<p>You wrote:</p>
<blockquote>
<p><strong>Definition</strong><span class="math-container">$\quad$</span> A matrix is stable if all of its characteristic values have negative real parts.</p>
<p>I really lack the knowledge about systems of differential equations, stable matrices, and perhaps linear dynamical systems [...]</p>
<ol>
<li>How come it is the case that, when <span class="math-container">$M$</span> is a stable matrix, the solution of the system of differential equations <span class="math-container">$\frac{dx(t)}{dt}=Mx(t)$</span> will converge to zero as <span class="math-container">$t\to\infty$</span> for <em>any</em> initial position <span class="math-container">$x(0)$</span>?</li>
</ol>
</blockquote>
<p>The answer to your question is linked to the theory of  <em><strong>linear systems of ordinary differential equations</strong></em>, and to the study of the <em><strong>stability</strong></em> 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.</p>
<p>Anyway, I try to give a <strong>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,</strong> hoping this brief outline  could be  useful to you and to some possible readers.</p>
<p>We can consider a  system of two differential linear equations, and show possible cases of solutions <strong>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 <em>stable</em> (in a sense to be specified): hence the name of <em><strong>stable matrix</strong></em>.</strong></p>
<h2>Linear ordinary differential equations systems with constant coefficients</h2>
<p>Consider a system of two <em>linear ordinary differential equations of the first</em> <em>order</em>, <em>with constant coefficients</em>:</p>
<p><span class="math-container">$$\begin{cases} {dx\over dt} = ax(t)+by(t) \\ {dy\over dt} = bx(t)+cy(t) \end{cases} \tag{1.1}$$</span></p>
<p>where <span class="math-container">$x(t)$</span> and <span class="math-container">$y(t)$</span> are given differentiable functions from <span class="math-container">$\mathbb{R}$</span> to <span class="math-container">$\mathbb{R}$</span>, and <span class="math-container">$a,b,c,d$</span> are real parameters.</p>
<p>We can associate to the system  two initial conditions
<span class="math-container">$$\begin{cases} x(0)=x_0 \\ y(0)=y_0 \end{cases} \tag{1.2}$$</span></p>
<p>The system <span class="math-container">$(1.1)$</span> together with the initial conditions <span class="math-container">$(1.2)$</span> is called <strong>an initial value problem</strong> (or <em>Cauchy's problem</em>).</p>
<p>The initial value problem can be written in a more compact form as:</p>
<p><span class="math-container">$${dX\over dt}= A X(t)\tag{2.1}$$</span>
<span class="math-container">$$X(0)=X_0,\tag{2.2}$$</span></p>
<p>where the matrix <span class="math-container">$A$</span> is the matrix of the coefficients of the system:</p>
<p><span class="math-container">$$A=\begin{pmatrix} a  &amp; b \\ c  &amp; d \end{pmatrix},  $$</span></p>
<p>and <span class="math-container">$$X(t)=\begin{pmatrix} x(t) \\ y(t) \end{pmatrix}, \;\; X_0=\begin{pmatrix} x_0 \\ y_0 \end{pmatrix}.$$</span></p>
<p><strong>The solution of this initial value problem is linked to the eigenvalues and the eigenvectors  of <span class="math-container">$A$</span>.</strong> In particular, the stability of an equilibrium solution is linked to the properties of the eigenvalues of <span class="math-container">$A$</span>.</p>
<p>Below, we will show, <strong>in particular, that if the eigenvalues of <span class="math-container">$A$</span> have a negative real part, the equilibrium solution is  stable</strong>.</p>
<p>Moreover, we will see that any trajectory (solution), for any initial condition, of the  system tends to <span class="math-container">$(0,0)$</span> as <span class="math-container">$t\rightarrow +\infty$</span>.</p>
<h2>Equilibrium and stability</h2>
<p>A state of the system is called an <em><strong>equilibrium</strong></em> or <em><strong>fixed point</strong></em> if, beginning from this state,  the state doesn’t change in time: this means that <span class="math-container">$dx(t)/dt=0, dy(t)/dt=0$</span>. In formal terms, an equilibrium is a <em>constant solution</em> of the system of differential equations.</p>
<p>It can be shown that <strong>in the case of a linear system as <span class="math-container">$(1.1)$</span></strong> there is a <strong>unique equilibrium solution, <span class="math-container">$y\equiv 0, y(t)\equiv 0$</span>,</strong> that is the constant solution given by the functions <span class="math-container">$x(t)=0$</span> and <span class="math-container">$y(t)=0\;\forall t$</span>.</p>
