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

2

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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

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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 17 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.
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$begingroup$

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 17 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.
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edited 17 mins ago

northerlywind

asked 1 hour ago

northerlywindnortherlywind

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edited 17 mins ago

northerlywind

asked 1 hour ago

northerlywindnortherlywind

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asked 1 hour ago

northerlywindnortherlywind

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northerlywind is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 1 hour ago

northerlywindnortherlywind

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2 Answers
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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.

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edited 46 mins ago

answered 55 mins ago

aghostinthefiguresaghostinthefigures

1,3441217

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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.

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answered 51 mins ago

amdamd

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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 46 mins ago

answered 55 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 46 mins ago

answered 55 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 46 mins ago

answered 55 mins ago

aghostinthefiguresaghostinthefigures

1,3441217

$endgroup$

share|cite|improve this answer

edited 46 mins ago

answered 55 mins ago

aghostinthefiguresaghostinthefigures

1,3441217

answered 55 mins ago

aghostinthefiguresaghostinthefigures

1,3441217

answered 55 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 51 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 51 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 51 mins ago

amdamd

32.3k21053

$endgroup$

share|cite|improve this answer

answered 51 mins ago

amdamd

32.3k21053

answered 51 mins ago

amdamd

32.3k21053

answered 51 mins ago

amdamd

32.3k21053