每週問題 December 23, 2013

這是關於冪零矩陣的問題。

Let A and B be n\times n matrices. Does ABAB=0 imply BABA=0?

 
參考解答:

因為 (AB)^2=ABAB=0AB 是冪零 (nilpotent) 矩陣,AB 的所有特徵值皆為零。由於 ABBA 有相同的特徵值,可知 BA 的特徵值也為零。若 n=1,命題明顯為真。若 n=2,設 BA 的 Jordan 典型形式為 BA=SJS^{-1},其中 J=\begin{bmatrix}  0&1\\  0&0  \end{bmatrix}J=\begin{bmatrix}  0&0\\  0&0  \end{bmatrix},因此 BABA=(BA)^2=SJ^2S^{-1}=S0S^{-1}=0。若 n\ge 3ABAB=0 不能推論 BABA=0。下為一例:

A=\begin{bmatrix}  0&0&1\\  0&0&0\\  0&1&0  \end{bmatrix},~~B=\begin{bmatrix}  0&0&1\\  1&0&0\\  0&0&0  \end{bmatrix}

計算可得 (AB)^2=0,但

(BA)^2=\begin{bmatrix}  0&0&1\\  0&0&0\\  0&0&0  \end{bmatrix}

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