6265689

证明：由于${A^2} = A$，且$r\left( A \right) = r$，则存在可逆阵$P$，使得

\[{P^{ - 1}}AP = \left( {\begin{array}{*{20}{c}}
{{E_r}}&{}\\
{}&0
\end{array}} \right)\]

即${P^{ - 1}}AP = \left( {\begin{array}{*{20}{c}}
{{E_r}}&{}\\
{}&0
\end{array}} \right)$，令$B = P\left( {\begin{array}{*{20}{c}}
{{0_r}}&{}&{}\\
{}&J&{}\\
{}&{}&0
\end{array}} \right){P^{ - 1}}$，则命题得证

其中$J$为$s$阶若当块，对角线全为$0$

$\bf注：$若$A$的零化多项式无重根，则$A$可相似对角化