How to interpret complex eigenvectors of the Jacobian matrix of a (linear) dynamical system?












5












$begingroup$


Consider a linear ODE system of the following form: $$ frac {dx} {dt} = Ax $$
In case $A$ has real eigenvectors, I can interpret them as the directions in which the system will move, if the initial value is already a point on the eigenvector. The eigenvalues corresponding to each eigenvector would tell me whether the system approaches the fixed point (negative eigenvalue) or moves away from it (pos. eigenvalue). The plot below shows an example vectorfield for the case
$ A = left( begin{array}{cc} 2 & 3\ 1 & 4 end{array} right) $, in which the Jacobian $A$ has eigenvalues $1$ and $5$ and eigenvectors $left( begin{array}{c} -3 \ 1 end{array} right)$ (blue) and $left( begin{array}{c} 1 \ 1 end{array} right)$ (green).



Plot of real eigenvectors:
Plot real eigenvectors



My question is how to interpret the eigenvectors in case they are complex? If the Jacobian has complex eigenvectors, this means the eigenvalues are also complex, so the system will oscillate. Do the complex eigenvectors tell me something about the rotational axis of these oscillations?



I have plotted the vectorfield of several linear oscillatory systems and to me it seems like the real part of eigenvector points in the direction of the ellipse in which the system oscillates. Here is the vectorfield plot for the system
$$A = left( begin{array}{cc} 0 & -3\ 1 & -2 end{array} right)$$
( eigenvalues are $-1 pm sqrt{-2}$). The real part of the eigenvector is shown in blue.



Plot of complex eigenvectors:
Plot complex eigenvectors



If this is correct, how can I interpret the imaginary parts of the eigenvectors? Could you maybe point me out an article or a book in which this is explained?










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$endgroup$












  • $begingroup$
    Please define "the direction of the ellipse in which the system oscillates."
    $endgroup$
    – Did
    Jun 1 '15 at 18:30
















5












$begingroup$


Consider a linear ODE system of the following form: $$ frac {dx} {dt} = Ax $$
In case $A$ has real eigenvectors, I can interpret them as the directions in which the system will move, if the initial value is already a point on the eigenvector. The eigenvalues corresponding to each eigenvector would tell me whether the system approaches the fixed point (negative eigenvalue) or moves away from it (pos. eigenvalue). The plot below shows an example vectorfield for the case
$ A = left( begin{array}{cc} 2 & 3\ 1 & 4 end{array} right) $, in which the Jacobian $A$ has eigenvalues $1$ and $5$ and eigenvectors $left( begin{array}{c} -3 \ 1 end{array} right)$ (blue) and $left( begin{array}{c} 1 \ 1 end{array} right)$ (green).



Plot of real eigenvectors:
Plot real eigenvectors



My question is how to interpret the eigenvectors in case they are complex? If the Jacobian has complex eigenvectors, this means the eigenvalues are also complex, so the system will oscillate. Do the complex eigenvectors tell me something about the rotational axis of these oscillations?



I have plotted the vectorfield of several linear oscillatory systems and to me it seems like the real part of eigenvector points in the direction of the ellipse in which the system oscillates. Here is the vectorfield plot for the system
$$A = left( begin{array}{cc} 0 & -3\ 1 & -2 end{array} right)$$
( eigenvalues are $-1 pm sqrt{-2}$). The real part of the eigenvector is shown in blue.



Plot of complex eigenvectors:
Plot complex eigenvectors



If this is correct, how can I interpret the imaginary parts of the eigenvectors? Could you maybe point me out an article or a book in which this is explained?










share|cite|improve this question











$endgroup$












  • $begingroup$
    Please define "the direction of the ellipse in which the system oscillates."
    $endgroup$
    – Did
    Jun 1 '15 at 18:30














5












5








5


4



$begingroup$


Consider a linear ODE system of the following form: $$ frac {dx} {dt} = Ax $$
In case $A$ has real eigenvectors, I can interpret them as the directions in which the system will move, if the initial value is already a point on the eigenvector. The eigenvalues corresponding to each eigenvector would tell me whether the system approaches the fixed point (negative eigenvalue) or moves away from it (pos. eigenvalue). The plot below shows an example vectorfield for the case
$ A = left( begin{array}{cc} 2 & 3\ 1 & 4 end{array} right) $, in which the Jacobian $A$ has eigenvalues $1$ and $5$ and eigenvectors $left( begin{array}{c} -3 \ 1 end{array} right)$ (blue) and $left( begin{array}{c} 1 \ 1 end{array} right)$ (green).



