每週問題 January 7, 2013

這是矩陣乘法定義的線性變換的行列式計算問題。

Let A be an n\times n matrix and define T(X)=AX for every n\times n matrix X. Show that \det T=\det(A^n).

 
參考解答:

\lambda_1,\ldots,\lambda_nA 的特徵值,對應特徵向量 \mathbf{x}_1,\ldots,\mathbf{x}_n。對於特徵值 \lambda_i,令 X_i(j) 為一 n\times n 階矩陣 X_i(j)=\begin{bmatrix}  \mathbf{0}&\cdots&\mathbf{x}_i&\cdots&\mathbf{0}  \end{bmatrix},其中第 j 行等於 \mathbf{x}_i,其餘行等於 \mathbf{0},則

\begin{aligned}  T(X_i(j))&=AX_i(j)\\  &=A\begin{bmatrix}  \mathbf{0}&\cdots&\mathbf{x}_i&\cdots&\mathbf{0}  \end{bmatrix}\\  &=\begin{bmatrix}  \mathbf{0}&\cdots&\lambda_i\mathbf{x}_i&\cdots&\mathbf{0}  \end{bmatrix}\\  &=\lambda_i\begin{bmatrix}  \mathbf{0}&\cdots&\mathbf{x}_i&\cdots&\mathbf{0}  \end{bmatrix}\\  &=\lambda_iX_i(j).\end{aligned}

因此,X_i(j)j=1,\ldots,n,是 T 的特徵向量,對應特徵值 \lambda_i,也就是說,T 的特徵值 \lambda_i 的相重數為 n (但 \lambda_i 未必相異)。所以,\det T=\lambda_1^n\cdots\lambda_n^n=\det(A^n)

PowSol-Jan-7-13

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