每週問題 June 5, 2017

Posted on 06/05/2017 by ccjou

證明 A+A^T 是不可逆矩陣的一個充分條件。

Let A and B be n\times n matrices, where n is an odd number. Prove that if AB=0 then at least one of the matrices A+A^T and B+B^T is singular.

 
參考解答：

假設 A 與 B 是 2m+1 階方陣。使用 Sylvester 不等式，

\hbox{rank}A+\hbox{rank}B\le\hbox{rank} (AB)+2m+1=2m+1。

因此，\hbox{rank}A\le m 或 \hbox{rank}B\le m。若 \hbox{rank}A\le m，則 \hbox{rank}A^T=\hbox{rank}A\le m。使用不等式 \hbox{rank}(X+Y)\le \hbox{rank}X+\hbox{rank}Y，可得

\hbox{rank}(A+A^T)\le \hbox{rank}A+\hbox{rank}A^T\le 2m<2m+1，

證明 A+A^T 是不可逆的。

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