Plot of real eigenvectors:
Plot real eigenvectors



My question is how to interpret the eigenvectors in case they are complex? If the Jacobian has complex eigenvectors, this means the eigenvalues are also complex, so the system will oscillate. Do the complex eigenvectors tell me something about the rotational axis of these oscillations?



I have plotted the vectorfield of several linear oscillatory systems and to me it seems like the real part of eigenvector points in the direction of the ellipse in which the system oscillates. Here is the vectorfield plot for the system
$$A = left( begin{array}{cc} 0 & -3\ 1 & -2 end{array} right)$$
( eigenvalues are $-1 pm sqrt{-2}$). The real part of the eigenvector is shown in blue.



Plot of complex eigenvectors:
Plot complex eigenvectors



If this is correct, how can I interpret the imaginary parts of the eigenvectors? Could you maybe point me out an article or a book in which this is explained?










share|cite|improve this question











$endgroup$




Consider a linear ODE system of the following form: $$ frac {dx} {dt} = Ax $$
In case $A$ has real eigenvectors, I can interpret them as the directions in which the system will move, if the initial value is already a point on the eigenvector. The eigenvalues corresponding to each eigenvector would tell me whether the system approaches the fixed point (negative eigenvalue) or moves away from it (pos. eigenvalue). The plot below shows an example vectorfield for the case
$ A = left( begin{array}{cc} 2 & 3\ 1 & 4 end{array} right) $, in which the Jacobian $A$ has eigenvalues $1$ and $5$ and eigenvectors $left( begin{array}{c} -3 \ 1 end{array} right)$ (blue) and $left( begin{array}{c} 1 \ 1 end{array} right)$ (green).



Plot of real eigenvectors:
Plot real eigenvectors



My question is how to interpret the eigenvectors in case they are complex? If the Jacobian has complex eigenvectors, this means the eigenvalues are also complex, so the system will oscillate. Do the complex eigenvectors tell me something about the rotational axis of these oscillations?



I have plotted the vectorfield of several linear oscillatory systems and to me it seems like the real part of eigenvector points in the direction of the ellipse in which the system oscillates. Here is the vectorfield plot for the system
$$A = left( begin{array}{cc} 0 & -3\ 1 & -2 end{array} right)$$
( eigenvalues are $-1 pm sqrt{-2}$). The real part of the eigenvector is shown in blue.



Plot of complex eigenvectors:
Plot complex eigenvectors



If this is correct, how can I interpret the imaginary parts of the eigenvectors? Could you maybe point me out an article or a book in which this is explained?







ordinary-differential-equations dynamical-systems stability-in-odes






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edited Jul 29 '15 at 3:49







user147263

















asked Jun 1 '15 at 17:51









JannisJannis

2912




2912












  • $begingroup$
    Please define "the direction of the ellipse in which the system oscillates."
    $endgroup$
    – Did
    Jun 1 '15 at 18:30


















  • $begingroup$
    Please define "the direction of the ellipse in which the system oscillates."
    $endgroup$
    – Did
    Jun 1 '15 at 18:30
















$begingroup$
Please define "the direction of the ellipse in which the system oscillates."
$endgroup$
– Did
Jun 1 '15 at 18:30




$begingroup$
Please define "the direction of the ellipse in which the system oscillates."
$endgroup$
– Did
Jun 1 '15 at 18:30










3 Answers
3






active

oldest

votes


















4












$begingroup$

Intuitively, when the system has real eigenvalues and real eigenvectors, the eigenvectors are the directions in which the flow $vec{x}$ does not change direction.



When the eigenvalues have imaginary part, hence the eigenvectors contain imaginary part, there are no directions in which the flow $vec{x}$ does not change direction.



