Why do real positive eigenvalues result in an unstable system? What about eigenvalues between 0 and 1? or 1?Classifying local behavior of fixed points using eigenvalues from linear stability analysis of 3D systemUse of Routh-Hurwitz if you have the eigenvalues?Dynamics of a three dimensional systemWhat to call the eigenvalues that don't correspond to a conserved quantityDetermining the stability of a system with a zero eigenvalueStability of zero solution of ODE system when largest real part of eigenvalues is $0$.Stable limit cycleFixed point stability of non-linear ode systemNumerically determine eigenvalues of real non-symmetric matrix known to have positive eigenvaluesHow to rotate a coordinate system to find the unstable manifold.
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Why do real positive eigenvalues result in an unstable system? What about eigenvalues between 0 and 1? or 1?
Classifying local behavior of fixed points using eigenvalues from linear stability analysis of 3D systemUse of Routh-Hurwitz if you have the eigenvalues?Dynamics of a three dimensional systemWhat to call the eigenvalues that don't correspond to a conserved quantityDetermining the stability of a system with a zero eigenvalueStability of zero solution of ODE system when largest real part of eigenvalues is $0$.Stable limit cycleFixed point stability of non-linear ode systemNumerically determine eigenvalues of real non-symmetric matrix known to have positive eigenvaluesHow to rotate a coordinate system to find the unstable manifold.
$begingroup$
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
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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
asked 8 hours ago
questionmarkquestionmark
456
New contributor
questionmark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
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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
asked 8 hours ago
questionmarkquestionmark
456
New contributor
questionmark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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
linear-algebra eigenvalues-eigenvectors dynamical-systems stability-theory
asked 8 hours ago
questionmarkquestionmark
456
New contributor
questionmark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
questionmarkquestionmark
456
New contributor
questionmark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 6 hours ago
questionmark
asked 8 hours ago
questionmarkquestionmark
456
New contributor
questionmark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 8 hours ago
questionmarkquestionmark
456
asked 8 hours ago
questionmarkquestionmark
456
456
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questionmark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
questionmark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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.
answered 7 hours ago
aghostinthefiguresaghostinthefigures
1,3941217
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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.
answered 7 hours ago
amdamd
32.3k21054
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2 Answers
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2 Answers
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$begingroup$
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.
answered 7 hours ago
aghostinthefiguresaghostinthefigures
1,3941217
$endgroup$
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.
answered 7 hours ago
aghostinthefiguresaghostinthefigures
1,3941217
$endgroup$
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.
answered 7 hours ago
aghostinthefiguresaghostinthefigures
1,3941217
$endgroup$
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.
answered 7 hours ago
aghostinthefiguresaghostinthefigures
1,3941217
edited 7 hours ago
answered 7 hours ago
aghostinthefiguresaghostinthefigures
1,3941217
answered 7 hours ago
aghostinthefiguresaghostinthefigures
1,3941217
answered 7 hours ago
aghostinthefiguresaghostinthefigures
1,3941217
1,3941217
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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.
answered 7 hours ago
amdamd
32.3k21054
$endgroup$
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.
answered 7 hours ago
amdamd
32.3k21054
$endgroup$
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.
answered 7 hours ago
amdamd
32.3k21054
$endgroup$
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.
answered 7 hours ago
amdamd
32.3k21054
answered 7 hours ago
amdamd
32.3k21054
answered 7 hours ago
amdamd
32.3k21054
answered 7 hours ago
amdamd
32.3k21054
32.3k21054
add a comment |
add a comment |
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