Let $X$ be a separable Hilbert space, and let $X_n$ be the subspace spanned by orthonormal vectors $(e_i)_{i=1}^n$. Consider the operator $I_n:\mathbb{R}^n\to X$ be such that $I_n u=\sum_{i=1}^n u_i e_i$ for all $u=(u_i)_{i=1}^n$. The adjoint $I_n^*:X\to \mathbb{R}^n$ is then given by $I^*_n x= (\langle x, e_i\rangle_X)_i$.
Let $M$ be a symmetric positive semidefinite operator on $X$, and $\lambda>0$. Is it possible to represent $$ \tag{1} (I_n^* MI_n+\lambda \operatorname{id}_{\mathbb{R}^n})^{-1} $$ or $$ \tag{2} I_n(I_n^* MI_n+\lambda \operatorname{id}_{\mathbb{R}^n})^{-1}I_n^* $$ with $( M +\lambda \operatorname{id}_{X})^{-1} $ together with some projection operators?
The matrix $I_n^* MI_n$ in equations (1) and (2) has a matrix representation $(\langle Me_i,e_j\rangle_X)_{ij}$. By using $I_n^*\operatorname{id}_{X} I_n=\operatorname{id}_{\mathbb{R}^n}$, we can rewrite (1) as $(I_n^* (M+\lambda\operatorname{id}_{X})I_n)^{-1}$. But it is unclear to me how to handle the inverse operator, essentially when $M$ does not preserve $X_n$. The question is motivated by the weak convergence of (2), as described here. This question asks a similar question but assumes the operator preserves the subspace.
