Let matrix $Y \in \mathbb{S}^{n}$ be symmetric and positive definite (thus invertible) and $X \in \mathbb{R}^{n \times n}$ such that $Y - X X^T \succ 0$. I think that the following identity should hold and I am trying to prove it (it works for $n=1$):
\begin{equation} \bigg(X^T \left( Y - X X^T \right)^{-1}X + I \bigg)^{-1}= I - X^T Y^{-1}X. \end{equation}
Do you have any idea or properties that could help me to prove it?
I observe that
$$I - X^T Y^{-1}X = \begin{pmatrix} Y^{1/2}&-X^T \end{pmatrix} \begin{pmatrix} Y^{-1} & 0 \\ 0 & Y^{-1}\end{pmatrix} \begin{pmatrix}Y^{1/2}\\ -X\end{pmatrix}$$
but I am not sure it can help since I do not know how to invert such a "product of block matrices".
Edit : My observation is wrong, the right-handside yields $I + X^T Y^{-1}X$.
Edit: My problem is a particular case of the "Woodbury matrix identity".