设 ${\bf X},{\bf Y}$ 分别为 $m\times n$ 与 $n\times m$ 阵, 且 $$\bex {\bf Y}{\bf X}={\bf E}_n,\quad {\bf A}={\bf E}_m+{\bf X}{\bf Y}. \eex$$ 证明: ${\bf A}$ 相似于对角阵.
证明: 由 ${\bf Y}{\bf X}={\bf E}_n$ 知 $$\bex n=\rank({\bf Y}{\bf X})\leq \min\sed{\rank({\bf Y}),\rank({\bf X})}\leq \min\sed{m,n}\leq n. \eex$$ 于是 $$\bex \rank({\bf Y})=\rank({\bf X})=n\leq m. \eex$$ 设 ${\bf X}=({\bf x}_1,{\bf x}_2,\cdots,{\bf x}_n)$, 而 ${\bf Y}{\bf x}={\bf 0}$ 的基础解系为 ${\bf x}_{n+1},\cdots,{\bf x}_m$ (由 $\rank({\bf Y})=n$), 则 (1) 断言: ${\bf x}_1,\cdots,{\bf x}_n,{\bf x}_{n+1},\cdots,{\bf x}_m$ 线性无关. 事实上, $$\beex \bea &\quad l_1{\bf x}_1+\cdots+l_n{\bf x}_n+l_{n+1}{\bf x}_{n+1}+\cdots+l_m{\bf x}_m={\bf 0}\\ &\ra l_1{\bf e}_1+\cdots+l_n{\bf e}_n={\bf 0}\quad\sex{\mbox{用 }{\bf Y}\mbox{ 作用后利用 }{\bf Y}{\bf X}={\bf E}_n}\\ &\ra l_1=\cdots=l_n=0\\ &\ra l_{n+1}{\bf x}_{n+1}+\cdots+l_m{\bf x}_m={\bf 0}\\ &\ra l_{n+1}=\cdots=l_m=0. \eea \eeex$$ (2) 由 $$\beex \bea {\bf A}({\bf x}_1,\cdots,{\bf x}_m) &=({\bf x}_1,\cdots,{\bf x}_m) +{\bf X}({\bf e}_1,\cdots,{\bf e}_n,{\bf 0},\cdots,{\bf 0})\\ &=({\bf x}_1,\cdots,{\bf x}_m) +({\bf x}_1,\cdots,{\bf x}_n,{\bf 0},\cdots,{\bf 0})\\ &=(2{\bf x}_1,\cdots,2{\bf x}_n,{\bf x}_{n+1},\cdots,{\bf x}_m)\\ &=({\bf x}_1,\cdots,{\bf x}_m)\sex{\ba{cc} 2{\bf E}_n&\\ &{\bf E}_{m-n} \ea} \eea \eeex$$ 知 $$\bex {\bf P}^{-1}{\bf A}{\bf P}=\sex{\ba{cc} 2{\bf E}_n&\\ &{\bf E}_{m-n} \ea},{\bf P}=({\bf x}_1,\cdots,{\bf x}_m). \eex$$
参考: 参考资料中的例4.
