每週問題 April 17, 2017

這是網友范智忠提供的問題

Let A and B be n\times n matrices. If A^2B+BA^2=2ABA, show that (AB-BA)^n=0.

 
參考解答:

改寫給定的條件,

\begin{aligned} A^2B+BA^2=2ABA &\Rightarrow A^2B-ABA=ABA-BA^2\\ &\Rightarrow A(AB-BA)=(AB-BA)A, \end{aligned}

可知 A 與交換子 (commutator) AB-BA 是可交換矩陣。接下來我們要證明 AB-BA 是一個冪零 (nilpotent) 矩陣,滿足 (AB-BA)^n=0。交換子 AB-BA 是冪零矩陣的一個充要條件為 \hbox{trace}((AB-BA)^k)=0k=1,\ldots,n。若 k=1,跡數的循環不變性立得 \hbox{trace}(AB-BA)=\hbox{trace}(AB)-\hbox{trace}(BA)=0。若 k > 1,使用 (AB-BA)^{k-1}A=A(AB-BA)^{k-1},可得

\begin{aligned} \hbox{trace}((AB-BA)^{k})&=\hbox{trace}((AB-BA)^{k-1}(AB-BA))\\ &=\hbox{trace}((AB-BA)^{k-1}AB)-\hbox{trace}((AB-BA)^{k-1}BA)\\ &=\hbox{trace}(A(AB-BA)^{k-1}B)-\hbox{trace}((AB-BA)^{k-1}BA)\\ &=\hbox{trace}((AB-BA)^{k-1}BA)-\hbox{trace}((AB-BA)^{k-1}BA)\\ &=0. \end{aligned}

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