<p>An equilibrium is said <em><strong>stable</strong></em> if, after a small perturbation,  the system stays 'near' the equilibrium. In more formal terms, an equilibrium  is <em>stable</em> if, given some starting value <span class="math-container">$(x_0,y_0)$</span> ‘close' to the equilibrium, i.e., within some distance<span class="math-container">$^1$</span> <span class="math-container">$\delta$</span>, the trajectory stays close to the equilibrium , i.e., within some distance <span class="math-container">$\epsilon &gt;\delta$</span>.</p>
<p><a href="https://i.sstatic.net/QMaHX.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/QMaHX.png" alt="enter image description here" /></a>
<span class="math-container">$\; Fig. 1 - Stability$</span></p>
<p>An equilibrium  is said <em><strong>asymptotically stable</strong></em>  if it is stable and if the solutions tend to it as <span class="math-container">$t \rightarrow +\infty$</span> (notice that the fact that the solutions tend to the equilibrium in time is not enough, it must be stable).</p>
<p>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 <em><strong>globally asymptotically stable.</strong></em> In linear systems as  <span class="math-container">$(1.1)$</span> we have global stability.</p>
<h2>Solutions and stability of equilibrium in relation to eigenvalues</h2>
<p>Let <span class="math-container">$\lambda_1$</span> and <span class="math-container">$\lambda_2$</span> be the eigenvalues of <span class="math-container">$A$</span>.</p>
<p>We can distinguish various cases, according to the properties of the eigenvalues of <span class="math-container">$A$</span>, which establish the stability or instability of the equilibrium position.</p>
<ol>
<li><strong>Negative real distinct eigenvalues</strong>: <span class="math-container">$\lambda_1\neq \lambda_2$</span> and <span class="math-container">$\lambda_1&lt;0$</span>,  <span class="math-container">$\lambda_2&lt;0$</span>.</li>
</ol>
<p>Let <span class="math-container">$(\xi_1,\eta_1)$</span> and <span class="math-container">$(\xi_2,\eta_2)$</span> the corresponding eigenvectors. It can be proved that the <em>general</em> solution of the system <span class="math-container">$(1.1)$</span> is</p>
<p><span class="math-container">$$ \begin{pmatrix} x(t) \\ y(t) \end{pmatrix}=c_1e^{\lambda_1 t}  \begin{pmatrix} \xi_1 \\ \eta_1 \end{pmatrix} + c_2 e^{\lambda_2 t}\begin{pmatrix} \xi_2 \\ \eta_2 \end{pmatrix}    \tag{3}        $$</span></p>
<p>where <span class="math-container">$c_1$</span> and <span class="math-container">$c_2$</span> are arbitrary constant. Therefore <span class="math-container">$(3)$</span> is the set of the infinite solutions of system <span class="math-container">$(1.1)$</span>: varying <span class="math-container">$c_1$</span> and <span class="math-container">$c_2$</span>,  the formula gives all the solutions of the system <span class="math-container">$(1.1)$</span>.</p>
<p>As in formula <span class="math-container">$(3)$</span> <strong>the negative eigenvalues appear as exponents of an exponential, the solutions tend to <span class="math-container">$(0,0)$</span> as <span class="math-container">$t$</span> tends to <span class="math-container">$+\infty$</span>, for any choice of the constants <span class="math-container">$c_1$</span> and <span class="math-container">$c_2$</span>, that is for any value of the initial conditions.</strong></p>
<p>(remember that the <strong>unique</strong><span class="math-container">$^2$</span> solution of an initial value problem as <span class="math-container">$(1.1)-(1.2)$</span> requires to determine the constants <span class="math-container">$c_1$</span> and <span class="math-container">$c_2$</span> in the general solution <span class="math-container">$(3)$</span>, using the given initial conditions).</p>
<p>If the solutions of the system are represented in a <em><strong>phase plane</strong></em><span class="math-container">$^3$</span>, with <span class="math-container">$x(t)$</span> and <span class="math-container">$y(t)$</span> on the axes, we have the following picture:</p>
<p><a href="https://i.sstatic.net/8KFzw.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/8KFzw.png" alt="enter image description here" /></a></p>
<p><span class="math-container">$\quad Fig. 2- Stable\; node$</span></p>
