<p>You got to the quadratic equation
<span class="math-container">$$
\lambda^2 - (\rho - n)\lambda + \frac{c^\ast f''(k^\ast)}{\varepsilon}
$$</span>
The discriminant is given by:
<span class="math-container">$$
\Delta = (\rho - n)^2 - 4 \frac{c^\ast f''(k^\ast)}{\varepsilon}
$$</span>
So the two roots are:
<span class="math-container">$$
\lambda_1, \lambda_2 = \frac{(\rho - n) \pm \sqrt{(\rho - n)^2 - 4 \frac{c^\ast f''(k^\ast)}{2}}}{2}
$$</span>
As <span class="math-container">$f''(k) &lt; 0$</span> and <span class="math-container">$\rho &gt; n$</span>  we see that both roots are real, that one of them is larger than <span class="math-container">$0$</span> and one is smaller than zero, so we have a saddle path.</p>
<p>The eigenvectors <span class="math-container">$V_1$</span> and <span class="math-container">$V_2$</span> satisfy <span class="math-container">$A V_1 = \lambda_1 V_1$</span> and <span class="math-container">$A V_2 = \lambda_2 V_2$</span>, so for the first eigenvector:
<span class="math-container">$$
\begin{bmatrix} \rho - n &amp; -1 \\ \frac{c^\ast f''(k^\ast)}{\varepsilon} &amp; 0\end{bmatrix}\begin{bmatrix}v_1\\ 1\end{bmatrix} = \lambda_1 \begin{bmatrix} v_1\\ 1\end{bmatrix}
$$</span>
Here we normalized the second component to 1 which we can do as eigenvectors are only given up to scale. So:
<span class="math-container">$$
\begin{align*}
&amp;(\rho - n) v_1 - 1 = \frac{\left[(\rho - n) + \sqrt{(\rho - n)^2 - 4 \frac{c^\ast f''(k^\ast)}{2}}\right]}{2}v_1\\
\leftrightarrow &amp;v_1 = \frac{2}{(\rho - n) - \sqrt{(\rho - n)^2) - 4 \frac{c^\ast f''k)}{\varepsilon}}}
\end{align*}
$$</span>
Equivalently, we can also use the second condition:
<span class="math-container">$$
v_1 =\frac{\varepsilon}{c^\ast f''(k^\ast)}\frac{\left[(\rho - n) + \sqrt{(\rho - n)^2 - 4 \frac{c^\ast f''(k^\ast)}{2}}\right]}{2}
$$</span>
It is easy to see that these two are the same.</p>
<p>For the second eigenvector we have:
<span class="math-container">$$
\begin{bmatrix} \rho - n &amp; -1 \\ \frac{c^\ast f''(k^\ast)}{\varepsilon} &amp; 0\end{bmatrix}\begin{bmatrix}v_2\\ 1\end{bmatrix} = \lambda_2 \begin{bmatrix} v_2\\ 1\end{bmatrix}
$$</span>
So:
<span class="math-container">$$
\begin{align*}
&amp;(\rho - n) v_2 - 1 = \frac{\left[(\rho - n) - \sqrt{(\rho - n)^2 - 4 \frac{c^\ast f''(k^\ast)}{2}}\right]}{2}v_2\\
\leftrightarrow &amp;v_2 = \frac{2}{(\rho - n) + \sqrt{(\rho - n)^2) - 4 \frac{c^\ast f''k)}{\varepsilon}}}
\end{align*}
$$</span></p>

