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 '16 at 8:57
  • $\begingroup$ Thank you for your suggestion. I have changed the notation. $\endgroup$ – Wuchen Apr 22 '16 at 18:19

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 $p_1=\cdots=p_k=1$.

| cite | improve this answer | |
  • $\begingroup$ 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 '16 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. $\endgroup$ – Liviu Nicolaescu Apr 23 '16 at 15:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.