Systems with Zero as an Eigenvalue
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
For
. The eigenvalues are
and
. Let us find the associated eigenvectors.
, set
The equation
translates into
The two equations are the same. So we have y = 2x. 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
infinity. But when
passing by (0,0) and having
. When
, the trajectory converge to the equilibrium point on the line of equilibrium points (that is
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
if
is an eigenvector associated to 0 and
We have two cases, whether
If
, then
, the trajectory goes to
or
as eigenvalues. Hence
an eigenvector associated to , then the general solution is
.
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
.