$\bf命题：$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${\alpha ^T}{A^{ - 1}}\alpha
< 1$
原文地址：http://www.cnblogs.com/ly758241/p/3706312.html
