I have a system of equations where the relationship between input and output is derived from a pixel lattice:
\begin{equation} x_i(k+1) = \sum_j \alpha e^{ \frac{\left(dr_{ji}^2 + dc_{ji}^2 \right)}{2 \sigma^2}} \, x_j(k) \end{equation} where dr_{ji} and dc_{ji} stand for the distance between row and column of pixel $i$ and $j$
In matrix form, this results in \begin{equation} x(k+1) = \alpha \, A x(k), \end{equation} where, for example, for a 3x3 lattice, A is:
\begin{equation} A = \begin{array}{ccccccccc} e^\frac{0}{2\sigma^2} & e^\frac{1}{2\sigma^2} & e^\frac{4}{2\sigma^2} & e^\frac{1}{2\sigma^2} & e^\frac{2}{2\sigma^2} & e^\frac{5}{2\sigma^2} & e^\frac{4}{2\sigma^2} & e^\frac{5}{2\sigma^2} & e^\frac{8}{2\sigma^2}\\ e^\frac{1}{2\sigma^2} & e^\frac{0}{2\sigma^2} & e^\frac{1}{2\sigma^2} & e^\frac{2}{2\sigma^2} & e^\frac{1}{2\sigma^2} & e^\frac{2}{2\sigma^2} & e^\frac{5}{2\sigma^2} & e^\frac{4}{2\sigma^2} & e^\frac{5}{2\sigma^2}\\ e^\frac{4}{2\sigma^2} & e^\frac{1}{2\sigma^2} & e^\frac{0}{2\sigma^2} & e^\frac{5}{2\sigma^2} & e^\frac{2}{2\sigma^2} & e^\frac{1}{2\sigma^2} & e^\frac{8}{2\sigma^2} & e^\frac{5}{2\sigma^2} & e^\frac{4}{2\sigma^2}\\ e^\frac{1}{2\sigma^2} & e^\frac{2}{2\sigma^2} & e^\frac{5}{2\sigma^2} & e^\frac{0}{2\sigma^2} & e^\frac{1}{2\sigma^2} & e^\frac{4}{2\sigma^2} & e^\frac{1}{2\sigma^2} & e^\frac{2}{2\sigma^2} & e^\frac{5}{2\sigma^2}\\ e^\frac{2}{2\sigma^2} & e^\frac{1}{2\sigma^2} & e^\frac{2}{2\sigma^2} & e^\frac{1}{2\sigma^2} & e^\frac{0}{2\sigma^2} & e^\frac{1}{2\sigma^2} & e^\frac{2}{2\sigma^2} & e^\frac{1}{2\sigma^2} & e^\frac{2}{2\sigma^2}\\ e^\frac{5}{2\sigma^2} & e^\frac{2}{2\sigma^2} & e^\frac{1}{2\sigma^2} & e^\frac{4}{2\sigma^2} & e^\frac{1}{2\sigma^2} & e^\frac{0}{2\sigma^2} & e^\frac{5}{2\sigma^2} & e^\frac{2}{2\sigma^2} & e^\frac{1}{2\sigma^2}\\ e^\frac{4}{2\sigma^2} & e^\frac{5}{2\sigma^2} & e^\frac{8}{2\sigma^2} & e^\frac{1}{2\sigma^2} & e^\frac{2}{2\sigma^2} & e^\frac{5}{2\sigma^2} & e^\frac{0}{2\sigma^2} & e^\frac{1}{2\sigma^2} & e^\frac{4}{2\sigma^2}\\ e^\frac{5}{2\sigma^2} & e^\frac{4}{2\sigma^2} & e^\frac{5}{2\sigma^2} & e^\frac{2}{2\sigma^2} & e^\frac{1}{2\sigma^2} & e^\frac{2}{2\sigma^2} & e^\frac{1}{2\sigma^2} & e^\frac{0}{2\sigma^2} & e^\frac{1}{2\sigma^2}\\ e^\frac{8}{2\sigma^2} & e^\frac{5}{2\sigma^2} & e^\frac{4}{2\sigma^2} & e^\frac{5}{2\sigma^2} & e^\frac{2}{2\sigma^2} & e^\frac{1}{2\sigma^2} & e^\frac{4}{2\sigma^2} & e^\frac{1}{2\sigma^2} & e^\frac{0}{2\sigma^2} \end{array} \end{equation}
I need to find $\alpha$ and $\sigma$ for known x(0) and x(K) vectors that solve the linear system in K steps. This is, perform this linear regression:
\begin{equation} x(k) = \alpha \, A^K x(0), \end{equation}
I thought about applying some natural logarithms, but it is not straightforward. Any ideas?
Best