<p>Straight-Line Solutions, Eigenvalues and Eigenvectors </p><p>In this section we will discuss the problem of finding two linearly independent solutions for the homogeneous linear system </p><p>Let us first start with an example to illustrate the technique we will be developping. </p><p>Example: Draw the direction field of the linear system </p><p>Answer: The following is the direction field: </p><p>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.</p><p>Straight-Line Solutions Consider the homogeneous linear system (in the matricial notation) </p><p>A straight-line solution is a vector function of the form </p><p>, </p><p> 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. </p><p>Remark: Note that if Y(t) is a straight-line solution, then is also a straight-line solution. </p><p>Clearly, we have </p><p>. </p><p>Therefore, we have </p><p>. </p><p>Since and are constant vectors, we deduce that is a constant function. Denote it by </p><p>Clearly, this is a first order differential equation which is linear as well as separable. Its solution is </p><p>, where C is an arbitrary constant. So, if a straight-line solution exists, it must be of the form </p><p>, where C is an arbitrary constant, and is a non-zero constant vector which satisfies </p><p>Note that we don't have to keep the constant C (read the above remark). </p><p>Let us illustrate the above ideas with an example.</p><p>Example: Find any straight-line solution to the system </p><p>Answer: First, let us find the constant vector </p><p> such that for some . Easy computations give </p><p>We have two cases: </p><p>Case 1. If , then (since is not the zero vector). The first equation gives . In this case, the second equation forces ; therefore we have </p><p>. </p><p>We may ignore the constant (see the above remark). Therefore, the solution is a straight-line solution to the system. </p><p>Case 2. If , then from the second equation we get . The first equation reduces to</p><p>, or equivalently . Therefore, we have </p><p>. </p><p>We may again ignore the constant . Hence, the solution </p><p> is a straight-line solution to the system. </p><p>So, we have found two straight-line solutions </p><p>Are these the only straight-lines? The answer is: "yes," but this will be discussed later.</p><p>Theorem: Straight-Line Solutions</p><p>Consider the homogeneous linear system </p><p>Any straight-line solution may be found in the form </p><p>, 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. </p><p>Computation of Eiegenvalues</p><p>Consider the matrix </p><p> 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 </p><p> which is equivalent to the system </p><p>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, </p><p>, which reduces to the algebraic equation </p><p>. Note that the above equation is independent of the vector . This equation is called the Characteristic Polynomial of the system.</p><p>Example: Find the characteristic polynomial and the eigenvalues of the matrix </p><p>Answer: The characteristic polynomial is given by </p><p>. </p><p>This is a quadratic equation. Its only roots are and . These are the eigenvalues of the matrix.</p><p>Computation of Eiegenvectors</p><p>Assume is an eigenvalue of the matrix A. An eigenvector associated to is given by the matricial equation </p><p>. </p><p>Set . Then, the above matricial equation reduces to the algebraic system </p><p> which is equivalent to the system </p><p>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. </p><p>Example: Consider the matrix . </p><p>Find all the eigenvectors associated to the eigenvalue .</p><p>Answer: In the above example we checked that in fact is an eigenvalue of the given matrix. Let</p><p> be an eigenvector associated to the eigenvalue . Set . Then we must have </p><p> which reduces to the only equation </p><p>, which yields . Therefore, we have </p><p>Note that we have all of the eigenvectors associated to the eigenvalue .</p><p>Conclusion</p><p>In order to find the straight-line solution to the homogeneous linear system </p><p>, perform the following steps: </p><p>First, we look for the eigenvalues through the characteristic polynomial </p><p>. </p><p>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 </p><p>. </p><p>If you find a parameter factorized in front of , there will be no need to keep it; </p><p>For an eigenvalue and an associated eigenvector , a straight-line solution will be given by </p><p>. </p><p>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 </p><p>. </p><p>Set . Then the second order differential equation is equivalent to the first order system </p><p>The matrix coefficient of the system is </p><p>. </p><p>The characteristic polynomial is , which is equivalent to the equation </p><p>. </p><p>We recognize the characteristic equation associated to the second order differential equation. </p>