If an eigenvector is $vec{v}_x+ivec{v}_y$, with corresponding eigenvalue $L_x+iL_y$, then $$A(vec{v}_x+ivec{v}_y)=(L_x+iL_y)(vec{v}_x+ivec{v}_y)\
=(L_xvec{v}_x-L_yvec{v}_y)+i(L_xvec{v}_y+L_yvec{v}_x)$$



Separate the real and imaginary parts:
$$Avec{v}_x=L_xvec{v}_x-L_yvec{v}_y\
Avec{v}_y=L_xvec{v}_y+L_yvec{v}_x$$



So on the real plane, you can see $vec{v}_x, vec{v}_y$ are two axis that the flow rotate around. Intuitively, they are major and minor axis of the ellipse or circle when you have a center as fixed point, or spiral. This is not very rigorous argument, but may give you some insight.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you, this is what I wanted to know. In the meantime I also have found a similar post which also includes a nice illustration.
    $endgroup$
    – Jannis
    Jun 2 '15 at 12:04












  • $begingroup$
    @Jannis: Thanks for the reference. That's a nice picture.
    $endgroup$
    – KittyL
    Jun 2 '15 at 14:12










  • $begingroup$
    Nitpick: L_y appears to change into L_b. Please let me know if I missed something. On a more substantial note, I still can't quite visualize how rotation in the complex plane translates into oscillation of the original system.
    $endgroup$
    – Ben Ogorek
    Dec 31 '18 at 0:59








  • 1




    $begingroup$
    @BenOgorek: Fixed! Thanks. "Rotation" means the vector $(x,y)$ going along the ellipse following the arrows. (See the OP's last picture) "Oscillation" refers to the change of values of $x$, or the change of values of $y$. Consider the change of values of $x$ in that picture. Suppose we start at the point $(6,0)$, where $x=6$. Following the arrows, it could go to $(0,-6)$, where $x=0$. Continuing following the arrows, the value of $x$ changes back to $6$, and so on. Same for the $y$ values.
    $endgroup$
    – KittyL
    Dec 31 '18 at 12:31






  • 1




    $begingroup$
    @BenOgorek: To simplify the matter, let's consider $x''(t)+x(t)=0$. This differential equation has solution $c_1e^{it}+c_2e^{-it}$ which can be written as $(c_1+c_2)cos t+(c_1i-c_2i)sin t$. These of course contain many solutions in the complex plane. But we want to pick the real solutions. Hence we let $d_1=c_1+c_2$ and $d_2=i(c_1-c_2)$ and only use the real values of $d_i$. This gives us all real solutions of the original equation. For the system, the reason is similar.
    $endgroup$
    – KittyL
    Dec 31 '18 at 20:52



















1












$begingroup$

Usually complex eigenvalues correspond to circular motion (not dissimilarly to the way that rotation matrices have complex eigenvalues/vectors. If there is a good reason for why rotations should correspond to complex eigenvectors, I don't know it. The real part of the eigenvalue indicates whether the motion is circular (or elliptical) or if it is spiralling in or out. The imaginary part and the eigenvector give you the direction in which this spiralling happens (clockwise/anti-clockwise).






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    0












    $begingroup$

    I scetched you the different cases.



    enter image description here



    edit: if one eigenvalue is zero, then there exists a line consisting of fixed points.






    share|cite|improve this answer











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      3 Answers
      3






      active

      oldest

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      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      Intuitively, when the system has real eigenvalues and real eigenvectors, the eigenvectors are the directions in which the flow $vec{x}$ does not change direction.



      When the eigenvalues have imaginary part, hence the eigenvectors contain imaginary part, there are no directions in which the flow $vec{x}$ does not change direction.



      If an eigenvector is $vec{v}_x+ivec{v}_y$, with corresponding eigenvalue $L_x+iL_y$, then $$A(vec{v}_x+ivec{v}_y)=(L_x+iL_y)(vec{v}_x+ivec{v}_y)\
      =(L_xvec{v}_x-L_yvec{v}_y)+i(L_xvec{v}_y+L_yvec{v}_x)$$



      Separate the real and imaginary parts:
      $$Avec{v}_x=L_xvec{v}_x-L_yvec{v}_y\
      Avec{v}_y=L_xvec{v}_y+L_yvec{v}_x$$



      So on the real plane, you can see $vec{v}_x, vec{v}_y$ are two axis that the flow rotate around. Intuitively, they are major and minor axis of the ellipse or circle when you have a center as fixed point, or spiral. This is not very rigorous argument, but may give you some insight.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Thank you, this is what I wanted to know. In the meantime I also have found a similar post which also includes a nice illustration.
        $endgroup$
        – Jannis
        Jun 2 '15 at 12:04












