Based off of http://en.wikipedia.org/wiki/Determinant#Block_matrices, I'm trying to find the formula for $\det(M)$ when $M = \left(\begin{matrix} A & B \\ B^T & C \end{matrix}\right)$. It is known that $A$ and $C$ are diagonal matrices in my use case.
I don't even know where to begin proving this. Any proofs or hints in the right direction would be appreciated.
If $A$ is invertible, then the factorization $$M:=\left(\begin{array}{cc} A & B \\ B^T & C\end{array}\right)= \left(\begin{array}{cc} A & 0 \\ B^T & 1\end{array}\right) \left(\begin{array}{cc} 1 & A^{-1}B \\ 0 & C-B^TA^{-1}B\end{array}\right)$$ implies that $$\operatorname{det}M=\operatorname{det}A\cdot \operatorname{det}\left(C-B^TA^{-1}B\right).$$ Similarly, if $C$ is invertible, then $$\operatorname{det}M=\operatorname{det}C\cdot \operatorname{det}\left(A-BC^{-1}B^T\right).$$ If both $A$ and $C$ are invertible, one can write the result in a symmetric form: \begin{align*}\operatorname{det}M&=\operatorname{det}A\cdot\operatorname{det}C\cdot \operatorname{det}\left(1-A^{-1}BC^{-1}B^T\right)=\\ &=\operatorname{det}A\cdot\operatorname{det}C\cdot \operatorname{det}\left(1-C^{-1}B^TA^{-1}B\right) \end{align*} The first formula is preferable when the size of $A$ is much smaller than the size of $C$ and vice versa. The diagonal form of $A,C$ does not simplify the result and its derivation.