Geometrical interpretation of complex eigenvectors in a system of differential equations 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?
 A: 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 $\pmb{a}+i\pmb{b}$.)
A: The other posts here are good, but I wanted to give an explicit solution in the 2D case and see exactly how the eigenvectors/eigenvalues relate to the shape of the trajectories.
I'll do this for the attractor spiral, repellor spiral, and center cases all at once, but I'm mainly concerned with the center case.

Setup and Calculations
Suppose we have a 2D linear system $\dot{\vec{x}} = M\vec{x}$ where $M$ is a nondegenerate $2\times 2$ real matrix.
For the sake of the problem, assume $M$ has non-real eigenvalues $\lambda_{1}, \lambda_{2}$ and non-real eigenvectors $\vec{v}_{1}, \vec{v}_{2}$ (equivalently, this amounts to assuming $(\operatorname{tr} M)^{2}/4 < \det M)$.
Since $M$ is a real matrix, one can show that $\lambda_{1} = \lambda_{2}^{*}$ and $\vec{v}_{1} = \vec{v}_{2}^{*}$.
Let us write everything in terms of real and imaginary components:
$$ \lambda_{1} = a + ib, \qquad \lambda_{2} = a - ib  $$
and
$$ \vec{v}_{1} = \vec{v}_{R} + i\vec{v}_{I}, \qquad \vec{v}_{2} = \vec{v}_{R} - i\vec{v}_{I}. $$
The general solution is
$$ \vec{x}(t) = A\vec{v}_{1}e^{at + ibt} + B\vec{v}_{2}e^{at - ibt} $$
where $A, B$ are complex constants.
When we expand everything out, we get the following:
\begin{align*}
 \vec{x}(t) &= (A_{1} + iA_{2})(\vec{v}_{R} + i\vec{v}_{I})e^{at+ibt} + (B_{1} + iB_{2})(\vec{v}_{R} - i\vec{v}_{I})e^{at-ibt} \\[1.5ex]
 &= \Big[ A_{1}\vec{v}_{R} - A_{2}\vec{v}_{I} + iA_{2}\vec{v}_{R} + iA_{1}\vec{v}_{I} \Big]e^{at+ibt} + \Big[ B_{1}\vec{v}_{R} + B_{2}\vec{v}_{I} + iB_{2}\vec{v}_{R} - iB_{1}\vec{v}_{I} \Big]e^{at-ibt} \\[1.5ex]
 &= \Big[ A_{1}\vec{v}_{R} - A_{2}\vec{v}_{I} + iA_{2}\vec{v}_{R} + iA_{1}\vec{v}_{I} \Big]\Big[ \cos bt + i\sin bt \Big]e^{at} \\
 &\qquad + \Big[ B_{1}\vec{v}_{R} + B_{2}\vec{v}_{I} + iB_{2}\vec{v}_{R} - iB_{1}\vec{v}_{I} \Big]\Big[ \cos bt - i\sin bt \Big]e^{at} \\[1.5ex]
 &= e^{at}\vec{v}_{R}\Big[ A_{1}\cos bt + \color{purple}{iA_{1}\sin bt} + \color{purple}{iA_{2}\cos bt} - A_{2}\sin bt \\
 &\qquad\qquad +B_{1}\cos bt - \color{purple}{iB_{1}\sin bt} + \color{purple}{iB_{2}\cos bt} + B_{2}\sin bt \Big] \\
 &\qquad + e^{at}\vec{v}_{I}\Big[ -A_{2}\cos bt - \color{purple}{iA_{2}\sin bt} + \color{purple}{iA_{1}\cos bt} - A_{1}\sin bt \\
 &\qquad\qquad + B_{2}\cos bt - \color{purple}{iB_{2}\sin bt} - \color{purple} {iB_{1}\cos bt} - B_{1}\sin bt \Big].
\end{align*}
Since we are concerned with real solutions only, we want the imaginary terms to cancel out.
For this, we need
$$ B_{1} = A_{1} \qquad\text{ and }\qquad B_{2} = -A_{2}. $$
This finally yields
\begin{align*}
 \vec{x}(t) &= 2e^{at}\vec{v}_{R} [A_{1}\cos bt - A_{2}\sin bt] - 2e^{at}\vec{v}_{I} [A_{2}\cos bt + A_{1}\sin bt] \\[1.5ex]
 &= 2e^{at}\vec{v}_{R} [A_{1}\cos bt - A_{2}\sin bt] - 2e^{at}\vec{v}_{I} [A_{1}\cos(bt - \tfrac{\pi}{2}) - A_{2}\sin(bt - \tfrac{\pi}{2})] \\[0.5ex]
 &= 2e^{at}\vec{v}_{R}\sqrt{A_{1}^{2}+A_{2}^{2}}\cos(bt + \delta) - 2e^{at}\vec{v}_{I}\sqrt{A_{1}^{2}+A_{2}^{2}}\cos(bt + \delta - \tfrac{\pi}{2}) \\[0.8ex]
 &= Ce^{at}[\vec{v}_{R}\cos(bt + \delta) - \vec{v}_{I}\sin(bt + \delta)]
\end{align*}
where $C$ and $\delta$ are some real constants.

