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
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.
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:
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Author: Mohamed Amine Khamsi