Discrete system with only real eigenvalues oscillates anyway

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I am trying to produce random linear discrete systems with random matrices. I produce a random matrix (real entries 0 to 1) and then to produce stable systems I normalize the eigenvalues. Then I remove certain eigenvalues e.g. imaginary part above certain threshold) to produce low frequency systems. I came across the following (pathological?) system:
A = [1.92163913525676 0.237219436248567;...
-11.5006034164096 -1.93377947537634]
Who has only real eigenvalues:
>> [V,D] = eig(A)
V =
0.2462 -0.0809
-0.9692 0.9967
D =
0.9879 0
0 -1.0000
But simulated as:
nT = 1000;
X = nan(nT,n);
X(1,:) = rand(1,n);
for i = 2:size(X,1)
X(i,:) = A*X(i-1,:)';
end
Produces Oscillations Oscillations
What am I doing wrong? to my understanding having real, less than 1 eigenvalues implies that the system is only decaying? Is this a numerical effect?
matrices discrete-mathematics eigenvalues-eigenvectors
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up vote
0
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I am trying to produce random linear discrete systems with random matrices. I produce a random matrix (real entries 0 to 1) and then to produce stable systems I normalize the eigenvalues. Then I remove certain eigenvalues e.g. imaginary part above certain threshold) to produce low frequency systems. I came across the following (pathological?) system:
A = [1.92163913525676 0.237219436248567;...
-11.5006034164096 -1.93377947537634]
Who has only real eigenvalues:
>> [V,D] = eig(A)
V =
0.2462 -0.0809
-0.9692 0.9967
D =
0.9879 0
0 -1.0000
But simulated as:
nT = 1000;
X = nan(nT,n);
X(1,:) = rand(1,n);
for i = 2:size(X,1)
X(i,:) = A*X(i-1,:)';
end
Produces Oscillations Oscillations
What am I doing wrong? to my understanding having real, less than 1 eigenvalues implies that the system is only decaying? Is this a numerical effect?
matrices discrete-mathematics eigenvalues-eigenvectors
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to produce random linear discrete systems with random matrices. I produce a random matrix (real entries 0 to 1) and then to produce stable systems I normalize the eigenvalues. Then I remove certain eigenvalues e.g. imaginary part above certain threshold) to produce low frequency systems. I came across the following (pathological?) system:
A = [1.92163913525676 0.237219436248567;...
-11.5006034164096 -1.93377947537634]
Who has only real eigenvalues:
>> [V,D] = eig(A)
V =
0.2462 -0.0809
-0.9692 0.9967
D =
0.9879 0
0 -1.0000
But simulated as:
nT = 1000;
X = nan(nT,n);
X(1,:) = rand(1,n);
for i = 2:size(X,1)
X(i,:) = A*X(i-1,:)';
end
Produces Oscillations Oscillations
What am I doing wrong? to my understanding having real, less than 1 eigenvalues implies that the system is only decaying? Is this a numerical effect?
matrices discrete-mathematics eigenvalues-eigenvectors
I am trying to produce random linear discrete systems with random matrices. I produce a random matrix (real entries 0 to 1) and then to produce stable systems I normalize the eigenvalues. Then I remove certain eigenvalues e.g. imaginary part above certain threshold) to produce low frequency systems. I came across the following (pathological?) system:
A = [1.92163913525676 0.237219436248567;...
-11.5006034164096 -1.93377947537634]
Who has only real eigenvalues:
>> [V,D] = eig(A)
V =
0.2462 -0.0809
-0.9692 0.9967
D =
0.9879 0
0 -1.0000
But simulated as:
nT = 1000;
X = nan(nT,n);
X(1,:) = rand(1,n);
for i = 2:size(X,1)
X(i,:) = A*X(i-1,:)';
end
Produces Oscillations Oscillations
What am I doing wrong? to my understanding having real, less than 1 eigenvalues implies that the system is only decaying? Is this a numerical effect?
matrices discrete-mathematics eigenvalues-eigenvectors
asked Aug 9 at 15:21
pepgma
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31
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1 Answer
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The dynamical systems you are considering are discrete-time systems.
The condition for a dynamical system to have only real eigenvalues in order for it to have no oscillating modes applies to continuous-time systems.
For discrete-time dynamical systems it translates to "only positive real eigenvalues".
That particular dynamical system in your question has the highest oscillating behaviour possible because one of its eigenvalues is $-1$: that is, signal vectors
have their component along the eigenvector corresponding to the eigenvalue $-1$, constant in absolute value, but continuously switched in sign at every iteration.
