Ttyjukl

Why does Arg'[1. + I] return -0.5?

Has a Nobel Peace laureate ever been accused of war crimes?

Why do games have consumables?

English or Hindi translation of Vyasa Smriti

Is there really no use for MD5 anymore?

Check if a string is entirely made of the same substring

Why do real positive eigenvalues result in an unstable system? What about eigenvalues between 0 and 1? or 1?

Retract an already submitted recommendation letter (written for an undergrad student)

How important is it that $TERM is correct?

Prove that the countable union of countable sets is also countable

Crossed out red box fitting tightly around image

view price of already bought and installed apps on play sotr

What is purpose of DB Browser(dbbrowser.aspx) under admin tool?

What does a straight horizontal line above a few notes, after a changed tempo mean?

A Paper Record is What I Hamper

Sharepoint Designer Discontinuation - software to modify existing workflows

Ramp colors through selected symbols in QGIS?

Combinatorics problem, right solution?

What is the best way to deal with NPC-NPC combat?

std::unique_ptr of base class holding reference of derived class does not show warning in gcc compiler while naked pointer shows it. Why?

Will I lose my paid in full property

Unable to completely uninstall Zoom meeting app

What is this word supposed to be?

I preordered a game on my Xbox while on the home screen of my friend's account. Which of us owns the game?

Why do real positive eigenvalues result in an unstable system? What about eigenvalues between 0 and 1? or 1?

2

$begingroup$

I'm confused why, if all eigenvalues of a linear system are real and positive, this entails an unstable system. For example, if eigenvalues are between 0 and 1, surely this means the system is gradually shrinking? And even more so, doesn't an eigenvalue of 1 mean the system is "staying put"? Why is a negative eigenvalue instead related to stability?

I'm confused because in the case of eigenvalues of a Markov matrix, it seemed an eigenvalue of 1 meant stability, and 0 < λ < 1 meant it was shrinking. But in both those cases the eigenvalue is positive.

linear-algebra eigenvalues-eigenvectors dynamical-systems stability-theory

share|cite|improve this question

edited 18 mins ago

northerlywind

asked 1 hour ago

northerlywindnortherlywind

355

New contributor

northerlywind is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

2

$begingroup$

I'm confused why, if all eigenvalues of a linear system are real and positive, this entails an unstable system. For example, if eigenvalues are between 0 and 1, surely this means the system is gradually shrinking? And even more so, doesn't an eigenvalue of 1 mean the system is "staying put"? Why is a negative eigenvalue instead related to stability?

I'm confused because in the case of eigenvalues of a Markov matrix, it seemed an eigenvalue of 1 meant stability, and 0 < λ < 1 meant it was shrinking. But in both those cases the eigenvalue is positive.

linear-algebra eigenvalues-eigenvectors dynamical-systems stability-theory

share|cite|improve this question

edited 18 mins ago

northerlywind

asked 1 hour ago

northerlywindnortherlywind

355

New contributor

northerlywind is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

I'm confused why, if all eigenvalues of a linear system are real and positive, this entails an unstable system. For example, if eigenvalues are between 0 and 1, surely this means the system is gradually shrinking? And even more so, doesn't an eigenvalue of 1 mean the system is "staying put"? Why is a negative eigenvalue instead related to stability?

I'm confused because in the case of eigenvalues of a Markov matrix, it seemed an eigenvalue of 1 meant stability, and 0 < λ < 1 meant it was shrinking. But in both those cases the eigenvalue is positive.

linear-algebra eigenvalues-eigenvectors dynamical-systems stability-theory

share|cite|improve this question

edited 18 mins ago

northerlywind

asked 1 hour ago

northerlywindnortherlywind

355

New contributor

northerlywind is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 18 mins ago

northerlywind

asked 1 hour ago

northerlywindnortherlywind

355

New contributor

northerlywind is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 18 mins ago

northerlywind

asked 1 hour ago

northerlywindnortherlywind

355

New contributor

northerlywind is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 1 hour ago

northerlywindnortherlywind

355

New contributor

northerlywind is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 1 hour ago

northerlywindnortherlywind

355

2 Answers
2

active

oldest

votes

4

$begingroup$

The key difference lies in the fact that the maps defined by a Markov matrix are discrete-time whereas the dynamical systems you refer to are continuous-time systems. In either case, a "location" on the system is stable or unstable depending on whether or not a perturbation off of the location grows or shrinks, which is measured differently in each system.

