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I am self-studying the mathematical appendix of Microeconomic Theory (MWG). They defined a stable matrix as follows:

Definition$\quad$ A matrix is stable if all of its characteristic values have negative real parts.

Then, they made the following remark:

Remark$\quad$ This terminology is motivated by the fact that in this case the solution of the system of differential equations $\frac{dx(t)}{dt}=Mx(t)$ will converge to zero as $t\to\infty$ for any initial position $x(0)$.

Moreover, they also applied the following theorem about stable matrices:

Theorem$\quad$ A matrix $A$ is stable if and only if there is a symmetric positive definite matrix $E$ such that $EA$ is negative definite.

I really lack the knowledge about systems of differential equations, stable matrices, and perhaps linear dynamical systems. Could someone please help with the following questions:

  1. How come it is the case that, when $M$ is a stable matrix, the solution of the system of differential equations $\frac{dx(t)}{dt}=Mx(t)$ will converge to zero as $t\to\infty$ for any initial position $x(0)$?
  2. How to prove the theorem above?

Any help would be really appreciate! It would also be very helpful if someone could provide some references (e.g., textbooks, monographs, research papers, lecture notes, and so forth) that systematically study the topics related to this post!

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    $\begingroup$ For the second part: see here $\endgroup$
    – tdm
    Mar 4 at 11:25
  • $\begingroup$ For part 1, this is a standard result of the theory of stability of systems of ordinary differential equation. The treatment of the subject and the proofs can vary, among the many books concerning differential equations. A treatment of the theorem and a sketch of the proof can be found also in Simon e Blume, Mathematics for Economists. $\endgroup$ Mar 4 at 17:37

2 Answers 2

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This answer shows asymptotic convergence to origin/equilibrium if the matrix of coefficients in the homogeneous linear system is diagonalizable. (Continuous time.)


Notation I will use the notation $\triangleq$ when introducing new variables.

I will usually not specify the time variable for trajectories, e.g. I will write $x$ and not $x(t)$.

I will use the dot above character operator to denote differential w.r.t. time, e.g. $$\dot{x} = \frac{\text{d}x}{\text{d}t}$$


Lemma 1 The dynamic system $$ \dot{x} = \lambda x $$ has a solution of the form $$ x(t) = x_0e^{\lambda t}, $$ where $x_0$ is the initial value of $x$.

Proof Take the time derivative of $x e^{-\lambda t}$: $$ \frac{\text{d}(x e^{-\lambda t})}{\text{d}t} = \dot{x} e^{-\lambda t} + x(-\lambda) e^{-\lambda t} = \lambda x e^{-\lambda t} + x(-\lambda) e^{-\lambda t} = 0. $$ This means that $x e^{-\lambda t}$ was constant w.r.t. time, $x e^{-\lambda t} = c$, hence $x = c e^{\lambda t}$. Combining this with the initial value condition we get $x_0 = c e^{\lambda 0} = c$. $\blacksquare$


Corollary Suppose that the linear dynamic system $$ \dot{\mathbf{x}} = \mathbf{D} \mathbf{x} $$ is such that $\mathbf{D}$ is diagonal, with eigenvalues $\lambda_1,...\lambda_n$. Then the system has a solution of the form $$ \mathbf{x} = \begin{bmatrix} x_{10} e^{\lambda_1 t} \\ \vdots \\ x_{n0} e^{\lambda_n t} \end{bmatrix} $$ where $x_{i0}$ are the initial values of $\mathbf{x}$ in dimension $i$.


The next lemma only covers diagonalizable matrices. A generic matrix is diagonizable, but some matrices are not. A very similar result holds, but the proof is more finicky. First we need to introduce some notation.

Let $\mathbf{M}$ be a diagonalizable matrix, with eigenvalues $\lambda_1,...\lambda_n$. We will denote the matrix with these values in the main diagonal and zeroes everywhere else by $\mathbf{D}$, i.e. $$ \mathbf{D} \triangleq \begin{bmatrix} \lambda_1 & 0 & \dots & 0 \\ 0 & \ddots & & \vdots \\ \vdots & & \ddots & 0 \\ 0 & \dots & 0 & \lambda_n \end{bmatrix} $$ We will denote the eigenvectors of $\mathbf{M}$ by $\mathbf{v_1}$, ... $\mathbf{v_n}$.