EDIT:
The language of the eigenvectors is a way to simplify the complexity of a linear problem by reducing its dimension. Chosing a basis of eigenvectors an $n$-dimensional linear problem is split into $n$ independent $1$-dimensional problem. Now in your case you would have $2$ independent $1$-dimensional dynamical systems, one of which as the (scalar) equation $xi_1(t+1)=-xi_1(t)$ whose output, you see, is indefinitely "flipping" (the other system has equation $xi_2(t+1) = 0.9879xi_2(t)$ that is damping). So it is the negativity of the real eigenvalue that gives the oscillating behaviour.
There are lots of references: start with this from MIT where the picture at page $19-1$ compares the two worlds (continuous vs discrete).
Great. Thanks. I did know about the difference from discrete to continuous systems for stability but assumed the oscillatory properties would be the same. I am trying to imagine what is the reason for this. In the continuous case it is easy as the solutions can easily be verified. I kind of understand from your comment that the signal vectors are determined by the eigenvectors but what produces the flip? I see that A has negative entries which implies a potential flip. Do you know a good reference for the behavior of these systems? Most of the literature online is for continuous systems.
â pepgma
Aug 10 at 12:05
edited to add the reply to your comment.
â trying
Aug 10 at 15:08
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The dynamical systems you are considering are discrete-time systems.
The condition for a dynamical system to have only real eigenvalues in order for it to have no oscillating modes applies to continuous-time systems.
For discrete-time dynamical systems it translates to "only positive real eigenvalues".
That particular dynamical system in your question has the highest oscillating behaviour possible because one of its eigenvalues is $-1$: that is, signal vectors
have their component along the eigenvector corresponding to the eigenvalue $-1$, constant in absolute value, but continuously switched in sign at every iteration.
EDIT:
The language of the eigenvectors is a way to simplify the complexity of a linear problem by reducing its dimension. Chosing a basis of eigenvectors an $n$-dimensional linear problem is split into $n$ independent $1$-dimensional problem. Now in your case you would have $2$ independent $1$-dimensional dynamical systems, one of which as the (scalar) equation $xi_1(t+1)=-xi_1(t)$ whose output, you see, is indefinitely "flipping" (the other system has equation $xi_2(t+1) = 0.9879xi_2(t)$ that is damping). So it is the negativity of the real eigenvalue that gives the oscillating behaviour.
There are lots of references: start with this from MIT where the picture at page $19-1$ compares the two worlds (continuous vs discrete).
Great. Thanks. I did know about the difference from discrete to continuous systems for stability but assumed the oscillatory properties would be the same. I am trying to imagine what is the reason for this. In the continuous case it is easy as the solutions can easily be verified. I kind of understand from your comment that the signal vectors are determined by the eigenvectors but what produces the flip? I see that A has negative entries which implies a potential flip. Do you know a good reference for the behavior of these systems? Most of the literature online is for continuous systems.
â pepgma
Aug 10 at 12:05
edited to add the reply to your comment.
â trying
Aug 10 at 15:08
add a comment |Â
up vote
0
down vote
accepted
The dynamical systems you are considering are discrete-time systems.
The condition for a dynamical system to have only real eigenvalues in order for it to have no oscillating modes applies to continuous-time systems.
For discrete-time dynamical systems it translates to "only positive real eigenvalues".
That particular dynamical system in your question has the highest oscillating behaviour possible because one of its eigenvalues is $-1$: that is, signal vectors
have their component along the eigenvector corresponding to the eigenvalue $-1$, constant in absolute value, but continuously switched in sign at every iteration.
EDIT:
The language of the eigenvectors is a way to simplify the complexity of a linear problem by reducing its dimension. Chosing a basis of eigenvectors an $n$-dimensional linear problem is split into $n$ independent $1$-dimensional problem. Now in your case you would have $2$ independent $1$-dimensional dynamical systems, one of which as the (scalar) equation $xi_1(t+1)=-xi_1(t)$ whose output, you see, is indefinitely "flipping" (the other system has equation $xi_2(t+1) = 0.9879xi_2(t)$ that is damping). So it is the negativity of the real eigenvalue that gives the oscillating behaviour.
There are lots of references: start with this from MIT where the picture at page $19-1$ compares the two worlds (continuous vs discrete).
Great. Thanks. I did know about the difference from discrete to continuous systems for stability but assumed the oscillatory properties would be the same. I am trying to imagine what is the reason for this. In the continuous case it is easy as the solutions can easily be verified. I kind of understand from your comment that the signal vectors are determined by the eigenvectors but what produces the flip? I see that A has negative entries which implies a potential flip. Do you know a good reference for the behavior of these systems? Most of the literature online is for continuous systems.
â pepgma
Aug 10 at 12:05
edited to add the reply to your comment.
â trying
Aug 10 at 15:08
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The dynamical systems you are considering are discrete-time systems.
