# (Asymmetric) matrix power series in closed form: $\sum_{i=0}^{\infty} A^i \left(A^i\right)^{\top}={?}$

Let $$A\in \mathbb{S}^{N\times N}$$ be a symmetric, real and stable matrix, i.e., $$\rho(A)<1$$, where $$\rho(A)$$ stands for the spectral radius of $$A$$. Then, $$\sum\limits_{i=0}^{\infty} A^{2i}=\left( I-A^2 \right)^{-1}.\tag{\star}\label{star}$$

Now, let $$A$$ be asymmetric. It is not clear whether the power series $$F(A):=\sum_{i=0}^{\infty} A^i (A^i)^{\top}$$ admits a compact closed-form expression as in equation \eqref{star} (when it converges). If the matrix $$A$$ is normal — i.e., it commutes with its transpose — and stable, then we have that $$F(A)=\left(I-AA^{\top}\right)^{-1}$$. In general, that is not the case.

Framework. Let $$A$$ be a random matrix generated as follows. Let $$\widetilde{A}\in \mathcal{G}(N,p)$$ be the adjacency matrix associated with the realization of a random graph over $$N$$ nodes with probability $$p$$ of arrow drawing, i.e., the entries $$\widetilde{A}_{ij}$$ of $$\widetilde{A}$$ are i.i.d. $$\mathsf{Bernoulli}(p)$$ for all $$i\neq j$$ — this is often referred to as a binomial random graph. Set $$\widetilde{A}_{ii}=0$$ for all $$i$$. Let $$A$$ be obtained from $$\widetilde{A}$$ by: i) [off-diagonal] normalizing the entries of $$\widetilde{A}$$ as $$A_{ij}:=\alpha_1 \widetilde{A}_{ij}/d_{\max}$$, for all $$i\neq j$$, where $$d_{\max}$$ is the maximum in-flow degree of the random graph and $$0<\alpha_1<1$$ is some constant; ii) [diagonal] setting the diagonal terms as $$A_{ii}:=\alpha-\sum_{k\neq i} \widetilde{A}_{ik}$$, where $$0<\alpha_1\leq\alpha<1$$. In other words, the rows of $$A$$ sum to $$\alpha$$ and its support is given by the realization of the binomial random graph on $$N$$ nodes.

Question. Is it true that $$\mathbb{P}\left(\lVert\sum_{i=0}^{\infty}A^i \left(A^i\right)^{\top} - \left(I-AA^{\top}\right)^{-1}\rVert_\text{max}>\varepsilon\right)\overset{N\rightarrow \infty}\longrightarrow 0?$$

Is there any known result in this direction? Any reference would be appreciated. The specifics of the random nature of $$A$$ are not important other than that of $$A$$ being stable and randomly generated (so that it is more natural to consider the limit $$N\rightarrow\infty$$).

Some numerical experiments are pointing in the direction of a concentration of $$\sum_{i=0}^{\infty} A^i \left(A^i\right)^{\top}$$ about $$\left(I-AA^{\top}\right)^{-1}$$ as the dimension $$N$$ grows large (I may attach them later, if necessary) under the referred framework.

Context. $$F(A)$$ is the covariance matrix associated with the time-series reflecting the state-evolution of certain linear stochastic dynamical systems. Having a closed form expression for the covariance in terms of $$A$$ is helpful for model identification purposes.

• I do not have much to say on the topic but I just want to thank you for such a well written (and interesting) question. This rarely happens when people post for the first time! Apr 21, 2022 at 13:56
• Thank you, @VladimirDotsenko. Apr 21, 2022 at 18:15

Q: Does $$F=\sum_{i=0}^{\infty} A^i (A^i)^{\top}$$ for non-symmetric $$A$$ have a closed-form solution analogous to the solution $$\sum_{i=0}^{\infty} A^{2i}=\left( I-A^2 \right)^{-1}$$ for a symmetric $$A$$?
A: Yes, in terms of the vectorization operation: $$\operatorname{vec}(F)=(I-{A} \otimes A)^{-1}\cdot\operatorname{vec}(I).\tag{\ast}\label{ast}$$ This does require the inversion of an $$N^2\times N^2$$ matrix, rather than the inversion of an $$N\times N$$ matrix as in the case of a symmetric $$A$$.
The vectorization formula follows because $$F$$ solves the Lyapunov equation $$AFA^\top=F-I.$$
I do not have an answer to the second question, whether \eqref{ast} converges for large $$N$$ to $$(I-AA^\top)^{-1}$$.