# Solution of coupled non-linear matrix difference equations encountered in calculating the determinat of a block partitioned matrix.

The determinant of the following $$n N \times n N$$ block partitioned complex symmetric matrix ($$N \times N$$ blocks) $$\begin{bmatrix}\mathbb{A} & \mathbb{B} &\cdots &\mathbb{B} \\ \mathbb{B}_{}^{T} & \ddots &\ddots &\vdots \\ \vdots & \ddots & \ddots & \mathbb{B}\\ \mathbb{B}_{}^{T}& \cdots & \mathbb{B}_{}^{T}& \mathbb{A}\end{bmatrix}$$ is given as $$\prod_{i=1}^{N}\det\left[\mathbb{A}_{i}^{}\right].$$ Here $$\mathbb{A}_{}^{}$$, $$\mathbb{B}_{}^{}$$ are $$n \times n$$ complex matrices with $$\mathbb{A}_{}^{T}=\mathbb{A}_{}^{}$$.

$$\mathbb{A}_{i}^{}$$ are $$n \times n$$ matrices defined by the following recurrence equations (difference equations) $$\mathbb{A}_{k}^{}=\mathbb{A}_{k-1}^{}-\mathbb{B}_{k-1}^{T}\mathbb{A}_{k-1}^{-1}\mathbb{B}_{k-1}^{}$$ $$\mathbb{B}_{k}^{}=\mathbb{B}_{k-1}^{}-\mathbb{B}_{k-1}^{T}\mathbb{A}_{k-1}^{-1}\mathbb{B}_{k-1}^{}$$ with the initial conditions $$\mathbb{A}_{1}^{}=\mathbb{A}_{}^{}$$ $$\mathbb{B}_{1}^{}=\mathbb{B}_{}^{}$$ Is there a method to find the solutions of the above recurrence equations? Is there a closed form expression for the determinant of the above block partitioned matrix?

$$\textbf{Progress achieved:}$$ After i have posted this question, i could make some progress as follows:

(i) If $$\mathbb{A}_{}^{T}=\mathbb{A}_{}^{}$$, then $$\mathbb{A}_{k}^{T}=\mathbb{A}_{k}^{}$$.

(ii) $$\mathbb{B}_{k}^{T}-\mathbb{B}_{k}^{}=\mathbb{B}_{}^{T}-\mathbb{B}_{}^{}$$.

(iii) $$\mathbb{A}_{k}^{}-\mathbb{B}_{k}^{}=\mathbb{A}_{}^{}-\mathbb{B}_{}^{}$$.

(iv) Using (iii), the above two recurrence equations can be decoupled and one only needs to solve for the following $$\mathbb{A}_{k}^{}=\left[\mathbb{S}_{}^{T}+\mathbb{S}_{}^{}\right]-\mathbb{S}_{}^{T}\mathbb{A}_{k-1}^{-1}\mathbb{S}_{}^{}$$ where $$\mathbb{S}_{}^{}=\left(\mathbb{A}_{}^{}-\mathbb{B}_{}^{}\right)$$ and the initial condition is $$\mathbb{A}_{1}^{}=\mathbb{A}_{}^{}$$.

This recurrence equation seems to be related to discrete time algebraic riccatti equation (someone please confirm this). If this is true, can analytical solution (not a simple iterative solution in terms of repeated fractions but some closed form solution) be achieved for the above recurrence equation?

Here's some semi-justified black magic. Perhaps it helps arrive at an answer

$$A_{k}^{}=\left[S^{T} + S \right] - S^{T} A_{k-1}^{-1} S$$

$$\mathbb{1} = A_{k}^{-1} \left[S^{T} + S \right] - A_{k}^{-1} S^{T} A_{k-1}^{-1} S$$

Let $$Z_k = A_k^{-1}S^T$$ and $$C = (S^T)^{-1} S$$, then

$$\mathbb{1} = Z_k \left[\mathbb{1} + C \right] - Z_k Z_{k-1} C$$ $$\mathbb{1} = Z_k \left[\mathbb{1} - (Z_{k-1} - \mathbb{1})C \right]$$

Subtract the bracket from both sides

$$(Z_{k-1} - \mathbb{1})C = (Z_k - \mathbb{1}) \left[\mathbb{1} - (Z_{k-1} - \mathbb{1})C \right]$$

Let $$Y_k = Z_k - \mathbb{1}$$, then

$$Y_{k-1}C = Y_k(\mathbb{1} - Y_{k-1}C)$$

$$Y_{k-1} = Y_k(C^{-1} - Y_{k-1})$$

$$Y_{k}^{-1} = (C^{-1}Y_{k-1}^{-1} - \mathbb{1})$$

Finally with $$W_k = Y_k^{-1}$$ and $$B = C^{-1}$$ we get

$$W_k =B W_{k-1} - \mathbb{1} = B^k W_0 - \sum_{i=0}^{k-1} B^i$$

Substituting back the expressions for $$W_k$$ and $$B$$ via $$A_k$$ and $$S$$ we get an analytical solution for $$A_k$$.

Of course, it all depends on the assumption that a bunch of matrices are invertible, but relaxing these assumptions would produce a lot of cases to consider. Hopefully this is sufficient to get you going

• Thanks for the answer. I liked the transformations you did to arrive at linear equation. I have to check your answer carefully though. I had solved these recurrece equations by converting to riccatti difference equation which can be solved using axillary linear system. Commented Aug 21, 2019 at 10:09
• Sorry, I have so far only heard of Ricatti, not used it before. But if you have already solved it, it's great. I'd be happy to see your solution to learn too Commented Aug 21, 2019 at 10:11
• Will try to post the solution soon. Commented Aug 21, 2019 at 10:12