I've been given the following second order difference equation, and a set of questions following, but I'm not certain I've done it correctly, verification and hints are appreciated.
Given $x_{t+2}=-\frac{1}{2}x_{t+1}+\frac{3}{2} x_t$:
- Write this as a system of first order difference equations and find all equilibria of the system $z_{t+1} = Az_t$.
For this problem I used a change of variable to say $y_{t+1}=-\frac{1}{2}y_t+\frac{3}{2} x_t$, and $y_t=x_{t+1}$, which when I converted to a matrix (forthcoming in 2) and subtracted the identity matrix, produced infinitely many equilibria of the form x=y=c, where c is any real number.
- Find the solution to this system, that is, find a matrix $D$ such that $z_{t+1} = D^tz_t$.
starting with the matrix from the system in 1, I have $$A=\begin{bmatrix}-\frac{1}{2} & \frac{3}{2} \\ 1 & 0 \end{bmatrix}$$by subtracting λI from this I get $$A-λI=\begin{bmatrix}-\frac{1}{2}-λ & \frac{3}{2} \\ 1 & -λ \end{bmatrix}$$ taking the determinant I have $λ^2+λ\frac{1}{2}-\frac{3}{2}=0$ which produces the roots $λ_1=1$ and $λ_2=\frac{-3}{2}$. The next step towards finding $D$ from my understanding is to plug the two values for lambda into $(A-λ_iI)v_i=0$ and solve for $v_1$ and $v_2$. At this point I get the following $$v_1=\begin{bmatrix}1 \\ 1 \end{bmatrix} v_2=\begin{bmatrix}1 \\ \frac{-2}{3} \end{bmatrix}$$ Since these are linearly independent I get matrix $P$ composed of $v_1,v_2$, $$P=\begin{bmatrix}1 & 1 \\ 1 & -\frac{2}{3} \end{bmatrix}$$ which I find the inverse of: $$P^{-1}=\frac{-5}{3}\begin{bmatrix}-\frac{2}{3} & -1 \\ -1 & 1 \end{bmatrix}$$ and finally our D matrix is $$D=\begin{bmatrix}1 & 0 \\ 0 & \frac{-3}{2}\end{bmatrix}$$ which from my understanding I can say that $z_{t+1} = PD^tP^{-1}z_t$.
- (last part) Are the equilibria you found in part 1 stable?
I've been digging into the definition of stability for this question because my professor stated that the above eigenvalues are all I needed, and that if their moduli are 1 or greater then the equilibria are not asymptotically stable. I've also seen definitions involving epsilon neighborhoods, and limits of n as it goes to infinity. Based on the moduli statement my answer is no the equilibria are not stable($|\frac{-3}{2}|>|1|=1$), but I have no idea of the intuition behind this, since any initial conditions with $x_1=x_0=c$ will return c infinitely.