We discussed the case of system with two distinct real eigenvalues,
repeated (nonzero) eigenvalue, and complex eigenvalues. But we did not
discuss the case when one of the eigenvalues is zero. In fact, it is
easy to see that this happen if and only if we have more than one
equilibrium point (which is (0,0)). In this case, we will have a
line of equilibrium points (the direction vector for this line is the
eigenvector associated to the eigenvalue zero).

**Example.** Find the general solution to

**Answer.** The characteristic polynomial of this system is

which reduces to . The eigenvalues are and . Let us find the associated eigenvectors.

- For , set
The equation translates into

The two equations are the same. So we have

*y*= 2*x*. Hence an eigenvector is - For , set
The equation translates into

The two equations are the same (as -

*x*-*y*=0). So we have*y*= -*x*. Hence an eigenvector is

Therefore the general solution is

Note that all the solutions are line parallel to the vector
.
When , the trajectory goes to infinity. But
when , the trajectory converge to the
equilibrium point on the line of equilibrium points (that is passing
by (0,0) and having as a direction vector). The picture below
explains more what is happening.

The general case is very similar to this example. Indeed, assume that a system has 0 and as eigenvalues. Hence if is an eigenvector associated to 0 and an eigenvector associated to , then the general solution is

We have two cases, whether or .

- If , then is an equilibrium point.
- If , then the solution is a line parallel
to the vector . Moreover, we have when
- if , the solution tends away from the
line of equilibrium;
- if , the solution tends to the
equilibrium point along a line parallel to .

**
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