We will denote the matrix we get by writing the eigenvectors side by side as column vectors by $\mathbf{B}$, i.e. $$ \mathbf{B} \triangleq \begin{bmatrix} \mathbf{v_1} & \dots & \mathbf{v_n} \end{bmatrix}. $$

From these definitions, we have $$ \mathbf{MB} = \begin{bmatrix} \lambda_1\mathbf{v_1} & \dots & \lambda_n\mathbf{v_n} \end{bmatrix} = \mathbf{BD}. $$ The inverse $\mathbf{B}^{-1}$ exists because $\mathbf{M}$ is diagonizable, so we also have $$ \mathbf{B^{-1}MB} = \mathbf{D}. $$


Lemma 2 Suppose that the linear dynamic system $$ \dot{\mathbf{x}} = \mathbf{M} \mathbf{x} $$ is such that $\mathbf{M}$ is diagonalizable. Then the system has a solution of the form $$ \mathbf{x} = \mathbf{B} \begin{bmatrix} c_1 e^{\lambda_1 t} \\ \vdots \\ c_2 e^{\lambda_n t} \end{bmatrix} $$ where $c_i$ are constants.


Proof Let us introduce a new vector variable $$ \mathbf{z} \triangleq \mathbf{B}^{-1} \mathbf{x}. $$ Since $\mathbf{B}^{-1}$ is a linear operator $$ \dot{\mathbf{z}} = \mathbf{B}^{-1} \dot{\mathbf{x}} $$ We will also use the fact that $$ \mathbf{B}\mathbf{z} = \mathbf{x}. $$ We are ready: $$ \begin{eqnarray*} \dot{\mathbf{x}} & = & \mathbf{M} \mathbf{x} \\ \\ \mathbf{B}^{-1}\dot{\mathbf{x}} & = & \mathbf{B}^{-1}\mathbf{M} \mathbf{x} \\ \\ \dot{\mathbf{z}} & = & \mathbf{B}^{-1}\mathbf{M} \mathbf{B}\mathbf{z} \\ \\ \dot{\mathbf{z}} & = & \mathbf{D}\mathbf{z} \end{eqnarray*} $$ Using the corollary of Lemma 1 we know that the form of $\mathbf{z}$'s trajectory is $$ \mathbf{z} = \begin{bmatrix} z_{10} e^{\lambda_1 t} \\ \vdots \\ z_{n0} e^{\lambda_n t} \end{bmatrix} $$ and $\mathbf{x} = \mathbf{B}\mathbf{z}$. $\blacksquare$


What does all this have to do with global asymptotic stability?

  1. Since the system is linear and homogeneous, the equilibrium point is the origin.
  2. The trajectory $$ \mathbf{z} = \begin{bmatrix} z_{10} e^{\lambda_1 t} \\ \vdots \\ z_{n0} e^{\lambda_n t} \end{bmatrix} $$ will converge to the origin if the real parts of the eigenvalues are negative. (We do not show here that the imaginary parts 'cancel out' and this is in fact a real trajectory. For such discussions consult your local linear algebrist.)
  3. The matrix $\mathbf{B}$ has a finite norm (its largest eigenvalue), so $\mathbf{x} = \mathbf{B}\mathbf{z}$ will also converge to origin. If one is not familiar with matrix norms, I think one can also use that if $\mathbf{B}$ is a linear operator in a finite dimensional euclidean space then $$ \lim_{t \to \infty} \mathbf{B}\mathbf{z}(t) = \mathbf{B} \lim_{t \to \infty}\mathbf{z}(t). $$

Thus the trajectory $\mathbf{x}$ will also converge to origin, and it will never be further than $\max_i \lambda_i ||\mathbf{x_0}||$.