CC BY-SA 4.0
2022-01-11T15:46:57.357Z
2022-01-11T15:46:57.357Z
tdm
<p>I'm currently struggling to proof the <strong>saddle-path stability</strong> in the <strong>Ramsey-Cass-Koopmans model</strong>. Based on the books <em>Introduction to Modern Economic Growth</em> by D. Acemoglu (chapter 7 and 8) and <em>Mathematics for Economists</em> by Blume and Simon (chapter 25.4) , the following was given:</p>
<p><em>Consider the boundary value problem, corresponding to the equilibrium conditions, of the
baseline neoclassical (Ramsey) growth model</em></p>
<p><span class="math-container">$ \dot k(t) =f(k(t))-(\delta+n)k(t)-c(t) \\ \dot c(t) =\frac{c(t)}{\epsilon_u(c(t))}(f'(k(t))-\delta-\rho) $</span></p>
<p><em>subject to <span class="math-container">$ k(t) , ~ c(t) ≥ 0 $</span>, the initial value condition <span class="math-container">$ k(0) = k_0 &gt; 0 $</span> and the transversality condition</em></p>
<p><span class="math-container">$ \lim_{t \rightarrow \infty} \big[k(t)e^{-\int_0^t(f'(k(s))-\delta-n)ds}\big]=0 $</span></p>
<p><em>where</em></p>
<p><span class="math-container">$ \epsilon_u(c(t)) \equiv - \frac{u''(c(t))~c(t)}{u'(c(t))} $</span></p>
<p><em>is the elasticity of marginal utility of consumption, and <span class="math-container">$ ρ &gt; 0 $</span> and <span class="math-container">$ 0 &lt; δ &lt; 1.$</span></em> <em>We assume that the production function <span class="math-container">$ f (·) $</span> satisfies <span class="math-container">$ f (0) = 0,~ f ′(k) &gt; 0,~ f ′′(k) &lt; 0 $</span> for all <span class="math-container">$ k &gt; 0 $</span> (so it features diminishing marginal returns to capital), and the two Inada conditions, <span class="math-container">$ \lim_{k→0} f ′(k) = ∞ $</span> and <span class="math-container">$\lim_{k→∞} f ′(k) = 0 $</span>.</em></p>
<p>Here's my attempt:</p>
<p>First, I linearized the two differential equations <span class="math-container">$ \dot c(t) $</span> and <span class="math-container">$ \dot k(t)$</span>. Then, I did a  first-order Taylor expansion around the steady-state <span class="math-container">$(k^∗, c^∗) $</span> which gives</p>
<p><span class="math-container">$ \begin{bmatrix}
        \dot k(t)\\ ~ \\
        \dot c(t)
    \end{bmatrix} 
     \approx
    \begin{bmatrix}
        \frac{\partial \dot k(t)}{\partial k(t)} \big\rvert_{k(t) = k^*,c(t)=c^*} &amp; \frac{\partial \dot k(t)}{\partial c(t)} \big\rvert_{k(t) = k^*,c(t)=c^*} \\ ~ \\
        \frac{\partial \dot c(t)}{\partial k(t)}\big\rvert_{k(t) = k^*,c(t)=c^*} &amp; \frac{\partial \dot c(t)}{\partial c(t)} \big\rvert_{k(t) = k^*,c(t)=c^*}
    \end{bmatrix}
    \begin{bmatrix}
        k(t)- k^* \\ ~ \\
        c(t)- c^*
    \end{bmatrix} \\ ~ \\
     = 
    \begin{bmatrix}
        f'(k^*) - (\delta + n) &amp; - 1 \\ ~ \\
        \frac{c^*}{\epsilon_u(c^*)} f''(k^*) &amp;  \frac{1}{\epsilon_u(c^*)} \big(f'(k^*)-\delta - \rho \big)
    \end{bmatrix}
    \begin{bmatrix}
        k(t)- k^* \\ ~ \\
        c(t)- c^*
    \end{bmatrix} \\ ~ \\ $</span></p>
<p>Since we know</p>
<p><span class="math-container">$ \dot k(t)=0 \implies c^*=f(k^*)-(\delta+n)k^* \\
\dot c(t)=0 \implies f'(k^*)=\delta + \rho 
     $</span></p>
<p>the linearized model simplifies to</p>
<p><span class="math-container">$ \begin{bmatrix}
        \dot k(t)\\ ~ \\
        \dot c(t)
    \end{bmatrix} 
 \approx
 \begin{bmatrix}
        \rho - n &amp; - 1 \\ ~ \\
        \frac{c^*}{\epsilon_u(c^*)} f''(k^*) &amp;  0
    \end{bmatrix}
    \begin{bmatrix}
        k(t)- k^* \\ ~ \\
        c(t)- c^*
    \end{bmatrix} \\ ~ \\ $</span></p>
