656

$f命题3：$设$A$实对称正定，$B$实对称半正定，则$trleft( {B{A^{ - 1}}} ight)trleft( A ight) ge trleft( B ight)$

方法一：同时合同对角化

由题可知，存在可逆阵$R$，使得[R'AR = E，R'BR = diagleft( {{lambda _1}, cdots ,{lambda _n}} ight)]
其中${lambda _i} ge 0,i = 1, cdots ,n$，则
egin{align*}
R'B{A^{ - 1}}{{R'}^{ - 1}}& = R'BR cdot {R^{ - 1}}{A^{ - 1}}{{R'}^{ - 1}}\&
= diagleft( {{lambda _1}, cdots ,{lambda _n}} ight)
end{align*}
即[trleft( {B{A^{ - 1}}} ight) = sumlimits_{i = 1}^n {{lambda _i}} ]
从而可知egin{align*}
trleft( B ight) &= trleft( {diagleft( {{lambda _1}, cdots ,{lambda _n}} ight)C} ight)\&
= sumlimits_{j = 1}^n {{lambda _j}{c_j}} le sumlimits_{i = 1}^n {{lambda _i}} cdot sumlimits_{j = 1}^n {{c_j}} \&
= trleft( {B{A^{ - 1}}} ight) cdot trleft( C ight)
end{align*}
其中$C = {left( {R'R} ight)^{ - 1}},{c_j}left( {j = 1, cdots ,n} ight)$为$C$的对角元；而$trleft( A ight) = trleft( C ight)$，故结论成立