      • $begingroup$
        @Jannis: Thanks for the reference. That's a nice picture.
        $endgroup$
        – KittyL
        Jun 2 '15 at 14:12










      • $begingroup$
        Nitpick: L_y appears to change into L_b. Please let me know if I missed something. On a more substantial note, I still can't quite visualize how rotation in the complex plane translates into oscillation of the original system.
        $endgroup$
        – Ben Ogorek
        Dec 31 '18 at 0:59








      • 1




        $begingroup$
        @BenOgorek: Fixed! Thanks. "Rotation" means the vector $(x,y)$ going along the ellipse following the arrows. (See the OP's last picture) "Oscillation" refers to the change of values of $x$, or the change of values of $y$. Consider the change of values of $x$ in that picture. Suppose we start at the point $(6,0)$, where $x=6$. Following the arrows, it could go to $(0,-6)$, where $x=0$. Continuing following the arrows, the value of $x$ changes back to $6$, and so on. Same for the $y$ values.
        $endgroup$
        – KittyL
        Dec 31 '18 at 12:31






      • 1




        $begingroup$
        @BenOgorek: To simplify the matter, let's consider $x''(t)+x(t)=0$. This differential equation has solution $c_1e^{it}+c_2e^{-it}$ which can be written as $(c_1+c_2)cos t+(c_1i-c_2i)sin t$. These of course contain many solutions in the complex plane. But we want to pick the real solutions. Hence we let $d_1=c_1+c_2$ and $d_2=i(c_1-c_2)$ and only use the real values of $d_i$. This gives us all real solutions of the original equation. For the system, the reason is similar.
        $endgroup$
        – KittyL
        Dec 31 '18 at 20:52
















      4












      $begingroup$

      Intuitively, when the system has real eigenvalues and real eigenvectors, the eigenvectors are the directions in which the flow $vec{x}$ does not change direction.



      When the eigenvalues have imaginary part, hence the eigenvectors contain imaginary part, there are no directions in which the flow $vec{x}$ does not change direction.



      If an eigenvector is $vec{v}_x+ivec{v}_y$, with corresponding eigenvalue $L_x+iL_y$, then $$A(vec{v}_x+ivec{v}_y)=(L_x+iL_y)(vec{v}_x+ivec{v}_y)\
      =(L_xvec{v}_x-L_yvec{v}_y)+i(L_xvec{v}_y+L_yvec{v}_x)$$



      Separate the real and imaginary parts:
      $$Avec{v}_x=L_xvec{v}_x-L_yvec{v}_y\
      Avec{v}_y=L_xvec{v}_y+L_yvec{v}_x$$



      So on the real plane, you can see $vec{v}_x, vec{v}_y$ are two axis that the flow rotate around. Intuitively, they are major and minor axis of the ellipse or circle when you have a center as fixed point, or spiral. This is not very rigorous argument, but may give you some insight.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Thank you, this is what I wanted to know. In the meantime I also have found a similar post which also includes a nice illustration.
        $endgroup$
        – Jannis
        Jun 2 '15 at 12:04












      • $begingroup$
        @Jannis: Thanks for the reference. That's a nice picture.
        $endgroup$
        – KittyL
        Jun 2 '15 at 14:12










      • $begingroup$
        Nitpick: L_y appears to change into L_b. Please let me know if I missed something. On a more substantial note, I still can't quite visualize how rotation in the complex plane translates into oscillation of the original system.
        $endgroup$
        – Ben Ogorek
        Dec 31 '18 at 0:59








      • 1




        $begingroup$
        @BenOgorek: Fixed! Thanks. "Rotation" means the vector $(x,y)$ going along the ellipse following the arrows. (See the OP's last picture) "Oscillation" refers to the change of values of $x$, or the change of values of $y$. Consider the change of values of $x$ in that picture. Suppose we start at the point $(6,0)$, where $x=6$. Following the arrows, it could go to $(0,-6)$, where $x=0$. Continuing following the arrows, the value of $x$ changes back to $6$, and so on. Same for the $y$ values.
        $endgroup$
        – KittyL
        Dec 31 '18 at 12:31






      • 1




        $begingroup$
        @BenOgorek: To simplify the matter, let's consider $x''(t)+x(t)=0$. This differential equation has solution $c_1e^{it}+c_2e^{-it}$ which can be written as $(c_1+c_2)cos t+(c_1i-c_2i)sin t$. These of course contain many solutions in the complex plane. But we want to pick the real solutions. Hence we let $d_1=c_1+c_2$ and $d_2=i(c_1-c_2)$ and only use the real values of $d_i$. This gives us all real solutions of the original equation. For the system, the reason is similar.
        $endgroup$
        – KittyL
        Dec 31 '18 at 20:52














      4












      4








      4





      $begingroup$

      Intuitively, when the system has real eigenvalues and real eigenvectors, the eigenvectors are the directions in which the flow $vec{x}$ does not change direction.



