I need to understand the behavior of the following series as $n$ grows very large, where $A$ is an $n\times n$ matrix, $B$ is a $n\times 1$ vector, and $y_0$ is any vector in $\mathbb{R}^n$ $$ y_n = A^ny_0 + \sum_{k=0}^{n-1}A^kB $$ Doing a little bit of analysis on the matrix I have (which is a really complicated matrix thats a function of some small value $s$ which is why I left it out) I was able to see that the largest eigenvalue of $A$ is $1$, all other values are $0\leq \lambda < 1$
Unfortunately that means that the geometric matrix series formula $$ \sum_{k=0}^{n-1}A^kB = (I-A)^{-1}(I-A^n)B $$ won't work because $(I-A)$ is singular. Does anyone know how to get around this issue?
supposing that $\lambda_1 =1$ is simple, then you should be able to use a common technique from Markov Chains. In particular, with $W := \mathbf v_1\mathbf u_1^*$ where these are the right and left eigenvectors associated with $\lambda_1$ and $\text{trace}(W)=1$, then
$\Big(I+A+A^2+....+ A^{n-1}\Big)\Big(I-A+W\Big) = I -A^n +nW$
and invert $\Big(I-A+W\Big)$
this is exercise 16 on p. 468 of Grinstead and Snell's free intro to probability book
https://math.dartmouth.edu/~prob/prob/prob.pdf