768

$\bf命题：$设$A$为$n$阶实对称阵，$\alpha $为$n$维实向量，$\left( {\begin{array}{*{20}{c}}A&\alpha \\{{\alpha ^T}}&1\end{array}} \right)$为正定阵，证明：$A$正定且${\alpha ^T}{A^{ - 1}}\alpha < 1$

证明：作合同变换
\[{\rm{ }}\left( {\begin{array}{*{20}{c}}
E&0\\
{ - {\alpha ^T}{A^{ - 1}}}&E
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
A&\alpha \\
{{\alpha ^T}}&1
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
E&{ - {A^{ - 1}}\alpha }\\
0&E
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
A&0\\
0&{1 - {\alpha ^T}{A^{ - 1}}\alpha }
\end{array}} \right)\]
而合同变换保持正定性，故$A$正定且${\alpha ^T}{A^{ - 1}}\alpha < 1$

$\bf注：$由命题我们容易得出下面考研题

$(03中科院七)$设$Q$为$n$阶正定阵，$x$为$n$维实向量，则${x^T}{\left( {Q + x{x^T}} \right)^{ - 1}}x < 1$