# Thread: Limit of Sums of matrices

1. ## Limit of Sums of matrices

Hello,

I have no idea how to solve this:

$\displaystyle x = (1-\alpha) \sum_{r=0}^\infty \alpha^r W^r Y$

where $\displaystyle x \in \mathbb{R}^n, 0 < \alpha < 1, W \in \mathbb{R}^{n x n}$ symmetric ,$\displaystyle Y \in \mathbb{R}^n$.

Derive x in the limit $\displaystyle \alpha \rightarrow 0$.

We are just interested in the sign of $\displaystyle x_i$. It is helpful to expand the sum in $\displaystyle x_i$. What happens as $\displaystyle \alpha \rightarrow 0$? Which terms in the sum are non-zero? Which terms dominate as $\displaystyle \alpha \rightarrow 0$?
Thank You!

2. I am not sure what is intended by the suggestion "to expand the sum in $\displaystyle x_i$."

The series $\displaystyle \sum_{r=0}^\infty \alpha^r W^r$ converges in the operator norm to $\displaystyle (I-\alpha W)^{-1}$ whenever $\displaystyle 0<\alpha<\|W\|^{-1}$. Therefore $\displaystyle x = (1-\alpha)(I-\alpha W)^{-1}Y\to Y$ as $\displaystyle \alpha\to0$.

Expanding in $\displaystyle x_i$ means solving this term with repect to the $\displaystyle i$-th component of $\displaystyle x$.
Consider the limit $\displaystyle \alpha \rightarrow 0$. Setting $\displaystyle \alpha = 0$ will yield a different result!