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9625

2024-07-02 20:26:34   224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

PT(A+B)P=diag(Er,0)
<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
PTAP=(A1A2′A2A3)
<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
PTBP=(Er?A1?A2′?A2?A3)
<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

QTA1Q=diag(λ1,?,λr)=C
<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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