TITLE: Inverse of a small submatrix
QUESTION [11 upvotes]: Let $A$ be a large matrix (say, $1000 \times 1000$), and let $\mathcal I = \{2,3,5\}$ be a set of row/column indices. Let $(A^{-1})_{\cal I \times I}$ denote the submatrix of $A^{-1}$ that consists of the $\{2,3,5\}$ rows and columns of $A^{-1}$.
Is there an efficient way of computing the following $3 \times 3$ matrix inverse $((A^{-1})_{\cal I \times \cal I})^{-1}$ without inverting the large matrix $A$?

REPLY [11 votes]: One way to go about this is as follows:
For $i,j \in \mathcal{I}$ Compute $e_i^TA^{-1}e_j$ by using the approach based on Gaussian quadrature; see for instance, a precise algorithm and analysis in our paper "Gauss quadrature for matrix inverse forms with applications." 
Now that you have $[A^{-1}]_{\mathcal{I},\mathcal{I}}$, getting its inverse is an easy matter since $|\mathcal{I}|$ is small.