每週問題 December 9, 2013

這是關於等價標準型和奇異值分解的應用問題。

Let A be an n\times n matrix of rank r<n. Prove that there exists an n\times n matrix B of rank n-r such that A+B is invertible.

 
參考解答:

使用等價標準型,A=U\begin{bmatrix}  I_r&0\\  0&0  \end{bmatrix}V,其中 UV 是可逆矩陣。令 B=U\begin{bmatrix}  0&0\\  0&I_{n-r}  \end{bmatrix}V,其中 n-r=\hbox{rank}B。計算可得

\displaystyle  A+B=U\begin{bmatrix}  I_r&0\\  0&0  \end{bmatrix}V+U\begin{bmatrix}  0&0\\  0&I_{n-r}  \end{bmatrix}V=UI_nV=UV

證明 A+B 是可逆矩陣。另一個證法採用奇異值分解。令 A 的奇異值分解為 A=U\Sigma V^\ast,其中 \Sigma=\hbox{diag}(\sigma_1,\ldots,\sigma_r,0,\ldots,0)\sigma_1\ge\cdots\ge\sigma_r>0UV 是么正 (unitary) 矩陣,U^\ast=U^{-1}V^\ast=V^{-1}。令 B=U\begin{bmatrix}  0&0\\  0&I_{n-r}  \end{bmatrix}V^\ast。因此,

\displaystyle   A+B=U\hbox{diag}(\sigma_1,\ldots,\sigma_r,\underbrace{1,\ldots,1}_{n-r})V^\ast

是可逆矩陣。

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