I have a problem I've been having quite some trouble to solve. Here it is.
We are given that $y_{t+1} = -\frac{3}{2}y_{t}+y_{t-1}$ and $x_t = y_{t+1}-y_t$
a) Find a matrix $A$ s.t $$ \left(\begin{matrix} x_t \\ y_t \\ \end{matrix}\right) = A \left(\begin{matrix} x_{t-1} \\ y_{t-1} \\ \end{matrix}\right)$$
My solution: I found that $A = \left(\begin{matrix} -5/2 & -3/2 \\ 1&1 \\ \end{matrix}\right)$
b) Derive an expression for $\left(\begin{matrix} x_t \\ y_t \\ \end{matrix}\right)$ as a function of the starting values $\left(\begin{matrix} x_0 \\ y_0 \\ \end{matrix}\right)$
My solution: $\left(\begin{matrix} x_t \\ y_t \\ \end{matrix}\right) = A^t\left(\begin{matrix} x_0 \\ y_0 \\ \end{matrix}\right)$
c) Determine the eigenvalues of the matrix $A$ and write down the matrix decomposition for $A$
My solution: $\lambda_1 = 1/2$ and $\lambda_2 = -2$ and $A = \left(\begin{matrix} -1/2 & -3 \\ 1&1 \\ \end{matrix}\right) \left(\begin{matrix} 1/2 & 0 \\ 0&-2 \\ \end{matrix}\right) \left(\begin{matrix} 2/5 & 6/5 \\ -2/5&-1/5 \\ \end{matrix}\right)$
I won't go into details to explain how I found all this, as it was pretty straight forward. However, it is the next question I'm having trouble solving.
d) Using the decomposed matrix $A$, find an explicit solution for $y_t$ subject to the condition $\lim_{t\to\infty} y_t = 0$ as a function of time $t$ only.
I have no idea how to tackle this question. If you could help, that would be nice.
What might help you is that once you have the Eigenvalue-Decomposition of $A$ (namely $U \Sigma U^{-1}$), you can compute $A^t$ very efficently: If $A = U \Sigma U^{-1}$, then $A^t = (U \Sigma U^{-1})^t = U \Sigma^t U^{-1}$. For example $A^2 = U \Sigma U^{-1} U \Sigma U^{-1} = U \Sigma^2 U^{-1}$. Now $\Sigma^t$ is very easy to compute, because it is a diagonal matrix. You have $(x_t, y_t) = A^t * (x_0, y_0)^T = U \Sigma^t U^{-1} (x_0, y_0)^T$. Computing the righthand side will give you a non-recursive solution to $y_t$ in terms of $y_0$ and $x_0$. From there on you will probably be able to solve it. Hope this helps.