<p>Each branch with arrows represents a solution of specific initial values  problems (it is a <em>trajectory</em>), with different initial conditions. The origin <span class="math-container">$(0,0))$</span> represents  the equilibrium solution, that is the constant solution <span class="math-container">$y(t)=0, x(t)=0 \;\;\forall t$</span>. The arrows, which depend on the sign of the derivatives of <span class="math-container">$y(t)$</span>, and <span class="math-container">$x(t)$</span>, represent the direction of the motion.</p>
<p>Such a configuration of the solutions is called <em><strong>stable node</strong></em>: 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.</p>
<ol start="2">
<li><strong>Complex conjugate eigenvalues with negative real part</strong>. <span class="math-container">$\lambda_1=\alpha +i\beta, \lambda_2= \alpha -i\beta,\;\beta\neq 0$</span>.</li>
</ol>
<p>In this case it can be proved that the general solution of the system <span class="math-container">$(1.1)$</span> is given by the formula:</p>
<p><span class="math-container">$$\begin{pmatrix} x(t) \\ y(t) \end{pmatrix}=e^{\alpha t}\begin{pmatrix} C_1 \cos (\beta t - \gamma_1) \\ C_2 \cos (\beta t - \gamma_2) \end{pmatrix}\tag {4}$$</span></p>
<p>whith <span class="math-container">$C_1\geq 0,\;C_2\geq 0, \gamma_1, \gamma_2$</span> constants.</p>
<p>(a) <span class="math-container">$\alpha&lt;0$</span>.  The solutions tend to <span class="math-container">$(0,0)$</span> as <span class="math-container">$t\rightarrow +\infty$</span>. The phase portrait becomes a spiral with arrows pointing to the origin, as in <em>Fig. 3 a)</em> below. This configuration takes the name of <em><strong>stable focus</strong></em>.</p>
<p>(b) <span class="math-container">$\alpha&gt;0$</span>. The solutions don't tend to <span class="math-container">$(0,0)$</span>, 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 <em><strong>unstable focus</strong></em></p>
<p>(c) <span class="math-container">$\alpha=0$</span>. The exponential disappears, the solutions become periodic functions, closed orbits around <span class="math-container">$(0,0)$</span>, as in Fig. 3 c). This configuration in the phase plane is called <em><strong>center</strong></em>.</p>
<p><a href="https://i.sstatic.net/bNgTn.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/bNgTn.png" alt="enter image description here" /></a></p>
<p><span class="math-container">$ \qquad \qquad Fig. 3 - a)\; \alpha&lt;0, stable \;focus\; b)\;\alpha&gt;0, unstable\; focus\; c)\; \alpha=0,center$</span></p>
<p>The analysis can be completed <strong>examining analogously the remaining cases</strong> relative to the eigenvalues of the matrix <span class="math-container">$A$</span>.</p>
<p>The conclusion will be that <strong>the equilibrium <span class="math-container">$(0,0)$</span> is asymptotically stable if the real parts of the eigenvalues of <span class="math-container">$A$</span>  are negative, it is not asymptotically stable in the other cases.</strong></p>
<p><span class="math-container">$$***$$</span></p>
<p>As for references, there are a lot of books about ordinary differential equations.</p>
<p>If you are looking for a treatment aimed at economic applications, you can consult some book on economic dynamics, as</p>
<p>Gandolfo G. <em>Economic Dynamics</em>, Fourth Ed., Springer,2009.</p>
<p>Shone R., <em>Economic Dynamics</em>, Second Ed., Cambridge University Press, 2002.</p>
<p>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</p>
<p>Braun M, <em>Differential Equations and their Applications</em>, Fourth Ed., Springer, 1993.</p>
<p>This is not a book for economists, but it is interesting  for economists too.</p>
<hr />
<p><span class="math-container">$^1$</span> <sup>This definition requires a mathematical definition of <em>distance</em>, in general this theory is carried out in  <em>normed vector spaces</em>, where a notion of distance is defined. </sup></p>
<p><span class="math-container">$^2$</span> <sup>This a consequence of the fundamental <em>theorem of existence and uniqueness for Cauchy problems</em>.</sup></p>
<p><span class="math-container">$^3$</span> <sup>The geometric representation in the phase plane could be not easy, in general. It requires the use of the theory of <em>curves</em> in <span class="math-container">$\mathbb{R}^2$</span>. A <em>curve</em> (in <em>parametric form</em>)  in <span class="math-container">$\mathbb{R}^2$</span>  is a continuous map <span class="math-container">$\phi \equiv (\phi_1, \phi_2):I \rightarrow \mathbb{R}^2$</span>, from an interval <span class="math-container">$I$</span> of <span class="math-container">$\mathbb{R}$</span> taking values in <span class="math-container">$\mathbb{R}^2$</span>.</sup></p>

