast B)^{2k+1}=(C^\ast C)^{2k+1}。Hermitian 半正定矩陣 A^\ast A 存在唯一的 2k+1 次方根,即 B^\ast B=C^\ast C。再者,Hermitian 半正定矩陣 B^\ast B 的特徵值滿足 \lambda_i\ge 0i=1,\ldots,n,並有完整的單範正交 (orthonormal) 特徵向量集 \{\mathbf{x}_1,\ldots,\mathbf{x}_n\},可當作 \mathbb{C}^n 的一組基底。如果我們能證明 B\mathbf{x}_i=C\mathbf{x}_ii=1,\ldots,n,也就證得 B=C。使用這個性質:B^\ast BB 有相同的零空間 (nullspace),C^\ast CC 有相同的零空間,故 B^\ast B, B, C 有相同的零空間。若 \lambda_i=0,則 B^\ast B\mathbf{x}_i=B\mathbf{x}_i=C\mathbf{x}_i=\mathbf{0}。若 \lambda_i\neq 0,則 A\mathbf{x}_i=B(B^\ast B)^k\mathbf{x}_i=\lambda_i^kB\mathbf{x}_iA\mathbf{x}_i=C(C^\ast C)^k\mathbf{x}_i=\lambda_i^kC\mathbf{x}_i,就有 B\mathbf{x}_i=C\mathbf{x}_i

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