99

$f命题：$设$A in {M_{m imes n}}left( F ight),B in {M_{n imes m}}left( F ight),m ge n,lambda e 0$，则

[{ m{ }}left| {lambda {E_m} - AB} ight| = {lambda ^{m - n}}left| {lambda {E_n} - BA} ight|]

方法二：等价标准形

由$rleft( A ight) = r$知，存在可逆阵$P,Q$，使得
[PAQ = left( {egin{array}{*{20}{c}}
{{E_r}}&0\
0&0
end{array}} ight)]令[{Q^{ - 1}}B{P^{ - 1}} = left( {egin{array}{*{20}{c}}
{{B_1}}&{{B_3}}\
{{B_4}}&{{B_2}}
end{array}} ight)]
其中${B_1}$为$r imes r$矩阵，于是有
[PAB{P^{ - 1}} = PAQ cdot {Q^{ - 1}}B{P^{ - 1}} = left( {egin{array}{*{20}{c}}
{{B_1}}&{{B_3}}\
0&0
end{array}} ight)]
且[{Q^{ - 1}}BAQ = {Q^{ - 1}}B{P^{ - 1}} cdot PAQ = left( {egin{array}{*{20}{c}}
{{B_1}}&0\
{{B_4}}&0
end{array}} ight)]
从而可知
[left| {lambda {E_m} - AB} ight| = left| {egin{array}{*{20}{c}}
{lambda {E_r} - {B_1}}&{ - {B_3}}\
0&{lambda {E_{m - r}}}
end{array}} ight| = {lambda ^{m - r}}left| {lambda {E_r} - {B_1}} ight|]
且[left| {lambda {E_n} - BA} ight| = left| {egin{array}{*{20}{c}}
{lambda {E_r} - {B_1}}&0\
{ - {B_4}}&{lambda {E_{n - r}}}
end{array}} ight| = {lambda ^{n - r}}left| {lambda {E_r} - {B_1}} ight|]

故结论成立

$f注1：$$left| {lambda {E_m} - AB} ight| = left| {{P^{ - 1}}} ight| cdot left| {lambda {E_m} - AB} ight| cdot left| P ight| = left| {{P^{ - 1}}lambda {E_m}P - {P^{ - 1}}ABP} ight| = left| {lambda {E_m} - {P^{ - 1}}ABP} ight|$