每週問題 February 24, 2014

這是關於 A 和轉置 A^T 的特徵方程表達問題。

Let \lambda be an eigenvalue of A=[a_{ij}] and let \mathbf{x}=(x_1,\ldots,x_n)^T and \mathbf{y}=(y_1,\ldots,y_n)^T be eigenvectors corresponding to \lambda of A and A^T, respectively. Show that

\displaystyle  \lambda\sum_{i=1}^nx_i=\sum_{j=1}^nx_jc_j

and

\displaystyle  \lambda\sum_{j=1}^ny_j=\sum_{i=1}^ny_ir_i,

where c_j=\sum_{i=1}^na_{ij} and r_i=\sum_{j=1}^na_{ij}, 1\le i,j\le n.

 
參考解答:

寫出特徵方程 A\mathbf{x}=\lambda\mathbf{x} 的顯式表達,如下:

\displaystyle  \sum_{j=1}^na_{ij}x_j=\lambda x_i,~~i=1,\ldots,n

將上面 n 個等式加總在一起,

\displaystyle\begin{aligned}  \lambda\sum_{i=1}^nx_i&=\sum_{i=1}^n\sum_{j=1}^na_{ij}x_j\\  &=\sum_{j=1}^nx_{j}\sum_{i=1}^na_{ij}\\  &=\sum_{j=1}^nx_jc_j.  \end{aligned}

同樣地,寫出 A^T\mathbf{y}=\lambda\mathbf{y} 的顯式表達:

\displaystyle  \sum_{i=1}^na_{ij}y_i=\lambda y_j,~~j=1,\ldots,n

加總上面 n 個等式,

\displaystyle\begin{aligned}  \lambda\sum_{j=1}^ny_j&=\sum_{j=1}^n\sum_{i=1}^na_{ij}y_i\\  &=\sum_{i=1}^ny_{i}\sum_{j=1}^na_{ij}\\  &=\sum_{i=1}^ny_ir_i.  \end{aligned}

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