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?







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    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?







    share|cite|improve this question






















      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?







      share|cite|improve this question












      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?









      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Aug 9 at 15:21









      pepgma

      31




      31




















          1 Answer
          1






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          oldest

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          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).






          share|cite|improve this answer






















          • 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.
            – pepgma
            Aug 10 at 12:05










          • edited to add the reply to your comment.
            – trying
            Aug 10 at 15:08










          Your Answer




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          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).






          share|cite|improve this answer






















          • 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.
            – pepgma
            Aug 10 at 12:05










          • edited to add the reply to your comment.
            – trying
            Aug 10 at 15:08














          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).






          share|cite|improve this answer






















          • 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.
            – pepgma
            Aug 10 at 12:05










          • edited to add the reply to your comment.
            – trying
            Aug 10 at 15:08












          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).






          share|cite|improve this answer














          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).







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          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.
            – pepgma
            Aug 10 at 12:05










          • edited to add the reply to your comment.
            – 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.
            – pepgma
            Aug 10 at 12:05










          • edited to add the reply to your comment.
            – 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.
          – 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.
          – pepgma
          Aug 10 at 12:05












          edited to add the reply to your comment.
          – trying
          Aug 10 at 15:08




          edited to add the reply to your comment.
          – trying
          Aug 10 at 15:08












           

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