Let $y,s \in \mathbb{R}^n$ be two vectors such that $y^\top s = 1$. To prove $$X = I + yy^\top - \frac{ss^\top}{s^\top s} $$ is a positive definite matrix, I found a source which shows: $$X = \left( I + \frac{ys^\top}{\sqrt{s^\top s}} - \frac{ss^\top}{s^\top s} \right) \left( I + \frac{ys^\top}{\sqrt{s^\top s}} - \frac{ss^\top}{s^\top s} \right) ^\top$$ hence $X \succ0$. How can I obtain this decomposition? I can, of course, reverse engineer the result, but I want to start from the first definition and decompose myself.
Edit: Ugly and meaningless reverse engineering: $$\begin{align*} X^* &= I - \frac{1}{s^\top s} ss^\top + yy^\top \\ & = I + \frac{s^\top y}{\sqrt{s^\top s}} - \frac{1}{s^\top s} ss^\top + yy^\top -\frac{s^\top y}{\sqrt{s^\top s}} \\ & = \left( I + \frac{s^\top y}{\sqrt{s^\top s}} - \frac{1}{s^\top s} ss^\top \right) + \left( \frac{ys^\top}{\sqrt{s^\top s}} + yy^\top - \frac{ys^\top}{\sqrt{s^\top s}} \right) - \left( \frac{ss^\top }{s^\top s} + \frac{sy^\top}{\sqrt{s^\top s}} - \frac{ss^\top}{s^\top s} \right) \\ & = \left( I + \frac{s^\top y}{\sqrt{s^\top s}} - \frac{ss^\top}{s^\top s} \right) + \left( \frac{ys^\top}{\sqrt{s^\top s}} + \frac{ys^\top s y^\top}{s^\top s} - \frac{ys^\top s s^\top}{s^\top s \sqrt{s^\top s}} \right) - \left( \frac{ss^\top }{s^\top s} + \frac{s s^\top s y^\top}{s^\top s \sqrt{s^\top s}} - \frac{s s^\top s s^\top}{s^\top s s^\top s} \right) \\ & = \left( I + \frac{ys^\top}{\sqrt{s^\top s}} - \frac{ss^\top}{s^\top s} \right) \left( I + \frac{sy^\top}{\sqrt{s^\top s}} - \frac{ss^\top}{s^\top s} \right) \\ & = \left( I + \frac{ys^\top}{\sqrt{s^\top s}} - \frac{ss^\top}{s^\top s} \right) \left( I + \frac{ys^\top}{\sqrt{s^\top s}} - \frac{ss^\top}{s^\top s} \right)^\top. \end{align*} $$
if you really want the decomposition, then I suppose this isn't the answer you want, but there is a simpler and very general approach.
Consider
$P:= I - \frac{\mathbf{ss}^T}{\mathbf s^T \mathbf s} \succeq 0$
you can in fact check that this is real symmetric and $P^2 = P$ so it is a projector and rank is given by its trace =$n-1$. (You can also do this with diagonalization of a symmetric rank one matrix and then shift all eigenvalues by one). So for any $\big\Vert \mathbf v\big \Vert_2 = 1$, we have
$\mathbf v^T P\mathbf v \geq 0$ with equality iff $\mathbf v \propto \mathbf s$
now consider $X = P + \mathbf {yy}^T$
for any unit length $\mathbf v\in \mathbb R^n$,
$\mathbf v^T X \mathbf v = \mathbf v^T P \mathbf v + \mathbf v^T \mathbf{yy}^T \mathbf v \gt 0$
and so $X \succ 0$
i.e. if $\mathbf v \propto \mathbf s$ then this reads as $0+1\gt 0$
and otherwise this reads $\alpha + \gamma\gt 0$ where $\alpha \gt 0$ and $\gamma \geq 0$
underlying idea:
the addition of a positive semidefinite matrix and a positive definite matrix gives a positive definite matrix. This idea turns out to be useful in many different areas and is rather simple unlike some decompositions.