每週問題 December 28, 2015

計算 \begin{bmatrix}A&A\\A&A\end{bmatrix} 的特徵值。

Let A be an n\times n matrix. Find the eigenvalues of \begin{bmatrix}A&A\\A&A\end{bmatrix} in terms of those of A.

 
參考解答:

解法1:令 \lambda_1,\ldots,\lambda_nA 的特徵值。使用分塊矩陣的行列式公式,

\begin{aligned} \begin{vmatrix}A-\lambda I&A\\A&A-\lambda I\end{vmatrix}&=\det(A-\lambda I+A)\det(A-\lambda I-A)\\&=\det(2A-\lambda I)\det(-\lambda I).\end{aligned}

所以,\begin{bmatrix}A&A\\A&A\end{bmatrix} 的特徵值由 2A 與零矩陣的特徵值組成,即 2\lambda_1,\ldots,2\lambda_n,以及 n0

解法2:考慮分塊矩陣乘法

\begin{bmatrix} I&0\\ -I&I\end{bmatrix}\begin{bmatrix} A&A\\A&A \end{bmatrix} \begin{bmatrix} I&0\\I&I\end{bmatrix}=\begin{bmatrix}2A&A\\0&0\end{bmatrix}

立知 \begin{bmatrix}A&A\\A&A\end{bmatrix} 相似於分塊上三角矩陣 \begin{bmatrix}2A&A\\0&0\end{bmatrix},故特徵值為 2\lambda_1,\ldots,2\lambda_n,以及 n0

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