# Limit of Sums of matrices

• Jan 26th 2009, 10:30 AM
Kyle_Katarn
Limit of Sums of matrices
Hello,

I have no idea how to solve this:

Quote:

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

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!
• Jan 26th 2009, 10:56 AM
Opalg
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 $\displaystyle \alpha\to0$.
• Jan 26th 2009, 11:20 AM
Kyle_Katarn

Expanding in $\displaystyle x_i$ means solving this term with repect to the $\displaystyle i$-th component of $\displaystyle x$.

Your post looks plausible to me, but are we allowed to do this? I forgot to mention the following hint in the description:

Quote:

Consider the limit $\displaystyle \alpha \rightarrow 0$. Setting $\displaystyle \alpha = 0$ will yield a different result!
• Jan 26th 2009, 11:54 AM
Opalg
I'm puzzled by that. I'd be interested to see the intended answer, because I don't see how it could be different from the one I suggested. Are you sure that the problem is correctly stated?
• Jan 26th 2009, 12:26 PM
Kyle_Katarn
The problem should be correct. I will write again when I have found something.