9656

$f命题1:$设$alpha $,$eta $为实$n$维非零列向量,求$alpha eta '{ m{ + }}eta alpha '$的正负惯性指数

方法一:由$alpha $是非零列向量知,存在可逆阵$P$,使得[Palpha = {left( {1,0, cdots ,0} ight)^prime }]
从而可设[Peta = {left( {{b_1},{b_2}, cdots ,{b_n}} ight)^prime }]
则[Pleft( {alpha eta '{ m{ + }}eta alpha '} ight)P' = left( {egin{array}{*{20}{c}}
{2{b_1}}&{gamma '}\
gamma &0
end{array}} ight)]
其中$gamma = {left( {{b_2}, cdots ,{b_n}} ight)^prime } $

$left( 1 ight)$若${b_i} = 0left( {i = 2, cdots ,n} ight)$,则由$eta e0$知${b_1} e 0$,从而

由合同标准形理论知,存在可逆阵$M$,使得[MPleft( {alpha eta '{ m{ + }}eta alpha '} ight)P'M' = left( {egin{array}{*{20}{c}}
varepsilon &0\
0&0
end{array}} ight)]
其中$varepsilon = pm 1$,即正负惯性指数分别为$1,0$或$0,1$

$left( 2 ight)$若存在$i$,使得${b_i} e 0$$left( {i = 2, cdots ,n} ight)$,则

由合同标准形理论知,存在可逆阵$N$,使得[NPleft( {alpha eta '{ m{ + }}eta alpha '} ight)P'N' = diagleft( {1, - 1,0} ight)]
此时正负惯性指数均为$1$

原文地址:https://www.cnblogs.com/ly758241/p/3706383.html