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    $\begingroup$ Well, that's a sunday morning I'm never getting back :/ $\endgroup$
    – Giskard
    Mar 10 at 6:52
  • $\begingroup$ You should add a little specification. Without further specification, the last formula giving the solutions z, with $z_{10}$ etc. is incorrect, and can suggest a confusion between Cauchy problems and general solution of a system. The notation with $0$ and the reference to the Corollary of Lemma 1 suggests that the $z_{i0}$ are initial conditions, and this is not true, they are arbitrary constants. And, when calculating them for specific initial conditions, the constants in general don’t result equal to the initial conditions. $\endgroup$ Mar 11 at 14:06
  • $\begingroup$ I am reluctant to add further preamble/notation to the statement of Lemma 2, and I think the current wording is precise; but for future readers those constants are given by $\mathbf{z_0}=\mathbf{B^{-1}x_0}$. $\endgroup$
    – Giskard
    Mar 11 at 16:52
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You wrote:

Definition$\quad$ A matrix is stable if all of its characteristic values have negative real parts.

I really lack the knowledge about systems of differential equations, stable matrices, and perhaps linear dynamical systems [...]

  1. How come it is the case that, when $M$ is a stable matrix, the solution of the system of differential equations $\frac{dx(t)}{dt}=Mx(t)$ will converge to zero as $t\to\infty$ for any initial position $x(0)$?

The answer to your question is linked to the theory of linear systems of ordinary differential equations, and to the study of the stability of the solutions. A detailed and rigorous explanation, of course, can't be contained in an answer, as it requires many concepts about ordinary differential equations.

Anyway, I try to give a concise (as far as possible, it is not easy to summarize) and informal explanation, limited to the case of a system of two differential equations, hoping this brief outline could be useful to you and to some possible readers.

We can consider a system of two differential linear equations, and show possible cases of solutions when the matrix of the coefficients of the system has eigenvalues (characteristic values) with negative real parts. We will see that in this case the equilibrium solutions are stable (in a sense to be specified): hence the name of stable matrix.

Linear ordinary differential equations systems with constant coefficients

Consider a system of two linear ordinary differential equations of the first order, with constant coefficients:

$$\begin{cases} {dx\over dt} = ax(t)+by(t) \\ {dy\over dt} = bx(t)+cy(t) \end{cases} \tag{1.1}$$

where $x(t)$ and $y(t)$ are given differentiable functions from $\mathbb{R}$ to $\mathbb{R}$, and $a,b,c,d$ are real parameters.

We can associate to the system two initial conditions $$\begin{cases} x(0)=x_0 \\ y(0)=y_0 \end{cases} \tag{1.2}$$

The system $(1.1)$ together with the initial conditions $(1.2)$ is called an initial value problem (or Cauchy's problem).

The initial value problem can be written in a more compact form as:

$${dX\over dt}= A X(t)\tag{2.1}$$ $$X(0)=X_0,\tag{2.2}$$

where the matrix $A$ is the matrix of the coefficients of the system:

$$A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}, $$

and $$X(t)=\begin{pmatrix} x(t) \\ y(t) \end{pmatrix}, \;\; X_0=\begin{pmatrix} x_0 \\ y_0 \end{pmatrix}.$$

The solution of this initial value problem is linked to the eigenvalues and the eigenvectors of $A$. In particular, the stability of an equilibrium solution is linked to the properties of the eigenvalues of $A$.

Below, we will show, in particular, that if the eigenvalues of $A$ have a negative real part, the equilibrium solution is stable.

Moreover, we will see that any trajectory (solution), for any initial condition, of the system tends to $(0,0)$ as $t\rightarrow +\infty$.

Equilibrium and stability

A state of the system is called an equilibrium or fixed point if, beginning from this state, the state doesn’t change in time: this means that $dx(t)/dt=0, dy(t)/dt=0$. In formal terms, an equilibrium is a constant solution of the system of differential equations.

It can be shown that in the case of a linear system as $(1.1)$ there is a unique equilibrium solution, $y\equiv 0, y(t)\equiv 0$, that is the constant solution given by the functions $x(t)=0$ and $y(t)=0\;\forall t$.

