每週問題 April 27, 2015

這是利用行列式證明一特殊矩陣型態必定可逆的問題。

For any n\times n matrix A, show that there exists a matrix J=\text{diag}(\pm 1, \ldots,\pm 1) such that A+J is nonsingular.

 
參考解答:

我們採用歸納法來證明。若 n=1,命題明顯成立。假設對於 (n-1)\times (n-1) 階矩陣,命題成立。考慮 n\times n 階矩陣 A=\begin{bmatrix} A_1&A_2\\ A_3&a \end{bmatrix},其中 A_1(n-1)\times(n-1) 階分塊。根據歸納假設,存在一 (n-1)\times(n-1) 階矩陣 J_1=\text{diag}(\pm 1, \ldots,\pm 1) 使得 \det (A_1+J_1)\neq 0。運用行列式餘因子公式,以最末行或最末列化簡,可得

\displaystyle\begin{aligned} \begin{vmatrix} A_1+J_1&A_2\\ A_3&a+1 \end{vmatrix}-\begin{vmatrix} A_1+J_1&A_2\\ A_3&a-1 \end{vmatrix}&=(a+1)\det(A_1+J_1)-(a-1)\det(A_1+J_1)\\ &=2\det(A_1+J_1)\neq 0. \end{aligned}

上式等號左邊的兩個行列式至少有一個不為零,因此得證。

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