This is the geometric series sum we all know and love: $$ \sum^\infty_{n=0} a^n=\frac{1}{1-a} $$ By moving things around a little bit we get: $$ \sum^\infty_{n=0} (1-a)^n=\frac{1}{a} $$ This works fine when a is between 0 and 2, by adding an extra factor called k I can include a bigger range $$ k\sum^\infty_{n=0} (1-ak)^n=\frac{1}{a} $$ If k is set to 0.1 I can let a be between 0 and 20. This works perfectly fine for normal numbers, like scalars.
When I plug in matrices A instead of a scalar a and use the identity matrix instead of 1 I can't seem to get anything remotely close to the matrix inverse. I've tried to set k to very small values but it always converges to something that is similar to n times the identity matrix, even though the matrix inverse to a is something totally different.
In other words, I keep getting something incorrect along these lines: $$ k\sum^\infty_{n=0} (I-Ak)^n \color{red}{\rightarrow \infty I} $$ Why won't I get something that resembles the inverse?
The reason for why I'm trying to calculate the matrix inverse in this way is mostly just for fun and to learn and see when math "breaks" or becomes very troublesome/problematic as the case is here.
Both Jose Pérez Cano and a math teacher I contacted said that it should work assuming that the operator norm of $(I-Ak)$ is below 1.
Given a 2x2 matrix it seemed to converge correctly every time the operator norm was below 1, but for 3x3 matrices and above it failed to converge every time unless the matrix had mostly diagonal elements. So perhaps the k variable has to be a matrix for a more general case to converge to the inverse.
