656

$\bf命题3：$设$A$实对称正定，$B$实对称半正定，则$tr\left( {B{A^{ - 1}}} \right)tr\left( A \right) \ge tr\left( B \right)$

方法一：同时合同对角化

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