<p>where <span class="math-container">$ c^*f''(k^*)/\epsilon_u(c^*) &lt; 0$</span> and <span class="math-container">$\rho - n &gt; 0$</span>.</p>
<p>Then I defined</p>
<p><span class="math-container">$ A \equiv \begin{bmatrix}
        \rho - n &amp; - 1 \\ ~ \\
        \frac{c^*}{\epsilon_u(c^*)} f''(k^*) &amp;  0
    \end{bmatrix} $</span></p>
<p>Thus, we get</p>
<p><span class="math-container">$ \begin{equation}
    det(A)  =\frac{c^*}{\epsilon_u(c^*)}f''(k^*) &lt; 0 \\
    tr(A)  = \rho - n &gt; 0 
\end{equation}$</span></p>
<p>Then, the characteristic polynomial is</p>
<p><span class="math-container">$ \begin{equation}
    det(A-\lambda I_2) = 0 \iff \lambda^2 - tr(A)\lambda + det(A) \overset{!}{=} 0 \iff \lambda^2 - (\rho-n)\lambda + \frac{c^*f''(k^*)}{\epsilon_u(c^*)} \overset{!}{=} 0 
 \end{equation} $</span></p>
<p>This is how far I've come. The next step would be to compute the eigenvalues <span class="math-container">$λ_{1,2}$</span> and the eigenvectors <span class="math-container">$V_1,~V_2$</span>. We were told that if <span class="math-container">$ tr(A) = λ_1 + λ_2 &gt; 0 $</span>, then the system is unstable. The general solution of the linearized system should look like this</p>
<p><span class="math-container">$ \begin{equation}
    \begin{bmatrix}
        k(t)\\c(t)
    \end{bmatrix} 
     =
    \begin{bmatrix}
        k^*\\ c^*
    \end{bmatrix} 
    + \gamma_1 ~ e^{\lambda_1 t} ~V_1 + \gamma_2 ~ e^{\lambda_2 t} ~ V_2 \end{equation}$</span></p>
<p>(cf. Simon, Blume), where <span class="math-container">$\lambda_1,~ \lambda_2$</span> are the coefficients of the roots <span class="math-container">$λ_1,~ λ_2$</span>. We were also told the following:</p>
<p>&quot;Recall that the goal was to solve the system of the two differential equations in
the neighborhood of the steady-state. If the eigenvalues of <span class="math-container">$A$</span> are smaller than
one in absolute values, the system is (globally) asymptotically stable. If, on
the other hand, all eigenvalues of <span class="math-container">$A$</span> are greater than one in absolute values, the
system can be thought of as “completely unstable,” since all trajectories explode
when they start from an initial position other than the steady-state itself. The
interesting case occurs when, as here, some of the eigenvalues of <span class="math-container">$A$</span> are greater
and some are smaller than one in absolute values. In order for the system to
be saddle-path stable, we require <span class="math-container">$|λ_1| &gt; 1$</span> and <span class="math-container">$|λ_2| &lt; 1$</span>. That means there
exist solution trajectories that, starting arbitrarily close to the steady-state, get
arbitrarily far from it as time goes to infinity. However, there exists a region
such that any trajectory that starts in the region (or is found in this region at
some point in time) converges to the steady-state. In that case the trajectory
remains inside this region while converging to the steady-state.</p>
<p>In case of <span class="math-container">$|λ_2| &lt; 1$</span>, the saddle-path will correspond to the eigenvector <span class="math-container">$V_2$</span> associated with <span class="math-container">$λ_2$</span>, allowing us to ignore the unstable path corresponding to the
eigenvector associated with <span class="math-container">$λ_1$</span>. This means that the coefficient <span class="math-container">$γ_1$</span> of the explosive (unstable) root, namely <span class="math-container">$|λ_1| &gt; 1$</span>, must be set to <span class="math-container">$γ_1 = 0$</span>.&quot;</p>
<p>In order to finish the proof, I need to find the solution to the (simplified) linearized system.</p>