For a continuously differentiable dynamical system, a local perturbation around the evolution of the system near a fixed point is described by an exponential $e^{lambda t}$ since (taking liberal approximations) $x(t+Delta t) approx x(t) + lambda x(t) = (1 + lambda)x(t)$; the function grows if $lambda > 0$ and shrinks if the opposite.

For a discrete-time system like a Markov chain, local perturbations grow depending on whether or not the "next" step of the system is larger than the previous one, a.k.a. if $x_{n+1} = lambda x_{n} > x_n$. Hence, $lambda > 1$ indicates an unstable fixed point.

share|cite|improve this answer

edited 47 mins ago

answered 56 mins ago

aghostinthefiguresaghostinthefigures

1,3441217

$endgroup$

add a comment | 

2

$begingroup$

With discrete-time Markov processes, you’re examining the behavior of successively larger powers of a fixed matrix, which depends on successively larger powers of its eigenvalues. If there are any with modulus greater than one, those contributions to the matrix power will grow without bound, any with modulus less than one will tend toward zero, and any with modulus exactly one will neither grow nor shrink.

If the “system” you’re asking about is a system of differential equations, then instead of the behavior of successive powers of a matrix $A$ we’re interested in its exponential $e^{tA}$, which in turn depends on exponentials $e^{lambda t}$of its eigenvalues. Here, it’s eigenvalues with a positive real part whose contributions grow without bound, while the contributions of ones with a negative real part tend to zero.

share|cite|improve this answer

answered 52 mins ago

amdamd

32.3k21053

$endgroup$

add a comment | 

Your Answer

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

northerlywind is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3202164%2fwhy-do-real-positive-eigenvalues-result-in-an-unstable-system-what-about-eigenv%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

4

$begingroup$

The key difference lies in the fact that the maps defined by a Markov matrix are discrete-time whereas the dynamical systems you refer to are continuous-time systems. In either case, a "location" on the system is stable or unstable depending on whether or not a perturbation off of the location grows or shrinks, which is measured differently in each system.

For a continuously differentiable dynamical system, a local perturbation around the evolution of the system near a fixed point is described by an exponential $e^{lambda t}$ since (taking liberal approximations) $x(t+Delta t) approx x(t) + lambda x(t) = (1 + lambda)x(t)$; the function grows if $lambda > 0$ and shrinks if the opposite.

For a discrete-time system like a Markov chain, local perturbations grow depending on whether or not the "next" step of the system is larger than the previous one, a.k.a. if $x_{n+1} = lambda x_{n} > x_n$. Hence, $lambda > 1$ indicates an unstable fixed point.

share|cite|improve this answer

edited 47 mins ago

answered 56 mins ago

aghostinthefiguresaghostinthefigures

1,3441217

$endgroup$

add a comment | 

4

$begingroup$

The key difference lies in the fact that the maps defined by a Markov matrix are discrete-time whereas the dynamical systems you refer to are continuous-time systems. In either case, a "location" on the system is stable or unstable depending on whether or not a perturbation off of the location grows or shrinks, which is measured differently in each system.

For a continuously differentiable dynamical system, a local perturbation around the evolution of the system near a fixed point is described by an exponential $e^{lambda t}$ since (taking liberal approximations) $x(t+Delta t) approx x(t) + lambda x(t) = (1 + lambda)x(t)$; the function grows if $lambda > 0$ and shrinks if the opposite.

