证明：由于${A^2} = A$，且$rleft( A ight) = r$，则存在可逆阵$P$，使得
[{P^{ - 1}}AP = left( {egin{array}{*{20}{c}}
{{E_r}}&{}\
{}&0
end{array}}
ight)]
即${P^{ - 1}}AP = left( {egin{array}{*{20}{c}}
{{E_r}}&{}\
{}&0
end{array}}
ight)$，令$B = Pleft( {egin{array}{*{20}{c}}
{{0_r}}&{}&{}\
{}&J&{}\
{}&{}&0
end{array}}
ight){P^{ - 1}}$，则命题得证
其中$J$为$s$阶若当块，对角线全为$0$
$f注：$若$A$的零化多项式无重根，则$A$可相似对角化