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$\bf命题1：$设$A$，$B$均为实对称半正定阵，则$tr\left( {AB} \right) \le tr\left( A \right) \cdot tr\left( B \right)$

证明：由$A$实对称知，存在正交阵$Q$，使得

$$A = Qdiag\left( {{\lambda _1}, \cdots ,{\lambda _n}} \right){Q^T}$$
其中${{\lambda _1}}$为$A$的最大特征值，则
$$\begin{align*} tr\left( {AB} \right) &= tr\left( {Qdiag\left( {{\lambda _1}, \cdots ,{\lambda _n}} \right){Q^T}B} \right)\\& {\rm{ }} = tr\left( {diag\left( {{\lambda _1}, \cdots ,{\lambda _n}} \right){Q^T}BQ} \right)\\& {\rm{ }} = \sum\limits_{i = 1}^n {{\lambda _i}{b_{ii}}} \le {\lambda _1}tr\left( B \right)\\& {\rm{ }} \le \left( {\sum\limits_{i = 1}^n {{\lambda _i}} } \right) \cdot tr\left( B \right) = tr\left( A \right) \cdot tr\left( B \right) \end{align*}$$
其中${b_{11}}, \cdots ,{b_{nn}}$为${{Q^T}BQ}$的对角元

$\bf注：$矩阵的迹的性质$tr\left( {MN} \right) = tr\left( {NM} \right)$

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