Consider a homogeneous constant-coefficient linear system:
The characteristic polynomial is
.
If is a root of this, we get a solution
with
, i.e.
Look at it from another point of view. This solution is of the form
where is a constant vector and is a constant
number. Suppose we look for solutions of that form.
Recall that we write the system in matrix and vector language as
where
is the matrix of coefficients.
If for some nonzero vector and some number , then we say that is an eigenvalue of the matrix and is an eigenvector of for that eigenvalue.
If is an eigenvector of for an eigenvalue , then is a solution of the homogeneous system . |
Now, how do we find the eigenvalues and their corresponding
eigenvectors? For systems we get the eigenvalues
from the characteristic polynomial of the corresponding
second-order differential equation:
.
This can be written in terms of determinants.
The determinant of the matrix is .
The characteristic polynomial of is .
The eigenvectors of are the roots of the characteristic polynomial. |
Once you have the eigenvalue , you can find an eigenvector by looking at one of the equations in the system . This will only determine the ratio of the entries and : since a multiple of an eigenvector is also an eigenvector, there is nothing else to be determined.
Example: . The characteristic polynomial of is . Its roots, which are the eigenvalues, are and .
For the eigenvalue , the system
says
Note that the two equations say the same thing (one is a multiple of the other). That must be the case if is an eigenvalue: otherwise the two equations together would require , which we don't allow for an eigenvector.
We can take any convenient value for one of the variables,
and determine the other
from one of the equations: e.g. take and then .
Thus
is an eigenvector for eigenvalue ,
and one solution of our system is
.
Similarly, an eigenvector for the other eigenvalue, 5, is
, corresponding to a solution
. Thus the general solution
of our homogeneous system is
In some cases the equation may say that one of the variables is 0,
so we take any convenient value (say 1) for the other
variable.
Also, one of the two equations may just say
, so we look at the other equation.
Both of these occur (for different eigenvalues) with the matrix
. The eigenvalues are 2 and 3.
For eigenvalue 2 we have
It is also possible for both of the equations to say , in which case all nonzero vectors are eigenvectors. This occurs with the matrix , whose only eigenvalue is 2.
We need two solutions of a homogeneous system to form a fundamental set. If there are two different eigenvalues, the solutions obtained using an eigenvector for each do form a fundamental set. If there is only one eigenvalue and all nonzero vectors are eigenvectors (as in the last paragraph), we can use any two eigenvectors that are not multiples of each other (e.g. and ) and obtain a fundamental set. However, if there is only one eigenvalue from which we get only one solution, we are missing a solution. In that case we can fall back on the reduction to a second-order equation, which will supply a second solution involving as well as .
Example: Let
.
The characteristic
polynomial is
,
so there is only one eigenvalue . The eigenvalue equations say
The next example has complex eigenvalues.
Example: Let .
The characteristic polynomial is
,
which has
complex roots
.
The eigenvector equations for the eigenvalue are
It may not be obvious that these equations say the same thing, but they must if we haven't made a mistake (in fact, the second is times the first). To avoid fractions, take and then , i.e. one solution is .
Now we could repeat this procedure for the second eigenvalue , but it's not really necessary: the result (if we again choose ) would be the complex conjugate of this one. If we're interested in real solutions (which we often are), we can use an extension of the Principle of Equating Real Parts:
If the matrix has real entries, the real and imaginary parts of a solution of are solutions. |
Writing
,
the real and imaginary parts of the
solution above are
and
respectively. These form a fundamental set. Thus when has
real entries we need only look at one eigenvalue from a pair of complex
conjugate eigenvalues, and take real and imaginary parts of the
solution obtained from that eigenvalue.