CC BY-SA 4.0
2024-03-10T05:42:05.147Z
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.
Giskard
CC BY-SA 4.0
2024-03-10T10:27:32.880Z
@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.
BakerStreet
CC BY-SA 4.0
2024-03-10T10:42:19.537Z
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.
BakerStreet
CC BY-SA 4.0
2024-03-11T09:27:48.450Z
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.
BakerStreet
CC BY-SA 4.0
2024-03-11T10:06:01.850Z
As I wrote above I enjoyed reading your answer, I just think the word "show" is perhaps not the best fit here. As for a proof taking up way too much space, I will immodestly recommend [my answer](https://economics.stackexchange.com/a/57867/1601), which is slightly shorter than yours, and is perhaps a reasonably rigorous proof of the statement.
Giskard
CC BY-SA 4.0
2024-03-11T13:29:24.060Z
@Giskard I said 'show' exactly to not say 'prove'. My answer + proofs would be impossible. And I prefere a conceptual answer(and it is rigorous, I have some lecture notes of analytic machanics that follow these lines)  that illustrates the key concepts to a person who doesn't know differential equations. In my opinion a proof as in your answer is too advanvced, implies advanced concept not explained (operators, norms, bounded operators ,complex exponential and Euler formula), and less useful to a beginner of the subject.
BakerStreet
CC BY-SA 4.0
2024-03-11T13:32:29.970Z
Anyway, the overall resut in this trhead is fine, our answers maybe are complementary.:-)
BakerStreet
CC BY-SA 4.0
2024-03-09T17:32:58.047Z
2024-03-11T15:48:28.767Z
2024-03-11T15:48:28.767Z
BakerStreet
<p>This answer shows asymptotic convergence to origin/equilibrium if the matrix of coefficients in the homogeneous linear system is diagonalizable. (Continuous time.)</p>
<hr />
<p><strong>Notation</strong> I will use the notation <span class="math-container">$\triangleq$</span> when introducing new variables.</p>
<p>I will usually not specify the time variable for trajectories, e.g. I will write <span class="math-container">$x$</span> and not <span class="math-container">$x(t)$</span>.</p>
<p>I will use the dot above character operator to denote differential w.r.t. time, e.g.
<span class="math-container">$$\dot{x} = \frac{\text{d}x}{\text{d}t}$$</span></p>
<hr />
<p><strong>Lemma 1</strong> The dynamic system
<span class="math-container">$$
\dot{x} = \lambda x
$$</span>
has a solution of the form
<span class="math-container">$$
x(t) = x_0e^{\lambda t},
$$</span>
where <span class="math-container">$x_0$</span> is the initial value of <span class="math-container">$x$</span>.</p>
<p><strong>Proof</strong> Take the time derivative of <span class="math-container">$x e^{-\lambda t}$</span>:
<span class="math-container">$$
\frac{\text{d}(x e^{-\lambda t})}{\text{d}t} = \dot{x} e^{-\lambda t} + x(-\lambda) e^{-\lambda t} = \lambda x e^{-\lambda t} + x(-\lambda) e^{-\lambda t} = 0.
$$</span>
This means that <span class="math-container">$x e^{-\lambda t}$</span> was constant w.r.t. time, <span class="math-container">$x e^{-\lambda t} = c$</span>, hence <span class="math-container">$x = c e^{\lambda t}$</span>. Combining this with the initial value condition we get <span class="math-container">$x_0 = c e^{\lambda 0} = c$</span>. <span class="math-container">$\blacksquare$</span></p>
<hr />
<p><strong>Corollary</strong> Suppose that the linear dynamic system
<span class="math-container">$$
\dot{\mathbf{x}} = \mathbf{D} \mathbf{x}
$$</span>
is such that <span class="math-container">$\mathbf{D}$</span> is diagonal, with eigenvalues <span class="math-container">$\lambda_1,...\lambda_n$</span>. Then the system has a solution of the form
<span class="math-container">$$
\mathbf{x} = \begin{bmatrix}
 x_{10} e^{\lambda_1 t} \\ \vdots \\ x_{n0} e^{\lambda_n t}
\end{bmatrix}
$$</span>
where <span class="math-container">$x_{i0}$</span> are the initial values of <span class="math-container">$\mathbf{x}$</span> in dimension <span class="math-container">$i$</span>.</p>
<hr />
<p>The next lemma only covers <em>diagonalizable</em> matrices. A <em>generic</em> 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.</p>
