每週問題 June 24, 2013

本週問題是關於 Jordan 典型形式的相似問題。

Let

A=\begin{bmatrix}  \lambda&1&0\\  0&\lambda&1\\  0&0&\lambda  \end{bmatrix}~~~\textrm{and}~~~B=\begin{bmatrix}  \lambda&c&0\\  0&\lambda&c\\  0&0&\lambda  \end{bmatrix},

where c\neq 0. Show that A is similar to B.

 
參考解答:

P=\begin{bmatrix}  c^2&0&0\\  0&c&0\\  0&0&1  \end{bmatrix}

因為 c\neq 0P 是一可逆矩陣。計算可得

\begin{aligned}  PAP^{-1}&=\begin{bmatrix}  c^2&0&0\\  0&c&0\\  0&0&1  \end{bmatrix}\begin{bmatrix}  \lambda&1&0\\  0&\lambda&1\\  0&0&\lambda  \end{bmatrix}\begin{bmatrix}  c^{-2}&0&0\\  0&c^{-1}&0\\  0&0&1  \end{bmatrix}\\  &=\begin{bmatrix}  c^2&0&0\\  0&c&0\\  0&0&1  \end{bmatrix}\begin{bmatrix}  c^{-2}\lambda&c^{-1}&0\\  0&c^{-1}\lambda&1\\  0&0&\lambda  \end{bmatrix}\\  &=\begin{bmatrix}  \lambda&c&0\\  0&\lambda&c\\  0&0&\lambda  \end{bmatrix}=B,\end{aligned}

因此證明 A 相似於 B

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This entry was posted in pow 典型形式, 每週問題 and tagged , . Bookmark the permalink.

4 Responses to 每週問題 June 24, 2013

  1. levinc417 says:

    阿,這一份沒有放PDF檔~

    • ccjou says:

      前一陣子更換主機,LaTeX排版軟體尚未重灌。等安裝妥,近期內會再貼上。

  2. levinc417 says:

    這個P怎麼想出來的啊!?

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