证明:$(1)$设$rleft( A ight) = r$,则由$Jordan$标准形理论知,存在可逆阵$P$,使得
[{P^{ - 1}}AP = left( {egin{array}{*{20}{c}}
0&{{E_r}}\
0&0
end{array}}
ight)]
从而可知
[{P^{ - 1}}{A^{k - 1}}P = {left( {{P^{ - 1}}AP}
ight)^{k - 1}} = {left( {egin{array}{*{20}{c}}
0&{{E_r}}\
0&0
end{array}}
ight)^{k - 1}}]
由于${A^{k - 1}} e 0$,则${P^{ - 1}}{A^{k - 1}}P e 0$,从而$rleft( A ight) = r ge k - 1$,而同理由${A^k} = 0$知,$rleft( A ight) = r < k < k + 1 < cdots < n$,
所以有$rleft( A ight) $的最小值为$k-1$,即$Ax=0$的解空间维数的最大值为$n-(k-1)$
$(2)$当$A$的秩最大时,$Ax=0$解空间的维数最小。此时若$pmid n$,则维数为$frac{n}{p}$;若$p mid n$,则维数为$left[ {frac{n}{p}} ight] + 1$