A generic $k \times k$ block symmetric matrix $\Sigma$ is denoted as
\begin{align}
\Sigma = \begin{bmatrix}\Sigma\_{11} & \Sigma\_{12} & \ldots & \Sigma\_{1k} \\ \Sigma\_{21} & \Sigma\_{22} & \ldots & \Sigma\_{2k} \\ \ldots & \ldots & \ldots & \ldots \\ \Sigma\_{k1} & \Sigma\_{k2} & \ldots & \Sigma\_{kk}\end{bmatrix}.
\end{align}
It has the following structures:


1. For all $i$, $\Sigma\_{ii} \in \mathbb{S}^{p\_i \times p\_i}$ and it is positive definite .
2. For $i \neq j$, $\Sigma\_{ij} \in \mathbb{R}^{p\_i \times p\_j} $ and $\Sigma\_{ij} = \lambda \Sigma\_{ii} \theta\_{i} \theta\_{j}^\top \Sigma\_{jj}$, where $\lambda \in \mathbb{R}$, $\theta\_{i} \in \mathbb{R}^{p\_i}$ and $\theta\_{i}^\top \Sigma\_{ii} \theta\_{i} = 1$ for all $i$.


Fix $\Sigma\_{ii}$ and $\lambda$, we can see that $\Sigma$ is fully characterized by $\theta = (\theta\_1, \ldots, \theta\_k)$, thus we can use $\Sigma\_{\theta}$ to denote the block symmetric matrix characterized by $\theta$. 


We consider the following three matrices. Let $\Sigma\_{u}$ denote the matrix characterized by $u = (u\_1, \ldots, u\_k)$ and $\Sigma\_{v}$ denote the matrix characterized by $v = (v\_1, \ldots, v\_k)$ and let $\Sigma\_{0}$ be the block diagonal matrix of $\{\Sigma\_{ii}\}$. We want to compute the following determinant


\begin{align}
\text{det}(\Sigma\_{u}^{-1} + \Sigma\_{v}^{-1} - \Sigma\_{0}^{-1}).
\end{align}


I have calculated this determinant for $k = 2$, where the answer is 


\begin{align}
\left(\frac{1 - \lambda^2 a b}{1 - \lambda^2}\right)^2 \frac{1}{\text{det}(\Sigma\_{11})\text{det}(\Sigma\_{22})},
\end{align}


with $a = u\_1^\top \Sigma\_{11} v\_1$ and $b = u\_2^\top \Sigma\_{22} v\_2$.


How could I compute the determinant for the general $k$?


Thank you!



Recursively calculate the determinant
Perhaps you should write this more geometrically. Denote by $(-,-)\_i$ the canonical Euclidean inner product on $\newcommand{\bR}{\mathbb{R}}$ $\bR^{p\_i}$. Then we can identify $\Sigma\_{ii}$ with a symmetric positive operator. $\Sigma\_{ij}: \bR^{p\_j}\to\bR^{p\_i}$ is the operator that has the invariant description


$$\Sigma\_{ij}=\lambda \Sigma\_{ii}\circ \theta\_i\otimes \theta\_j^T\circ \Sigma\_{jj}. $$


Now fix $(-,-)\_i$ orthonormal bases in $\bR^{p\_i}$ such that all the operators $\Sigma\_{ii}$ are diagonal. Perform the computations in this case and express the result in invariant terms.


Try to understand what happens in the "simplest" case $p\_1=\cdots=p\_k=1$.


