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    $f命题1:$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$trleft( {MN} ight) = trleft( {NM} ight)$

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  • 原文地址:https://www.cnblogs.com/ly758241/p/3706392.html
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