I'm having trouble proving that the trace of the following matrix is positive: $$A_+ :=A - \frac{As^{T}sA}{s^{T}As}$$ where $A \in \mathbb{R}^{n \times n}$ is a symmetric positive definite matrix and $s \in \mathbb{R}^n$, $s \neq 0$. This matrix is almost the BFGS update matrix.
Applying the properties of the trace of a matrix, I got to the following expression: $$\text{trace}(A_+) = \text{trace}(A) - \frac{||As||^2}{s^T A s}.$$
How should I conclude the statement that $\text{trace}(A_+) > 0$? I also tried proving that $A_+$ is a positive definite matrix but couldn't get anywhere.
Hint: Assume that $As \neq 0$ (if it were, then the denominator of $\frac{As^TsA}{s^TAs}$ would be zero). Let $u_1$ denote the unit vector in the direction of $As$, and extend this vector into an orthonormal basis $\{u_1,u_2,\dots,u_n\}$ of $\Bbb R^n$. Note that $$ \operatorname{Tr}(A_+) = \sum_{i=1}^n u_i^TA_+u_i \geq u_1 ^TA_+u_1. $$
Regarding the proof that $A_+$ is positive semidefinite: note that any $x \in \Bbb R^n$ can be written as $x = x_1 + x_2$ where $x_1$ is parallel to $u_1$ and $x_2$ is orthogonal to $u_1$. Write $$ x^TA_+x = (x_1 + x_2)^TA_+(x_1 + x_2) = x_1^TA_+x_1 + x_1^TA_+x_2 + x_2^TA_+x_1 + x_2^TA_+x_2. $$ Of course, if you prove this statement first, then you get the statement about the trace for free.