      When the eigenvalues have imaginary part, hence the eigenvectors contain imaginary part, there are no directions in which the flow $vec{x}$ does not change direction.



      If an eigenvector is $vec{v}_x+ivec{v}_y$, with corresponding eigenvalue $L_x+iL_y$, then $$A(vec{v}_x+ivec{v}_y)=(L_x+iL_y)(vec{v}_x+ivec{v}_y)\
      =(L_xvec{v}_x-L_yvec{v}_y)+i(L_xvec{v}_y+L_yvec{v}_x)$$



      Separate the real and imaginary parts:
      $$Avec{v}_x=L_xvec{v}_x-L_yvec{v}_y\
      Avec{v}_y=L_xvec{v}_y+L_yvec{v}_x$$



      So on the real plane, you can see $vec{v}_x, vec{v}_y$ are two axis that the flow rotate around. Intuitively, they are major and minor axis of the ellipse or circle when you have a center as fixed point, or spiral. This is not very rigorous argument, but may give you some insight.






      share|cite|improve this answer











      $endgroup$



      Intuitively, when the system has real eigenvalues and real eigenvectors, the eigenvectors are the directions in which the flow $vec{x}$ does not change direction.



      When the eigenvalues have imaginary part, hence the eigenvectors contain imaginary part, there are no directions in which the flow $vec{x}$ does not change direction.



      If an eigenvector is $vec{v}_x+ivec{v}_y$, with corresponding eigenvalue $L_x+iL_y$, then $$A(vec{v}_x+ivec{v}_y)=(L_x+iL_y)(vec{v}_x+ivec{v}_y)\
      =(L_xvec{v}_x-L_yvec{v}_y)+i(L_xvec{v}_y+L_yvec{v}_x)$$



      Separate the real and imaginary parts:
      $$Avec{v}_x=L_xvec{v}_x-L_yvec{v}_y\
      Avec{v}_y=L_xvec{v}_y+L_yvec{v}_x$$



      So on the real plane, you can see $vec{v}_x, vec{v}_y$ are two axis that the flow rotate around. Intuitively, they are major and minor axis of the ellipse or circle when you have a center as fixed point, or spiral. This is not very rigorous argument, but may give you some insight.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 31 '18 at 12:25

























      answered Jun 1 '15 at 18:46









      KittyLKittyL

      13.8k31534




      13.8k31534












      • $begingroup$
        Thank you, this is what I wanted to know. In the meantime I also have found a similar post which also includes a nice illustration.
        $endgroup$
        – Jannis
        Jun 2 '15 at 12:04












      • $begingroup$
        @Jannis: Thanks for the reference. That's a nice picture.
        $endgroup$
        – KittyL
        Jun 2 '15 at 14:12










      • $begingroup$
        Nitpick: L_y appears to change into L_b. Please let me know if I missed something. On a more substantial note, I still can't quite visualize how rotation in the complex plane translates into oscillation of the original system.
        $endgroup$
        – Ben Ogorek
        Dec 31 '18 at 0:59








      • 1




        $begingroup$
        @BenOgorek: Fixed! Thanks. "Rotation" means the vector $(x,y)$ going along the ellipse following the arrows. (See the OP's last picture) "Oscillation" refers to the change of values of $x$, or the change of values of $y$. Consider the change of values of $x$ in that picture. Suppose we start at the point $(6,0)$, where $x=6$. Following the arrows, it could go to $(0,-6)$, where $x=0$. Continuing following the arrows, the value of $x$ changes back to $6$, and so on. Same for the $y$ values.
        $endgroup$
        – KittyL
        Dec 31 '18 at 12:31






