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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?

Jannis
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3 Answers3

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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 $\vec{v}_x+i\vec{v}_y$, with corresponding eigenvalue $L_x+iL_y$, then $$A(\vec{v}_x+i\vec{v}_y)=(L_x+iL_y)(\vec{v}_x+i\vec{v}_y)\\ =(L_x\vec{v}_x-L_y\vec{v}_y)+i(L_x\vec{v}_y+L_y\vec{v}_x)$$

Separate the real and imaginary parts: $$A\vec{v}_x=L_x\vec{v}_x-L_y\vec{v}_y\\ A\vec{v}_y=L_x\vec{v}_y+L_y\vec{v}_x$$

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.

KittyL
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  • 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. – Jannis Jun 02 '15 at 12:04
  • @Jannis: Thanks for the reference. That's a nice picture. – KittyL Jun 02 '15 at 14:12
  • 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. – Ben Ogorek Dec 31 '18 at 00:59
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    @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. – KittyL Dec 31 '18 at 12:31
  • Regarding the solution, it's almost like the eigenvectors start you off in a different plane than physical reality, a place in which there are directions where the system will stay in. But it's not reality. There's a similar question here. You can see I still don't quite understand how the grounded-in-reality solution makes so literal use of the imaginary component (just drop the i and multiply by C_2!). – Ben Ogorek Dec 31 '18 at 17:51
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    @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. – KittyL Dec 31 '18 at 20:52
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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.

the.polo
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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).

Dan Robertson
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