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Unformatted text preview: Chap . 8 Linear Algebra : Matrix Eigenvalue Problems Sec . 8 . 1 Eigenvalues , Eigenvectors Eigenvectors of a (square!) matrix A are vectors x , not zero vectors, such that if you multiply them by A , you get a vector y u Ax that is proportional to x , say, y u u x . The factor of proportionality u is called an eigenvalue of A . How to find the eigenvalues? You have to solve the “characteristic equation” (4) of A , which is a quadratic equation in u if A is a 2 U 2 matrix, a cubic equation for a 3 U 3 matrix, and so on. Hence in practice you may need a root-finding method or one of the interesting iterative methods to be discussed in numerics in Chap. 20. Once you have found an eigenvalue, to find a corresponding eigenvector is simpler. You can find it by solving a linear system of equations, as explained on p. 337 and later in this chapter. Or you get an eigenvector along with an eigenvalue by using those iterative methods just mentioned. Problem Set 8 . 1 . Page 338 1 . Eigenvalues and eigenvectors . For a diagonal matrix the eigenvalues are the main diagonal entries because the characteristic equation is det u A u u I U u a 11 u u a 22 u u u u a 11 u u Uu a 22 u u U u 0. For the given matrix you obtain from this ¡ 1 u u 2, ¡ 2 u 0.4. Now determine an eigenvector of A corresponding to ¡ 1 u u 2. In components, u A u ¡ 1 I U x u is u a 11 u u 1 U x 1 ¢ a 12 x 2 u u u 2 u u u 2 UU x 1 ¢ x 2 u a 21 x 1 ¢ u a 22 u u 1 U x 2 u x 1 ¢ u 0.4 ¢ 2 U x 2 u 0. The first equation gives no condition. The second gives x 2 u 0. Hence an eigenvector of A corresponding to ¡ 1 u u 2 is ¡ x 1 ¢ T . Since an eigenvector is determined only up to a nonzero constant, you can simply take ¡ 1 ¢ T as an eigenvector. For u 2 u 0.4 the procedure is similar and leads to ¡ 1 ¢ T . 5 . Eigenvalues and eigenvectors . Problem 1 concerned a diagonal matrix, a case in which you could see the eigenvalues almost immediately. For a general 2 U 2 matrix the determination of eigenvalues and eigenvectors follows the same pattern. Example 1 on p. 334 illustrates this. For Prob. 5 the matrix is A u 5 u 2 9 u 6 . Calculate the characteristic equation 5 u u u 2 9 u 6 u u u u 5 u u Uu u 6 u u U ¢ 2 U 9 u u 2 ¢ u u 30 ¢ 18 u u u ¢ 4 Uu u u 3 U and from it the eigenvalues u 1 u u 4, u 2 u 3. Then find eigenvectors. For u 1 u u 4 obtain the system (2), p. 335, 80 Linear Algebra, Vector Calculus Part B u 5 u 4 U x 1 u 2 x 2 U say, x 1 U 2, x 2 U 9 9...
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This note was uploaded on 04/08/2012 for the course MT 423 taught by Professor M.lee during the Spring '12 term at National Taiwan University.
- Spring '12