每週問題 August 29, 2016

證明 Sherman-Morrison-Woodbury 公式的一個特例。

Let A be an m\times m matrix, B be an n\times m matrix and C be an n\times n matrix. If A and C are symmetric positive definite, show the following identities.
(a) (A^{-1}+B^TC^{-1}B)^{-1}B^TC^{-1}=AB^T(BAB^T+C)^{-1}
(b) (A^{-1}+B^TC^{-1}B)^{-1}=A-AB^T(BAB^T+C)^{-1}BA

 
參考解答:

(a) 考慮

\begin{aligned} B^TC^{-1}BAB^T+B^T&=B^TC^{-1}(BAB^T+C),\\ B^T+B^TC^{-1}BAB^T&=(A^{-1}+B^TC^{-1}B)AB^T.\end{aligned}

上面兩式等號右邊左乘 (A^{-1}+B^TC^{-1}B)^{-1},右乘 (BAB^T+C)^{-1},可得

(A^{-1}+B^TC^{-1}B)^{-1}B^TC^{-1}=AB^T(BAB^T+C)^{-1}

(b) 使用 (a),

\begin{aligned}(A^{-1}+B^TC^{-1}B)^{-1}&=(A^{-1}+B^TC^{-1}B)^{-1}(I+B^TC^{-1}BA-B^TC^{-1}BA)\\ &=(A^{-1}+B^TC^{-1}B)^{-1}((A^{-1}+B^TC^{-1}B)A-B^TC^{-1}BA)\\ &=A-(A^{-1}+B^TC^{-1}B)^{-1}B^TC^{-1}BA\\ &=A-AB^T(BAB^T+C)^{-1}BA.\end{aligned}

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This entry was posted in pow 線性方程與矩陣代數, 每週問題 and tagged . Bookmark the permalink.

1 Response to 每週問題 August 29, 2016

  1. Meiyue Shao says:

    我觉得SMW公式最好是讲思想,单纯的代数演算既不利于理解,也不容易记住。

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