Solution
So now we have the solution in terms of real and imaginary parts of one of our eigenvectors:
$$ \boxed{\vec{x}(t) =  Ce^{at}[\vec{v}_{R}\cos(bt + \delta) - \vec{v}_{I}\sin(bt + \delta)]} $$
where $C, \delta$ are real constants.
(The minus sign here seems strange, but I don't think it's a mistake.)
There are many observations we can make that can give us useful information.

Meanings of Real and Imaginary Parts of Eigenvalues
From our form of the solution above we can interpret the meanings of $a$ and $b$ here:

*

*$\operatorname{sign}(a)$ determines whether we are dealing with a spiral sink, center, or spiral repellor,

*$|a|$ determines how fast every point is repelled or attracted,

*$\operatorname{sign}(b)$ determines whether the solutions go clockwise or counterclockwise (an explanation here is needed because it matters what $\vec{v}_{R}$ and $\vec{v}_{I}$ are), and

*$|b|$ is the angular frequency of the trajectories as they go around the origin.


Clockwise vs Counterclockwise
Let us define the 2D cross-product as
$$ \vec{x}\times\vec{y} = x_{1}y_{2} - x_{2}y_{1}. $$
This tells us the orientation of $(\vec{x}, \vec{y})$ in the plane.
If the above is positive, then $(\vec{x}, \vec{y})$ is positively oriented.
If the above is negative, then $(\vec{x}, \vec{y})$ is positively oriented.
With the 2D cross-product, we can tell whether the solution goes clockwise or counterclockwise.
We calculate
\begin{align*}
\vec{x}\times\dot{\vec{x}} &= Ce^{at}[\vec{v}_{R}\cos(bt + \delta) - \vec{v}_{I}\sin(bt + \delta)] \times Ce^{at}b [-\vec{v}_{R}\sin(bt + \delta) - \vec{v}_{I}\cos(bt + \delta)] \\[1.5ex]
&= C^{2}e^{2at}b\, [\vec{v}_{R}\cos(bt + \delta) - \vec{v}_{I}\sin(bt + \delta)] \times [-\vec{v}_{R}\sin(bt + \delta) - \vec{v}_{I}\cos(bt + \delta)] \\[1.5ex]
&= C^{2}e^{2at}b\, (0 - \vec{v}_{R}\times\vec{v}_{I}\cos^{2}(bt+\delta) + \vec{v}_{I}\times\vec{v}_{R}\sin^{2}(bt+\delta) + 0) \\[1.5ex]
&= C^{2}e^{2at}b\, (0 - \vec{v}_{R}\times\vec{v}_{I}\cos^{2}(bt+\delta) - \vec{v}_{R}\times\vec{v}_{I}\sin^{2}(bt+\delta) + 0) \\[1.5ex]
&= C^{2}e^{2at} b\, (-\vec{v}_{R}\times\vec{v}_{I})
\end{align*}
Now since $C$ and $a$ are real, the factor $C^{2}e^{2at}$ is positive and has no bearing on the sign of $\vec{x}\times\dot{\vec{x}}$.
Then we have the following implications:

*

*$\vec{x}\times\dot{\vec{x}} > 0$ $\iff$ $-b(\vec{v}_{R}\times\vec{v}_{I}) > 0$ $\iff$ the solutions are counterclockwise,

*$\vec{x}\times\dot{\vec{x}} < 0$ $\iff$ $-b(\vec{v}_{R}\times\vec{v}_{I}) < 0$ $\iff$ the solutions are clockwise.

This gives us the relationship between the eigenvalues, eigenvectors, and the direction in which the solutions go.
As we can see, it's not an obvious relationship, but it does exist.
(Of course, there is a much more efficient way of telling whether the solutions go clockwise or counterclockwise: consider $M\binom{1}{0} = \binom{M_{11}}{M_{21}}$, and just look at the sign of entry $M_{21}$.
Nonetheless, the above relationship between the eigenvalues, eigenvectors, and the direction of solutions is still interesting.)

Choice of Eigenvalue/Eigenvector
Notice that in my calculations I did not assume that $b$ is a positive quantity (although I could have if I wanted to).
Instead, the only requirement used was that $\vec{v}_{1} = \vec{v}_{R} + i\vec{v}_{I}$ is an eigenvector associated with eigenvalue $\lambda_{1} = a+ib$.
If we switch $\vec{v}_{I}$ for $-\vec{v}_{I}$ in our calculations, then we must also switch $b$ for $-b$ (so that the eigenvector corresponds to the correct eigenvalue).
As one can see, this kind of change preserves the sign of $-b(\vec{v}_{R}\times\vec{v}_{I})$.
Thus, it doesn't matter which eigenvector is which as long as we don't mix up which eigenvalue is associated with which eigenvector (in particular, which eigenvector
corresponds to $\operatorname{Im}\lambda > 0$ and which eigenvector corresponds to $\operatorname{Im}\lambda < 0$).

