Block-matrix inverse of $\left(\begin{array}{ccc} I_N & - A & - B\\ - A^{\top} & NI_M & 0\\ - B^{\top} & 0 & NI_K \end{array}\right)$ Consider a matrix $M$ written in block-form as follows:
$$M = \left(\begin{array}{ccc}
  I_N & - A & - B\\
  - A^{\top} & NI_M & 0\\
  - B^{\top} & 0 & NI_K
\end{array}\right)$$
where $A\in\mathbb{R}^{N\times M}$, $B\in\mathbb{R}^{N\times K}$ are rectangular matrices, and $I_N$ is the $N\times N$ identity matrix.
I want to find the inverse matrix $M^{-1}$. I suspect that I should be able to write down some expression for $M^{-1}$ exploiting the block structure of $M$.
Can we invert $M$?
 A: I will assume for the purpose of this answer that $N < M + K$, but even without this assumption the same approach can be taken with appropriate adjustments.
As the comment on your question notes, we can write $M$ in the form
$$
M = \pmatrix{I_N&-F\\ -F^T&NI_{M+K}},
$$
and from there it is indeed possible to use the Schur complement to obtain a formula for the inverse. However, the fact that the diagonal entries are both multiples of the identity matrix leads to a nice alternative approach using the SVD. Suppose that $F$ has singular value decomposition $F = U\Sigma V^T$ where $U$ is orthogonal and $N \times N$, $V$ is orthogonal and $(M+K)\times(M+K)$ and
$$
\Sigma = \pmatrix{\Sigma_0&0}
$$
where $\Sigma$ has the same shape as $F$ and $\Sigma_0$ is square and diagonal with non-negative entries. We can write $M$ in the form $M = WPW^T$, where
$$
W = \pmatrix{U & 0\\0 & V}, \quad P = \pmatrix{I_N & \Sigma \\ \Sigma^T & N I_{M + K}}.
$$
We can write
$$
P = 
\left( \begin{array}{cc|c}
I_N & \Sigma_0 & 0\\ 
\Sigma_0 & N \cdot I_N & 0\\
\hline 
0 & 0 & N \cdot I_{M + K - N}
\end{array}\right).
$$
It now suffices to take the inverse of each block. It is easy to verify that
$$
\pmatrix{I & \Sigma_0\\ \Sigma_0 & N\cdot I}\pmatrix{N \cdot I & -\Sigma_0\\ -\Sigma_0 & I} = \pmatrix{N\cdot I - \Sigma_0^2 & 0\\0 & N \cdot I - \Sigma_0^2} \implies\\
\pmatrix{I & \Sigma_0\\ \Sigma_0 & N\cdot I}^{-1} = 
\pmatrix{N \cdot I & -\Sigma_0\\ -\Sigma_0 & I} 
\pmatrix{N\cdot I - \Sigma_0^2 & 0\\0 & N \cdot I - \Sigma_0^2}^{-1} = \\
\pmatrix{N(N\cdot I - \Sigma_0^2)^{-1} & -(N\cdot I - \Sigma_0^2)^{-1}\Sigma_0\\
(N\cdot I - \Sigma_0^2)^{-1} \Sigma_0 & (N\cdot I - \Sigma_0^2)^{-1}}.
$$
From there, we get
$$
P^{-1} = \pmatrix{N(N\cdot I - \Sigma_0^2)^{-1} & -(N\cdot I - \Sigma_0^2)^{-1}\Sigma_0 & 0\\
(N\cdot I - \Sigma_0^2)^{-1} \Sigma_0 & (N\cdot I - \Sigma_0^2)^{-1} & 0\\
0 & 0 & \frac 1{N} I_{M + K - N}}.
$$
Finally,
$$
M^{-1} = [WPW^T]^{-1} = W P^{-1}W^T.
$$
