768

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

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

$f注：$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${x^T}{left( {Q + x{x^T}} ight)^{ - 1}}x < 1$