Semi-major/Semi-minor Axes
Lastly, I'll look at how at how the eigenvectors can be used to find the semi-major and semi-minor axes of the ellipses traced out in the center case ($a=0$).
It's important to note that $\vec{v}_{R}, \vec{v}_{I}$ are not (necessarily) proportional to the semi-major/semi-minor axes of the ellipses traced out by our solutions.
In fact, the two vectors don't even have to be orthogonal to each other.
Despite this, we will find a way to relate these to the semi-major/semi-minor axes of our ellipses.
Notice that if $\vec{v}$ is an eigenvector of $M$, so is any complex scalar multiple of $\vec{v}$.
This means we have some degrees of freedom in choosing what $\vec{v}_{1}$ and $\vec{v}_{2}$ could be.
In particular, any vector of the form $\vec{w}_{1} = ce^{i\phi}\vec{v}_{1}$ where $c>0$ and $\phi\in\mathbb{R}$ would be an eigenvector of $M$ with eigenvalue $\lambda_{1}$.
For convenience, let us set the second eigenvector associated with $\lambda_{2}$ to be  $\vec{w}_{2} = \vec{w}_{1}^{*}$.
From this, we get
\begin{align*}
\vec{w}_{1} &= ce^{i\phi}\vec{v}_{1} = c(\cos\phi + i\sin\phi)(\vec{v}_{R}+i\vec{v}_{I}) = c(\vec{v}_{R}\cos\phi - \vec{v}_{I}\sin\phi) + ic(\vec{v}_{R}\sin\phi + \vec{v}_{I}\cos\phi)
\end{align*}
and
\begin{align*}
\vec{w}_{2} &= ce^{-i\phi}\vec{v}_{2} = c(\cos\phi - i\sin\phi)(\vec{v}_{R}-i\vec{v}_{I}) = c(\vec{v}_{R}\cos\phi - \vec{v}_{I}\sin\phi) - ic(\vec{v}_{R}\sin\phi + \vec{v}_{I}\cos\phi).
\end{align*}
This yields
$$ \vec{w}_{R} = c(\vec{v}_{R}\cos\phi - \vec{v}_{I}\sin\phi) \qquad\text{ and }\qquad \vec{w}_{I} = c(\vec{v}_{R}\sin\phi + \vec{v}_{I}\cos\phi) $$
with solution
$$ \vec{x}(t) = C[\vec{w}_{R}\cos(bt + \delta) - \vec{w}_{I}\sin(bt + \delta)] $$
to our system.
Scaling $\vec{w}_{R}, \vec{w}_{I}$ by a positive constant doesn't do anything interesting, and the constant $c>0$ would be absorbed by $C$ in our solution anyways.
For our sake we'll set $c=1$.
If we want $\vec{w}_{R}, \vec{w}_{I}$ to be proportional to the semi-major/semi-minor axes of an ellipse, we would clearly want the two vectors to be orthogonal to each other.
It turns out this alone is enough for them to be proportional to the semi-major/semi-minor axes by virtue of the form of our solution for $\vec{x}(t)$ above.
We want
\begin{align*}
0 &= \vec{w}_{I}\cdot\vec{w}_{R} \\
&= (\vec{v}_{R}\cos\phi - \vec{v}_{I}\sin\phi)\cdot (\vec{v}_{R}\sin\phi + \vec{v}_{I}\cos\phi) \\
&= \vec{v}_{R}\cdot\vec{v}_{I}(\cos^{2}\phi - \sin^{2}\phi) + \vec{v}_{R}^{2}\sin\phi\cos\phi - \vec{v}_{I}^{2}\sin\phi\cos\phi \\
&= \vec{v}_{R}\cdot\vec{v}_{I} \cos(2\phi) + (\vec{v}_{R}^{2} - \vec{v}_{I}^{2})\sin(2\phi)/2.
\end{align*}
By doing some algebra, the criterion can be rewritten as
\begin{align*}
\cot 2\phi = \frac{\vec{v}_{I}^{2} - \vec{v}_{R}^{2}}{2\vec{v}_{R}\cdot\vec{v}_{I}}.
\end{align*}
(If the denominator is zero, then we already have our desired vectors that are orthogonal to each other.)
From the above formula, we can obtain a value of $\phi$ (but the value might not be unique) and get the appropriate vectors
$$ \vec{w}_{R} = \vec{v}_{R}\cos\phi - \vec{v}_{I}\sin\phi \qquad\text{ and }\qquad \vec{w}_{I} = \vec{v}_{R}\sin\phi + \vec{v}_{I}\cos\phi, $$
which would be proportional to the semi-major/semi-minor axes.
A: 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.