<p>Let <span class="math-container">$\mathbf{M}$</span> be a diagonalizable matrix, with eigenvalues <span class="math-container">$\lambda_1,...\lambda_n$</span>. We will denote the matrix with these values in the main diagonal and zeroes everywhere else by <span class="math-container">$\mathbf{D}$</span>, i.e.
<span class="math-container">$$
\mathbf{D} \triangleq 
\begin{bmatrix}
\lambda_1 &amp; 0 &amp; \dots &amp; 0 \\
0 &amp; \ddots &amp; &amp; \vdots   \\
\vdots &amp;   &amp; \ddots &amp; 0
\\
0 &amp; \dots &amp; 0 &amp; \lambda_n
\end{bmatrix}
$$</span>
We will denote the eigenvectors of <span class="math-container">$\mathbf{M}$</span> by <span class="math-container">$\mathbf{v_1}$</span>, ... <span class="math-container">$\mathbf{v_n}$</span>.</p>
<p>We will denote the matrix we get by writing the eigenvectors side by side as column vectors by <span class="math-container">$\mathbf{B}$</span>, i.e.
<span class="math-container">$$
\mathbf{B} \triangleq 
\begin{bmatrix}
\mathbf{v_1} &amp; \dots &amp; \mathbf{v_n}
\end{bmatrix}.
$$</span></p>
<p>From these definitions, we have
<span class="math-container">$$
\mathbf{MB} = 
\begin{bmatrix}
\lambda_1\mathbf{v_1} &amp; \dots &amp; \lambda_n\mathbf{v_n}
\end{bmatrix}
=
\mathbf{BD}.
$$</span>
The inverse <span class="math-container">$\mathbf{B}^{-1}$</span> exists because <span class="math-container">$\mathbf{M}$</span> is diagonizable, so we also have
<span class="math-container">$$
\mathbf{B^{-1}MB} = 
\mathbf{D}.
$$</span></p>
<hr />
<p><strong>Lemma 2</strong> Suppose that the linear dynamic system
<span class="math-container">$$
\dot{\mathbf{x}} = \mathbf{M} \mathbf{x}
$$</span>
is such that <span class="math-container">$\mathbf{M}$</span> is diagonalizable. Then the system has a solution of the form
<span class="math-container">$$
\mathbf{x} = \mathbf{B}
\begin{bmatrix}
c_1 e^{\lambda_1 t} \\ \vdots \\ c_2 e^{\lambda_n t}
\end{bmatrix}
$$</span>
where <span class="math-container">$c_i$</span> are constants.</p>
<hr />
<p><strong>Proof</strong> Let us introduce a new vector variable
<span class="math-container">$$
\mathbf{z} \triangleq \mathbf{B}^{-1} \mathbf{x}.
$$</span>
Since <span class="math-container">$\mathbf{B}^{-1}$</span> is a linear operator
<span class="math-container">$$
\dot{\mathbf{z}} = \mathbf{B}^{-1} \dot{\mathbf{x}}
$$</span>
We will also use the fact that
<span class="math-container">$$
\mathbf{B}\mathbf{z} = \mathbf{x}.
$$</span>
We are ready:
<span class="math-container">$$
\begin{eqnarray*}
\dot{\mathbf{x}} &amp; = &amp; \mathbf{M} \mathbf{x} \\
\\
\mathbf{B}^{-1}\dot{\mathbf{x}} &amp; = &amp; \mathbf{B}^{-1}\mathbf{M} \mathbf{x} \\
\\
\dot{\mathbf{z}} &amp; = &amp; \mathbf{B}^{-1}\mathbf{M} \mathbf{B}\mathbf{z} \\
\\
\dot{\mathbf{z}} &amp; = &amp; \mathbf{D}\mathbf{z}
\end{eqnarray*}
$$</span>
Using the corollary of <strong>Lemma 1</strong> we know that the form of <span class="math-container">$\mathbf{z}$</span>'s trajectory is
<span class="math-container">$$
\mathbf{z} = \begin{bmatrix}
 z_{10} e^{\lambda_1 t} \\ \vdots \\ z_{n0} e^{\lambda_n t}
\end{bmatrix}
$$</span>
and <span class="math-container">$\mathbf{x} = \mathbf{B}\mathbf{z}$</span>. <span class="math-container">$\blacksquare$</span></p>
<hr />
<p>What does all this have to do with global asymptotic stability?</p>
<ol>
<li>Since the system is linear and homogeneous, the equilibrium point is the origin.</li>
<li>The trajectory
<span class="math-container">$$
\mathbf{z} = \begin{bmatrix}
 z_{10} e^{\lambda_1 t} \\ \vdots \\ z_{n0} e^{\lambda_n t}
\end{bmatrix}
$$</span>
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.)</li>
<li>The matrix <span class="math-container">$\mathbf{B}$</span> has a finite norm (its largest eigenvalue), so
<span class="math-container">$\mathbf{x} = \mathbf{B}\mathbf{z}$</span> will also converge to origin. If one is not familiar with matrix norms, I think one can also use that if <span class="math-container">$\mathbf{B}$</span> is a linear operator in a finite dimensional euclidean space then
<span class="math-container">$$
\lim_{t \to \infty} \mathbf{B}\mathbf{z}(t) =
\mathbf{B} \lim_{t \to \infty}\mathbf{z}(t).
$$</span></li>
</ol>
<p>Thus the trajectory <span class="math-container">$\mathbf{x}$</span> will also converge to origin, and it will never be further than <span class="math-container">$\max_i \lambda_i ||\mathbf{x_0}||$</span>.</p>

