<p>$\newcommand{\vect}[1]{{\bf #1}}$</p>

<p>Note that you can write the system in the form</p>

<p>$$
\dot{\vect{x}} = A \vect{x} + \vect{b}
$$</p>

<p>where </p>

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

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

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

<p>Now define the variable</p>

<p>$$
\vect{z} = \vect{x} - \vect{x}^*
$$</p>

<p>and note that </p>

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

<p>That is,</p>

<p>$$
\dot{\vect{z}} = A \vect{z}
$$</p>

<p>What you need to do now is to find the eigen-decomposition of $A$</p>

CC BY-SA 3.0
2018-01-02T16:40:19.883Z
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
CC BY-SA 3.0
2018-01-02T16:45:45.637Z
@Edie The variable ${\bf z}$ represent deviations from the fixed point ${\bf x}^*$. The characteristic roots are the eigenvalues of $A$
caverac
CC BY-SA 3.0
2018-01-02T15:32:36.400Z
2018-01-02T15:32:36.400Z
caverac
<blockquote>
  <p>For the system
  \begin{align}
\dot{x} = x + y + 1\\
\dot{y} = 2x - y + 5
\end{align}</p>
  
  <p>(i) Find the fixed point.</p>
  
  <p>(ii) Transform the system into deviations <strong>from the fixed point</strong>. What are the characteristic roots of the transformed <strong>system</strong>?</p>
  
  <p>(iii) Derive the equations for the stable and unstable <strong>arms</strong>.</p>
  
  <p>(iv) Set the model up on a spreadsheet and plot, <strong>on the same graph</strong>, the following trajectories:</p>
  
  <p>(a) The trajectory from <strong>initial</strong> point <strong>(-2,4)</strong></p>
  
  <p>(b) The unstable arm <strong>passing through</strong> the point at which $x=2$</p>
  
  <p><em>4th exercise page 172 in the "An Introduction to Economic Dynamics book"</em></p>
</blockquote>

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

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

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

CC BY-SA 3.0
2018-01-02T13:55:51.800Z
Your English seems fine. Do you know what a the concepts of fixed point and characteristic roots mean?
Giskard
CC BY-SA 3.0
2018-01-02T14:15:53.513Z
I'm afraid not, if you don't mind explaining that as well, I would appreciate it.
Edie
CC BY-SA 3.0
2018-01-02T15:01:14.470Z
How did you solve (i) then? What book are you studying from?
Giskard
CC BY-SA 3.0
2018-01-02T15:31:25.893Z
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
<rational-expectations>
Saddles and Rational Expectations. Language barrier preventing me to solve this exercise
Edie
# Saddles and Rational Expectations. Language barrier preventing me to solve this exercise

(Asked by: username_0 on 2018-01-02)

> 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.

**Tags:** rational-expectations
> **Comments:**
> **username_1**: Your English seems fine. Do you know what a the concepts of fixed point and characteristic roots mean?  
> **username_0**: I'm afraid not, if you don't mind explaining that as well, I would appreciate it.  
> **username_1**: How did you solve (i) then? What book are you studying from?  
> **username_0**: 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.  

---

## 1 answers

### Answer by username_2 on 2018-01-02. Score: 1 | Accepted answer

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

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 $A$

> **Comments:**
> **username_0**: 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 ?  
> **username_2**: @Edie The variable ${\bf z}$ represent deviations from the fixed point ${\bf x}^*$. The characteristic roots are the eigenvalues of $A$  

---