CC BY-SA 4.0
2022-01-11T14:42:05.280Z
What is your question exactly? You have a quadratic equation in $\lambda$ which gives you two roots. These will correspond to the eigenvalues you are looking for. Also, for a continuous system, you want to look for eigenvalues with positive or negative sign (not for ones where the absolute values are greater or less than one).
tdm
CC BY-SA 4.0
2022-01-11T14:51:29.870Z
I have problems with the calculation of the eigenvalues and vectors. How do I calculate them in this particular case?
freakonomist
<macroeconomics><mathematical-economics><economic-growth><ramsey>
Proof of Saddle-Path Stablility in Ramsey-Cass-Koopmans model
freakonomist
# Proof of Saddle-Path Stablility in Ramsey-Cass-Koopmans model

(Asked by: username_0 on 2022-01-11)

I'm currently struggling to proof the **saddle-path stability** in the **Ramsey-Cass-Koopmans model**. Based on the books *Introduction to Modern Economic Growth* by D. Acemoglu (chapter 7 and 8) and *Mathematics for Economists* by Blume and Simon (chapter 25.4) , the following was given:

*Consider the boundary value problem, corresponding to the equilibrium conditions, of the
baseline neoclassical (Ramsey) growth model*

$ \dot k(t) =f(k(t))-(\delta+n)k(t)-c(t) \\ \dot c(t) =\frac{c(t)}{\epsilon\_u(c(t))}(f'(k(t))-\delta-\rho) $

*subject to $ k(t) , ~ c(t) ≥ 0 $, the initial value condition $ k(0) = k\_0 > 0 $ and the transversality condition*

$ \lim\_{t \rightarrow \infty} \big[k(t)e^{-\int\_0^t(f'(k(s))-\delta-n)ds}\big]=0 $

*where*

$ \epsilon\_u(c(t)) \equiv - \frac{u''(c(t))~c(t)}{u'(c(t))} $

*is the elasticity of marginal utility of consumption, and $ ρ > 0 $ and $ 0 < δ < 1.$* *We assume that the production function $ f (·) $ satisfies $ f (0) = 0,~ f ′(k) > 0,~ f ′′(k) < 0 $ for all $ k > 0 $ (so it features diminishing marginal returns to capital), and the two Inada conditions, $ \lim\_{k→0} f ′(k) = ∞ $ and $\lim\_{k→∞} f ′(k) = 0 $.*

Here's my attempt:

First, I linearized the two differential equations $ \dot c(t) $ and $ \dot k(t)$. Then, I did a first-order Taylor expansion around the steady-state $(k^∗, c^∗) $ which gives

$ \begin{bmatrix}
\dot k(t)\\ ~ \\
\dot c(t)
\end{bmatrix}
\approx
\begin{bmatrix}
\frac{\partial \dot k(t)}{\partial k(t)} \big\rvert\_{k(t) = k^\*,c(t)=c^\*} & \frac{\partial \dot k(t)}{\partial c(t)} \big\rvert\_{k(t) = k^\*,c(t)=c^\*} \\ ~ \\
\frac{\partial \dot c(t)}{\partial k(t)}\big\rvert\_{k(t) = k^\*,c(t)=c^\*} & \frac{\partial \dot c(t)}{\partial c(t)} \big\rvert\_{k(t) = k^\*,c(t)=c^\*}
\end{bmatrix}
\begin{bmatrix}
k(t)- k^\* \\ ~ \\
c(t)- c^\*
\end{bmatrix} \\ ~ \\
=
\begin{bmatrix}
f'(k^\*) - (\delta + n) & - 1 \\ ~ \\
\frac{c^\*}{\epsilon\_u(c^\*)} f''(k^\*) & \frac{1}{\epsilon\_u(c^\*)} \big(f'(k^\*)-\delta - \rho \big)
\end{bmatrix}
\begin{bmatrix}
k(t)- k^\* \\ ~ \\
c(t)- c^\*
\end{bmatrix} \\ ~ \\ $