CC BY-SA 4.0
2024-03-10T06:52:49.530Z
Well, that's a sunday morning I'm never getting back :/
Giskard
CC BY-SA 4.0
2024-03-11T14:06:16.847Z
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.
BakerStreet
CC BY-SA 4.0
2024-03-11T16:52:58.537Z
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 $\mathbf{z_0}=\mathbf{B^{-1}x_0}$.
Giskard
CC BY-SA 4.0
2024-03-10T06:52:18.933Z
2024-03-10T18:22:02.220Z
2024-03-10T18:22:02.220Z
Giskard
<p>I am self-studying the mathematical appendix of <em>Microeconomic Theory</em> (MWG). They defined a stable matrix as follows:</p>
<blockquote>
<p><strong>Definition</strong><span class="math-container">$\quad$</span> A matrix is stable if all of its characteristic values have negative real parts.</p>
</blockquote>
<p>Then, they made the following remark:</p>
<blockquote>
<p><strong>Remark</strong><span class="math-container">$\quad$</span> This terminology is motivated by the fact that in this case the solution of the system of differential equations <span class="math-container">$\frac{dx(t)}{dt}=Mx(t)$</span> will converge to zero as <span class="math-container">$t\to\infty$</span> for <em>any</em> initial position <span class="math-container">$x(0)$</span>.</p>
</blockquote>
<p>Moreover, they also applied the following theorem about stable matrices:</p>
<blockquote>
<p><strong>Theorem</strong><span class="math-container">$\quad$</span> A matrix <span class="math-container">$A$</span> is stable if and only if there is a symmetric positive definite matrix <span class="math-container">$E$</span> such that <span class="math-container">$EA$</span> is negative definite.</p>
</blockquote>
<p>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:</p>
<ol>
<li>How come it is the case that, when <span class="math-container">$M$</span> is a stable matrix, the solution of the system of differential equations <span class="math-container">$\frac{dx(t)}{dt}=Mx(t)$</span> will converge to zero as <span class="math-container">$t\to\infty$</span> for <em>any</em> initial position <span class="math-container">$x(0)$</span>?</li>
<li>How to prove the theorem above?</li>
</ol>
<p>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!</p>

