NUMERICAL METHODS with COMPUTER APPLICATIONS NEE 411 – EE1 NUMERICAL METHODS Boundary Value and Eigen Value Problems of Ordinary Differential Equations Introduction of Eigen Values and Eigen Vectors Diagonalization Definition 1: Let L: V → V be a linear transformation of an n-dimensional vector space V into itself. We say that L is diagonizable or can be diagonalized if there exists a basis S for V such that L is represented with respect to S by a diagonal matrix D. 1 0 D = 0 0 1 0 ... 0 n 0 NUMERICAL METHODS Introduction of Eigen Values and Eigen Vectors Diagonalization Definition 2: Let L: V → V be a linear transformation of an n-dimensional vector space V into itself (a linear operator on V). The real number is called an eigenvalue of L if there exists a nonzero vector (x) in V such that L(x) = x. Every nonzero vector x satisfying this equation is then called an eigenvector of L associated with the eigenvalue . Eigenvalues are also called proper, characteristic, or latent values, and eigenvectors are also called proper, characteristic, or latent vectors. Note that x = 0V always satisfies (1), but 0V is not an eigenvector, since we insist that an eigenvector be a nonzero vector. NUMERICAL METHODS Introduction of Eigen Values and Eigen Vectors Diagonalization Theorem 1: Let L: V → V be a linear transformation of an n-dimensional vector space V into itself. Then L is diagonalizable if and only if V has a basis S of eigenvectors of L. Moreover, if D is the diagonal matrix representing L with respect to S, then the entries on the main diagonal D are the eigenvalues of L. NUMERICAL METHODS Introduction of Eigen Values and Eigen Vectors Diagonalization Example 1: Let L: V→V be the linear operator defined by L(x) = 2x. We can see that the only eigenvalue of L is = 2 and that every nonzero vector in V is an eigenvector of L associated with the eigenvalue = 2. NUMERICAL METHODS Introduction of Eigen Values and Eigen Vectors Diagonalization Example 2: Let L: R2→R2 be the linear transformation defined by a a L 1 = 1 a2 a2 Then we can see that a a L = 1 a a and a Thus any vector of the form a a a L = −1 −a −a , where a is any nonzero real number, NUMERICAL METHODS Introduction of Eigen Values and Eigen Vectors Diagonalization for example, 1 x1 = 1 , is an eigenvector of L associated with the a eigenvalue = 1; any vector of the form , where a is any nonzero −a 1 real number, such as x2 = , is an eigenvector of L associated with −1 the eigenvalue = -1. NUMERICAL METHODS Diagonalization An eigenvalue can be associated with it many different eigenvectors. In fact, if x is an eigenvector of L associated with the eigenvalue [i.e., L(x) = x], then L(rx) = rL(x) = r(x) = (rx), for any real number r. Thus, if r 0, then rx is also an eigenvector of L associated with so that eigenvectors are never unique. NUMERICAL METHODS Diagonalization Theorem 2: An n x n matrix A is similar to a diagonal matrix D if and only if Rn has a basis of eigenvectors of A. Moreover, the elements on the main diagonal D are the eigenvalues of A. To use Theorem 2, we need only to show that there is a set of n eigenvectors of A that are linearly independent, since n linearly independent vectors in Rn form a basis for Rn. NUMERICAL METHODS Example 3: Let 1 A= −2 1 4 . We wish to find the eigenvalues of A and their associated eigenvectors. Thus we wish to find all scalars and all nonzero vectors a1 x= a2 Solution: 1 −2 a1 + a2 = a1 −2a1 + 4a2 = a2 or 1 a1 a1 = → (1) 4 a2 a2 ( − 1)a1 − a2 = 0 2a1 + ( − 4)a2 = 0 NUMERICAL METHODS which gives us system of two equations in two unknowns has a nontrivial solution if and only if the determinant of the coefficient matrix is zero. Thus −1 − 1 =0 2 − 4 which gives us the characteristic equation of 2 − 5 + 6 = 0 = ( − 3)( − 2) therefore, = 2 and = 3 are the eigenvalues of A. To find all the eigenvectors of A associated with =2, we substitute = 2 in equation (1) (2 − 1)a1 − a2 = 0 2a1 + (2 − 4)a2 = 0 a1 − a2 = 0 or 2a1 − 2a2 = 0 NUMERICAL METHODS which gives use all the eigenvectors associated with = 2. a1 = a2 a2 = any real number r. hence all eigenvectors associated with the eigenvalue 1 = 2 are given by r x1 = r and r is any nonzero real number. In particular, for r = 1, 1 x1 = 1 NUMERICAL METHODS To find all the eigenvectors of A associated with =3, we substitute = 3 in equation (1) (3 − 1)a1 − a2 = 0 2a1 + (3 − 4)a2 = 0 2a1 − a2 = 0 or 2a1 − a2 = 0 which gives use all the eigenvectors associated with = 3. a1 = 1 a2 2 a2 = any real number r. hence all eigenvectors associated with the eigenvalue 2 = 3 are given by r/2 x2 = and r is any nonzero real number. In particular, for r = 2, r 1 x2 = 2 NUMERICAL METHODS Problem: Find the characteristic polynomial, the eigenvalues and their associated eigenvectors for each of the following matrices. Answer: 1 (a) A = 2 -1 4 2 (b) A = 0 0 -2 3 -1 3 -2 2 2 (c) A = 1 2 2 2 -2 3 1 1 (a) f( ) = 2 - 5 + 6; = 2 and = 3 1 1 x1 = and x2 = -1 -2 (b) f( ) = 3 - 7 2 + 14 - 8; 1 = 1, 2 = 2 and 3 = 4 -1 1 7 x1 = 1 , x 2 = 0 and x 3 = -4 1 0 2 (c) f( ) = 3 - 5 2 + 2 + 8; 1 = -1, 2 = 2 and 3 = 4 1 -2 8 x1 = 0 , x 2 = -3 and x 3 = 5 -1 2 2 NUMERICAL METHODS Problem: Find the characteristic polynomial, the eigenvalues and their associated eigenvectors for each of the following matrices. 4 (d) A = 2 0 -4 -2 (f) A = 2 -1 2 1 -2 -3 -6 0 6 (g) A = 2 -2 2 5 0 -2 0 7 2 (h) A = 0 -2 0 0 -2 -2 -2 1 1 (i) A = 2 6 0 4 4 0 0 2 5 (e) A = 9 -2 -6