# 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} &= \frac{1}{1+\frac{1}{\rho}} - \frac{1}{1+\lambda}\\ \dot{\lambda} &= \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,& \frac{1}{\left(\frac{1}{\rho}+1\right)^2}\\ 1,& \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 ($\mu_1>0$) and the second one is always negative ($\mu_2<0$). 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 $k<1$ (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. • Perhaps you could include the equations describing the dynamics? A guess: You are in a discrete time model, not a continuous one. The stability criterion for discrete time is that the absolute value of the eigenvalues are smaller than one. See en.wikipedia.org/wiki/… – Giskard Apr 19 '15 at 9:58
• You've nailed it. – clueless Apr 19 '15 at 10:30
• Dont delete the question, there's valuable information here that might be useful for future visitors - and that's all StackExchange is about, generating a stock of information. Perhaps instead just accept the answer and move on :) – FooBar Apr 21 '15 at 15:22

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

• It looks like $E$ should be the Identity matrix, $I$. – Alecos Papadopoulos Apr 19 '15 at 13:36
• You are right. I learned linear algebra in another language, and our notation is sometimes different. Corrected. – Giskard Apr 19 '15 at 18:46
• The system equations are formulated in a related question link – clueless Apr 21 '15 at 9:43

I find that the system is saddle-path stable.

Setting $z\equiv 1/k$ we get

\begin{eqnarray*} \dot{z} & = & -z^2\left(\frac{\rho}{1+\rho} - \frac{1}{1+\lambda}\right) \\ \\ \dot{\lambda} & = & \rho \lambda - z \end{eqnarray*}

The fixed point is $E=\{z^*, \lambda^*\} = \{1, 1/\rho\}$

The Jacobian of this system evaluated at the steady state is

\begin{align} J_E = \begin{bmatrix}0& -\frac{\rho^2}{(1+\rho)^2}\\ -1& \rho\end{bmatrix} \end{align}

The determinant is

$${\rm det}(J_E) = 0-\frac{\rho^2}{(1+\rho)^2} < 0$$

and when the determinant of the Jaconbian in a two-by-two system is negative, the system is saddle-path stable (and so mathematically speaking, unstable), irrespective of whether the trace of the Jacobian (here equal to $\rho>0$) is positive, negative, or zero.

• Thanks. Say, I don't know whether I have a real or complex eigenvalue. But I know $\det(J_E)<0$ and is real. Is that sufficient to inference on a saddle and/or real eigenvalues? From the link you posted I'd guess so, since $\det(J_E)<0$ cannot hold for complex eigenvalues?! – clueless Apr 21 '15 at 16:14
• The point is, that I want to save space and do not want to write down those long eigenvalues and show that they are real and opposite in sign for all $\rho$. And the $\det(J_E)<0$ is way shorter and smoother. I actually do have an additional parameter which is implicity set to unity here, but would occur as well. – clueless Apr 21 '15 at 16:23