An equilibrium is said stable if, after a small perturbation, the system stays 'near' the equilibrium. In more formal terms, an equilibrium is stable if, given some starting value $(x_0,y_0)$ ‘close' to the equilibrium, i.e., within some distance$^1$ $\delta$, the trajectory stays close to the equilibrium , i.e., within some distance $\epsilon >\delta$.

enter image description here $\; Fig. 1 - Stability$

An equilibrium is said asymptotically stable if it is stable and if the solutions tend to it as $t \rightarrow +\infty$ (notice that the fact that the solutions tend to the equilibrium in time is not enough, it must be stable).

If asymptotic stability holds for all initial conditions, that is if every trajectory approaches the equilibrium (i.e. , not only for points near the equilibrium, but also for points far away from equilibrium) the equilibrium is said to be globally asymptotically stable. In linear systems as $(1.1)$ we have global stability.

Solutions and stability of equilibrium in relation to eigenvalues

Let $\lambda_1$ and $\lambda_2$ be the eigenvalues of $A$.

We can distinguish various cases, according to the properties of the eigenvalues of $A$, which establish the stability or instability of the equilibrium position.

  1. Negative real distinct eigenvalues: $\lambda_1\neq \lambda_2$ and $\lambda_1<0$, $\lambda_2<0$.

Let $(\xi_1,\eta_1)$ and $(\xi_2,\eta_2)$ the corresponding eigenvectors. It can be proved that the general solution of the system $(1.1)$ is

$$ \begin{pmatrix} x(t) \\ y(t) \end{pmatrix}=c_1e^{\lambda_1 t} \begin{pmatrix} \xi_1 \\ \eta_1 \end{pmatrix} + c_2 e^{\lambda_2 t}\begin{pmatrix} \xi_2 \\ \eta_2 \end{pmatrix} \tag{3} $$

where $c_1$ and $c_2$ are arbitrary constant. Therefore $(3)$ is the set of the infinite solutions of system $(1.1)$: varying $c_1$ and $c_2$, the formula gives all the solutions of the system $(1.1)$.

As in formula $(3)$ the negative eigenvalues appear as exponents of an exponential, the solutions tend to $(0,0)$ as $t$ tends to $+\infty$, for any choice of the constants $c_1$ and $c_2$, that is for any value of the initial conditions.

(remember that the unique$^2$ solution of an initial value problem as $(1.1)-(1.2)$ requires to determine the constants $c_1$ and $c_2$ in the general solution $(3)$, using the given initial conditions).

If the solutions of the system are represented in a phase plane$^3$, with $x(t)$ and $y(t)$ on the axes, we have the following picture:

enter image description here

$\quad Fig. 2- Stable\; node$

Each branch with arrows represents a solution of specific initial values problems (it is a trajectory), with different initial conditions. The origin $(0,0))$ represents the equilibrium solution, that is the constant solution $y(t)=0, x(t)=0 \;\;\forall t$. The arrows, which depend on the sign of the derivatives of $y(t)$, and $x(t)$, represent the direction of the motion.

Such a configuration of the solutions is called stable node: we can see from the arrows in the picture that each solution, represented by a trajectory in the plane, approaches the origin, the equilibrium, as time goes on.

  1. Complex conjugate eigenvalues with negative real part. $\lambda_1=\alpha +i\beta, \lambda_2= \alpha -i\beta,\;\beta\neq 0$.

In this case it can be proved that the general solution of the system $(1.1)$ is given by the formula:

$$\begin{pmatrix} x(t) \\ y(t) \end{pmatrix}=e^{\alpha t}\begin{pmatrix} C_1 \cos (\beta t - \gamma_1) \\ C_2 \cos (\beta t - \gamma_2) \end{pmatrix}\tag {4}$$

whith $C_1\geq 0,\;C_2\geq 0, \gamma_1, \gamma_2$ constants.

(a) $\alpha<0$. The solutions tend to $(0,0)$ as $t\rightarrow +\infty$. The phase portrait becomes a spiral with arrows pointing to the origin, as in Fig. 3 a) below. This configuration takes the name of stable focus.

