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!

  • $\begingroup$ For better legibility, I'd suggest that you replace $\theta^{1} = (\theta^{1}_1, \ldots, \theta^{1}_k)$ by e.g. $V = (v_1, \ldots, v_k)$ and $\theta^{2} = (\theta^{2}_1, \ldots, \theta^{2}_k)$ by $W = (w_1, \ldots, w_k)$ where the $v_i$ and $w_i$ are vectors of length $p_i$. Using upper indexes is not a good idea when powers are also involved! So you'd have $a =v_1^{\top} \Sigma_{11} w_1$ (I suppose not $w_1^{\top}$) etc. $\endgroup$
    – Wolfgang
    Apr 22, 2016 at 8:57
  • $\begingroup$ Thank you for your suggestion. I have changed the notation. $\endgroup$
    – Wuchen
    Apr 22, 2016 at 18:19

1 Answer 1


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 $p_1=\cdots=p_k=1$.

  • $\begingroup$ Can you explain more on that? I'm not familiar with the terms you used, especially the "big" equation you used. Thank you very much! $\endgroup$
    – Wuchen
    Apr 22, 2016 at 18:20
  • $\begingroup$ Don't worry about the meaning of $\otimes$. What I wrote is equivalent with your definition. My not-really-an-answer makes two points: 1). it suffices to assume the matrices $\Sigma_{ii}$ are diagonal; 2) even the simplest case $p_1=\cdots =p_k=1$ is nontrivial. $\endgroup$ Apr 23, 2016 at 15:42

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