Since we know

$ \dot k(t)=0 \implies c^\*=f(k^\*)-(\delta+n)k^\* \\
\dot c(t)=0 \implies f'(k^\*)=\delta + \rho
$

the linearized model simplifies to

$ \begin{bmatrix}
\dot k(t)\\ ~ \\
\dot c(t)
\end{bmatrix}
\approx
\begin{bmatrix}
\rho - n & - 1 \\ ~ \\
\frac{c^\*}{\epsilon\_u(c^\*)} f''(k^\*) & 0
\end{bmatrix}
\begin{bmatrix}
k(t)- k^\* \\ ~ \\
c(t)- c^\*
\end{bmatrix} \\ ~ \\ $

where $ c^\*f''(k^\*)/\epsilon\_u(c^\*) < 0$ and $\rho - n > 0$.

Then I defined

$ A \equiv \begin{bmatrix}
\rho - n & - 1 \\ ~ \\
\frac{c^\*}{\epsilon\_u(c^\*)} f''(k^\*) & 0
\end{bmatrix} $

Thus, we get

$ \begin{equation}
det(A) =\frac{c^\*}{\epsilon\_u(c^\*)}f''(k^\*) < 0 \\
tr(A) = \rho - n > 0
\end{equation}$

Then, the characteristic polynomial is

$ \begin{equation}
det(A-\lambda I\_2) = 0 \iff \lambda^2 - tr(A)\lambda + det(A) \overset{!}{=} 0 \iff \lambda^2 - (\rho-n)\lambda + \frac{c^\*f''(k^\*)}{\epsilon\_u(c^\*)} \overset{!}{=} 0
\end{equation} $

This is how far I've come. The next step would be to compute the eigenvalues $λ\_{1,2}$ and the eigenvectors $V\_1,~V\_2$. We were told that if $ tr(A) = λ\_1 + λ\_2 > 0 $, then the system is unstable. The general solution of the linearized system should look like this

$ \begin{equation}
\begin{bmatrix}
k(t)\\c(t)
\end{bmatrix}
=
\begin{bmatrix}
k^\*\\ c^\*
\end{bmatrix}
+ \gamma\_1 ~ e^{\lambda\_1 t} ~V\_1 + \gamma\_2 ~ e^{\lambda\_2 t} ~ V\_2 \end{equation}$

(cf. Simon, Blume), where $\lambda\_1,~ \lambda\_2$ are the coefficients of the roots $λ\_1,~ λ\_2$. We were also told the following:

"Recall that the goal was to solve the system of the two differential equations in
the neighborhood of the steady-state. If the eigenvalues of $A$ are smaller than
one in absolute values, the system is (globally) asymptotically stable. If, on
the other hand, all eigenvalues of $A$ are greater than one in absolute values, the
system can be thought of as “completely unstable,” since all trajectories explode
when they start from an initial position other than the steady-state itself. The
interesting case occurs when, as here, some of the eigenvalues of $A$ are greater
and some are smaller than one in absolute values. In order for the system to
be saddle-path stable, we require $|λ\_1| > 1$ and $|λ\_2| < 1$. That means there
exist solution trajectories that, starting arbitrarily close to the steady-state, get
arbitrarily far from it as time goes to infinity. However, there exists a region
such that any trajectory that starts in the region (or is found in this region at
some point in time) converges to the steady-state. In that case the trajectory
remains inside this region while converging to the steady-state.

