## 每週問題 February 16, 2015

Let $\{\mathbf{x}_1,\ldots,\mathbf{x}_k\}$ be a set of linearly independent eigenvectors for an $n\times n$ matrix $A$ corresponding to respective eigenvalues $\lambda_1,\ldots,\lambda_k$. Let $X$ be any $n\times (n-k)$ matrix such that $S=\begin{bmatrix} \mathbf{x}_1&\cdots&\mathbf{x}_k&X \end{bmatrix}$ is nonsingular. Show that if $S^{-1}=\begin{bmatrix} \mathbf{y}_1^\ast\\ \vdots\\ \mathbf{y}_k^\ast\\ Y^\ast \end{bmatrix}$, where $\mathbf{y}_i^\ast$ are $1\times n$ and $Y^\ast$ is $(n-k)\times n$, then $\{\mathbf{y}_1,\ldots,\mathbf{y}_k\}$ is a set of linearly independent left eigenvectors associated with $\lambda_1,\ldots,\lambda_k$, respectively, i.e., $\mathbf{y}_i^\ast A=\lambda_i\mathbf{y}_i^\ast$, $1\le i\le k$.

$\displaystyle \begin{bmatrix} \mathbf{y}_1^\ast\\ \vdots\\ \mathbf{y}_k^\ast\\ Y^\ast \end{bmatrix}\begin{bmatrix} \mathbf{x}_1&\cdots&\mathbf{x}_k&X \end{bmatrix}=\begin{bmatrix} \mathbf{y}_1^\ast\mathbf{x}_1&\cdots&\mathbf{y}_1^\ast\mathbf{x}_k&\mathbf{y}_1^\ast X\\ \vdots&\ddots&\vdots&\vdots\\ \mathbf{y}_k^\ast\mathbf{x}_1&\cdots&\mathbf{y}_k^\ast\mathbf{x}_k&\mathbf{y}_k^\ast X\\ Y^\ast\mathbf{x}_1&\cdots&Y^\ast\mathbf{x}_k&Y^\ast X\\ \end{bmatrix}=\begin{bmatrix} 1&&&\\ &\ddots&&\\ &&1&\\ &&&I_{n-k} \end{bmatrix}$

\displaystyle\begin{aligned} S^{-1}AS&=\begin{bmatrix} \mathbf{y}_1^\ast\\ \vdots\\ \mathbf{y}_k^\ast\\ Y^\ast \end{bmatrix}A\begin{bmatrix} \mathbf{x}_1&\cdots&\mathbf{x}_k&X \end{bmatrix}=\begin{bmatrix} \mathbf{y}_1^\ast\\ \vdots\\ \mathbf{y}_k^\ast\\ Y^\ast \end{bmatrix}\begin{bmatrix} A\mathbf{x}_1&\cdots&A\mathbf{x}_k&AX \end{bmatrix}\\ &=\begin{bmatrix} \mathbf{y}_1^\ast\\ \vdots\\ \mathbf{y}_k^\ast\\ Y^\ast \end{bmatrix}\begin{bmatrix} \lambda_1\mathbf{x}_1&\cdots&\lambda_k\mathbf{x}_k&AX \end{bmatrix}=\begin{bmatrix} \lambda_1&&&\\ &\ddots&&\\ &&\lambda_k&\\ &&&Y^\ast AX \end{bmatrix}=B. \end{aligned}

$\displaystyle S^{-1}A=\begin{bmatrix} \mathbf{y}_1^\ast\\ \vdots\\ \mathbf{y}_k^\ast\\ Y^\ast \end{bmatrix}A=\begin{bmatrix} \mathbf{y}_1^\ast A\\ \vdots\\ \mathbf{y}_k^\ast A\\ Y^\ast A \end{bmatrix}$

$\displaystyle BS^{-1}=\begin{bmatrix} \lambda_1&&&\\ &\ddots&&\\ &&\lambda_k&\\ &&&Y^\ast AX \end{bmatrix}\begin{bmatrix} \mathbf{y}_1^\ast\\ \vdots\\ \mathbf{y}_k^\ast\\ Y^\ast \end{bmatrix}=\begin{bmatrix} \lambda_1\mathbf{y}_1^\ast\\ \vdots\\ \lambda_k\mathbf{y}_k^\ast\\ Y^\ast AXY^\ast \end{bmatrix}$

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