      • 1




        $begingroup$
        @BenOgorek: To simplify the matter, let's consider $x''(t)+x(t)=0$. This differential equation has solution $c_1e^{it}+c_2e^{-it}$ which can be written as $(c_1+c_2)cos t+(c_1i-c_2i)sin t$. These of course contain many solutions in the complex plane. But we want to pick the real solutions. Hence we let $d_1=c_1+c_2$ and $d_2=i(c_1-c_2)$ and only use the real values of $d_i$. This gives us all real solutions of the original equation. For the system, the reason is similar.
        $endgroup$
        – KittyL
        Dec 31 '18 at 20:52


















      • $begingroup$
        Thank you, this is what I wanted to know. In the meantime I also have found a similar post which also includes a nice illustration.
        $endgroup$
        – Jannis
        Jun 2 '15 at 12:04












      • $begingroup$
        @Jannis: Thanks for the reference. That's a nice picture.
        $endgroup$
        – KittyL
        Jun 2 '15 at 14:12










      • $begingroup$
        Nitpick: L_y appears to change into L_b. Please let me know if I missed something. On a more substantial note, I still can't quite visualize how rotation in the complex plane translates into oscillation of the original system.
        $endgroup$
        – Ben Ogorek
        Dec 31 '18 at 0:59








      • 1




        $begingroup$
        @BenOgorek: Fixed! Thanks. "Rotation" means the vector $(x,y)$ going along the ellipse following the arrows. (See the OP's last picture) "Oscillation" refers to the change of values of $x$, or the change of values of $y$. Consider the change of values of $x$ in that picture. Suppose we start at the point $(6,0)$, where $x=6$. Following the arrows, it could go to $(0,-6)$, where $x=0$. Continuing following the arrows, the value of $x$ changes back to $6$, and so on. Same for the $y$ values.
        $endgroup$
        – KittyL
        Dec 31 '18 at 12:31






      • 1




        $begingroup$
        @BenOgorek: To simplify the matter, let's consider $x''(t)+x(t)=0$. This differential equation has solution $c_1e^{it}+c_2e^{-it}$ which can be written as $(c_1+c_2)cos t+(c_1i-c_2i)sin t$. These of course contain many solutions in the complex plane. But we want to pick the real solutions. Hence we let $d_1=c_1+c_2$ and $d_2=i(c_1-c_2)$ and only use the real values of $d_i$. This gives us all real solutions of the original equation. For the system, the reason is similar.
        $endgroup$
        – KittyL
        Dec 31 '18 at 20:52
















      $begingroup$
      Thank you, this is what I wanted to know. In the meantime I also have found a similar post which also includes a nice illustration.
      $endgroup$
      – Jannis
      Jun 2 '15 at 12:04






      $begingroup$
      Thank you, this is what I wanted to know. In the meantime I also have found a similar post which also includes a nice illustration.
      $endgroup$
      – Jannis
      Jun 2 '15 at 12:04














      $begingroup$
      @Jannis: Thanks for the reference. That's a nice picture.
      $endgroup$
      – KittyL
      Jun 2 '15 at 14:12




      $begingroup$
      @Jannis: Thanks for the reference. That's a nice picture.
      $endgroup$
      – KittyL
      Jun 2 '15 at 14:12












      $begingroup$
      Nitpick: L_y appears to change into L_b. Please let me know if I missed something. On a more substantial note, I still can't quite visualize how rotation in the complex plane translates into oscillation of the original system.
      $endgroup$
      – Ben Ogorek
      Dec 31 '18 at 0:59






      $begingroup$
      Nitpick: L_y appears to change into L_b. Please let me know if I missed something. On a more substantial note, I still can't quite visualize how rotation in the complex plane translates into oscillation of the original system.
      $endgroup$
      – Ben Ogorek
      Dec 31 '18 at 0:59






      1




      1




      $begingroup$
      @BenOgorek: Fixed! Thanks. "Rotation" means the vector $(x,y)$ going along the ellipse following the arrows. (See the OP's last picture) "Oscillation" refers to the change of values of $x$, or the change of values of $y$. Consider the change of values of $x$ in that picture. Suppose we start at the point $(6,0)$, where $x=6$. Following the arrows, it could go to $(0,-6)$, where $x=0$. Continuing following the arrows, the value of $x$ changes back to $6$, and so on. Same for the $y$ values.
      $endgroup$
      – KittyL
      Dec 31 '18 at 12:31




