(2014-04-18 from 352558840@qq.com [南开大学 2014 年高等代数考研试题]反对称矩阵的组合) 设 ${\bf A},{\bf B}$ 都是反对称矩阵, 且 ${\bf A}$ 可逆, 则 $|{\bf A}^2-{\bf B}|>0$.
证明: 由 ${\bf A}^T=-{\bf A}$ 知 $$\bex |{\bf A}|=|{\bf A}^T|=(-1)^n |{\bf A}|. \eex$$ 故 $n$ 为偶数 (否则, $|{\bf A}|=0$, ${\bf A}$ 不可逆). 又 ${\bf A}$ 可逆, ${\bf A}^T{\bf A}$ 正定, 而存在可逆阵 ${\bf P}$, 使得 ${\bf P}^T{\bf A}^T{\bf A}{\bf P}={\bf E}$. 于是 $$\beex \bea |{\bf P}^T|\cdot|{\bf A}^2-{\bf B}|\cdot|{\bf P}| &=|{\bf P}^T|\cdot|-{\bf A}^T{\bf A}-{\bf B}|\cdot|{\bf P}|\\ &=|{\bf P}^T|\cdot|{\bf A}^T{\bf A}+{\bf B}|\cdot|{\bf P}|\quad\sex{n\mbox{ 为偶数}}\\ &=|{\bf E}+{\bf P}^T{\bf B}{\bf P}|. \eea \eeex$$ 既然 ${\bf P}^T{\bf B}{\bf P}$ 也是反对称矩阵, 而存在正交阵 ${\bf Q}$, 使得 (参考文献) $$\bex {\bf Q}^T{\bf P}^T{\bf B}{\bf P}{\bf Q} =\sex{\ba{cccc} {\bf D}&&&\\ &\ddots&&\\ &&{\bf D}&\\ &&&{\bf 0} \ea},\quad{\bf D}=\sex{\ba{cc} 0&1\\ -1&0 \ea}. \eex$$ 于是 $$\beex \bea |{\bf Q}^T\cdot{\bf P}^T|\cdot|{\bf A}^2-{\bf B}|\cdot|{\bf P}\cdot{\bf Q}| &=\sev{{\bf E}+\sex{\ba{cccc} {\bf D}&&&\\ &\ddots&&\\ &&{\bf D}&\\ &&&{\bf 0} \ea}}\\ &=2^r\quad\sex{r\mbox{ 为 }\sex{\ba{cccc} {\bf D}&&&\\ &\ddots&&\\ &&{\bf D}&\\ &&&{\bf 0} \ea}\mbox{ 中 }{\bf D}\mbox{ 的个数}}\\ &>0. \eea \eeex$$
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