# Complex Eigenvalues

Consider the linear homogeneous system

The Characteristic polynomial is

In this section, we consider the case when the above quadratic equation has complex roots (that is if ). The roots (eigenvalues) are

where

In this case, the difficulty lies with the definition of

In order to get around this difficulty we use Euler's formula

Therefore, we have

In this case, the eigenvector associated to will have complex components.

Example. Find the eigenvalues and eigenvectors of the matrix

Answer. The characteristic polynomial is

Its roots are

Set . The associated eigenvector V is given by the equation . Set

The equation translates into

Since , then the two equations are the same (which should have been expected, do you see why?). Hence we have which implies that an eigenvector is

We leave it to the reader to show that for the eigenvalue , the eigenvector is

Let us go back to the system

with complex eigenvalues . Note that if V, where

is an eigenvector associated to , then the vector

(where is the conjugate of v) is an eigenvector associated to . On the other hand, we have seen that

are solutions. Note that these solutions are complex functions. In order to find real solutions, we used the above remarks. Set

then we have

which gives

Similarly we have

Putting everything together we get

Clearly this implies where

It is easy to see that we have

Since the sum and difference of solutions lead to another solution, then both and are solutions of the system. These are real solutions. It is very easy to check in fact that they are linearly independent. Let us summarize the above technique.

Summary (of the complex case). Consider the system

Write down the characteristic polynomial

and find its roots

we are assuming that . Note that at this step, you need to know and . The common mistake is to forget to divide by 2.

Find an eigenvector V associated to the eigenvalue . Write down the eigenvector as

Two linearly independent solutions are given by the formulas

The general solution is

where and are arbitrary numbers. Note that in this case, we have

Example. Consider the harmonic oscillator

Find the general solution using the system technique.

Answer. First we rewrite the second order equation into the system

The matrix coefficient of this system is

We have already found the eigenvalues and eigenvectors of this matrix. Indeed the eigenvalues are

Hence we have

The eigenvector associated to is

Next we write down the two linearly independent solutions

and

The general solution of the equivalent system is

or

Below we draw some solutions. Notice how the solutions spiral and dye at the origin (see the discussion below)

Since we are looking for the general solution of the differential equation, we only consider the first component. Therefore we have

You may want to check that the second component is just the derivative of y.
Below we draw some solutions for the differential equation

Qualitative Analysis of Systems with Complex Eigenvalues.

Recall that in this case, the general solution is given by

The behavior of the solutions in the phase plane depends on the real part . Indeed, we have three cases:

the case: . The solutions tend to the origin (when ) while spiraling. In this case, the equilibrium point is called a spiral sink.

The case: The solutions explode or get away from the origin (when ) while spiraling. In this case, the equilibrium point is called a spiral source.

The case: The solutions are periodic. This means that the trajectories are closed curves or cycles. In this case, the equilibrium point is called a center.

### If you would like more practice, click on Example.

[Differential Equations] [First Order D.E.]
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Author: Mohamed Amine Khamsi

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