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 . 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 = 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 . 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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