Considering a symmetric positive matrix $A$:
$M=\begin{bmatrix} A & B \\ B' & C \end{bmatrix} $,
is there any mathematical relation between the inverses/determinant of the blocks $A$ and $C$
($M$ being similar to a covariance matrix)?
Particularly I am trying to relate the inverse of $M$ to the smaller and faster-to-compute inverses of $A$ and $C$.
Let $$L = \begin{bmatrix} I_{n} & 0 \\ -C^{-1}B & I_m \end{bmatrix}. $$ Then, we have $$ ML = \begin{bmatrix} A & B \\ B' & C \end{bmatrix} \begin{bmatrix} I_{n} & 0 \\ -C^{-1}B & I_m \end{bmatrix} = \begin{bmatrix} A-BC^{-1}B' & B \\ 0 & C \end{bmatrix} $$ Hence, by taking the inverse in both sides, we shall have
$$ L^{-1}M^{-1} = \begin{bmatrix} (A-BC^{-1}B')^{-1} & (A-BC^{-1}B')^{-1}BC^{-1} \\ 0 & C^{-1} \end{bmatrix} $$ and thus the inverse of M can be computed using the following formula: $$ M^{-1} = L\begin{bmatrix} (A-BC^{-1}B')^{-1} & (A-BC^{-1}B')^{-1}BC^{-1} \\ 0 & C^{-1} \end{bmatrix} $$
You can procedure by doing the following steps:
1) Compute $C^{-1}$.
2) Compute $(A-BC^{-1}B')^{-1}$ (Which is a problem of the same size of computing $~A^{-1}$).
3) Compute $L$ and use the result to compute $M^{-1}$.
I hope that this answer your question.