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Let's consider a system of differential equations in the form

$$\dot{X} = M X$$

in two dimensions ($X = (x(t), y(t))$).

In the case that $M$ has real values, it is easy to give a geometric interpretation of the eigenvectors in the $(x, y)$ plane: they are the directions along which the dynamical system is "sucked" or "expelled" from the stable point.

If the eigenvalues are complex, then the eigenvectors are complex too. Let's say the eigenvalues are purely imaginary, so that the trajectory is an ellipse.

Can I draw anything in the $(x, y)$ plane that is related to the eigenvectors? In particular, do the eigenvectors have any simple relation to the rotation and eccentricity of the ellipse?

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This is a wild guess, but it is possible that you something like the angular velocity vector? Also, this could be better asked at Math.SE –  Angel Joaniquet Tukiainen Apr 15 '13 at 18:16
    
@AngelJoaniquetTukiainen Hmm, well I think it might be appropriate because he's looking for a physical interpretation of it. –  Ataraxia Apr 15 '13 at 18:19
    
Related: physics.stackexchange.com/q/22873/2451 –  Qmechanic Apr 15 '13 at 19:13
    
Note that while purely imaginary eigenvalues imply elliptical trajectories, general complex eigenvalues give spiral trajectories. (Your third paragraph can be read as implying this, but it has other readings, too, so I thought I'd clarify.) –  Ted Pudlik May 15 '13 at 23:32
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2 Answers 2

With a short straightforward calculation, I came to this picture:

That is, if the ellipse semi-major and semi-minor axes are given by vectors $\pmb{a}$ and $\pmb{b}$, then the eigenverctors are proportional to $\pmb{a}\pm i\pmb{b}$ (with maybe some complex factors), and their order would give the direction of rotation: from the $\mathrm{Im}\,\lambda>0$ eigenvector to the $\mathrm{Im}\,\lambda<0$ eigenvector. (On the figure, from $\pmb{a}-i\pmb{b}$ to $\pmb{a}+i\pmb{b}$.)

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could you please provide a sketch of this calculation? It seems interesting. –  Lorenzo Pistone May 17 '13 at 0:58
    
First I considered the simplest paraxial ellipse and the $M$ of the form $\left(\begin{smallmatrix}0&-a\\1/a&0\end{smallmatrix}\right)$. Then I applied an arbitrary rotation to the plane. Of course, the semi-major and semi-minor axes are interchangeable. –  firtree May 17 '13 at 8:01
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If you have real matrix $M$, its eigenvalues and eigen vectors come in conjugate pairs -- for $MX=\lambda X$ you always have $M\bar{X}=\bar{\lambda}\bar{X}$. You can convince yourself that a general solution to $\dot{Y}=MY$ in 2D is $$ Y(t)=Re\left\{a\exp(\lambda t) X\right\},\,a\in\mathbb{C}. $$ In general, in higher dimensions, real eigenvalues correspond to invariant 1-d subspaces -- lines, while conjugate pairs correspond to invariant 2-d planes. For example, for any 3d-rotation you have an invariant plane and an invariant line.

As for the relation between eigenvectors and ellipse parametrs, it surely exists, but I wasn't able to find any simple form for it. The reason is that all above can be seen as statements about operators, without explicit coordinate system. It is the coordinate system (actually, the standard dot product that comes with it) that gives sense to such notion as "semiaxis", "eccentricity". So you have to deal with coordinates of $X$, which not as nice as $X$ in its integrity.

For example, take the basis to be $Re X,\,Im X$, and then the trajectory is just a circle (for pure imaginary eigenvalue). Note that these two vectors are always linearly independent, because otherwise it follows that $\lambda$ is real.

Update

Take $A=Re X,\, B=Im X,\, \gamma=Re\lambda,\,\omega=Im\lambda,\,a=C\exp(i\phi)$. Then the solution is: $$ Y(t)=Ce^{\gamma t}\left(\cos(\omega t')A-\sin(\omega t')B\right), $$ where $t'=t+\phi/\omega$.

Now recall that once upon a time the solution coincides with one of the semiaxes, and after a time $\pi/2\omega$ it coincides with the second one. Two positions are orthgonal. Let the semiaxes be $S_1,\,S_2$: $$ S_1=\cos(\alpha)A-\sin(\alpha)B\\ S_2=-\sin(\alpha)A-\cos(\alpha)B\\ $$ Then $$ 0=-(S_1,S_2)=\cos(2\alpha)(A,B)+\sin(2\alpha)(A^2-B^2)/2\\ \tan(2\alpha)=\frac{2(A,B)}{B^2-A^2} $$ This gives you $\alpha=\omega t'$ the time when the solution goes trough $S_1$, and coordinates of $S_1$ and $S_2$. Now it is straight forward to obtain their lengthes and so on.

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well, now i see an error, i am working on it –  Peter Kravchuk Apr 15 '13 at 20:20
    
now looks ok, finally –  Peter Kravchuk Apr 15 '13 at 20:26
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