Real Positive Eigenvalue, but Stable Dynamics UPDATE
I was not thinking straight anymore and got totally confused after working hours on my equations. The point is, I have an unstable system, but I force it on the stable path. After realizing that crucial point everything made perfect sense.

Problem

(I fixed some major errors)

I'm having a two dimensional system with dynamics
\begin{align}
\dot{k} &amp;= \frac{1}{1+\frac{1}{\rho}} - \frac{1}{1+\lambda}\\
\dot{\lambda} &amp;= \rho\lambda - \frac{1}{k}.
\end{align}
Where $k\in[0,2]$ is the state, and $\lambda$ the costate. There is a (symmetric) fixed point $E(\tilde{k},\tilde{\lambda})$ at $\tilde{\lambda} = \frac{1}{k\rho}$ which yields $\tilde{k}=1$. The Jacobian at the fixed point is given by
\begin{align}
J_E = \begin{bmatrix}0,&amp; \frac{1}{\left(\frac{1}{\rho}+1\right)^2}\\
1,&amp; \rho\end{bmatrix}_E
\end{align}
I get two eigenvalues which are opposite in sign
\begin{align}
(\mu_1,\mu_2)=\left(\frac{p(p + \sqrt{p^2 + 2p + 5}+ 1}{2(p + 1)},\frac{p(p - \sqrt{p^2 + 2p + 5}+ 1}{2(p + 1)}\right)
\end{align}
where $\rho\in\mathbb{R}_{++}$ (time preference rate). The first eigenvalue is always positive ($\mu_1&gt;0$) and the second one is always negative ($\mu_2&lt;0$). So it is an unstable saddle,right? Nonetheless the system is stable. How is that possible, cause I used to think that it must be unstable?  The image shows the evolution of the state and the control which is a function of the costate. ($\rho = 0.05$)


For reference I add figures with $\rho=2,5$. 
The first one seems to converge to a different fixed point $k&lt;1$ (I didn't solve for that one, cause I deal with a symmetric situation; I think there are three in total).

 

And the second picture shows a strange attractor? Which I actually quite like cause of its chaotic stability. For $k\in[0,3]$ the policy function $\tau_1(k)$ is quite weird.
If you have are trying to discretize the continuous time model
$$
\dot{\textbf{x}} = A\textbf{x},
$$
then in discrete time you will have
$$
\textbf{x}_{t+1} = B \textbf{x}_t
$$
but $A\neq B$, since $A$ describes the change in $\textbf{x}$ while $B$ describes the next value of $\textbf{x}$, not just the change. However
$$
\Delta\textbf{x}_t = \textbf{x}_{t+1} - \textbf{x}_t = B \textbf{x}_t - \textbf{x}_t = (B-I) \textbf{x}_t
$$
This new matrix $B-I$ would correspond to $A$.

(From this you can also see why the stability criterion is different.)

Given your equations
\begin{eqnarray*}
\dot{k} &amp; = &amp;  \frac{1}{1+\frac{1}{\rho}} - \frac{1}{1+\lambda} \\
\\
\dot{\lambda} &amp; = &amp;  \rho \lambda - \frac{1}{k}
\end{eqnarray*}
the equations for the discrete modell would be
\begin{eqnarray*}
k_{t+1} &amp; = &amp;  k_t + \frac{1}{1+\frac{1}{\rho}} - \frac{1}{1+\lambda_t} \\
\\
\lambda_{t+1} &amp; = &amp;  \lambda_t + \rho \lambda_t - \frac{1}{k_t}.
\end{eqnarray*}
You will have to calculate the Jacobian of this system to determine stability.