The puzzling eigenvalues of a differential equation system

I have a question here, in order to have a stable solution of a differential equation, is that right we should have all eigenvalues less one?

But I have seen a incomprehensible situation in difference equation, for instance

\begin{align*} x_{t+1}&=a x_{t} -b y_t\\ y_t&= y_{t+1}-cx_{t+1} \end{align*} $a$, $b$ and $c$ are parameters, there are two eignenvalues, one has to be less than one, the other has to be larger than one in order to keep system stable. I completely do not grasp this idea. If anyone knows, please explain to me! Thanks!

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Notice this difference between difference and differential equations: $x_{n+1}=ax_n$ has solutions involving $a^n$, while $x'=ax$ has solutions involving $e^{at}$. For $a^n$, the big difference is between $a\gt1$ and $a\lt1$; for $e^{at}$, the big difference is between $a\gt0$ and $a\lt0$. – Gerry Myerson Jul 11 '12 at 12:41

Just rewrite the equations into the form \begin{align*} x_{t+1}&=a x_{t} -b y_t\\ y_{t+1}-cx_{t+1}&=y_t. \end{align*} This has the form $A\bar x_{t+1}=C\bar x_t$. Now, $A$ is not singular and so one has $\bar x_{t+1}=A^{-1}C\bar x_t$. Now put $$\bar x_t=\lambda^t\left(\begin{array}{c} x_1 \\ x_2\end{array}\right)=\lambda^t\bar x_0$$ and you are left with the eigenvalue problem after substitution
$$\lambda\bar x_0=A^{-1}C\bar x_0.$$
As stated in the comment area you should consider the case of the eigenvalues being $\lambda>1$ and $\lambda<1$.