(b) $\alpha>0$. The solutions don't tend to $(0,0)$, the phase portrait is again a spiral, but with reversed arrows pointing outside, as in Fig. 3 b): the equilibrium is not stable. This configuration is called unstable focus

(c) $\alpha=0$. The exponential disappears, the solutions become periodic functions, closed orbits around $(0,0)$, as in Fig. 3 c). This configuration in the phase plane is called center.

enter image description here

$ \qquad \qquad Fig. 3 - a)\; \alpha<0, stable \;focus\; b)\;\alpha>0, unstable\; focus\; c)\; \alpha=0,center$

The analysis can be completed examining analogously the remaining cases relative to the eigenvalues of the matrix $A$.

The conclusion will be that the equilibrium $(0,0)$ is asymptotically stable if the real parts of the eigenvalues of $A$ are negative, it is not asymptotically stable in the other cases.

$$***$$

As for references, there are a lot of books about ordinary differential equations.

If you are looking for a treatment aimed at economic applications, you can consult some book on economic dynamics, as

Gandolfo G. Economic Dynamics, Fourth Ed., Springer,2009.

Shone R., Economic Dynamics, Second Ed., Cambridge University Press, 2002.

If you are looking for a more general approach, a beautiful book, not advanced, as it requires basic mathematical analysis only, not too theorical, with many examples and applications is

Braun M, Differential Equations and their Applications, Fourth Ed., Springer, 1993.

This is not a book for economists, but it is interesting for economists too.


$^1$ This definition requires a mathematical definition of distance, in general this theory is carried out in normed vector spaces, where a notion of distance is defined.

$^2$ This a consequence of the fundamental theorem of existence and uniqueness for Cauchy problems.

$^3$ The geometric representation in the phase plane could be not easy, in general. It requires the use of the theory of curves in $\mathbb{R}^2$. A curve (in parametric form) in $\mathbb{R}^2$ is a continuous map $\phi \equiv (\phi_1, \phi_2):I \rightarrow \mathbb{R}^2$, from an interval $I$ of $\mathbb{R}$ taking values in $\mathbb{R}^2$.

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  • $\begingroup$ Great writeup, my one qualm is that you write "Below, we will show, in particular, that if the eigenvalues of A have a negative real part, the equilibrium solution is stable." but you never actually prove this, you list most of the cases and provide useful nomenclature and path attributes. $\endgroup$
    – Giskard
    Mar 10 at 5:42
  • $\begingroup$ @Giskard, look at the formula of the solutions! There is the negative real part eigenvalue as exponent! So the solution go to zero, I said that clearly twice! The proofs of the formulas would take too much room, and such subject are complicated, it is better the OP refers to a book. The OP said they has no knowledge of systems of differential equations and their stability, so it was in my opinion more important to explain the fundamental concepts and give a general idea. $\endgroup$ Mar 10 at 10:27
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    $\begingroup$ The usual general mathematical treatement and proofs involve the exponential matrix, and it is not the case and possible in an an answer to this question. For a more complete treatment, as the OP says that they wants a more in-depht study, it is necessary refer to books of differential equations. And consider that for the applications of the qualitative-geometric theory of differential equations, that is what my answer talks bout, the proofs can be not necessary, you can study a lot of classical mechanics without these proofs. $\endgroup$ Mar 10 at 10:42
  • $\begingroup$ A precisation, anyway. A proof of the formulas is immediate, one must just substitute them into the differential equations, needless to say. A different matter is to describe the procedure that leads to the formulas, without knowing them in advance. This would take up too much space, impossible in my answer, that focuses on concepts and applications of qualitative analysis with phase plane (an important, mainly for applications, and beatiful part of the theory of differential equations), and it would imply more theory, for which one must look at a book. $\endgroup$ Mar 11 at 9:27
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    $\begingroup$ Anyway, the overall resut in this trhead is fine, our answers maybe are complementary.:-) $\endgroup$ Mar 11 at 13:32

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