$f命题1：$设$A$，$B$均为实对称半正定阵，则$trleft( {AB} ight) le trleft( A ight) cdot trleft( B ight)$
证明：由$A$实对称知，存在正交阵$Q$，使得[A = Qdiagleft( {{lambda _1}, cdots ,{lambda _n}}
ight){Q^T}]
其中${{lambda _1}}$为$A$的最大特征值，则
egin{align*}
trleft( {AB}
ight) &= trleft( {Qdiagleft( {{lambda _1}, cdots ,{lambda _n}}
ight){Q^T}B}
ight)\&
{
m{ }} = trleft( {diagleft( {{lambda _1}, cdots ,{lambda _n}}
ight){Q^T}BQ}
ight)\&
{
m{ }} = sumlimits_{i = 1}^n {{lambda _i}{b_{ii}}} le {lambda _1}trleft( B
ight)\&
{
m{ }} le left( {sumlimits_{i = 1}^n {{lambda _i}} }
ight) cdot trleft( B
ight) = trleft( A
ight) cdot trleft( B
ight)
end{align*}
其中${b_{11}}, cdots ,{b_{nn}}$为${{Q^T}BQ}$的对角元
$f注：$矩阵的迹的性质$trleft( {MN} ight) = trleft( {NM} ight)$