CC BY-SA 4.0
2024-03-04T11:25:38.097Z
For the second part: see [here](https://math.stackexchange.com/questions/3269508/a-matrix-a-is-stable-iff-there-is-a-positive-definite-matrix-e-such-that-ea-is)
tdm
CC BY-SA 4.0
2024-03-04T17:37:48.967Z
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.
BakerStreet
<microeconomics><mathematical-economics><linear-algebra><dynamical-systems>
Question About Stable Matrices - MWG Mathematical Appendix M.D Matrices: Negative (Semi)Definiteness and Other Properties
Beerus
# Question About Stable Matrices - MWG Mathematical Appendix M.D Matrices: Negative (Semi)Definiteness and Other Properties

(Asked by: username_0 on 2024-03-03)

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!

**Tags:** microeconomics, mathematical-economics, linear-algebra, dynamical-systems
> **Comments:**
> **username_1**: For the second part: see [here](https://math.stackexchange.com/questions/3269508/a-matrix-a-is-stable-iff-there-is-a-positive-definite-matrix-e-such-that-ea-is)  
> **username_2**: 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.  

---

## 2 answers

### Answer by username_2 on 2024-03-09. Score: 3

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](https://i.sstatic.net/QMaHX.png)](https://i.sstatic.net/QMaHX.png)
$\; Fig. 1 - Stability$

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](https://i.sstatic.net/8KFzw.png)](https://i.sstatic.net/8KFzw.png)

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

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.

2. **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](https://i.sstatic.net/bNgTn.png)](https://i.sstatic.net/bNgTn.png)

$ \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 $\mathbb{R}^2$.

> **Comments:**
> **username_3**: 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.  
> **username_2**: @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.  
> **username_2**: 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.  
> **username_2**: 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.  
> **username_3**: As I wrote above I enjoyed reading your answer, I just think the word "show" is perhaps not the best fit here. As for a proof taking up way too much space, I will immodestly recommend [my answer](https://economics.stackexchange.com/a/57867/1601), which is slightly shorter than yours, and is perhaps a reasonably rigorous proof of the statement.  
> **username_2**: @Giskard I said 'show' exactly to not say 'prove'. My answer + proofs would be impossible. And I prefere a conceptual answer(and it is rigorous, I have some lecture notes of analytic machanics that follow these lines)  that illustrates the key concepts to a person who doesn't know differential equations. In my opinion a proof as in your answer is too advanvced, implies advanced concept not explained (operators, norms, bounded operators ,complex exponential and Euler formula), and less useful to a beginner of the subject.  
> **username_2**: Anyway, the overall resut in this trhead is fine, our answers maybe are complementary.:-)  

---

### Answer by username_3 on 2024-03-10. Score: 3

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 $\max\_i \lambda\_i ||\mathbf{x\_0}||$.

> **Comments:**
> **username_3**: Well, that's a sunday morning I'm never getting back :/  
> **username_2**: 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.  
> **username_3**: 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 $\mathbf{z_0}=\mathbf{B^{-1}x_0}$.  

---

