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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 $C$.

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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.

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    What exactly are the dimension m and n? I'm assuming that they are correspondent to the sizes of A and C. In my case, A and C are equal-sized. – 2017-01-16
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    They are correspondent to the sizes of A and C. – 2017-01-16
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    Does it answer your question, @joaocandre ? – 2017-01-16
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    While it supplies extremely relevant information, I was looking for a way to scale A^-1 and C^-1 without requiring B (which is probably not possible). I did not specify that in the question though, so I'll mark it as answered. – 2017-01-16
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    Wikipedia also has the final result: https://en.wikipedia.org/wiki/Invertible_matrix#Blockwise_inversion – 2017-01-16