First of all, this is my fourth question about dynamical systems in a week, sorry for that.

Considering a linear bidimensional dynamical (autonomous) system, the orbits can be plotted in the phase state as so: from wikipedia

Let me specify that I already managed to solve the dynamical equations in every case (regarding eigenvalues), and plot similar orbits.

My question is on the third image: there, there is one double (positive) eigenvalue. The dimension of the unstable manifold is therefore $2$. It is spanned by an eigenvector (here, $(1,0)$ for e.g.) and a generalized eigenvector.

My question is, is there a link between the shape of the trajectories and a generalized eigenvector? For the eigenvector the link is quite obvious (also in the first and last cases in the figure), but I am not able to show a generalized eigenvector from the orbits.

Thank you

edit: I wrote the dynamical equations for complex conjugate eigenvalues, and: - the (un)stable manifold is of obviously bidimensional (stable/unstable manifold theorem); - but nothing prevents me from trying to find a 1D-submanifold of the (un)stable manifold, by seeking a parametric curve of the form $(x,h(x))$ and writing that it is invariant ($h$ is written as its Taylor expansion up to a finite degree). The interesting thing is that there is no such curve!

As I have not read anything about this, I'd be willing to have your opinions: in some cases (1 and 4 in the figure), the (un)stable manifold is of dimension 2, and two submanifolds of dimension 1 can always be calculated. If the eigenvalues are complex (not real) (case 2 in above figure), the (un)stable manifold is of dimension 2, but this manifold cannot be "decomposed" in two (un)stable submanifolds. I'm not sure what I'm taking about when writing about decomposition in submanifolds... Highlight welcome!


For concreteness let's look at

$$x' = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} x.$$

Here we have

$$x(t) = \begin{bmatrix} e^t & t e^t \\ 0 & e^t \end{bmatrix} x(0).$$


$$x(0)=\begin{bmatrix} 0 \\ 1 \end{bmatrix},$$

that is, an initial condition aligned with a generalized eigenvector. Then we have

$$x(t) = \begin{bmatrix} t e^t \\ e^t \end{bmatrix}.$$

So the trajectory satisfies $x_1=tx_2$, that is, the component of $x(t)$ in the direction of $e_1$ is becoming more prominent than that in the direction of $e_2$. In particular, the ratio of these is growing linearly. By contrast, if we have distinct positive eigenvalues, then the ratio grows exponentially, and if we have a multiple nondefective eigenvalue (which, in two dimensions, only occurs for multiples of the identity), then the ratio remains constant. So what you're seeing in the trajectories in the third image is that the contribution from $e_1$ is becoming larger, but it does so rather slowly compared to the case of distinct eigenvalues.

I doubt it will be very easy to clearly see a generalized eigenvector in a plot of trajectories, simply because any defective matrix is arbitrarily close to a nondefective matrix. As a result, the trajectory of my example above will look (in a figure) essentially the same as the trajectory of a nondefective matrix.

As an example, the trajectories of the system I described above and

$$x' = \begin{bmatrix} 1 & 1 \\ 0 & 1 + 10^{-12} \end{bmatrix} x$$

will look very similar for small to moderate $t$, even though one is defective and the other is not. Indeed, look at $x(0)=\begin{bmatrix} 1 \\ 10^{-12} \end{bmatrix}$ in the first system:

$$x(t) = \begin{bmatrix} e^t \left ( 1 + 10^{-12} t \right ) \\ 10^{-12} e^t \end{bmatrix}.$$

This takes a very long time to stop looking like a multiple of $x(0)$, since we have to wait for $10^{-12} t$ to "catch up" with $1$.

  • $\begingroup$ Thanks a lot, it makes it clearer. Still I don't get what is the unstable (or stable) bidimensional manifold: the theorem states that it is tangent to the eigenvector at the origin, but how can it be tangent to a generalized eigenvector? $\endgroup$
    – anderstood
    Aug 5 '14 at 5:09

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