Stability conditions of discrete dynamic systems I have a system of 6 difference equations with three state variables. Let $X_t = [x_{1t}, x_{2t}, x_{3t}, x_{4t}, x_{5t}, x_{6t}]^T$. The general form of the problem is $g(X_{t+1}, X_t) = 0$. The log-linear form of this system around the steady state $X^*$ is:
\begin{equation}
X_{t+1} - X^* = M \cdot (X_t - X^*)
\end{equation}
The eigenvalues of $M$ are $-32.43$, $0.56 \pm 0.94i$, $0.98 \pm 0.12i$ and $0.94$. The stability conditions of discrete time system ($|\lambda|<1$), which imply that there are three stable roots and the system is saddle-point stable. 
If, however, I set up the system in a `differential form':
\begin{equation}
X_{t+1} - X_t = (M - I) \cdot (X_t - X^*) = N \cdot (X_t - X^*)
\end{equation}
The eigenvalues of $N$ are $-33.43$, $-0.44 \pm 0.94i$, $-0.02 \pm 0.12i$ and $-0.06$. Now, we can apply the stability conditions of differential system ($Re (\lambda)<0$), which imply that there are six stable roots and the system has multiple paths to the steady state. 
How do I explain the difference in the predictions of the two forms of the same model? Which is the right way to look at the problem?
 A: I will use zero for the eigenvector consisting of 0s.
We are going to assume in this discussion that X* and Y* are zero (which can be justified using a change of variables).
Now let us analyse the first equation: 
If you choose $X_0$ to be one of the nice eigenvectors (whose corresponding eigenvalue is $\lambda$ of modulus less than 1), it follows then $X_1= \lambda X_0$, $X_2= \lambda^2 X_0$ and so on and so forth. It follows in particular that $X_t$ goes to zero as $t$ goes to infinity.
It follows that $X_0$ is a linear combination of these three eigenvectors (whose corresponding eigenvalue is $\lambda$ of modulus less than $1$) then the corresponding $X_t$ goes to zero rapidly. However if there is even a tiny perturbation which takes $X_0$ outside this space then instead the solution goes to infinity.
Now let us consider the second equation:
Given $X_0$, I do not see a method of finding the solution other than converting the equation into the previous form. 
I think the confusion arose because differential and difference equations have been treated in the same way. The corresponding differential equation has a general solution of the form:
$X(t)= c e^{Nt} X_0$ where c is a constant. The eigenvalues of $e^N$ determine the number/stability of solutions. For instance it follows that if all the eigenvalues of $N$ have negative real part then all the eigenvalues of $e^N$ have modulus less than $1$ and $X(t)$ must go to zero irrespective of what $X(0)$ is. 
However I do not see a way of proving that there is such a solution in case we replace the differential equation by difference equations and hence can’t see how to use the eigenvalue condition given along with it.
I hope this helps!
Nishant.
