Eigenvalues and Eigenvectors Technique

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 tex2html_wrap_inline233 is a constant vector not equal to the zero vector tex2html_wrap_inline235 . The vector tex2html_wrap_inline233 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 tex2html_wrap_inline241 is also a straight-line solution.

Clearly, we have


Therefore, we have


Since tex2html_wrap_inline247 and tex2html_wrap_inline233 are constant vectors, we deduce that tex2html_wrap_inline251 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 tex2html_wrap_inline233 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 tex2html_wrap_inline273 for some tex2html_wrap_inline275 . Easy computations give


We have two cases:

Case 1. If tex2html_wrap_inline279 , then tex2html_wrap_inline281 (since tex2html_wrap_inline233 is not the zero vector). The first equation gives tex2html_wrap_inline285 . In this case, the second equation forces tex2html_wrap_inline279 ; therefore we have


We may ignore the constant tex2html_wrap_inline291 (see the above remark). Therefore, the solution


is a straight-line solution to the system.

Case 2. If tex2html_wrap_inline295 , then from the second equation we get tex2html_wrap_inline297 . The first equation reduces to tex2html_wrap_inline299 , or equivalently tex2html_wrap_inline301 . Therefore, we have


We may again ignore the constant tex2html_wrap_inline291 . Hence, the solution


is a straight-line solution to the system.

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 tex2html_wrap_inline233 is a non-zero constant vector which satisfies


The constant tex2html_wrap_inline275 is called an eigenvalue of the matrix A, and tex2html_wrap_inline233 is called an eigenvector associated to the eigenvalue tex2html_wrap_inline275 of the matrix A. Clearly, if tex2html_wrap_inline233 is an eigenvector associated to tex2html_wrap_inline275, then tex2html_wrap_inline333 is also an eigenvector associated to tex2html_wrap_inline275 . 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 tex2html_wrap_inline275 is an eigenvalue of A. Then there must exist a non-zero vector tex2html_wrap_inline343, such that tex2html_wrap_inline345 . This equation may be rewritten as the algebraic system


which is equivalent to the system


Since both tex2html_wrap_inline291 and tex2html_wrap_inline353 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 tex2html_wrap_inline233 . This equation is called the Characteristic Polynomial of the system.

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 tex2html_wrap_inline285 and tex2html_wrap_inline297 . These are the eigenvalues of the matrix.

Computation of Eiegenvectors

Assume tex2html_wrap_inline275 is an eigenvalue of the matrix A. An eigenvector tex2html_wrap_inline233 associated to tex2html_wrap_inline275 is given by the matricial equation


Set tex2html_wrap_inline343. Then, the above matricial equation reduces to the algebraic system


which is equivalent to the system


Since tex2html_wrap_inline275 is known, this is now a system of two equations and two unknowns. You must keep in mind that if tex2html_wrap_inline233 is an eigenvector, then tex2html_wrap_inline333 is also an eigenvector.

Example: Consider the matrix


Find all the eigenvectors associated to the eigenvalue tex2html_wrap_inline297 .

Answer: In the above example we checked that in fact tex2html_wrap_inline297 is an eigenvalue of the given matrix. Let tex2html_wrap_inline233 be an eigenvector associated to the eigenvalue tex2html_wrap_inline297 . Set tex2html_wrap_inline343 . Then we must have


which reduces to the only equation


which yields tex2html_wrap_inline301 . Therefore, we have


Note that we have all of the eigenvectors associated to the eigenvalue tex2html_wrap_inline297 .


In order to find the straight-line solution to the homogeneous linear system

displaymath413, 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 tex2html_wrap_inline275 is found from the characteristic polynomial, then we look for the eigenvectors tex2html_wrap_inline233 associated to it through the matricial equation


If you find a parameter factorized in front of tex2html_wrap_inline233 , there will be no need to keep it;

For an eigenvalue tex2html_wrap_inline275 and an associated eigenvector tex2html_wrap_inline233 , 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 tex2html_wrap_inline439 . 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.

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Author: Mohamed Amine Khamsi

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