# Infinite sum of product of vector elements

1. Jan 16, 2010

### diazona

I've been trying to simplify a formula that looks something like this:
$$X = \sum_{k=0}^{\infty} v_{x[k]} v_{y[k]}$$
where $v_{x[k]}$ and $v_{y[k]}$ are the horizontal and vertical components of a velocity vector. Both components are positive for all k, and
$$v_{x[k]} < v_{x[k-1]}$$
$$v_{y[k]} < v_{y[k-1]}$$
for all $k \ge 1$. I also know (or, I'm assuming) that there is a linear recurrence relation that expresses $v_{x[k]}$ and $v_{y[k]}$ in terms of $v_{x[k-1]}$ and $v_{y[k-1]}$:
$$\begin{pmatrix}v_{x[k]} \\ v_{y[k]}\end{pmatrix} = \begin{pmatrix}C_{xx} & C_{xy} \\ C_{yx} & C_{yy}\end{pmatrix}\begin{pmatrix}v_{x[k-1]} \\ v_{y[k-1]}\end{pmatrix}$$
Each of the elements of the matrix C satisfies $0 \le C_{ij} \le 1$. (More specifically: $0 < C_{xx} \le 1$, $0 < C_{yy} < 1$, I have reason to suspect that $C_{xx},C_{yy} > C_{xy},C_{yx}$, and it's quite possible that $C_{yx} = 0$.)

I managed to figure out without any trouble that if the matrix C is diagonal, the series sums to
$$X = \frac{v_{x[0]}v_{y[0]}}{1 - C_{xx}C_{yy}}$$
But when C has off-diagonal terms, I haven't been able to do anything meaningful with the sum. So far I've been playing with a structure like
$$X = \sum_{k=0}^{\infty} \vec{v}_{x[0]}^T(C^T)^k \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}C^k \vec{v}_{y[0]}$$
What I'd like, ideally, is to find some way to express the sum X in closed form when the matrix is not diagonal. Even if that's not possible (as I'm suspecting), anything I can say about the sum could potentially be useful. I wanted to see if someone who knows linear algebra better than I do could suggest anything.

(this is not homework)