In case of $|λ\_2| < 1$, the saddle-path will correspond to the eigenvector $V\_2$ associated with $λ\_2$, allowing us to ignore the unstable path corresponding to the
eigenvector associated with $λ\_1$. This means that the coefficient $γ\_1$ of the explosive (unstable) root, namely $|λ\_1| > 1$, must be set to $γ\_1 = 0$."

In order to finish the proof, I need to find the solution to the (simplified) linearized system.

**Tags:** macroeconomics, mathematical-economics, economic-growth, ramsey

> **Comments:**
> **username_1**: What is your question exactly? You have a quadratic equation in $\lambda$ which gives you two roots. These will correspond to the eigenvalues you are looking for. Also, for a continuous system, you want to look for eigenvalues with positive or negative sign (not for ones where the absolute values are greater or less than one).  
> **username_0**: I have problems with the calculation of the eigenvalues and vectors. How do I calculate them in this particular case?  

---

## 1 answers

### Answer by username_1 on 2022-01-11. Score: 6 | Accepted answer

You got to the quadratic equation
$$
\lambda^2 - (\rho - n)\lambda + \frac{c^\ast f''(k^\ast)}{\varepsilon}
$$
The discriminant is given by:
$$
\Delta = (\rho - n)^2 - 4 \frac{c^\ast f''(k^\ast)}{\varepsilon}
$$
So the two roots are:
$$
\lambda\_1, \lambda\_2 = \frac{(\rho - n) \pm \sqrt{(\rho - n)^2 - 4 \frac{c^\ast f''(k^\ast)}{2}}}{2}
$$
As $f''(k) < 0$ and $\rho > n$ we see that both roots are real, that one of them is larger than $0$ and one is smaller than zero, so we have a saddle path.

The eigenvectors $V\_1$ and $V\_2$ satisfy $A V\_1 = \lambda\_1 V\_1$ and $A V\_2 = \lambda\_2 V\_2$, so for the first eigenvector:
$$
\begin{bmatrix} \rho - n & -1 \\ \frac{c^\ast f''(k^\ast)}{\varepsilon} & 0\end{bmatrix}\begin{bmatrix}v\_1\\ 1\end{bmatrix} = \lambda\_1 \begin{bmatrix} v\_1\\ 1\end{bmatrix}
$$
Here we normalized the second component to 1 which we can do as eigenvectors are only given up to scale. So:
$$
\begin{align\*}
&(\rho - n) v\_1 - 1 = \frac{\left[(\rho - n) + \sqrt{(\rho - n)^2 - 4 \frac{c^\ast f''(k^\ast)}{2}}\right]}{2}v\_1\\
\leftrightarrow &v\_1 = \frac{2}{(\rho - n) - \sqrt{(\rho - n)^2) - 4 \frac{c^\ast f''k)}{\varepsilon}}}
\end{align\*}
$$
Equivalently, we can also use the second condition:
$$
v\_1 =\frac{\varepsilon}{c^\ast f''(k^\ast)}\frac{\left[(\rho - n) + \sqrt{(\rho - n)^2 - 4 \frac{c^\ast f''(k^\ast)}{2}}\right]}{2}
$$
It is easy to see that these two are the same.

For the second eigenvector we have:
$$
\begin{bmatrix} \rho - n & -1 \\ \frac{c^\ast f''(k^\ast)}{\varepsilon} & 0\end{bmatrix}\begin{bmatrix}v\_2\\ 1\end{bmatrix} = \lambda\_2 \begin{bmatrix} v\_2\\ 1\end{bmatrix}
$$
So:
$$
\begin{align\*}
&(\rho - n) v\_2 - 1 = \frac{\left[(\rho - n) - \sqrt{(\rho - n)^2 - 4 \frac{c^\ast f''(k^\ast)}{2}}\right]}{2}v\_2\\
\leftrightarrow &v\_2 = \frac{2}{(\rho - n) + \sqrt{(\rho - n)^2) - 4 \frac{c^\ast f''k)}{\varepsilon}}}
\end{align\*}
$$


---

