每週問題 February 24, 2014

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

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

\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}

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

\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}