Straight-Line Solutions, Eigenvalues and Eigenvectors
In this section we will discuss the problem of finding two linearly independent solutions for the
homogeneous linear system
Let us first start with an example to illustrate the technique we will be developping.
Example: Draw the direction field of the linear system
Answer: The following is the direction field:
Remark: From the above example we notice that some solutions lie on straight lines (can you spot
them?). So it is natural to investigate whether and when an homogeneous linear system has solutions
which are straight-lines.
Straight-Line Solutions
Consider the homogeneous linear system (in the matricial notation)
A straight-line solution is a vector function of the form
,
where is a constant vector not equal to the zero vector
. The vector is the direction vector
of the line on which the solution lives. Keep in mind that the solutions of the system may describe
trajectories of moving objects. So, in this case, we may think of it as an object moving along a straight
line.
Remark: Note that if Y(t) is a straight-line solution, then
is also a straight-line solution.
Clearly, we have
.
Therefore, we have
.
Since
and
are constant vectors, we deduce that
is a constant function. Denote it by
Clearly, this is a first order differential equation which is linear as well as separable. Its solution is
,
where C is an arbitrary constant. So, if a straight-line solution exists, it must be of the form
,
where C is an arbitrary constant, and
is a non-zero constant vector which satisfies
Note that we don't have to keep the constant C (read the above remark).
Let us illustrate the above ideas with an example.
Example: Find any straight-line solution to the system
Answer: First, let us find the constant vector
such that
for some . Easy computations give
We have two cases:
Case 1. If
, then
(since
this case, the second equation forces
is not the zero vector). The first equation gives
; therefore we have
.
We may ignore the constant
(see the above remark). Therefore, the solution
is a straight-line solution to the system.
Case 2. If
, then from the second equation we get
, or equivalently
. The first equation reduces to
. Therefore, we have
.
We may again ignore the constant
. Hence, the solution
is a straight-line solution to the system.
. In
So, we have found two straight-line solutions
Are these the only straight-lines? The answer is: "yes," but this will be discussed later.
Theorem: Straight-Line Solutions
Consider the homogeneous linear system
Any straight-line solution may be found in the form
,
where
is a non-zero constant vector which satisfies
The constant is called an eigenvalue of the matrix A, and
is called an eigenvector associated to
the eigenvalue of the matrix A. Clearly, if is an eigenvector associated to , then
is also an
eigenvector associated to . Our next target is to find out how to search for the eigenvalues and
eigenvectors of a matrix.
Computation of Eiegenvalues
Consider the matrix
and assume that is an eigenvalue of A. Then there must exist a non-zero vector
such that
. This equation may be rewritten as the algebraic system
which is equivalent to the system
,
Since both and can not be equal to zero at the same time, we must have the determinant of the
system equal to zero. That is,
,
which reduces to the algebraic equation
.
Note that the above equation is independent of the vector
Characteristic Polynomial of the system.
. This equation is called the
Example: Find the characteristic polynomial and the eigenvalues of the matrix
Answer: The characteristic polynomial is given by
.
This is a quadratic equation. Its only roots are
matrix.
and
. These are the eigenvalues of the
Computation of Eiegenvectors
Assume is an eigenvalue of the matrix A. An eigenvector
equation
associated to is given by the matricial
.
Set
. Then, the above matricial equation reduces to the algebraic system
which is equivalent to the system
Since is known, this is now a system of two equations and two unknowns. You must keep in mind
that if
is an eigenvector, then
is also an eigenvector.
Example: Consider the matrix
.
Find all the eigenvectors associated to the eigenvalue
Answer: In the above example we checked that in fact
Let
be an eigenvector associated to the eigenvalue
.
is an eigenvalue of the given matrix.
. Set
. Then we must have
which reduces to the only equation
,
which yields
. Therefore, we have
Note that we have all of the eigenvectors associated to the eigenvalue
.
Conclusion
In order to find the straight-line solution to the homogeneous linear system
, perform the following steps:
First, we look for the eigenvalues through the characteristic polynomial
.
This is a quadratic equation which has one double real root, or two distinct real roots, or two complex
roots.
Once an eigenvalue is found from the characteristic polynomial, then we look for the eigenvectors
associated to it through the matricial equation
.
If you find a parameter factorized in front of
, there will be no need to keep it;
For an eigenvalue and an associated eigenvector
, a straight-line solution will be given by
.
Remark: It is not hard to show that two straight-line solutions generated by two different eigenvalues
are in fact linearly independent. Combined with the results of the previous section we now see how
straight-lines may be used to help find the solutions of an homogeneous linear system. This
technique is also related to the case of second order differential equation with constant coefficients.
Indeed, consider the second order differential equation
.
Set
. Then the second order differential equation is equivalent to the first order system
The matrix coefficient of the system is
.
The characteristic polynomial is
,
which is equivalent to the equation
.
We recognize the characteristic equation associated to the second order differential equation.