设 ${\bf A}$ 为 $n$ 阶正定矩阵, ${\bf x}$, ${\bf y}$ 为 $n$ 维列向量且满足 ${\bf x}^t{\bf y}>0$. 证明矩阵 $$\bex {\bf M}={\bf A}+\cfrac{{\bf x}{\bf x}^t}{{\bf x}^t{\bf y}} -\cfrac{{\bf A}{\bf y}{\bf y}^t{\bf A}}{{\bf y}^t{\bf A}{\bf y}} \eex$$ 正定.
证明: （来自 chxp1234）易知$Y\neq0,$从而$Y^{T}AY>0.$$\forall Z\in R^{n},Z\neq0,$有 $$\begin{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}MZ\geq0.$$ 下证$Z^{T}MZ>0.$否则易知$Z^{T}MZ=0$的充要条件为 $$(Z,Z)(Y,Y)-(Z,Y)^{2}=0\mbox{且}X^{T}Z=0.$$ 而$(Z,Z)(Y,Y)-(Z,Y)^{2}=0$的充要条件为$Y,Z$线性相关,设为$Z=kY(k\in R,k\neq0),$此时 $$X^{T}Z=X^{T}(kY)=kX^{T}Y>0.$$ 从而$\forall Z\in R^{n},Z\neq0,$有$Z^{T}MZ>0.$ 又易知$M$是实对称的,从而$M$是正定矩阵.