Suppose that $H^+ = H - (\mathbf y^TH \mathbf y)^{-1} H\mathbf y \mathbf y^T H + (\mathbf y ^T \mathbf s )^{-1}\mathbf s \mathbf s^T $
where H is symmetric and positive definite.
Supposing that $\mathbf y ^T \mathbf s > 0$, show that $H^+ $ is also symmetric positive definite.
$H^+ $ just stands for the new H in this equation
I was given the following hint: Split $\mathbf z=a \mathbf s + \mathbf w$ where $\mathbf w^T \mathbf s = 0$ and show that $\mathbf z^T H^+ \mathbf z > 0$ for all $\mathbf z \neq 0$. Also, $(\mathbf y + b\mathbf x)^T H (\mathbf y + b \mathbf x) \geq 0$ for all b since H is positive definite; use this to show that $(\mathbf y^TH \mathbf y)(\mathbf x^T H \mathbf x)\geq (\mathbf y^TH \mathbf x)^2$ for any $\mathbf x$
I'm not exactly sure how to apply the hint- any suggestions would be appreciated!
Inner-product notation is good here: $x^{T}Hx=(Hx,y)$. Using this notation, $$ (H^{+}x,x)=(Hx,x)-\frac{1}{(Hy,y)}(Hx,y)^{2}+\frac{(x,s)^{2}}{(s,y)} $$ Using the hint, $(Hx,y)^{2}\le (Hx,x)(Hy,y)$ gives $$ (H^{+}x,x) \ge (Hx,x)-(Hx,x)+\frac{(x,s)^{2}}{(s,y)} = \frac{(x,s)^{2}}{(s,y)} \ge 0. $$