# Saddles and Rational Expectations. Language barrier preventing me to solve this exercise

For the system \begin{align} \dot{x} = x + y + 1\\ \dot{y} = 2x - y + 5 \end{align}

(i) Find the fixed point.

(ii) Transform the system into deviations from the fixed point. What are the characteristic roots of the transformed system?

(iii) Derive the equations for the stable and unstable arms.

(iv) Set the model up on a spreadsheet and plot, on the same graph, the following trajectories:

(a) The trajectory from initial point (-2,4)

(b) The unstable arm passing through the point at which $x=2$

4th exercise page 172 in the "An Introduction to Economic Dynamics book"

Can anyone explain to me how do we solve - (ii) Transform the system into deviations from the fixed point. What are the characteristic roots of this transformed system ?

I simply do not understand the point of the exercise. Anyone care to explain what it wants us to do?

I am a Slovakian student, and the exercises in English are getting me stuck. Any help would be appreciated.

• Your English seems fine. Do you know what a the concepts of fixed point and characteristic roots mean? Jan 2, 2018 at 13:55
• I'm afraid not, if you don't mind explaining that as well, I would appreciate it.
– Edie
Jan 2, 2018 at 14:15
• How did you solve (i) then? What book are you studying from? Jan 2, 2018 at 15:01
• Oh I understand now, let's say I know what is a fixed point and what do characteristic roots mean... I am really low on time though, so could you please explain the (ii) part ? I would be able to finish the exercise if you helped me with just the (ii). Just an explanation how and what to do. Thank you for your time anyway.
– Edie
Jan 2, 2018 at 15:31

$\newcommand{\vect}[1]{{\bf #1}}$

Note that you can write the system in the form

$$\dot{\vect{x}} = A \vect{x} + \vect{b}$$

where

$$\vect{x} = \left(\begin{array}{c} x \\ y\end{array}\right), ~~~~ A = \left(\begin{array}{cc} 1 & 1 \\ 2 & -1\end{array}\right), ~~~~~ \vect{b} = \left(\begin{array}{c} 1 \\ 5\end{array}\right)$$

A fixed point $\vect{x}^*$ is a point for which $A\vect{x}^* + \vect{b} = 0$, which in this case is

$$\vect{x}^* = \left(\begin{array}{c} -2 \\ 1\end{array}\right)$$

Now define the variable

$$\vect{z} = \vect{x} - \vect{x}^*$$

and note that

$$\dot{\vect{z}} = \dot{\vect{x}} = A\vect{x} + \vect{b} = A\vect{x} - A\vect{x}^* = A(\vect{x} - \vect{x}^*) = A\vect{z}$$

That is,

$$\dot{\vect{z}} = A \vect{z}$$

What you need to do now is to find the eigen-decomposition of $A$

• The question says "Transform the system into deviations from the fixed point"... but I don't understand which part of what you said, is that certain action ? Also how do I get the characteristic roots of this system ?
– Edie
Jan 2, 2018 at 16:40
• @Edie The variable ${\bf z}$ represent deviations from the fixed point ${\bf x}^*$. The characteristic roots are the eigenvalues of $A$ Jan 2, 2018 at 16:45