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2024-07-02 20:26:34 223人阅读
$\bf命题2:$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
<script id="MathJax-Element-1" type="math/tex; mode=display">{P^T}\left( {A + B} \right)P = diag\left( {{E_r},0} \right)</script>
设<script id="MathJax-Element-2" type="math/tex; mode=display">{P^T}AP = \left( {\begin{array}{*{20}{c}} {{A_1}}&{{A_2}}\\ {{A_2}^\prime }&{{A_3}} \end{array}} \right)</script>
则<script id="MathJax-Element-3" type="math/tex; mode=display">{P^T}BP = \left( {\begin{array}{*{20}{c}} {{E_r} - {A_1}}&{ - {A_2}}\\ { - {A_2}^\prime }&{ - {A_3}} \end{array}} \right)</script>
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
<script id="MathJax-Element-4" type="math/tex; mode=display">{Q^T}{A_1}Q = diag\left( {{\lambda _1}, \cdots ,{\lambda _r}} \right) = C</script>
从而可知
diag(QT,En?r)PTAPdiag(Q,En?r)=diag(C,0) <script id="MathJax-Element-5" type="math/tex; mode=display">diag\left( {{Q^T},{E_{n - r}}} \right){P^T}APdiag\left( {Q,{E_{n - r}}} \right) = diag\left( {C,0} \right)</script>
diag(QT,En?r)PTBPdiag(Q,En?r)=diag(Er?C,0) <script id="MathJax-Element-6" type="math/tex; mode=display">diag\left( {{Q^T},{E_{n - r}}} \right){P^T}BPdiag\left( {Q,{E_{n - r}}} \right) = diag\left( {{E_r} - C,0} \right)</script>
令$R = Pdiag\left( {Q,{E_{n - r}}} \right)$,则结论成立$\bf注1:$
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