The condition for a dynamical system to have only real eigenvalues in order for it to have no oscillating modes applies to continuous-time systems.
For discrete-time dynamical systems it translates to "only positive real eigenvalues".
That particular dynamical system in your question has the highest oscillating behaviour possible because one of its eigenvalues is $-1$: that is, signal vectors
have their component along the eigenvector corresponding to the eigenvalue $-1$, constant in absolute value, but continuously switched in sign at every iteration.
EDIT:
The language of the eigenvectors is a way to simplify the complexity of a linear problem by reducing its dimension. Chosing a basis of eigenvectors an $n$-dimensional linear problem is split into $n$ independent $1$-dimensional problem. Now in your case you would have $2$ independent $1$-dimensional dynamical systems, one of which as the (scalar) equation $xi_1(t+1)=-xi_1(t)$ whose output, you see, is indefinitely "flipping" (the other system has equation $xi_2(t+1) = 0.9879xi_2(t)$ that is damping). So it is the negativity of the real eigenvalue that gives the oscillating behaviour.
There are lots of references: start with this from MIT where the picture at page $19-1$ compares the two worlds (continuous vs discrete).
The dynamical systems you are considering are discrete-time systems.
The condition for a dynamical system to have only real eigenvalues in order for it to have no oscillating modes applies to continuous-time systems.
For discrete-time dynamical systems it translates to "only positive real eigenvalues".
That particular dynamical system in your question has the highest oscillating behaviour possible because one of its eigenvalues is $-1$: that is, signal vectors
have their component along the eigenvector corresponding to the eigenvalue $-1$, constant in absolute value, but continuously switched in sign at every iteration.
EDIT:
The language of the eigenvectors is a way to simplify the complexity of a linear problem by reducing its dimension. Chosing a basis of eigenvectors an $n$-dimensional linear problem is split into $n$ independent $1$-dimensional problem. Now in your case you would have $2$ independent $1$-dimensional dynamical systems, one of which as the (scalar) equation $xi_1(t+1)=-xi_1(t)$ whose output, you see, is indefinitely "flipping" (the other system has equation $xi_2(t+1) = 0.9879xi_2(t)$ that is damping). So it is the negativity of the real eigenvalue that gives the oscillating behaviour.
There are lots of references: start with this from MIT where the picture at page $19-1$ compares the two worlds (continuous vs discrete).
edited Aug 10 at 15:20
answered Aug 9 at 16:55
trying
4,1661722
4,1661722
Great. Thanks. I did know about the difference from discrete to continuous systems for stability but assumed the oscillatory properties would be the same. I am trying to imagine what is the reason for this. In the continuous case it is easy as the solutions can easily be verified. I kind of understand from your comment that the signal vectors are determined by the eigenvectors but what produces the flip? I see that A has negative entries which implies a potential flip. Do you know a good reference for the behavior of these systems? Most of the literature online is for continuous systems.
â pepgma
Aug 10 at 12:05
edited to add the reply to your comment.
â trying
Aug 10 at 15:08
add a comment |Â
Great. Thanks. I did know about the difference from discrete to continuous systems for stability but assumed the oscillatory properties would be the same. I am trying to imagine what is the reason for this. In the continuous case it is easy as the solutions can easily be verified. I kind of understand from your comment that the signal vectors are determined by the eigenvectors but what produces the flip? I see that A has negative entries which implies a potential flip. Do you know a good reference for the behavior of these systems? Most of the literature online is for continuous systems.
â pepgma
Aug 10 at 12:05
edited to add the reply to your comment.
â trying
Aug 10 at 15:08
Great. Thanks. I did know about the difference from discrete to continuous systems for stability but assumed the oscillatory properties would be the same. I am trying to imagine what is the reason for this. In the continuous case it is easy as the solutions can easily be verified. I kind of understand from your comment that the signal vectors are determined by the eigenvectors but what produces the flip? I see that A has negative entries which implies a potential flip. Do you know a good reference for the behavior of these systems? Most of the literature online is for continuous systems.
â pepgma
Aug 10 at 12:05
Great. Thanks. I did know about the difference from discrete to continuous systems for stability but assumed the oscillatory properties would be the same. I am trying to imagine what is the reason for this. In the continuous case it is easy as the solutions can easily be verified. I kind of understand from your comment that the signal vectors are determined by the eigenvectors but what produces the flip? I see that A has negative entries which implies a potential flip. Do you know a good reference for the behavior of these systems? Most of the literature online is for continuous systems.
â pepgma
Aug 10 at 12:05
edited to add the reply to your comment.
â trying
Aug 10 at 15:08
edited to add the reply to your comment.
â trying
Aug 10 at 15:08
add a comment |Â
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