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:
For all $i$, $\Sigma_{ii} \in \mathbb{S}^{p_i \times p_i}$ and it is positive definite .
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!
