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$\bf命题2：$设$A$，$B$均为实对称半正定阵，则$A$，$B$可同时合同对角化

证明：由$A,B$半正定知$A+B$半正定，则存在可逆阵$P$，使得
\[{P^T}\left( {A + B} \right)P = diag\left( {{E_r},0} \right)\]
设\[{P^T}AP = \left( {\begin{array}{*{20}{c}}
{{A_1}}&{{A_2}}\\
{{A_2}^\prime }&{{A_3}}
\end{array}} \right)\]
则\[{P^T}BP = \left( {\begin{array}{*{20}{c}}
{{E_r} - {A_1}}&{ - {A_2}}\\
{ - {A_2}^\prime }&{ - {A_3}}
\end{array}} \right)\]
由于${ - {A_3}}$半正定且${ {A_3}}$半正定，故${{A_3} = 0}$，从而可知${{A_2} = 0}$

又由${{A_1}}$半正定知，存在正交阵$Q$，使得
\[{Q^T}{A_1}Q = diag\left( {{\lambda _1}, \cdots ,{\lambda _r}} \right) = C\]
从而可知
\[diag\left( {{Q^T},{E_{n - r}}} \right){P^T}APdiag\left( {Q,{E_{n - r}}} \right) = diag\left( {C,0} \right)\]
\[diag\left( {{Q^T},{E_{n - r}}} \right){P^T}BPdiag\left( {Q,{E_{n - r}}} \right) = diag\left( {{E_r} - C,0} \right)\]
令$R = Pdiag\left( {Q,{E_{n - r}}} \right)$，则结论成立

$\bf注1：$