设 ${f A}$ 为 $n$ 阶正定矩阵, ${f x}$, ${f y}$ 为 $n$ 维列向量且满足 ${f x}^t{f y}>0$. 证明矩阵 $$ex {f M}={f A}+cfrac{{f x}{f x}^t}{{f x}^t{f y}} -cfrac{{f A}{f y}{f y}^t{f A}}{{f y}^t{f A}{f y}} eex$$ 正定.
证明: （来自 chxp1234）易知$Y eq0,$从而$Y^{T}AY>0.$$forall Zin R^{n},Z eq0,$有 $$egin{aligned} &Z^{T}MZ\ =&Z^{T}AZ+dfrac{Z^{T}XX^{T}Z}{X^{T}Y}-dfrac{Z^{T}AYY^{T}AZ}{Y^{T}AY}\ =&dfrac{X^{T}Y[(Z^{T}AZ)(Y^{T}AY)-(Z^{T}AY)^{2}]+(X^{T}Z)^{2}(Y^{T}AY)}{(X^{T}Y)(Y^{T}AY)} end{aligned}$$ 在内积$(X,Y)=X^{T}AY$下,$R^{n}$构成欧氏空间,于是由柯西不等式 $$(Z^{T}AZ)(Y^{T}AY)-(Z^{T}AY)^{2}=(Z,Z)(Y,Y)-(Z,Y)^{2}geq0,$$ 这样就有 $$Z^{T}MZgeq0.$$ 下证$Z^{T}MZ>0.$否则易知$Z^{T}MZ=0$的充要条件为 $$(Z,Z)(Y,Y)-(Z,Y)^{2}=0mbox{且}X^{T}Z=0.$$ 而$(Z,Z)(Y,Y)-(Z,Y)^{2}=0$的充要条件为$Y,Z$线性相关,设为$Z=kY(kin R,k eq0),$此时 $$X^{T}Z=X^{T}(kY)=kX^{T}Y>0.$$ 从而$forall Zin R^{n},Z eq0,$有$Z^{T}MZ>0.$ 又易知$M$是实对称的,从而$M$是正定矩阵.