      $begingroup$
      @BenOgorek: Fixed! Thanks. "Rotation" means the vector $(x,y)$ going along the ellipse following the arrows. (See the OP's last picture) "Oscillation" refers to the change of values of $x$, or the change of values of $y$. Consider the change of values of $x$ in that picture. Suppose we start at the point $(6,0)$, where $x=6$. Following the arrows, it could go to $(0,-6)$, where $x=0$. Continuing following the arrows, the value of $x$ changes back to $6$, and so on. Same for the $y$ values.
      $endgroup$
      – KittyL
      Dec 31 '18 at 12:31




      1




      1




      $begingroup$
      @BenOgorek: To simplify the matter, let's consider $x''(t)+x(t)=0$. This differential equation has solution $c_1e^{it}+c_2e^{-it}$ which can be written as $(c_1+c_2)cos t+(c_1i-c_2i)sin t$. These of course contain many solutions in the complex plane. But we want to pick the real solutions. Hence we let $d_1=c_1+c_2$ and $d_2=i(c_1-c_2)$ and only use the real values of $d_i$. This gives us all real solutions of the original equation. For the system, the reason is similar.
      $endgroup$
      – KittyL
      Dec 31 '18 at 20:52




      $begingroup$
      @BenOgorek: To simplify the matter, let's consider $x''(t)+x(t)=0$. This differential equation has solution $c_1e^{it}+c_2e^{-it}$ which can be written as $(c_1+c_2)cos t+(c_1i-c_2i)sin t$. These of course contain many solutions in the complex plane. But we want to pick the real solutions. Hence we let $d_1=c_1+c_2$ and $d_2=i(c_1-c_2)$ and only use the real values of $d_i$. This gives us all real solutions of the original equation. For the system, the reason is similar.
      $endgroup$
      – KittyL
      Dec 31 '18 at 20:52











      1












      $begingroup$

      Usually complex eigenvalues correspond to circular motion (not dissimilarly to the way that rotation matrices have complex eigenvalues/vectors. If there is a good reason for why rotations should correspond to complex eigenvectors, I don't know it. The real part of the eigenvalue indicates whether the motion is circular (or elliptical) or if it is spiralling in or out. The imaginary part and the eigenvector give you the direction in which this spiralling happens (clockwise/anti-clockwise).






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Usually complex eigenvalues correspond to circular motion (not dissimilarly to the way that rotation matrices have complex eigenvalues/vectors. If there is a good reason for why rotations should correspond to complex eigenvectors, I don't know it. The real part of the eigenvalue indicates whether the motion is circular (or elliptical) or if it is spiralling in or out. The imaginary part and the eigenvector give you the direction in which this spiralling happens (clockwise/anti-clockwise).






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Usually complex eigenvalues correspond to circular motion (not dissimilarly to the way that rotation matrices have complex eigenvalues/vectors. If there is a good reason for why rotations should correspond to complex eigenvectors, I don't know it. The real part of the eigenvalue indicates whether the motion is circular (or elliptical) or if it is spiralling in or out. The imaginary part and the eigenvector give you the direction in which this spiralling happens (clockwise/anti-clockwise).






          share|cite|improve this answer









          $endgroup$



          Usually complex eigenvalues correspond to circular motion (not dissimilarly to the way that rotation matrices have complex eigenvalues/vectors. If there is a good reason for why rotations should correspond to complex eigenvectors, I don't know it. The real part of the eigenvalue indicates whether the motion is circular (or elliptical) or if it is spiralling in or out. The imaginary part and the eigenvector give you the direction in which this spiralling happens (clockwise/anti-clockwise).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jun 1 '15 at 18:53









          Dan RobertsonDan Robertson

          2,581512




          2,581512























              0












              $begingroup$

              I scetched you the different cases.



              enter image description here



              edit: if one eigenvalue is zero, then there exists a line consisting of fixed points.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                I scetched you the different cases.



                enter image description here



                edit: if one eigenvalue is zero, then there exists a line consisting of fixed points.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  I scetched you the different cases.



                  enter image description here



                  edit: if one eigenvalue is zero, then there exists a line consisting of fixed points.






                  share|cite|improve this answer











                  $endgroup$



                  I scetched you the different cases.



                  enter image description here



                  edit: if one eigenvalue is zero, then there exists a line consisting of fixed points.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jun 1 '15 at 19:00

























                  answered Jun 1 '15 at 18:52









                  the.polothe.polo

                  510512




                  510512






























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