For a discrete-time system like a Markov chain, local perturbations grow depending on whether or not the "next" step of the system is larger than the previous one, a.k.a. if $x_{n+1} = lambda x_{n} > x_n$. Hence, $lambda > 1$ indicates an unstable fixed point.

share|cite|improve this answer

edited 47 mins ago

answered 56 mins ago

aghostinthefiguresaghostinthefigures

1,3441217

$endgroup$

add a comment | 

$begingroup$

The key difference lies in the fact that the maps defined by a Markov matrix are discrete-time whereas the dynamical systems you refer to are continuous-time systems. In either case, a "location" on the system is stable or unstable depending on whether or not a perturbation off of the location grows or shrinks, which is measured differently in each system.

For a continuously differentiable dynamical system, a local perturbation around the evolution of the system near a fixed point is described by an exponential $e^{lambda t}$ since (taking liberal approximations) $x(t+Delta t) approx x(t) + lambda x(t) = (1 + lambda)x(t)$; the function grows if $lambda > 0$ and shrinks if the opposite.

For a discrete-time system like a Markov chain, local perturbations grow depending on whether or not the "next" step of the system is larger than the previous one, a.k.a. if $x_{n+1} = lambda x_{n} > x_n$. Hence, $lambda > 1$ indicates an unstable fixed point.

share|cite|improve this answer

edited 47 mins ago

answered 56 mins ago

aghostinthefiguresaghostinthefigures

1,3441217

$endgroup$

share|cite|improve this answer

edited 47 mins ago

answered 56 mins ago

aghostinthefiguresaghostinthefigures

1,3441217

answered 56 mins ago

aghostinthefiguresaghostinthefigures

1,3441217

answered 56 mins ago

aghostinthefiguresaghostinthefigures

1,3441217

2

$begingroup$

With discrete-time Markov processes, you’re examining the behavior of successively larger powers of a fixed matrix, which depends on successively larger powers of its eigenvalues. If there are any with modulus greater than one, those contributions to the matrix power will grow without bound, any with modulus less than one will tend toward zero, and any with modulus exactly one will neither grow nor shrink.

If the “system” you’re asking about is a system of differential equations, then instead of the behavior of successive powers of a matrix $A$ we’re interested in its exponential $e^{tA}$, which in turn depends on exponentials $e^{lambda t}$of its eigenvalues. Here, it’s eigenvalues with a positive real part whose contributions grow without bound, while the contributions of ones with a negative real part tend to zero.

share|cite|improve this answer

answered 52 mins ago

amdamd

32.3k21053

$endgroup$

add a comment | 

2

$begingroup$

With discrete-time Markov processes, you’re examining the behavior of successively larger powers of a fixed matrix, which depends on successively larger powers of its eigenvalues. If there are any with modulus greater than one, those contributions to the matrix power will grow without bound, any with modulus less than one will tend toward zero, and any with modulus exactly one will neither grow nor shrink.

If the “system” you’re asking about is a system of differential equations, then instead of the behavior of successive powers of a matrix $A$ we’re interested in its exponential $e^{tA}$, which in turn depends on exponentials $e^{lambda t}$of its eigenvalues. Here, it’s eigenvalues with a positive real part whose contributions grow without bound, while the contributions of ones with a negative real part tend to zero.

share|cite|improve this answer

answered 52 mins ago

amdamd

32.3k21053

$endgroup$

add a comment | 

$begingroup$

With discrete-time Markov processes, you’re examining the behavior of successively larger powers of a fixed matrix, which depends on successively larger powers of its eigenvalues. If there are any with modulus greater than one, those contributions to the matrix power will grow without bound, any with modulus less than one will tend toward zero, and any with modulus exactly one will neither grow nor shrink.

If the “system” you’re asking about is a system of differential equations, then instead of the behavior of successive powers of a matrix $A$ we’re interested in its exponential $e^{tA}$, which in turn depends on exponentials $e^{lambda t}$of its eigenvalues. Here, it’s eigenvalues with a positive real part whose contributions grow without bound, while the contributions of ones with a negative real part tend to zero.

share|cite|improve this answer

answered 52 mins ago

amdamd

32.3k21053

$endgroup$

share|cite|improve this answer

answered 52 mins ago

amdamd

32.3k21053

answered 52 mins ago

amdamd

32.3k21053

answered 52 mins ago

amdamd

32.3k21053