# System of Difference Equations, Decoupling

I have the nonhomogenous systen of difference equations with $$0<\beta<1$$: $$\underbrace{\begin{bmatrix}d_{t+1}\\e_{t+1}\end{bmatrix}}_\text{X_{t+1}} =\underbrace{\begin{bmatrix}1&0\\-1&\frac{1}{\beta}\end{bmatrix}}_\text{\Phi}\underbrace{\begin{bmatrix}d_t\\e_t\end{bmatrix}}_\text{X_t}+\underbrace{\begin{bmatrix}0\\1\end{bmatrix}}_\text{=A}s_t$$ I want to decouple the system into: $$Y_{t+1}=\Lambda Y_t+B s_t$$ Where $$Q$$ are the eigenvectors of $$\Phi$$, $$Y_t=Q^{-1}X_t$$ and $$B=Q^{-1}A$$ and the eigenvalues:$$\Lambda=\begin{bmatrix}\lambda_1&0\\0&\lambda_2\end{bmatrix}$$ Then I want to solve the decoupled system with the larger eigenvalue.

EDIT:
I think I figured out a step: $$Y_{1,t+1}=\lambda_1 Y_{1,t}$$ $$Y_{2,t+1}=\lambda_2 Y_{2,t} + z_t$$ With $$\lambda_2=\frac{1}{\beta}$$ being larger I use: $$Y_{2,t+1}= \frac{1}{\beta}Y_{2,t} + z_t$$ But then how do I solve this, if it's still depending on two periods?

What I have thus far: $$\lambda_1=1,\lambda_2=\frac{1}{\beta}$$ $$Q=\begin{bmatrix}\frac{1-\beta}{\beta}&0\\1&1\end{bmatrix}$$ $$Q^{-1}=\begin{bmatrix}\frac{\beta}{1-\beta}&0 \\ -\frac{\beta}{1-\beta}&1\end{bmatrix}$$ And when I combine everything I get: $$Q^{-1}X_{t+1}=\begin{bmatrix}\frac{\beta}{1-\beta}&0 \\ -\frac{\beta}{1-\beta}&\frac{1}{\beta}\end{bmatrix}X_t+\begin{bmatrix}0 \\ 1\end{bmatrix}s_t$$

• Is this system already decoupled?
• How can I solve this with the larger eigenvalue (which is $$\lambda_2=\frac{1}{\beta}$$ since $$\beta<1$$)?
Both eigenvalues are still contained in the equation, so I just don't get how you could solve it without both of them.

I'm very gratefull for any help or advice :)

With

$$Q = \left( \begin{array}{cc} 0 & -\frac{\beta -1}{\beta } \\ 1 & 1 \\ \end{array} \right),\ \ \ \Lambda = \left( \begin{array}{cc} \frac{1}{\beta } & 0 \\ 0 & 1 \\ \end{array} \right),\ \ \ \delta_k = \left( \begin{array}{c} 0\\ 1 \end{array} \right)s_k$$

we have

$$y_{k+1} = Q\cdot\Lambda\cdot Q^{-1}y_k + \delta_k$$

or

$$Q^{-1}y_{k+1} = \Lambda\cdot Q^{-1}y_k + Q^{-1}\delta_k$$

or

$$Y_{k+1} = \Lambda\cdot Y_k + d_k$$

with solution

$$Y_k = \Lambda^{k-1}Y_0 + \Lambda^{k-1}\sum_{j=0}^{k-1}\Lambda^{-j}d_j$$

and finally

$$y_k = Q\cdot\left( \Lambda^{k-1}Q^{-1}y_0 + \Lambda^{k-1}\sum_{j=0}^{k-1}\Lambda^{-j}Q^{-1}\delta_j\right)$$

or

$$y_k = M^{k-1}y_0 + \sum_{j=0}^{k-1}M^{k-j-1}\delta_j$$